# Error Estimates for Binomial Approximations of Game Put Options

Authors: Yonatan Iron; Yuri Kifer
Journal: ISRN Probability and Statistics
 (2014)

---

## Abstract

A game or Israeli option is an American style option where both the writer and the holder have the right to terminate the contract before the expiration time. Kifer (2000) shows that the fair price for this option can be expressed as the value of a Dynkin game. In general, there are no explicit formulas for fair prices of American and game options and approximations are used for their computations. The paper by Lamberton (1998) provides error estimates for binomial approximation of American put options and here we extend the approach of Lamberton (1998) in order to obtain error estimates for binomial approximations of game put options which is more complicated as it requires us to deal with two free boundaries corresponding to the writer and to the holder of the game option.


---

## Body

## 1. Introduction

A put option on a stock can be interpreted as a contract between a holder and a writer which allows the former to claim from the latter at an exercise timet the amount (K-St)+, where K is a fixed amount called the option’s strike, St is the stock price at time t, and (x)+=max⁡(x,0). In the American options case its holder has the right to choose any exercise time before the contract matures, while in the game options case the contract writer also has the right to terminate it at any time before its maturity, but then he is required to pay a cancellation fee in addition to the payoff above.The fair price of American options and of game options is defined as the minimal amount the writer needs to construct a self-financing portfolio which covers his obligation to pay according to the option’s contract. It is well known that in the American options case the fair price can be obtained as a value of an appropriate optimal stopping problem, while for game options we have to deal with an optimal stopping (Dynkin) game (see [1]). For more information about results on Dynkin games and game options we refer the reader to the survey [2]. In general, both for American options and, even more so, for game options with finite maturity explicit formulas for their price are not available and approximation methods come into the picture, while estimates of their errors become important. One of the most easily implemented methods is the binomial approximation of stock prices modelled by the geometric Brownian motion and [3] provided corresponding error estimates for American put options. In the present paper we extend this approach in order to provide error estimates of binomial approximations for game put options. We observe that for perpetual game options some explicit formulas can be obtained (see [4]), but the finite maturity case studied here seems to be more realistic.Approximating the Brownian motion by appropriately normalized sums of Bernoulli random variables the paper [3] provided (error) estimates const·n-3/4 and const·n-2/3 for the difference between the price of an American put option and the price of its corresponding nth binomial model approximation. Using again the binomial approximation of the Brownian motion as above we construct in this paper two approximating procedures such that the difference between the price of a game put option and its nth approximation in the first procedure is between const·n-3/4 and const·n-1/2 and in the second procedure is between const·n-1/2 and const·n-2/3. The error estimates here are somewhat worse than in the case of American put options which is due to the lack of a smooth fit on the boundary of the writer’s stopping region which causes substantial difficulties in the study of regularity of payoff functions.We observe that specific properties of game put options had to be used in order to obtain error estimates with the above precision. For instance, when payoffs are path dependent (and not only dependent on the present value of the stock) [5] provides error estimates of similar binomial approximations only of order n-1/4(ln⁡n)3/4. Since price functions of game options can be represented as solutions of doubly reflected backward stochastic differential equations the results of [6] are also related to game options approximations. Nevertheless, approximations in [6] are not by binomial models, where computations can be done by means of the effective dynamical programming algorithm (see [5]), but by time discretizations, and so relevant probability space and σ-algebras remain infinite which prevents effective computations. Furthermore, error estimates in [6] applied to our situation are of order n-1/4; that is, they are worse than for binomial approximations which we construct here for the specific case of game put options.Our exposition proceeds as follows. In Section2 we provide basic results concerning game put option price functions, introduce our approximation processes, and formulate our main result Theorem 1. In Section 3 we show that the price function can be represented as a solution of a variational inequality problem closely related to the Stefan problem (see [7]). We then use this representation to study regularity properties of the price function near the free boundary of the option’s holder exercise region. In Section 4 we study the price function near the boundary of the exercise region of the writer. We use the information about this region from  [8] in order to represent the price function as an explicit solution of the heat equation. This representation enables us to understand better the behavior of the price function near the boundary. We estimate also the rate of decay of the price function when the initial stock price tends to infinity. Section 5 is devoted to the proof of Theorem 1. Finally, in Section 6 we exhibit some computations of the price functions and of the free boundaries.


## 2. Preliminaries and Main Results

The Black-Scholes (BS) model of a financial market consists of two assets among which one is nonrisky and the other one is risky. A nonrisky asset is called a bond and its priceBt at time t is given by the formula Bt=B0ert, where r is interpreted as the interest rate. A risky asset is called a stock and its price at time t is determined by a geometric Brownian motion
(1)St=S0exp⁡((r-κ22)t+κWt),
where κ>0 is called volatility and Wt,t≥0 is a standard Brownian motion defined on a complete probability space (Ω,ℱ,P). If S0=x we write also Stx for St. The fair price of an American put option at time t with a strike (price) K and a maturity (horizon) time T<∞ can now be written as a function FA(t,St) of time and the current stock price having the form (see, e.g., [9])
(2)FA(t,x)=sup⁡τ∈𝒯0,T-tEexp⁡(-rτ)(K-xexp⁡((r-κ22)τ+κWτ))+,
where 𝒯0,T-t denotes the set of all stopping times of the Brownian filtration with values in the interval [0,T-t] and E is the expectation with respect to the measure P. If we set ψ(x)=(K-ex)+, PA(t,x)=FA(t,ex), and μ=r-κ2/2 then we can rewrite (2) in the form
(3)PA(t,x)=sup⁡τ∈𝒯0,T-tEexp⁡(-rτ)ψ(x+μτ+κWτ).Relying on [1] (see also [4, 8, 10]) we can also write the fair price of a game put option at time t with a strike price K, a maturity time T, and a constant penalty δ>0 as a function F(t,St) of time and the current stock price in the form
(4)F(t,x)=inf⁡σ∈𝒯0,T-tsup⁡τ∈𝒯0,T-tEexp⁡(-rσ∧τ)R(σ,τ),
where R(s,t)=(K-Stx)++δ𝕀s<t and 𝕀Q is the indicator of an event Q. Using the functions P(t,x)=FA(t,ex) and ψ as above we can rewrite this formula in the form
(5)P(t,x)=inf⁡σ∈𝒯0,T-tsup⁡τ∈𝒯0,T-tEexp⁡(-rσ∧τ)×(ψ(x+μσ∧τ+κWσ∧τ)+δ𝕀σ<τ).
It follows also (see [1, 8, 10, 11]) that the saddle point (optimal) stopping times for the game value expressions (4) and (5) are given by
(6)σ*=inf⁡{s<T-t:F(t+s,Ssx)=(K-Ssx)++δ}∧T,τ*=inf⁡{s<T-t:F(t+s,Ssx)=(K-Ssx)+}∧T.Next, we introduce our binomial approximations of the Brownian motion(7)Wt(n)=Tn∑k=1[nt/T]‍ϵk,t∈[0,T],n=1,2,…,
where ϵk,k=1,2,… are independent identically distributed (i.i.d.) random variables taking on values 1 and −1 with probability 1/2 and [a] denotes the integral part of a number a. It is convenient to view {ϵk}k=1∞ as defined on the sequence space Ωϵ={-1,1}ℕ={ξ=(ξ1,ξ2,…):ξi=±1} by the formula ϵk(ξ)=ξk if ξ=(ξ1,ξ2,…). Then Wt(m) will be defined on the probability space (Ωϵ,ℱϵ,𝒫ϵ), where 𝒫ϵ={1/2,1/2}ℕ is the product measure and ℱϵ is generated by cylinder sets.Now setδ*=FA(0,K) which is the price of the American put option with a maturity T and a strike K provided the initial stock price is K. It is easy to see that if the penalty δ≥δ* then it does not make sense for the writer to cancel the corresponding game put option (see Lemma 3.1 in [10]), and so in this case the prices of American and game options are the same; that is, FA(0,K)=F(0,K). Since approximations of American options were studied in [3] we assume in this paper that δ<δ*. Observe that FA(t,K) is continuous in t and it is strictly decreasing to 0 as t increases to T, and so for each δ∈[0,δ*] there exists a unique tδ<T such that FA(tδ,K)=δ. Furthermore, we can define kn to be the minimal k∈ℕ such that δ≥FA(Tk/n,K) and set β(n)=Tkn/n. In order to define two sequences of functions P1(n) and P2(n),n=1,2,… which will approximate P(0,x) we set Xt(n)=x+κWt(n), h=T/n and introduce stopping times
(8)σ(n)(s)=inf⁡{t∈[0,s):ln⁡K-|μ|h-2κhcccccccccccccccccc<μh[th]+Xt(n)cccccccccccccccccc<ln⁡K+|μ|h+2κh}∧T,
where σ(n)=T if the infimum above is taken over the empty set and we set σ(n)=σ(n)(β(n)). Introduce a filtration {𝒢t=ℱ[t/h],t≥0}, where ℱ0 is the trivial σ-algebra and ℱk is generated by ϵ1,…,ϵk. Denote by 𝒯(n) the set of all stopping times with respect to the filtration {𝒢t} taking on value in the set {kh,k=0,1,…,n}. Then, clearly, σ(n)∈𝒯(n). Now, for x≤ln⁡K we define
(9)P1(n)(s,x)=sup⁡τ∈𝒯(n)E(e-rσ(n)(s)∧τ(ψ(μτ+Xτ(n))𝕀{τ≤σ(n)(s)}cccccccccccccccccccccc+δ𝕀{σ(n)(s)<τ})e-rσ(n)(s)∧τ),
and for x>ln⁡K we set
(10)P1(n)(s,x)=sup⁡τ∈𝒯(n)E(e-rσ(n)(s)∧τ(ψ(μτ+Xτ(n))𝕀{τ≤σ(n)(s)}cccccccccccccccccccccc+(δ-Ke(|μ|h+κh))ccccccccccccccccccccccc×𝕀{σ(n)(s)<τ})e-rσ(n)(s)∧τ).
The second approximation function is defined for all x by
(11)P2(n)(s,x)=sup⁡τ∈𝒯(n)E(e-rσ(n)(s)∧τ(ψ(μτ+Xτ(n))𝕀{τ≤σ(n)(s)}cccccccccccccccccccccc+(ψ(μσ(n)(s)+Xσ(n)(s)(n))+δ)cccccccccccccccccccccc×𝕀{σ(n)<τ})e-rσ(n)(s)∧τ).
Setting Pi(n)(x)=Pi(n)(β(n),x) we formulate now our main result.Theorem 1.
For eachx there exists C=C(x) such that for all n=1,2,…,
(12)-Cn1/2≤P1(n)(x)-P(0,x)≤Cn3/4,-Cn2/3≤P2(n)(x)-P(0,x)≤Cn1/2.Observe thatPi(n)(x),i=1,2 appearing in Theorem 1 is defined via β(n) and kn which can be obtained only knowing precise price function FA(t,K) of the American put option with the initial stock price equal to K. But from the computational point of view we can obtain FA only approximately using, for instance, the algorithm from [3]. One of the ways to overcome this difficulty is to proceed as follows. Let FA(n)(t,K) denote the nth binomial approximation of FA(n)(t,K) obtained in [3] which uniformly in t∈[0,T] satisfies
(13)-c~n2/3≤FA(n)(t,K)-FA(t,K)≤C~n3/4
for some c~,C~>0. Denote by mn the minimal m≤n such that δ≥FA(n)(Tm/n,K) taking mn=n if this inequality does not hold true for all m≤n. Set γ(n)=Tmn/n which unlike β(n) can be computed employing [3]. It is well known that ∂FA(t,K)/∂t exists (see, e.g., [3]) and, clearly, this derivative is nonpositive. In fact, it is possible to show that
(14)∂FA(t,K)∂t<0,d=inf⁡0<t<T|∂FA(t,K)∂t|>0.
This together with (13) yields that
(15)-C~dn3/4≤β(n)-γ(n)≤c~dn2/3.
From the definitions (9)–(11) it follows that for each x>0 there exists C~~=C~~(x)>0 independent of s,s~∈[0,T] such that
(16)|Pi(n)(s,x)-Pi(n)(s~,x)|≤C~~|s-s~|,i=1,2.
Now we obtain from Theorem 1 together with (15) and (16) the following.Corollary 2.
For eachx>0 there exists C=C(x)>0 such that for all n=1,2,…,
(17)-Cn1/2≤P1(n)(γ(n),x)-P(0,x)≤Cn2/3,-Cn2/3≤P1(n)(γ(n),x)-P(0,x)≤Cn1/2.In the following sections we will analyze regularity properties of the price functionP(t,x) of game put options and will complete the proof of Theorem 1 in Section 5 providing some computations in Section 6. The general strategy of the proof resembles that of [3], but the study of the price function of game put options is more complicated than in the American options case, in particular, because of appearance of two exercise boundaries (holder’s and writer’s) having different properties. Our proof will be based on regularity properties of solutions of parabolic partial differential equations with free boundary and of the corresponding variational inequalities and we will rely also on some prior results from [3, 4, 8].


## 3. Price Function Near the Holder’s Exercise Boundary

### 3.1. Some Previous Results

First, we state the following result from [8] (see also [10]) which we will use later on.Proposition 3.
(i) There exists an increasing functionb:[0,T)→ℝ such that lim⁡t→Tb(t)=K and F(t,x)=K-x for all (t,x) satisfying 0<x≤b(t).
(ii) There exists0<δ* such that for every 0≤δ≤δ* there is a β=β(δ)≥0 so that F(t,K)=δ for t∈[0,β] and for t≥β one has F(t,x)=FA(t,x) for all x≥0.
(iii) Furthermore,(18)Ft(t,x)+12κ2x2Fxx(t,x)+rxFx(t,x)-rF(t,x)=0,
for all
(19)(t,x)∈(0,T)×ℝ+∖({(t,x):x≤b(t)}∪[0,β)×{K}).
In particular,F(t,x) is of class C1,2, that is, continuously differentiable once with respect to t and twice with respect to x, and so, in fact, it is a smooth function there.
(iv) Finally,F(t,x) is convex and strictly decreasing in x and nonincreasing in t.Next, we introduce an operator𝒟 which acts on Borel functions u(t,x) on [0,T]×ℝ by
(20)𝒟u(t,x)=12[u(t+h,x+κh)+u(t+h,x-κh)]-u(t,x).
Clearly, (1/h)𝒟(t,x) can be viewed as a discretization of the differential operator (∂/∂t)+(κ2/2)(∂2/∂x2). We will rely on the following results from [3] concerning the operator 𝒟.Proposition 4.
For each Borel functionu on [0,T]×ℝ there exists a martingale (Mt)0≤t≤T with respect to the filtration 𝒢t,t≥0 such that M0=0 and for every t∈{0,h,2h,…,T},
(21)u(t,Xt(n))=u(0,x)+Mt+∑j=1nt/T‍𝒟u((j-1)h,X(j-1)h(n)).Proposition 5.
Let0≤t≤T-h and x∈ℝ. Assume that u is a C2 function on ([t,t+h]× [x+κh,x-κh]). Then
(22)𝒟u(t,x)=1κ∫0h‍dy∫-κyκy‍dz(z∂2u∂t∂x(t+y2,x+z)hhhhhhhhhhhhhhhh+δ(u)(t+y2,x+z)∂2u∂t∂x),
where
(23)δ(u)(t,x)=ut(t,x)+κ22uxx(t,x).We will need also the following result concerning the free boundarys(t)=ln⁡(b(t)) of the holder exercise region of our game put option which in the case of American options appears as Proposition 1 in [3] and it can be proved for game options in the same way.Proposition 6.
Let0≤t1<t2≤β and let x0=s(0)<x1=s(β)<ln⁡K; then (s(t1)-s(t2))2≤ sup⁡x0≤x≤x1|P(t1,x)-P(t2,x)|.We also observe that it follows from the Berry-Esseen estimate (see [12]) that, for some constant C1>0 independent of j,n≥1, and z∈ℝ,
(24)P{|Xjh(n)-z|≤κh}≤C1j.We will also rely on the following standard bounds on derivatives of solutions of 2nd-order parabolic equations with constant coefficients (see, e.g., [13, 14]).Proposition 7.
LetD=(0,T)×(0,1) and let w(t,x)∈C[D-] be a solution in D of the following parabolic equation:
(25)κ22wxx+μwx-rw=wt.
Suppose that w(0,x)=0 for all 0≤x≤1 and that there exists A>0 such that |w(t,x)|<A for all (x,t)∈D-. Then for every k,n and 0<a<b<1 there exists C=C(k,n,a,b,T,A) such that
(26)|∂k+nw∂kx∂nt(t,x)|<C∀(t,x)∈(0,T)×[a,b].


### 3.2. Price Function and Variational Inequalities

Next, we will show that the price function of the game put option can be represented as a solution of a variational inequality (v.i.) problem which is a generalization of the Stefan problem (see [7], VIII). This will enable us to derive certain regularity properties of this price function which we will use later on. Details of some of the proofs concerning the solutions of the v.i. problem below which are similar to the proofs in the case of the Stefan problem will not be given here. For the corresponding results in the American put option case we refer the reader to [3, 9] and to references there in.LetT′ be such that β<T′<T and set
(27)A=κ22∂2∂x2+μ∂∂x-r,whereμ=r-κ22.
Using the maximum principle, properties of price functions of American and game put options, and the fact that after time β the price functions of the game and American option are the same we obtain that for every x>s(T′) the time derivative Pt(T′,x)=PA,t(T′,x) is strictly negative and we can find a,b satisfying s(T′)<a<b<ln⁡K such that, for some constant c>0,
(28)-Pt(T′,x)>c∀x∈[a,b].
Relying on Proposition 3(iii) we also observe that for all (t,x)∈[0,T′]×(s(t),ln⁡K),
(29)∂P∂t(t,x)+AP(t,x)=0,P(t,x)>K-ex∀(t,x)∈[0,T′]×(s(t),ln⁡K),P(t,x)=K-ex∀t∈[0,T′],∀x≤s(t),Pt≤0.Leta0 be such that a0<s(0)<s(T′)<b. Introduce the domain D=(0,T′)×(a0,b) and for all (t,x) in the closure D- of D define the functions
(30)v(t,x)=P(T′-t,x)-P(T′,x),f(x)=AP(T′,x).
We obtain that
(31)f(x)={-Pt(T′,x),s(T′)<x≤b-rK,a0≤x≤s(T′)
and from the definition of v(t,x) it follows that for any (t,x)∈D-,
(32)Pt(T′-t,x)=-vt(t,x),Ptx(T′-t,x)=-vtx(t,x),Ptt(T′-t,x)=vtt(t,x),Px(T′-t,x)-Px(T′,x)=vx(t,x),Pxx(T′-t,x)-Pxx(T′,x)=vxx(t,x).
Since Px(T′,x) and Pxx(T′,x) are bounded we obtain that the integrability properties of the first- and second-order derivatives of P(t,x) and v(t,x) are the same in D-. Now set
(33)ψ(t)=v(t,b),g(t)=vt(t,b)for0≤t≤T′.
Then by (28) and (32),
(34)ψ(t)=∫0t‍g(τ)dτ=∫0t‍vt(τ,b)dτ≥0for0≤t≤T′.
It follows from (29) and (30)-(31) that on the set v>0,
(35)vt-Av-f=-Pt(T′-t,x)-AP(T′-t,x)+AP(T′,x)-f(x)=0,
and on the set v=0 we obtain
(36)vt-Av-f=rK>0.
Hence we arrive at the following (see [7]).Lemma 8.
The functionv is the unique solution of the following variational inequality problem.
v.i. Problem 1. Find v∈L2[0,T′;H2(a0,b)]∩H1[D] such that(i)
v,vt≥0,(ii)
(vt-Av)(w-v)≥f(w-v) a.s for every w∈L2[D],w≥0,(iii)
v(t,b)=ψ for 0≤t≤T′,x=b,(iv)
v(t,a0)=0 for 0≤t≤T′,x=a0,(v)
v(0,x)=0 for t=0,a0≤x≤b.Proof.
We will prove uniqueness and the fact thatv is a solution to v.i. Problem 1 follows from (30)–(36). Assume that v and v~ are two solutions of v.i. Problem 1. Since v~≥0 (property (i)) we can use the property (ii) of v and replace w by v~. Since both of them are solutions we obtain that
(37)(vt-Av)(v~-v)≥f(v~-v),(v~t-Av~)(v-v~)≥f(v-v~).
Define the parabolic boundary as the boundary of D without the interval {T′}×(a0,b) and let u=v-v~. Note that u is zero on the parabolic boundary and the sum of the two inequalities (37) is
(38)utu-κ22uxxu-μuxu+ru2=(ut-Au)u≤0.
Integrating both sides of (38) on (0,T′)×(a0,b) we obtain four terms on the left side. For the first term we have
(39)∫a0b‍∫0T′‍u(t,x)ut(t,x)dtdx=∫a0b‍12u2(T′,x)dx≥0.
Integration by parts of the second term and the fact that u=0 on the parabolic boundary yield
(40)-κ22∫0T′‍∫a0b‍uxx(t,x)u(t,x)dxdth=κ22∫0T′‍∫a0b‍ux2(t,x)dxdt≥0.
For the third term note that uxu=(1/2)(du2/dx) and that u(t,a0)=u(t,b0)=0 for every t, and so
(41)μ∫0T′‍∫a0b‍ux(t,x)u(t,x)dxdt=0.
The last term satisfies r∫0T′‍∫a0b‍u2(t,x)dxdt≥0 since r>0. We conclude that the left side of (38) cannot be negative and so it must be zero. Since all terms in the left-hand side of (38) are nonnegative and their sum is equal to 0 we obtain that r∫0T′‍∫a0b‍u2(t,x)dxdt=0, and so u=0 almost everywhere (a.e.). Hence, v=v~ a.e., and so there is only one continuous solution.Denote parts of the boundary ofD=(0,T′)×(a0,b) by
(42)Γ1=[0,T′]×{b},Γ2={0}×(a0,b),Γ3=[0,T′]×{a0},Γ=Γ1∪Γ2∪Γ3
and set
(43)L=∂∂t-A=∂∂t-κ22∂2∂x2-μ∂∂x+r.
Thus, Γ is a parabolic boundary of D. For every ε>0 we define following functions.(1)
A smooth functionf(ε)(x)≥f(x) on (a0,b) such that f(ε)(x)=f(x) for s(T′)<a<x≤b and for a0≤x≤a1, where a1 satisfies a0<a1<s(T′) and lim⁡ε→0f(ε)(x)=f(x) for a0≤x≤b.(2)
A smooth functionβ(ε)(v) satisfying
(44)β(ε)(v)=0∀v≥ε,β(ε)(0)=-1,βv(ε)(v)≥0,βvv(ε)(v)≤0.(3)
ψ(ε)(t)=ψ(t)+ε with ψ defined in (33).(4)
A smooth functionη(x) such that 0≤η(x)≤1 and for some a<a2<b,
(45)η(x)=1fora2≤x≤b,η(x)=0fora0≤x≤a.Set Fε(x,v)=f(ε)(x)-rKβ(ε)(v) which is a Lipschitz continuous function and for every constant C there is M0 such that C|F(ε)(x,v)|≤M whenever M≥M0 and |v|≤M. Let ϕ(ε) be a function on Γ satisfying
(46)ϕ(ε)|Γ1=ψ(ε)(t),ϕ(ε)|Γ2=εη(x),ϕ(ε)|Γ3=0,
and, moreover, relying on Chapter 3 in [14] we can choose ϕ(ε) so that(1)
ϕ(ε)∈C-2+δ[D] for some 0<δ<1 (in fact for each δ) and we refer the reader to Chapter 3 in [14] for the definition of C-2+δ[D] and for conditions yielding that a function defined only on the boundary Γ can be extended to a function from C-2+δ[D];(2)
Lϕ(ε)=F(ε)(x,ψ) at the points (0,b) and (0,a0).By the theory of semilinear parabolic equations (see [14]) there exists a function v(ε)∈C-2+γ[D] for some 0<γ<1 such that
(47)Lv(ε)=F(ε)(x,v(ε)),v(ε)|Γ=ϕ(ε).
In particular v(ε),vx(ε),vxx(ε),vt(ε) are continuous on D-.Letw=vt(ε). By differentiating with respect to t (47) and taking into account (33), (47), and the properties of ϕ(ε) we obtain that
(48)wt-κ22wxx-μwx+(r+rKβv(ε)(v(ε)))w=0,wherew(t,b)=g(t)∀0≤t≤T′,w(t,a0)=0∀0≤t≤T′w(0,x)=f(ε)(x)+ε(κ22ηx,x(x)+μηx(x)-rη(x))-rKβv(ε)(v(ε)(0,x))∀a0≤x≤b.
We see that in D the function w is a solution to a parabolic equation and since r+rKβv(ε)≥0 we can use the maximum principle
(49)min⁡(min⁡Γ(w),0)≤w(t,x)≤max⁡(max⁡Γ(w),0)∀(t,x)∈D.
Therefore in order to bound the function w we only need to bound its values on the parabolic boundary. First, we estimate the left hand side of (49). For a≤x≤b we have that v(ε)(0,x)=εη(x)≤ε, and so β(ε)(v(ε))≤0. In view of (28), (31), and the definition of f(ε) above there exists ε0>0 such that, for every 0<ε≤ε0,
(50)w(0,x)≥f(x)+ε(ηxx(x)+μηx(x)-rη(x))=-Pt(T′,x)+ε(κ22ηxx(x)+μηx(x)-rη(x))≥0∀a≤x≤b.
On the interval a0≤x≤a we have η=0 and since v(ε)(0,x)=εη(x)=0 we see that β(ε)(v(ε)(0,x))=-1. Since on this interval f(x)≥-rK we obtain
(51)w(0,x)=f(ε)(x)-rKβ(ε)(0)≥f(x)+rK≥0fora0≤x<a.
Hence, w≥0 on Γ2. We obtain next that
(52)w(t,b)=-∂P∂t(T′-t,b)≥0on⁡Γ1,w(t,b)=0on⁡Γ3.
It follows that min⁡(min⁡Γ(w),0)=0.Next, we estimate the right-hand side of (49). On Γ2 we have that
(53)w(0,x)≤|f(ε)(x)+ε(κ22ηxx(x)+μηx(x)-rη(x))-rKβ(ε)(v(ε)(0,x))κ22|≤sup⁡|f(ε)|+εsup⁡|κ22ηxx+μηx-rη|+rK≤C0,
where C0>0 is a constant independent of ε, and so
(54)0=min⁡⁡(min⁡Γ(w),0)≤w(t,x)≤max⁡(max⁡(-∂P∂t(T′-t,b),C0))=C1.
We conclude that there are some constants ε0 and C1 such that for every 0<ε≤ε0,
(55)0≤vt(ε)(t,x)≤C1.
Since vt(ε)≥0 and v(ε)(0,x)≥0 for a0≤x≤b we deduce that v(ε)(t,x)≥0 and because vt(ε) is uniformly bounded it follows that v(ε) is also uniformly bounded. By the properties of β(ε) we see that
(56)-1≤β(ε)(v(ε))≤0or∥β(ε)(v(ε))∥L∞[D]≤1.
Let D0=(0,T)×(a2,b) be an upper subrectangle of D, where a2 is the same as in the definition of the function η in (4). From the definition we have v(ε)(0,x)=εη(x)=ε in D-0 and since vt(ε) is nonnegative, we obtain that v(ε)(t,x)≥ε, and so β(ε)(v(ε)(t,x))=0.This means that onD0 the function v(ε) satisfies the parabolic equation
(57)Lv(ε)=f(ε).
For w=vt(ε) and 0<ε<ε0 we also have that
(58)Lw(t,x)=0∀(t,x)∈D0,w(0,x)=f(ε)(x)whena2≤x≤b.Next, lety(t,x) be a function on D0 such that
(59)Ly(t,x)=0∀(t,x)∈D0,y(0,x)=f(ε)(x)∀a2≤x≤b,
and all of its first- and second-order derivatives are bounded there. Such a function exists since we can choose a smooth function on the remaining part Γ0∖{0}×[a2,b] of the parabolic boundary Γ0 of D0 which extends f(ε)(x) as a smooth function to the whole D0 and then use Theorem 12 from Chapter 3 in [7]. For each ε<ε0 we define z(t,x)=w(t,x)-y(t,x) in the domain D0, where w(t,x)=vt(ε)(t,x). Then z(0,x)=w(0,x)-y(0,x)=0 for every a2≤x≤b. Fix x0∈(a2,b); then by Proposition 4.5 from Section 4.1 of [7] we obtain that |zt(t,x0)|,|ztt(t,x0)|<C for every 0≤t≤T′, where a constant C>0 is independent of ε. Since we assume that |yt(t,x0)|,|ytt(t,x0)|<C1 for 0≤t≤T′ it follows that |wt(t,x0)|,|wtt(t,x0)|<C1+C (for every ε). Let D1=(x0,b)×(0,T); then by Theorem 6 in Chapter 3 of [14] we obtain that for every ε>0 (and in fact every 0<α<1) v(ε),vt(ε)∈C-2+α[D1] and there is a constant C independent of ε such that
(60)|v(ε)vt(ε)|¯2+α+|vt(ε)|¯2+α<C.
For the definition of norms in (60) see Chapter 3 in [14].In particular, we get(61)∥vx(ε)∥L∞(D1)+∥vtx(ε)∥L∞(D1)+∥vtt(ε)∥L∞(D1)<C.
Considering again the whole region D we have
(62)-∫xb‍vxx(ε)(t,y)dy=2κ2∫xb‍(μvx(ε)(t,y)-vt(ε)(t,y)-rv(ε)(t,y)+β(ε)(v(ε)(t,y))+f(ε)(y))dy.
Hence,
(63)vx(ε)(t,x)=vx(ε)(t,b)+2κ2[μ(v(ε)(t,b)-v(ε)(t,x))+∫xb‍(-vt(ε)(t,y)-rv(ε)(t,y)ccccccccccc+β(ε)(v(ε)(t,y))+f(ε)(y))dy].
Since all terms in the right-hand side are uniformly bounded there is a constant C>0 independent of ε such that ∥vx(ε)∥<C for every 0<ε≤ε0. Now we see that in the equation
(64)vxx(ε)(t,y)=2κ2(2κ2μvx(ε)(t,y)-vt(ε)(t,y)-rv(ε)(t,y)hh+β(ε)(v(ε)(t,y))+f(ε)(y)2κ2)
all terms in the right hand side are uniformly bounded and therefore the term in the left is uniformly bounded, as well.We summarize this in the following lemma.Lemma 9.
There are constantsC>0,ε0>0 such that, for every ε≤ε0,
(65)∥vxx(ε)∥L∞[D]+∥vx(ε)∥L∞[D]+∥vt(ε)∥L∞[D]+∥v(ε)∥L∞[D]≤C.We now obtain the following (see [7]).Proposition 10.
For any1<p<∞ and t∈[0,T′], v(ε)→v as ε→0 weakly in W1,p(D). Furthermore, v(ε)→v uniformly on D and also vx(ε)→vx uniformly in x∈[0,K] for each t∈[0,T′]. The function v is the unique solution of v.i. Problem 1.Next, we analyze properties of second order derivatives starting with the following result.Lemma 11.
There is a constantC>0 such that, for any 0<ε≤ε0,
(66)∫0T′∫a0b‍(vtx(ε)(t,x))2dxdt<C.Proof.
Setv=v(ε),β=β(ε), and w=vt(ε). Multiply (48) by w to obtain
(67)wwt-κ22wwxx-μwwx+(r+rKβ′(v))w2=0.
Integrating this equation over (a0,b) and recalling that β′(v),t and K are nonnegative we obtain that, for any 0≤t≤T′,
(68)12ddt∫a0b‍w2(t,x)dx-κ22∫a0b‍w(t,x)wxx(t,x)dx-μ∫a0b‍12dw2dx(t,x)dx≤0.
By (47) and (55) we estimate the third term in (68):
(69)|μ∫a0b‍12dw2dx(t,x)dx|=|μ||w2(t,b)-w2(t,a0)|=|μ|w2(t,b)<C12.
For the second term in (68) we see that
(70)-κ22∫a0b‍w(t,x)wxx(t,x)dx=κ22(∫a0b‍wx2(t,x)dx-w(t,b)wx(t,b)+w(t,a0)wx(t,a0)∫a0b‍).
Since w(t,a0)=0 and the function w(t,x) is uniformly bounded in D we see in view of (61) that wx=vtx(ε) is uniformly bounded near the boundary [0,T′]×{b} and w(t,a0)wx(t,a0)=0, while |w(t,b)wx(t,b)|<C2 for some constant C2>0 independent of ε. Thus, we conclude from (68) that
(71)12ddt∫a0b‍w2(t,x)dx+κ22∫a0b‍wx2(t,x)dx≤C3
for some C3>0 independent of ε. Integrating the last equation over [0,T′] we obtain
(72)κ22∫0T′∫a0b‍wx2(t,x)dxdt+12∫a0b‍(w2(T,x)-w2(0,x))dx≤C3.
Since the function w is uniformly bounded it follows that there is C>0 independent of ε such that
(73)∫0T′‍∫a0b‍wx2(t,x)dxdt≤C.We will now deal with theL2 properties of the function vtt(ε)(t,x).Lemma 12.
There is a constantC>0 such that for any 0<ε≤ε0 and every 0<σ≤t≤T′,
(74)∫a0b‍(vtx(ε)(t,x))2dx+∫σt‍∫a0b‍(vtt(ε)(x,t))2dxdτ≤Cσ.Proof.
Setv=v(ε),β=β(ε), and w=vt(ε). Multiplying (48) by the function wt we have
(75)wt2-κ22wxxwt-μwxwt+(r+rKβ′(v))wwt=0
and an integration with respect to x over (a0,b) yields
(76)∫a0b‍wt2dx-κ22∫a0b‍wxxwtdx-μ∫a0b‍wxwtdx+∫a0b‍(r+rKβ′(v))wwtdx=0.
Fix some t∈[0,T]. Since wt(t,a0)=w(t,a0)=0 we see that
(77)κ22∫a0b‍wxx(t,x)wt(t,x)dx=κ22wx(t,b)wt(t,b)-κ24ddt∫a0b‍wx(t,x)2dx.
From (61) it follows that (κ2/2)|wx(t,b)wt(t,b)|≤C1 for some constant C1>0 independent of ε, and so we obtain
(78)κ24ddt∫a0b‍wx2(t,x)dx+∫a0b‍wt2dx+∫a0b‍(r+rKβ′(v))w(t,x)wt(t,x)dx≤μ∫a0b‍wx(t,x)wt(t,x)dx+C1.
Now we deal with the last term in (76). Since β′′(v)≤0 and v,w≥0 we obtain that
(79)∫a0b‍(r+rKβ′(v))wwtdx=12∫a0b‍(r+rKβ′(v))ddtw2dx=12ddt∫a0b‍(r+rKβ′(v))w2(t,x)dx-12∫a0b‍rKβ′′(v)w3dx≥12ddt∫a0b‍(r+rKβ′(v))w2dx.
We plug this inequality into (78) and obtain
(80)12ddt∫a0b‍[κ22wx2(t,x)+(r+rKβ′(v))w2(t,x)]dx+∫a0b‍wt2(t,x)dx≤μ∫a0b‍wx(t,x)wt(t,x)dx+C1.
Integrate the last inequality with respect to τ′ over the interval (τ,t) to obtain
(81)12∫a0b‍[κ22wx2(t,x)+(r+rKβ′(v))w2(t,x)]dx+∫τt‍∫a0b‍wt2(t,x)dxdτ′≤μ∫τt‍∫a0b‍|wx(t,x)||wt(t,x)|dxdτ′+C1(t-τ)+12∫a0b‍[κ22wx2(τ,x)+(r+rKβ′(v))w2(τ,x)]dx.
Next, integrating in τ over the interval (0,σ) for some 0<σ<t and taking into account that (r+rKβ′(v))w2≥0 by the property (2) of β we obtain that
(82)σ2∫a0b‍κ22wx2(t,x)dx+∫0σ‍∫τt‍∫a0b‍wt2(t,x)dxdτ′dτ≤C2+|μ|∫0σ‍∫τt‍∫a0b‍|wx(t,x)||wt(t,x)|dxdτ′dτ+12∫0σ‍∫a0b‍[κ22wx2(τ,x)+(r+rKβ′(v))w2(τ,x)]dxdτ.
Now, by (55), (56), and Lemma 11 together with the Cauchy–Schwarz inequality we estimate the right hand side of (82) by a constant C3>0 independent of ε. Hence,
(83)C3≥σκ24∫a0b‍wx2dx+∫0σ‍∫τt‍∫a0b‍wt2dxdτ′dτ≥σκ24∫a0b‍wx2dx+σ∫σt‍∫a0b‍wt2dxdτ′
and Lemma 12 follows.As a corollary of previous results we obtain the following.Proposition 13.
Letβ<σ<T and a<s(0)<s(σ)<b<ln⁡K. Define Dσ=(0,σ)×(a,b). Then
(84)P(t,x)∈H2(Dσ),
where by definition H2[U] is the set of all the functions in L2[U] with an L2 weak second order derivatives. Also there exists C>0 such that for every 0≤t≤T′,
(85)∫ab‍|∂2P∂x∂t(t,x)|2dx=∥∂2P∂x∂t(t,x)∥L2[a,b]<C.Proof.
From Lemmas12, 11, and 9 we obtain that {v(ε)}ε<ε0 are uniformly bounded in H2[Dσ] and so they have a weak limit v~∈H2[Dσ]. Since v(ε)→v uniformly we must have that v=v~, and so v∈H2[Dσ]. Since v is the solution of (36) we can apply Proposition 7 and using the fact that the constant C in (26) does not depend on t we can obtain in a similar way that for a fixed σ there is a constant C>0 such that, for every 0≤t≤σ,
(86)∥vx,t(t,·)∥L2[a0,b]<C.
From (32) we can deduce the same result for the function P(t,x).Corollary 14.
For each0≤t<T the function vt(t,x) is Hölder continuous with the exponent 1/2.Proof.
For every0<t<T Proposition 13 gives us that vt(t,x)∈H1[a0,b]. Hence, the result is a consequence of the Sobolev inequality.Corollary 15.
For every0≤t<T′ the functions Pt(t,x) and Pxx(t,x) as functions of x are continuous in the closed interval [s(t),b].Proof.
For the functionPt(t,x) the result follows from (32) and the previous corollary. Since P(t,x) is a solution of (29) in the interval {(t,x):s(t)<x<ln⁡K} and since the functions Px(t,x) and P(t,x) are continuous in the interval [s(t),b] we obtain the result for Pxx, as well.Corollary 16.
Letβ<σ<T and a<s(0)<s(σ)<b<lnK. Define E={(t,x): 0<t<β, a-μt<x<b-μt} and u(t,x)=e-rtP(t,x+μt). Then
(87)u(t,x)∈L2[E],
and there exists C>0 such that, for every 0≤t≤β,
(88)∫a-μtb-μt‍|∂2u∂x∂t(t,x)|2dx<C.Proof.
The assertion (87) follows from Proposition 13 and the definition of u(t,x). For (88) note that
(89)∂2u∂x∂t(t,x)=e-rt(-r∂P∂x(t,x+μt)+μ∂2P∂x2(t,x+μt)cccccccccc+∂2P∂x∂t(t,x+μt));
then use (85) and the fact that for (t,x)∈E the functions (∂2P/∂x2)(t,x+μt) and (∂P/∂x)(t,x+μt) are bounded.


## 3.1. Some Previous Results

First, we state the following result from [8] (see also [10]) which we will use later on.Proposition 3.
(i) There exists an increasing functionb:[0,T)→ℝ such that lim⁡t→Tb(t)=K and F(t,x)=K-x for all (t,x) satisfying 0<x≤b(t).
(ii) There exists0<δ* such that for every 0≤δ≤δ* there is a β=β(δ)≥0 so that F(t,K)=δ for t∈[0,β] and for t≥β one has F(t,x)=FA(t,x) for all x≥0.
(iii) Furthermore,(18)Ft(t,x)+12κ2x2Fxx(t,x)+rxFx(t,x)-rF(t,x)=0,
for all
(19)(t,x)∈(0,T)×ℝ+∖({(t,x):x≤b(t)}∪[0,β)×{K}).
In particular,F(t,x) is of class C1,2, that is, continuously differentiable once with respect to t and twice with respect to x, and so, in fact, it is a smooth function there.
(iv) Finally,F(t,x) is convex and strictly decreasing in x and nonincreasing in t.Next, we introduce an operator𝒟 which acts on Borel functions u(t,x) on [0,T]×ℝ by
(20)𝒟u(t,x)=12[u(t+h,x+κh)+u(t+h,x-κh)]-u(t,x).
Clearly, (1/h)𝒟(t,x) can be viewed as a discretization of the differential operator (∂/∂t)+(κ2/2)(∂2/∂x2). We will rely on the following results from [3] concerning the operator 𝒟.Proposition 4.
For each Borel functionu on [0,T]×ℝ there exists a martingale (Mt)0≤t≤T with respect to the filtration 𝒢t,t≥0 such that M0=0 and for every t∈{0,h,2h,…,T},
(21)u(t,Xt(n))=u(0,x)+Mt+∑j=1nt/T‍𝒟u((j-1)h,X(j-1)h(n)).Proposition 5.
Let0≤t≤T-h and x∈ℝ. Assume that u is a C2 function on ([t,t+h]× [x+κh,x-κh]). Then
(22)𝒟u(t,x)=1κ∫0h‍dy∫-κyκy‍dz(z∂2u∂t∂x(t+y2,x+z)hhhhhhhhhhhhhhhh+δ(u)(t+y2,x+z)∂2u∂t∂x),
where
(23)δ(u)(t,x)=ut(t,x)+κ22uxx(t,x).We will need also the following result concerning the free boundarys(t)=ln⁡(b(t)) of the holder exercise region of our game put option which in the case of American options appears as Proposition 1 in [3] and it can be proved for game options in the same way.Proposition 6.
Let0≤t1<t2≤β and let x0=s(0)<x1=s(β)<ln⁡K; then (s(t1)-s(t2))2≤ sup⁡x0≤x≤x1|P(t1,x)-P(t2,x)|.We also observe that it follows from the Berry-Esseen estimate (see [12]) that, for some constant C1>0 independent of j,n≥1, and z∈ℝ,
(24)P{|Xjh(n)-z|≤κh}≤C1j.We will also rely on the following standard bounds on derivatives of solutions of 2nd-order parabolic equations with constant coefficients (see, e.g., [13, 14]).Proposition 7.
LetD=(0,T)×(0,1) and let w(t,x)∈C[D-] be a solution in D of the following parabolic equation:
(25)κ22wxx+μwx-rw=wt.
Suppose that w(0,x)=0 for all 0≤x≤1 and that there exists A>0 such that |w(t,x)|<A for all (x,t)∈D-. Then for every k,n and 0<a<b<1 there exists C=C(k,n,a,b,T,A) such that
(26)|∂k+nw∂kx∂nt(t,x)|<C∀(t,x)∈(0,T)×[a,b].


## 3.2. Price Function and Variational Inequalities

Next, we will show that the price function of the game put option can be represented as a solution of a variational inequality (v.i.) problem which is a generalization of the Stefan problem (see [7], VIII). This will enable us to derive certain regularity properties of this price function which we will use later on. Details of some of the proofs concerning the solutions of the v.i. problem below which are similar to the proofs in the case of the Stefan problem will not be given here. For the corresponding results in the American put option case we refer the reader to [3, 9] and to references there in.LetT′ be such that β<T′<T and set
(27)A=κ22∂2∂x2+μ∂∂x-r,whereμ=r-κ22.
Using the maximum principle, properties of price functions of American and game put options, and the fact that after time β the price functions of the game and American option are the same we obtain that for every x>s(T′) the time derivative Pt(T′,x)=PA,t(T′,x) is strictly negative and we can find a,b satisfying s(T′)<a<b<ln⁡K such that, for some constant c>0,
(28)-Pt(T′,x)>c∀x∈[a,b].
Relying on Proposition 3(iii) we also observe that for all (t,x)∈[0,T′]×(s(t),ln⁡K),
(29)∂P∂t(t,x)+AP(t,x)=0,P(t,x)>K-ex∀(t,x)∈[0,T′]×(s(t),ln⁡K),P(t,x)=K-ex∀t∈[0,T′],∀x≤s(t),Pt≤0.Leta0 be such that a0<s(0)<s(T′)<b. Introduce the domain D=(0,T′)×(a0,b) and for all (t,x) in the closure D- of D define the functions
(30)v(t,x)=P(T′-t,x)-P(T′,x),f(x)=AP(T′,x).
We obtain that
(31)f(x)={-Pt(T′,x),s(T′)<x≤b-rK,a0≤x≤s(T′)
and from the definition of v(t,x) it follows that for any (t,x)∈D-,
(32)Pt(T′-t,x)=-vt(t,x),Ptx(T′-t,x)=-vtx(t,x),Ptt(T′-t,x)=vtt(t,x),Px(T′-t,x)-Px(T′,x)=vx(t,x),Pxx(T′-t,x)-Pxx(T′,x)=vxx(t,x).
Since Px(T′,x) and Pxx(T′,x) are bounded we obtain that the integrability properties of the first- and second-order derivatives of P(t,x) and v(t,x) are the same in D-. Now set
(33)ψ(t)=v(t,b),g(t)=vt(t,b)for0≤t≤T′.
Then by (28) and (32),
(34)ψ(t)=∫0t‍g(τ)dτ=∫0t‍vt(τ,b)dτ≥0for0≤t≤T′.
It follows from (29) and (30)-(31) that on the set v>0,
(35)vt-Av-f=-Pt(T′-t,x)-AP(T′-t,x)+AP(T′,x)-f(x)=0,
and on the set v=0 we obtain
(36)vt-Av-f=rK>0.
Hence we arrive at the following (see [7]).Lemma 8.
The functionv is the unique solution of the following variational inequality problem.
v.i. Problem 1. Find v∈L2[0,T′;H2(a0,b)]∩H1[D] such that(i)
v,vt≥0,(ii)
(vt-Av)(w-v)≥f(w-v) a.s for every w∈L2[D],w≥0,(iii)
v(t,b)=ψ for 0≤t≤T′,x=b,(iv)
v(t,a0)=0 for 0≤t≤T′,x=a0,(v)
v(0,x)=0 for t=0,a0≤x≤b.Proof.
We will prove uniqueness and the fact thatv is a solution to v.i. Problem 1 follows from (30)–(36). Assume that v and v~ are two solutions of v.i. Problem 1. Since v~≥0 (property (i)) we can use the property (ii) of v and replace w by v~. Since both of them are solutions we obtain that
(37)(vt-Av)(v~-v)≥f(v~-v),(v~t-Av~)(v-v~)≥f(v-v~).
Define the parabolic boundary as the boundary of D without the interval {T′}×(a0,b) and let u=v-v~. Note that u is zero on the parabolic boundary and the sum of the two inequalities (37) is
(38)utu-κ22uxxu-μuxu+ru2=(ut-Au)u≤0.
Integrating both sides of (38) on (0,T′)×(a0,b) we obtain four terms on the left side. For the first term we have
(39)∫a0b‍∫0T′‍u(t,x)ut(t,x)dtdx=∫a0b‍12u2(T′,x)dx≥0.
Integration by parts of the second term and the fact that u=0 on the parabolic boundary yield
(40)-κ22∫0T′‍∫a0b‍uxx(t,x)u(t,x)dxdth=κ22∫0T′‍∫a0b‍ux2(t,x)dxdt≥0.
For the third term note that uxu=(1/2)(du2/dx) and that u(t,a0)=u(t,b0)=0 for every t, and so
(41)μ∫0T′‍∫a0b‍ux(t,x)u(t,x)dxdt=0.
The last term satisfies r∫0T′‍∫a0b‍u2(t,x)dxdt≥0 since r>0. We conclude that the left side of (38) cannot be negative and so it must be zero. Since all terms in the left-hand side of (38) are nonnegative and their sum is equal to 0 we obtain that r∫0T′‍∫a0b‍u2(t,x)dxdt=0, and so u=0 almost everywhere (a.e.). Hence, v=v~ a.e., and so there is only one continuous solution.Denote parts of the boundary ofD=(0,T′)×(a0,b) by
(42)Γ1=[0,T′]×{b},Γ2={0}×(a0,b),Γ3=[0,T′]×{a0},Γ=Γ1∪Γ2∪Γ3
and set
(43)L=∂∂t-A=∂∂t-κ22∂2∂x2-μ∂∂x+r.
Thus, Γ is a parabolic boundary of D. For every ε>0 we define following functions.(1)
A smooth functionf(ε)(x)≥f(x) on (a0,b) such that f(ε)(x)=f(x) for s(T′)<a<x≤b and for a0≤x≤a1, where a1 satisfies a0<a1<s(T′) and lim⁡ε→0f(ε)(x)=f(x) for a0≤x≤b.(2)
A smooth functionβ(ε)(v) satisfying
(44)β(ε)(v)=0∀v≥ε,β(ε)(0)=-1,βv(ε)(v)≥0,βvv(ε)(v)≤0.(3)
ψ(ε)(t)=ψ(t)+ε with ψ defined in (33).(4)
A smooth functionη(x) such that 0≤η(x)≤1 and for some a<a2<b,
(45)η(x)=1fora2≤x≤b,η(x)=0fora0≤x≤a.Set Fε(x,v)=f(ε)(x)-rKβ(ε)(v) which is a Lipschitz continuous function and for every constant C there is M0 such that C|F(ε)(x,v)|≤M whenever M≥M0 and |v|≤M. Let ϕ(ε) be a function on Γ satisfying
(46)ϕ(ε)|Γ1=ψ(ε)(t),ϕ(ε)|Γ2=εη(x),ϕ(ε)|Γ3=0,
and, moreover, relying on Chapter 3 in [14] we can choose ϕ(ε) so that(1)
ϕ(ε)∈C-2+δ[D] for some 0<δ<1 (in fact for each δ) and we refer the reader to Chapter 3 in [14] for the definition of C-2+δ[D] and for conditions yielding that a function defined only on the boundary Γ can be extended to a function from C-2+δ[D];(2)
Lϕ(ε)=F(ε)(x,ψ) at the points (0,b) and (0,a0).By the theory of semilinear parabolic equations (see [14]) there exists a function v(ε)∈C-2+γ[D] for some 0<γ<1 such that
(47)Lv(ε)=F(ε)(x,v(ε)),v(ε)|Γ=ϕ(ε).
In particular v(ε),vx(ε),vxx(ε),vt(ε) are continuous on D-.Letw=vt(ε). By differentiating with respect to t (47) and taking into account (33), (47), and the properties of ϕ(ε) we obtain that
(48)wt-κ22wxx-μwx+(r+rKβv(ε)(v(ε)))w=0,wherew(t,b)=g(t)∀0≤t≤T′,w(t,a0)=0∀0≤t≤T′w(0,x)=f(ε)(x)+ε(κ22ηx,x(x)+μηx(x)-rη(x))-rKβv(ε)(v(ε)(0,x))∀a0≤x≤b.
We see that in D the function w is a solution to a parabolic equation and since r+rKβv(ε)≥0 we can use the maximum principle
(49)min⁡(min⁡Γ(w),0)≤w(t,x)≤max⁡(max⁡Γ(w),0)∀(t,x)∈D.
Therefore in order to bound the function w we only need to bound its values on the parabolic boundary. First, we estimate the left hand side of (49). For a≤x≤b we have that v(ε)(0,x)=εη(x)≤ε, and so β(ε)(v(ε))≤0. In view of (28), (31), and the definition of f(ε) above there exists ε0>0 such that, for every 0<ε≤ε0,
(50)w(0,x)≥f(x)+ε(ηxx(x)+μηx(x)-rη(x))=-Pt(T′,x)+ε(κ22ηxx(x)+μηx(x)-rη(x))≥0∀a≤x≤b.
On the interval a0≤x≤a we have η=0 and since v(ε)(0,x)=εη(x)=0 we see that β(ε)(v(ε)(0,x))=-1. Since on this interval f(x)≥-rK we obtain
(51)w(0,x)=f(ε)(x)-rKβ(ε)(0)≥f(x)+rK≥0fora0≤x<a.
Hence, w≥0 on Γ2. We obtain next that
(52)w(t,b)=-∂P∂t(T′-t,b)≥0on⁡Γ1,w(t,b)=0on⁡Γ3.
It follows that min⁡(min⁡Γ(w),0)=0.Next, we estimate the right-hand side of (49). On Γ2 we have that
(53)w(0,x)≤|f(ε)(x)+ε(κ22ηxx(x)+μηx(x)-rη(x))-rKβ(ε)(v(ε)(0,x))κ22|≤sup⁡|f(ε)|+εsup⁡|κ22ηxx+μηx-rη|+rK≤C0,
where C0>0 is a constant independent of ε, and so
(54)0=min⁡⁡(min⁡Γ(w),0)≤w(t,x)≤max⁡(max⁡(-∂P∂t(T′-t,b),C0))=C1.
We conclude that there are some constants ε0 and C1 such that for every 0<ε≤ε0,
(55)0≤vt(ε)(t,x)≤C1.
Since vt(ε)≥0 and v(ε)(0,x)≥0 for a0≤x≤b we deduce that v(ε)(t,x)≥0 and because vt(ε) is uniformly bounded it follows that v(ε) is also uniformly bounded. By the properties of β(ε) we see that
(56)-1≤β(ε)(v(ε))≤0or∥β(ε)(v(ε))∥L∞[D]≤1.
Let D0=(0,T)×(a2,b) be an upper subrectangle of D, where a2 is the same as in the definition of the function η in (4). From the definition we have v(ε)(0,x)=εη(x)=ε in D-0 and since vt(ε) is nonnegative, we obtain that v(ε)(t,x)≥ε, and so β(ε)(v(ε)(t,x))=0.This means that onD0 the function v(ε) satisfies the parabolic equation
(57)Lv(ε)=f(ε).
For w=vt(ε) and 0<ε<ε0 we also have that
(58)Lw(t,x)=0∀(t,x)∈D0,w(0,x)=f(ε)(x)whena2≤x≤b.Next, lety(t,x) be a function on D0 such that
(59)Ly(t,x)=0∀(t,x)∈D0,y(0,x)=f(ε)(x)∀a2≤x≤b,
and all of its first- and second-order derivatives are bounded there. Such a function exists since we can choose a smooth function on the remaining part Γ0∖{0}×[a2,b] of the parabolic boundary Γ0 of D0 which extends f(ε)(x) as a smooth function to the whole D0 and then use Theorem 12 from Chapter 3 in [7]. For each ε<ε0 we define z(t,x)=w(t,x)-y(t,x) in the domain D0, where w(t,x)=vt(ε)(t,x). Then z(0,x)=w(0,x)-y(0,x)=0 for every a2≤x≤b. Fix x0∈(a2,b); then by Proposition 4.5 from Section 4.1 of [7] we obtain that |zt(t,x0)|,|ztt(t,x0)|<C for every 0≤t≤T′, where a constant C>0 is independent of ε. Since we assume that |yt(t,x0)|,|ytt(t,x0)|<C1 for 0≤t≤T′ it follows that |wt(t,x0)|,|wtt(t,x0)|<C1+C (for every ε). Let D1=(x0,b)×(0,T); then by Theorem 6 in Chapter 3 of [14] we obtain that for every ε>0 (and in fact every 0<α<1) v(ε),vt(ε)∈C-2+α[D1] and there is a constant C independent of ε such that
(60)|v(ε)vt(ε)|¯2+α+|vt(ε)|¯2+α<C.
For the definition of norms in (60) see Chapter 3 in [14].In particular, we get(61)∥vx(ε)∥L∞(D1)+∥vtx(ε)∥L∞(D1)+∥vtt(ε)∥L∞(D1)<C.
Considering again the whole region D we have
(62)-∫xb‍vxx(ε)(t,y)dy=2κ2∫xb‍(μvx(ε)(t,y)-vt(ε)(t,y)-rv(ε)(t,y)+β(ε)(v(ε)(t,y))+f(ε)(y))dy.
Hence,
(63)vx(ε)(t,x)=vx(ε)(t,b)+2κ2[μ(v(ε)(t,b)-v(ε)(t,x))+∫xb‍(-vt(ε)(t,y)-rv(ε)(t,y)ccccccccccc+β(ε)(v(ε)(t,y))+f(ε)(y))dy].
Since all terms in the right-hand side are uniformly bounded there is a constant C>0 independent of ε such that ∥vx(ε)∥<C for every 0<ε≤ε0. Now we see that in the equation
(64)vxx(ε)(t,y)=2κ2(2κ2μvx(ε)(t,y)-vt(ε)(t,y)-rv(ε)(t,y)hh+β(ε)(v(ε)(t,y))+f(ε)(y)2κ2)
all terms in the right hand side are uniformly bounded and therefore the term in the left is uniformly bounded, as well.We summarize this in the following lemma.Lemma 9.
There are constantsC>0,ε0>0 such that, for every ε≤ε0,
(65)∥vxx(ε)∥L∞[D]+∥vx(ε)∥L∞[D]+∥vt(ε)∥L∞[D]+∥v(ε)∥L∞[D]≤C.We now obtain the following (see [7]).Proposition 10.
For any1<p<∞ and t∈[0,T′], v(ε)→v as ε→0 weakly in W1,p(D). Furthermore, v(ε)→v uniformly on D and also vx(ε)→vx uniformly in x∈[0,K] for each t∈[0,T′]. The function v is the unique solution of v.i. Problem 1.Next, we analyze properties of second order derivatives starting with the following result.Lemma 11.
There is a constantC>0 such that, for any 0<ε≤ε0,
(66)∫0T′∫a0b‍(vtx(ε)(t,x))2dxdt<C.Proof.
Setv=v(ε),β=β(ε), and w=vt(ε). Multiply (48) by w to obtain
(67)wwt-κ22wwxx-μwwx+(r+rKβ′(v))w2=0.
Integrating this equation over (a0,b) and recalling that β′(v),t and K are nonnegative we obtain that, for any 0≤t≤T′,
(68)12ddt∫a0b‍w2(t,x)dx-κ22∫a0b‍w(t,x)wxx(t,x)dx-μ∫a0b‍12dw2dx(t,x)dx≤0.
By (47) and (55) we estimate the third term in (68):
(69)|μ∫a0b‍12dw2dx(t,x)dx|=|μ||w2(t,b)-w2(t,a0)|=|μ|w2(t,b)<C12.
For the second term in (68) we see that
(70)-κ22∫a0b‍w(t,x)wxx(t,x)dx=κ22(∫a0b‍wx2(t,x)dx-w(t,b)wx(t,b)+w(t,a0)wx(t,a0)∫a0b‍).
Since w(t,a0)=0 and the function w(t,x) is uniformly bounded in D we see in view of (61) that wx=vtx(ε) is uniformly bounded near the boundary [0,T′]×{b} and w(t,a0)wx(t,a0)=0, while |w(t,b)wx(t,b)|<C2 for some constant C2>0 independent of ε. Thus, we conclude from (68) that
(71)12ddt∫a0b‍w2(t,x)dx+κ22∫a0b‍wx2(t,x)dx≤C3
for some C3>0 independent of ε. Integrating the last equation over [0,T′] we obtain
(72)κ22∫0T′∫a0b‍wx2(t,x)dxdt+12∫a0b‍(w2(T,x)-w2(0,x))dx≤C3.
Since the function w is uniformly bounded it follows that there is C>0 independent of ε such that
(73)∫0T′‍∫a0b‍wx2(t,x)dxdt≤C.We will now deal with theL2 properties of the function vtt(ε)(t,x).Lemma 12.
There is a constantC>0 such that for any 0<ε≤ε0 and every 0<σ≤t≤T′,
(74)∫a0b‍(vtx(ε)(t,x))2dx+∫σt‍∫a0b‍(vtt(ε)(x,t))2dxdτ≤Cσ.Proof.
Setv=v(ε),β=β(ε), and w=vt(ε). Multiplying (48) by the function wt we have
(75)wt2-κ22wxxwt-μwxwt+(r+rKβ′(v))wwt=0
and an integration with respect to x over (a0,b) yields
(76)∫a0b‍wt2dx-κ22∫a0b‍wxxwtdx-μ∫a0b‍wxwtdx+∫a0b‍(r+rKβ′(v))wwtdx=0.
Fix some t∈[0,T]. Since wt(t,a0)=w(t,a0)=0 we see that
(77)κ22∫a0b‍wxx(t,x)wt(t,x)dx=κ22wx(t,b)wt(t,b)-κ24ddt∫a0b‍wx(t,x)2dx.
From (61) it follows that (κ2/2)|wx(t,b)wt(t,b)|≤C1 for some constant C1>0 independent of ε, and so we obtain
(78)κ24ddt∫a0b‍wx2(t,x)dx+∫a0b‍wt2dx+∫a0b‍(r+rKβ′(v))w(t,x)wt(t,x)dx≤μ∫a0b‍wx(t,x)wt(t,x)dx+C1.
Now we deal with the last term in (76). Since β′′(v)≤0 and v,w≥0 we obtain that
(79)∫a0b‍(r+rKβ′(v))wwtdx=12∫a0b‍(r+rKβ′(v))ddtw2dx=12ddt∫a0b‍(r+rKβ′(v))w2(t,x)dx-12∫a0b‍rKβ′′(v)w3dx≥12ddt∫a0b‍(r+rKβ′(v))w2dx.
We plug this inequality into (78) and obtain
(80)12ddt∫a0b‍[κ22wx2(t,x)+(r+rKβ′(v))w2(t,x)]dx+∫a0b‍wt2(t,x)dx≤μ∫a0b‍wx(t,x)wt(t,x)dx+C1.
Integrate the last inequality with respect to τ′ over the interval (τ,t) to obtain
(81)12∫a0b‍[κ22wx2(t,x)+(r+rKβ′(v))w2(t,x)]dx+∫τt‍∫a0b‍wt2(t,x)dxdτ′≤μ∫τt‍∫a0b‍|wx(t,x)||wt(t,x)|dxdτ′+C1(t-τ)+12∫a0b‍[κ22wx2(τ,x)+(r+rKβ′(v))w2(τ,x)]dx.
Next, integrating in τ over the interval (0,σ) for some 0<σ<t and taking into account that (r+rKβ′(v))w2≥0 by the property (2) of β we obtain that
(82)σ2∫a0b‍κ22wx2(t,x)dx+∫0σ‍∫τt‍∫a0b‍wt2(t,x)dxdτ′dτ≤C2+|μ|∫0σ‍∫τt‍∫a0b‍|wx(t,x)||wt(t,x)|dxdτ′dτ+12∫0σ‍∫a0b‍[κ22wx2(τ,x)+(r+rKβ′(v))w2(τ,x)]dxdτ.
Now, by (55), (56), and Lemma 11 together with the Cauchy–Schwarz inequality we estimate the right hand side of (82) by a constant C3>0 independent of ε. Hence,
(83)C3≥σκ24∫a0b‍wx2dx+∫0σ‍∫τt‍∫a0b‍wt2dxdτ′dτ≥σκ24∫a0b‍wx2dx+σ∫σt‍∫a0b‍wt2dxdτ′
and Lemma 12 follows.As a corollary of previous results we obtain the following.Proposition 13.
Letβ<σ<T and a<s(0)<s(σ)<b<ln⁡K. Define Dσ=(0,σ)×(a,b). Then
(84)P(t,x)∈H2(Dσ),
where by definition H2[U] is the set of all the functions in L2[U] with an L2 weak second order derivatives. Also there exists C>0 such that for every 0≤t≤T′,
(85)∫ab‍|∂2P∂x∂t(t,x)|2dx=∥∂2P∂x∂t(t,x)∥L2[a,b]<C.Proof.
From Lemmas12, 11, and 9 we obtain that {v(ε)}ε<ε0 are uniformly bounded in H2[Dσ] and so they have a weak limit v~∈H2[Dσ]. Since v(ε)→v uniformly we must have that v=v~, and so v∈H2[Dσ]. Since v is the solution of (36) we can apply Proposition 7 and using the fact that the constant C in (26) does not depend on t we can obtain in a similar way that for a fixed σ there is a constant C>0 such that, for every 0≤t≤σ,
(86)∥vx,t(t,·)∥L2[a0,b]<C.
From (32) we can deduce the same result for the function P(t,x).Corollary 14.
For each0≤t<T the function vt(t,x) is Hölder continuous with the exponent 1/2.Proof.
For every0<t<T Proposition 13 gives us that vt(t,x)∈H1[a0,b]. Hence, the result is a consequence of the Sobolev inequality.Corollary 15.
For every0≤t<T′ the functions Pt(t,x) and Pxx(t,x) as functions of x are continuous in the closed interval [s(t),b].Proof.
For the functionPt(t,x) the result follows from (32) and the previous corollary. Since P(t,x) is a solution of (29) in the interval {(t,x):s(t)<x<ln⁡K} and since the functions Px(t,x) and P(t,x) are continuous in the interval [s(t),b] we obtain the result for Pxx, as well.Corollary 16.
Letβ<σ<T and a<s(0)<s(σ)<b<lnK. Define E={(t,x): 0<t<β, a-μt<x<b-μt} and u(t,x)=e-rtP(t,x+μt). Then
(87)u(t,x)∈L2[E],
and there exists C>0 such that, for every 0≤t≤β,
(88)∫a-μtb-μt‍|∂2u∂x∂t(t,x)|2dx<C.Proof.
The assertion (87) follows from Proposition 13 and the definition of u(t,x). For (88) note that
(89)∂2u∂x∂t(t,x)=e-rt(-r∂P∂x(t,x+μt)+μ∂2P∂x2(t,x+μt)cccccccccc+∂2P∂x∂t(t,x+μt));
then use (85) and the fact that for (t,x)∈E the functions (∂2P/∂x2)(t,x+μt) and (∂P/∂x)(t,x+μt) are bounded.


## 4. Price Function Near the Writer’s Exercise Boundary

### 4.1. Regularity Properties of Price Function

LetF(t,x) be the price function of the put game option (see Section 2). We begin this section by showing that near the writer’s exercise region Γ1={(t,K):0≤t≤β} the function ∂F/∂t is continuous. Let
(90)Yt[s,x]=(Yt1,[s,x],Yt2,[s,x])=(s+t,Stx),
which is a non homogeneous in time Markov process in ℝ+×ℝ, where Stx=xeμt+κBt and μ=r-(κ2/2). Let
(91)LY=∂∂t+κ2x22∂2∂x2+rx∂∂x-r,
which is the infinitesimal generator of Yt when considered on the space of all C2 functions. This is a parabolic operator with bounded smooth coefficients in the domain
(92)D=(0,β)×(k,K),
where k>0. Let P[s,x] and E[s,x] be the probability and the corresponding expectation for the Markov process Y starting at the point [s,x]. We will first show that for any t0∈[0,β),
(93)lim⁡(t,x)→(t0,K)P[t,x][Yτ∈Γ1]=1,
where τ=τ(Γ) and for any closed set Q⊂ℝ+×ℝ we set τ(Q) to be the arrival time at the set Q for a Markov process under consideration which is Yt here. Indeed, choosing an appropriate nonnegative function ϕ≤1 on the boundary Γ and relying on Chapter 3 in [14] we can choose u(t,x)∈C1,2(D) which solves the equation LYu=0 in D and equals 1 on the boundary part Γ1 for 0≤t≤t1<β while decaying smoothly to 0 when t grows to β. Then
(94)u(t,x)=E[t,x]ϕ(Yτ)≤P[t,x]{Yτ∈Γ1},
and so
(95)1≥liminf⁡(t,x)→(t0,K)P[t,x][Yτ∈Γ1]≥lim⁡(t,x)→(t0,K)u(t,x)=u(t0,K)=1.Next letf(x)=(K-x)+ and g(x)=f(x)+δ. Recall that the price of a put game option with an expiration time T and a constant penalty δ can be written in the form
(96)F(t,x)=sup⁡0≤τ≤T~inf⁡0≤σ≤T~J[t,x](f,g,σ,τ),
where T~=inf⁡{t:Yt1=T} and for any bounded Borel functions f^ and g^ we write
(97)J[t,x](f^,g^,σ,τ)=E[t,x][e-rσ∧τ(g^(Yσ2)𝕀{σ<τ}+f^(Yτ2)𝕀{τ≤σ})].
Set
(98)fs(x)=F(s,x)whenβ<s<T,Fs(t,x)=sup⁡0≤τ≤s~⁡inf⁡0≤σ≤s~⁡J[t,x](fs,g,σ,τ),
where s~=inf⁡{u:Yu(1)=s}. Let 〈σ*,τ*〉 and 〈σs*,τs*〉 be the two saddle points (see [1]) corresponding to the optimal stopping games with values F(t,x) and Fs(t,x), respectively, and so
(99)σ*=inf⁡{0≤t≤T~:F(Yt)=g(Yt2)},τ*=inf⁡{0≤t≤T~:F(Yt)=f(Yt2)},σs*=inf⁡{0≤t≤s~:Fs(Yt)=g(Yt2)},τs*=inf⁡{0≤t≤s~:Fs(Yt)=fs(Yt2)}.
Then
(100)F(t,x)=J[t,x](f,g,σ*,τ*),Fs(t,x)=J[t,x]s(fs,g,σs*,τs*).Lemma 17.
For all0≤t≤s<T and x>0, Fs(t,x)=F(t,x).Proof.
We have(101)Fs(t,x)=J[t,x](fs,g,σs*,τs*)≤J[t,x](fs,g,σ*,τs*)=E[t,x][e-rσ*∧τs*(fs(Yτs*2)𝕀{τs*≤σ*}hhhhhhhhhhhh+F(Yσ*)𝕀{σ*<τs*})e-rσ*∧τs*]≤E[t,x][e-rσ*∧τs*(F(Yτs*)𝕀{τs*≤σ*}hhhhhhhhhhhh+F(Yσ*)𝕀{σ*<τs*})e-rσ*∧τs*]=E[t,x][e-rσ*∧τs*F(Yσ*∧τs*)]≤F(t,x).
Indeed, the first inequality above follows by the saddle point property. The second inequality holds true since F is nonincreasing in the time variable, τs*≤s~=s-t for Y[t,x] and Y1,[t,x](τs*)≤s. The third inequality is satisfied since the process e-rYσ*∧u𝕀F(Yσ*∧u) is a continuous supermartingale in u with respect to P[t,x] (see [15]). For the other direction we have
(102)F(t,x)≤E[t,x][e-rs~∧σs*∧τ*F(Ys~∧σs*∧τ*)]=E[t,x][e-rs~∧σs*∧τ*(f(Yτ*2)𝕀τ*≤s~∧σs*+F(Ys~)𝕀s~<τ*∧σs*hhhhhhhhhhhhh+g(Yσs*2)𝕀σs*<s~∧τ*)e-rs~∧σs*∧τ*]≤E[t,x][e-rs~∧σs*∧τ*(fs(Yτ*∧s2)𝕀τ*∧s~≤σs*hhhhhhhhhhhhh+g(Yσs*2)𝕀σs*<s~∧τ*)e-rs~∧σs*∧τ*]=J[t,x](fs,g,σs*,τ*∧s~)≤J[t,x](fs,g,σs*,τs*)=Fs(t,x),
where we use the submartingale property of e-rYτ*∧u𝕀F(Yτ*∧u) in u.Now for any bounded Borel functionsf^ and g^ set
(103)Is(t,x,f^,g^)=sup⁡0≤τ≤s~inf⁡0≤σ≤s~J[t,x](f^,g^,σ,τ).
From the time homogeneity of the process Yt2=St we obtain that
(104)Is+h(t+h,x,f^,g)=Is(t,x,f^,g^).Proposition 18.
There is a constantC>0 such that for any (t,x)∈(0,β)×(k,K),
(105)0≤-∂F∂t(t,x)≤CP[t,x][τs*<σs+h*].Proof.
The left hand side of the above inequality follows from (iii) and (iv) of Proposition3. For the right hand side, let h>0 be such that β+h<T-h and t+h<β and let β<s<T-h. By [3] the price function of an American put option has a bounded derivative with respect to t in [0,s+h]×ℝ; that is, C=sup⁡(t,x)∈[0,s+h]×ℝ+|(∂FA(t,x))/∂t|<∞. This together with Proposition 3(ii) yields
(106)sup⁡β<s<T,x≥0|∂F(s,x)∂s|≤C.
Next, by Lemma 17 and the saddle point property,
(107)F(t,x)=Fs(t,x)=J[t,x](fs,g,σs*,τs*)≤J[t,x](fs,g,σs+h*,τs*).
By Lemma 17, (104), and the saddle point property,
(108)F(t+h,x)=Fs+h(t+h,x)=Is+h(t+h,x,fs+h,g)=Is(t,x,fs+h,g)=J[t,x](fs+h,g,σs+h*,τs+h*)≤J[t,x](fs+h,g,σs+h*,τs*).
Now, (98), (106), (107), and (108) yield that
(109)0≤1h(F(t,x)-F(t+h,x))≤1hE[t,x][e-rσs+h*∧τs*(fs(Yτs*2)-fs+h(Yτs*2))𝕀{τs*≤σs+h*}]≤1hE[e-rσs+h*∧τs*Ch𝕀{τs*≤σs+h*}]≤CP[t,x][τs*≤σs+h*].
Passing to the limit as h→0 we obtain the result.Corollary 19.
For every0≤t0<β, lim⁡(t,x)→(t0,K)(∂F/∂t)(t,x)=0, and so lim⁡(t,x)→(t0,ln⁡K)(∂P/∂t)(t,x)=0.Proof.
In view of Proposition18 we only have to show that for every 0≤t0<β,
(110)lim⁡(t,x)→(t0,K)P[t,x][τs*≤σs+h*]=0.
Let D be as in (92), Γ2={(β,x):k≤x≤K}, and Γ3={(t,k):0≤t≤β}. It follows from the definition of τs* and σs+h* that for every (x,t)∈D,
(111){τ(Γ1)<τ(Γ2∩Γ3)}⊂{σs+h*<τs*}withrespecttoP[t,x].
From (93) we obtain
(112)lim⁡(t,x)→(t0,K)P[t,x][σs+h*<τs*]=1,
and the result follows.Next, we deal with functionsP(t,x)=F(t,ex), and so it is natural to consider the domain D0=(0,β)×(k,ln⁡K) for some positive k<ln⁡K (which is, essentially, the same domain after the space coordinate change) and let
(113)c=PA,t(β,ln⁡K)=lim⁡x→log⁡KPA,t(β,x).
Let v(t,x) be a function solving the equation ((∂/∂t)+A)v(t,x)=0 with A defined by (27) and satisfying the boundary conditions
(114)v(t,ln⁡K)=c,v(t,k)=Pt(t,k)for0≤t≤β,v(β,x)=Pt(β,x)fork<x<ln⁡K.
Since these boundary conditions are continuous then (see [14]) they are satisfied by a unique solution in C1,2[D] of the above equation. Let w(t,x) be a function on D-0 such that
(115)Pt(t,x)=w(t,x)+v(t,x)∀(t,x)∈D-0∖(β,ln⁡K).
Thus, w(t,x)∈C1,2[D′] and it satisfies the same parabolic equation in D0 as (∂P/∂t)(t,x) and v(t,x). Its boundary values are
(116)w(t,ln⁡K)=-c,w(t,k)=0for0≤t≤β,w(β,x)=0fork<x<ln⁡K.
From the continuity of v(t,x) on D-0 we see that it is bounded there and since ∂P/∂t is also bounded there we obtain the same result for the function w as for v. Hence,
(117)w(t,x),v(t,x)∈C1,2[D0]∩L∞[D0].


### 4.2. Integrability ofwt(t,x) and wx(t,x)

Now we will analyze the functionw(t,x). Let Zt[u,x]=(u+t,Xtx) be the diffusion process in the plane whose infinitesimal generator is equal to L1=(∂/∂t)+A on the space of C2 functions. For each ε>0 define Dε=(0,β-ε)×(k+ε,ln⁡K-ε). Let Γε be the parabolic boundary of Dε. For every ε>0 which is sufficiently small we can find a smooth function w-(t,x) with compact support on the plane such that in D-ε it is equal to w(t,x). By the Dynkin formula we obtain that for every (u,x)∈Dε,
(118)E[u,x][w-(Zτ(Γε))]=w-(u,x)+E[u,x][∫0τ(Γε)‍L1w-(Zs)ds],
where τ(Q) denotes the arrival time to Q by the process Zt[u,x]. Note that since w(t,x)=w-(t,x) for (t,x)∈D-ε we can replace w- by w in the above formula and since Zs[u,x]∈D0 for s≤τ we obtain that L1w-(Zs)=0. It follows that for every ε>0,
(119)w(u,x)=E[u,x][w(Zτ(Γε))].
Now fix (u,x)∈D0 and a continuous path ω0. Let ℰ={Zτ(Γ1/n)[u,x](ω0)}n0<n⊂D-0, where n0 is such that (u,x)∈D1/n0. The sequence of times {τ(Γ1/n)(ω)}n>n0 is nondecreasing with respect to n and so it has a limit ρ≤T. Let γ be an accumulation point in ℰ; that is, lim⁡k→∞Zτ(Γ1/nk)[u,x](ω0)=γ for some subsequence nk. Define d(y,Γ0)=inf⁡{|y-x|:x∈Γ0} and note that this function is continuous on D-0 and it is 0 if and only if y∈Γ0. Since d(Yτ(Γ1/nk)[u,x](ω0),Γ0)≤1/nk for each k we conclude that γ∈Γ0 and since lim⁡k→∞τ(Γ1/nk)=ρ it follows that Zρ[u,x](ω0)=lim⁡k→∞Zτ(Γ1/nk)[u,x](ω0)=γ. Hence, τ(Γ0)(ω0)=ρ. By the definition w(t,x) is continuous except at the point (β,ln⁡K), but because P[u,x][Zτ(Γ0)=(β,ln⁡K)]=0 for every (u,x)∈D0 we can ignore paths that reach the point (β,ln⁡K), and so
(120)lim⁡ε→0w(Zτ(Γε)[u,x])=w(Zτ(Γ0)[u,x])P[u,x]a.s.Corollary 20.
For every(t,x)∈D0,
(121)w(t,x)=E[t,x][w(Zτ(Γ0))]=-cE[t,x][𝕀{τ(Γ01)<τ(Γ02∪Γ03)}]=-cP[t,x][τ(Γ01)<τ(Γ02∪Γ03)],
where Γ01={(t,ln⁡K):0≤t≤β}, Γ02={(β,x):k≤x≤ln⁡K}, and Γ03={(t,k):0≤t≤β}.Proof.
From (117) we know that the function w(t,x) is bounded and so we can use the Lebesgue bounded convergence theorem and from the boundary conditions on w(t,x) it follows that
(122)w(t,x)=lim⁡ε→0E[t,x][w(Zτ(Γε))]=E[t,x][lim⁡ε→0w(Zτ(Γε))]=E[t,x][w(Zτ(Γ0))]
which gives the first equality of the corollary, while the second equality follows from (116) and the third equality is obvious.Let(t,x),(t′,x)∈D0 and assume that t≤t′. Then it is not difficult to understand that
(123)P[t,x][τ(Γ01)<τ(Γ02∪Γ03)]hhhh≥P[t′,x][τ(Γ01)<τ(Γ02∪Γ03)],
and so w(t,x) is nonincreasing in t for every x which implies that
(124)∂w∂t(t,x)≥0∀(t,x)∈D0.
It is also easy to see that for 0≤t<T and 0≤x≤x′≤ln⁡K,
(125)P[t,x][τ(Γ01)<τ(Γ02∪Γ03)]hhhh≤P[t,x′][τ(Γ01)<τ(Γ02∪Γ03)],
and so
(126)∂w∂x≤0∀(t,x)∈D0.Lemma 21.
The functionswt and wx are in L1[D0].Proof.
We will use (124) in order to prove the result for wt(t,x). The case of wx(t,x) can be proven similarly by using (126). Using (116), (124), and the continuity of w(0,x) on {0}×[k,ln⁡K] we obtain that
(127)∫D0‍|∂w∂t|dtdx=∫kln⁡K‍∫0β‍∂w∂tdtdx=lim⁡ε→0∫kln⁡K-ε‍∫0β‍∂w∂tdtdx=lim⁡ε→0∫kln⁡K-ε‍(w(β,x)-w(0,x))dx=-lim⁡ε→0∫kln⁡K-ε‍w(0,x)dx=-∫kln⁡K‍w(0,x)dx<∞.
Using (126) in place of (124) the proof of integrability of wx is similar.


### 4.3. Integrability ofvt(t,x) and vx(t,x)

We continue this section by analyzing the functionv(t,x) solving the equation L1v=0 with the boundary conditions given by (114). Let C1,2[D-0] be the set of all functions which have one derivative in t and two derivatives in x both uniformly continuous in D0.Lemma 22.
There exists a functionz(t,x)∈C1,2[D-0] such that
(128)z(t,x)=v(t,x)∀(t,x)∈Γ0.Proof.
Recall thatPA,t(T,x)=Pt(T,k) for k≤x<ln⁡K and note that the functions PA,t(T,x),PA,t(t,x), and Pt(t,k) as function of (t,x) belong to the space C1,2[D-0]. Set
(129)z~(t,x)=ln⁡K-xln⁡K-k(Pt(t,k)+PA,t(T,x)-PA,t(T,k))+x-kln⁡K-kPA,t(t,x).
Then z~(t,x)∈C1,2[D-] since it is a linear combination of functions from this space. We also have
(130)z~(t,k)=ln⁡K-kln⁡K-k(Pt(t,k)+PA,t(T,k)-PA,t(T,k))h=Pt(t,x)∀0≤t≤βz~(t,ln⁡K)=PA,t(t,ln⁡K)when0≤t≤β,∀k≤x≤ln⁡K,z~(T,x)=ln⁡K-xln⁡K-k(Pt(T,k)+PA,t(T,x)-PA,t(T,k))hh+x-kln⁡K-kPA,t(T,x)=PA,t(T,x).
Thus, we obtain
(131)z(t,x)=ln⁡K-xln⁡K-kz~(t,x)+x-kln⁡K-kz~(T,x)∈C1,2[D-].
Since
(132)z(t,k)=z~(t,k)=Pt(t,x),z(t,ln⁡K)=z~(T,ln⁡K)=PA,t(T,ln⁡K)=c,z(T,x)=z~(T,x)=PA,t(T,x),
it follows that
(133)z(t,x)=v(t,x)∀(t,x)∈Γ.Next, definef(t,x)=-Lz(t,x). From Lemma 22 we obtain that f(x,t) is bounded in D0 and so it belongs to Lp[D0] for every 1≤p≤∞. Set v~(t,x)=v(t,x)-z(x,t) and observe that
(134)Lv~(t,x)=f(t,x),v~(t,x)=0∀(t,x)∈Γ0.
We conclude that the function v~(t,x) is the unique solution of the following problem (see [16]).Theorem 23.
Let1≤p<∞, then for any f(t,x)∈Lp[D0] there exists a unique function v~ such that(i)
v~∈Lp[0,T;W2,p(0,1)]∩Lp[0,T;W01,p(0,1)],(ii)
∂v~/∂t∈Lp[D0],(iii)
Lv~(t,x)=f(t,x) for every (t,x)∈D0,(iv)
v~|Γ0=0.From assertions (i) and (ii) of Theorem23 we obtain that the functions v~x(t,x) and v~t(t,x) are both in Lp[D] for every 0≤p<∞ and since z(t,x)∈C1,2[D-0] we obtain the following.Corollary 24.
For every1≤p<∞ the functions vt(t,x) and vx(t,x) belong to the space Lp[D0].We can now summarize most of the results of this section as follows.Proposition 25.
Lets(β)<k<ln⁡K<k′ and define
(135)D0=(0,β)×(k,ln⁡K),D0′=(0,β)×(ln⁡K,k′).
Then the function Pt(t,x) is continuous at every point in the domain D-0∖{(β,ln⁡K)}, and there exist two functions w(t,x) and v(t,x) on D0 such that
(136)Pt(t,x)=w(t,x)+v(t,x)forevery(t,x)∈D-0∖{(β,ln⁡K)},(137)w(t,x),v(t,x)∈C1,2(D0)∪L∞[D0],
and both functions are solutions of the parabolic equation L1u=0. Furthermore, w(t,x) is continuous in D0 and it satisfies
(138)w(t,ln⁡K)=PA,t(β,ln⁡K),w(t,b)=0when0≤t≤β,w(β,x)=0whenk<x<ln⁡K,(139)wt(t,x),wx(t,x)∈L1[D].
Finally, v(t,x)∈C(D-) and for every 1≤p<∞,
(140)vt(t,x),vx(t,x)∈Lp[D].
The same decomposition of Pt(t,x) with the same properties holds true in the domain D0′.Proof.
Taking the same functionsv and w as in (115) we see that (137) is actually the same as (117) and the fact that both v and w are solution of L1u=0 is clear from their definitions. Next we see that (138) is the same as (116), that (139) is the same as Lemma 21, and that (140) is, in fact, Corollary 24. Observe that we did not use in this section the fact that k<ln⁡K, so all the proofs are also applicable to the case k′>ln⁡K and the domain D0′.From (136), (139), and (140), and estimating Pxx via other derivatives in view of (29) we obtain the following.Corollary 26.
LetD~={(t,x):0<t<β,k-μt<x<ln⁡K-μt} and
(141)u(t,x)=e-rtP(t,x+μt).
Then
(142)∂2u∂t2∈L1[D~].


### 4.4. Price Function When Initial Stock Price Is Large

LetF(t,x),P(t,x), and u(t,x) be as above. Recall that in the domain (0,T)×(ln⁡K,∞) the function P(t,x) satisfies the equation L1P=0 and it is continuous in the closure of [0,T]×[ln⁡K,∞) and P(T,x)=(K-ex)+=0 for x>ln⁡K. Define
(143)v(t,x)=u(T-t,κ2x+ln⁡K+|μ|T),
where u is given by (141) and set G=(0,T)×(0,∞). It follows from Proposition 25 that(1)
v(t,x)∈C1,2[G]∪C[G-],(2)
vxx(t,x)=vt(t,x) for every (t,x)∈G,(3)
v(t,0)=u(T-t,ln⁡K+|μ|T) is continuous,(4)
v(0,x)=0 for every x>0,(5)
v(t,x) is bounded (since P(t,x) is).Since a bounded solution of the heat equation inG is unique (see [13]) then for every (t,x)∈G,
(144)v(t,x)=-2∫0t‍∂K∂x(t-τ,x)v(τ,0)dτ,whereK(t,x)=14πte-(x2/4t),
and so
(145)v(t,x)=14π∫0t‍xe-(x2/4(t-τ))v(τ,0)dτ(t-τ)3/2.
Differentiating v we obtain polynomials Qk,n(s,x) such that for all k,n∈ℕ,
(146)∂k+nv∂nt∂kx(t,x)=∫0t‍Qn,k((t-τ)-1/2,x)e-x2/4(t-τ)v(τ,0)dτ.
If N is large enough and c>0 then ((t-τ)N/xN)Qk,n((t-τ)-1/2,x) is a polynomial in (t-τ)1/2 and 1/x and it is bounded on (0,T)×(c,∞). Since supy≥0yNe-y<∞ for any N∈ℕ we can set y=x2/4(t-τ) deriving that, for any N∈ℕ and (t,x)∈(0,T)×(c,∞),
(147)∂k+nv∂nt∂kx(t,x)=∫0t‍4N(t-τ)Nx2NQn,k((t-τ)-1/2,x)yNe-yv(τ,0)dτ≤(4x)N∫0t‍((t-τ)NxNQn,k(x,(t-τ)-1/2))cccccccccccccccc×yNe-yv(τ,0)dτ≤CxN.
For some C=C(N)>0. Hence, the following results hold true.Corollary 27.
For anyk,n positive integers k,n, and c>0,
(148)∂k+nv(t,x)∂kt∂nx∈L2[(0,T)×(c,∞)].Corollary 28.
Let(2/κ)(ln⁡K+|μ|T)<k′ and G~={(t,x):0<t<β,k′-μ<x<∞}. Then
(149)∂2u∂t2(t,x)∈L2[G~].


## 4.1. Regularity Properties of Price Function

LetF(t,x) be the price function of the put game option (see Section 2). We begin this section by showing that near the writer’s exercise region Γ1={(t,K):0≤t≤β} the function ∂F/∂t is continuous. Let
(90)Yt[s,x]=(Yt1,[s,x],Yt2,[s,x])=(s+t,Stx),
which is a non homogeneous in time Markov process in ℝ+×ℝ, where Stx=xeμt+κBt and μ=r-(κ2/2). Let
(91)LY=∂∂t+κ2x22∂2∂x2+rx∂∂x-r,
which is the infinitesimal generator of Yt when considered on the space of all C2 functions. This is a parabolic operator with bounded smooth coefficients in the domain
(92)D=(0,β)×(k,K),
where k>0. Let P[s,x] and E[s,x] be the probability and the corresponding expectation for the Markov process Y starting at the point [s,x]. We will first show that for any t0∈[0,β),
(93)lim⁡(t,x)→(t0,K)P[t,x][Yτ∈Γ1]=1,
where τ=τ(Γ) and for any closed set Q⊂ℝ+×ℝ we set τ(Q) to be the arrival time at the set Q for a Markov process under consideration which is Yt here. Indeed, choosing an appropriate nonnegative function ϕ≤1 on the boundary Γ and relying on Chapter 3 in [14] we can choose u(t,x)∈C1,2(D) which solves the equation LYu=0 in D and equals 1 on the boundary part Γ1 for 0≤t≤t1<β while decaying smoothly to 0 when t grows to β. Then
(94)u(t,x)=E[t,x]ϕ(Yτ)≤P[t,x]{Yτ∈Γ1},
and so
(95)1≥liminf⁡(t,x)→(t0,K)P[t,x][Yτ∈Γ1]≥lim⁡(t,x)→(t0,K)u(t,x)=u(t0,K)=1.Next letf(x)=(K-x)+ and g(x)=f(x)+δ. Recall that the price of a put game option with an expiration time T and a constant penalty δ can be written in the form
(96)F(t,x)=sup⁡0≤τ≤T~inf⁡0≤σ≤T~J[t,x](f,g,σ,τ),
where T~=inf⁡{t:Yt1=T} and for any bounded Borel functions f^ and g^ we write
(97)J[t,x](f^,g^,σ,τ)=E[t,x][e-rσ∧τ(g^(Yσ2)𝕀{σ<τ}+f^(Yτ2)𝕀{τ≤σ})].
Set
(98)fs(x)=F(s,x)whenβ<s<T,Fs(t,x)=sup⁡0≤τ≤s~⁡inf⁡0≤σ≤s~⁡J[t,x](fs,g,σ,τ),
where s~=inf⁡{u:Yu(1)=s}. Let 〈σ*,τ*〉 and 〈σs*,τs*〉 be the two saddle points (see [1]) corresponding to the optimal stopping games with values F(t,x) and Fs(t,x), respectively, and so
(99)σ*=inf⁡{0≤t≤T~:F(Yt)=g(Yt2)},τ*=inf⁡{0≤t≤T~:F(Yt)=f(Yt2)},σs*=inf⁡{0≤t≤s~:Fs(Yt)=g(Yt2)},τs*=inf⁡{0≤t≤s~:Fs(Yt)=fs(Yt2)}.
Then
(100)F(t,x)=J[t,x](f,g,σ*,τ*),Fs(t,x)=J[t,x]s(fs,g,σs*,τs*).Lemma 17.
For all0≤t≤s<T and x>0, Fs(t,x)=F(t,x).Proof.
We have(101)Fs(t,x)=J[t,x](fs,g,σs*,τs*)≤J[t,x](fs,g,σ*,τs*)=E[t,x][e-rσ*∧τs*(fs(Yτs*2)𝕀{τs*≤σ*}hhhhhhhhhhhh+F(Yσ*)𝕀{σ*<τs*})e-rσ*∧τs*]≤E[t,x][e-rσ*∧τs*(F(Yτs*)𝕀{τs*≤σ*}hhhhhhhhhhhh+F(Yσ*)𝕀{σ*<τs*})e-rσ*∧τs*]=E[t,x][e-rσ*∧τs*F(Yσ*∧τs*)]≤F(t,x).
Indeed, the first inequality above follows by the saddle point property. The second inequality holds true since F is nonincreasing in the time variable, τs*≤s~=s-t for Y[t,x] and Y1,[t,x](τs*)≤s. The third inequality is satisfied since the process e-rYσ*∧u𝕀F(Yσ*∧u) is a continuous supermartingale in u with respect to P[t,x] (see [15]). For the other direction we have
(102)F(t,x)≤E[t,x][e-rs~∧σs*∧τ*F(Ys~∧σs*∧τ*)]=E[t,x][e-rs~∧σs*∧τ*(f(Yτ*2)𝕀τ*≤s~∧σs*+F(Ys~)𝕀s~<τ*∧σs*hhhhhhhhhhhhh+g(Yσs*2)𝕀σs*<s~∧τ*)e-rs~∧σs*∧τ*]≤E[t,x][e-rs~∧σs*∧τ*(fs(Yτ*∧s2)𝕀τ*∧s~≤σs*hhhhhhhhhhhhh+g(Yσs*2)𝕀σs*<s~∧τ*)e-rs~∧σs*∧τ*]=J[t,x](fs,g,σs*,τ*∧s~)≤J[t,x](fs,g,σs*,τs*)=Fs(t,x),
where we use the submartingale property of e-rYτ*∧u𝕀F(Yτ*∧u) in u.Now for any bounded Borel functionsf^ and g^ set
(103)Is(t,x,f^,g^)=sup⁡0≤τ≤s~inf⁡0≤σ≤s~J[t,x](f^,g^,σ,τ).
From the time homogeneity of the process Yt2=St we obtain that
(104)Is+h(t+h,x,f^,g)=Is(t,x,f^,g^).Proposition 18.
There is a constantC>0 such that for any (t,x)∈(0,β)×(k,K),
(105)0≤-∂F∂t(t,x)≤CP[t,x][τs*<σs+h*].Proof.
The left hand side of the above inequality follows from (iii) and (iv) of Proposition3. For the right hand side, let h>0 be such that β+h<T-h and t+h<β and let β<s<T-h. By [3] the price function of an American put option has a bounded derivative with respect to t in [0,s+h]×ℝ; that is, C=sup⁡(t,x)∈[0,s+h]×ℝ+|(∂FA(t,x))/∂t|<∞. This together with Proposition 3(ii) yields
(106)sup⁡β<s<T,x≥0|∂F(s,x)∂s|≤C.
Next, by Lemma 17 and the saddle point property,
(107)F(t,x)=Fs(t,x)=J[t,x](fs,g,σs*,τs*)≤J[t,x](fs,g,σs+h*,τs*).
By Lemma 17, (104), and the saddle point property,
(108)F(t+h,x)=Fs+h(t+h,x)=Is+h(t+h,x,fs+h,g)=Is(t,x,fs+h,g)=J[t,x](fs+h,g,σs+h*,τs+h*)≤J[t,x](fs+h,g,σs+h*,τs*).
Now, (98), (106), (107), and (108) yield that
(109)0≤1h(F(t,x)-F(t+h,x))≤1hE[t,x][e-rσs+h*∧τs*(fs(Yτs*2)-fs+h(Yτs*2))𝕀{τs*≤σs+h*}]≤1hE[e-rσs+h*∧τs*Ch𝕀{τs*≤σs+h*}]≤CP[t,x][τs*≤σs+h*].
Passing to the limit as h→0 we obtain the result.Corollary 19.
For every0≤t0<β, lim⁡(t,x)→(t0,K)(∂F/∂t)(t,x)=0, and so lim⁡(t,x)→(t0,ln⁡K)(∂P/∂t)(t,x)=0.Proof.
In view of Proposition18 we only have to show that for every 0≤t0<β,
(110)lim⁡(t,x)→(t0,K)P[t,x][τs*≤σs+h*]=0.
Let D be as in (92), Γ2={(β,x):k≤x≤K}, and Γ3={(t,k):0≤t≤β}. It follows from the definition of τs* and σs+h* that for every (x,t)∈D,
(111){τ(Γ1)<τ(Γ2∩Γ3)}⊂{σs+h*<τs*}withrespecttoP[t,x].
From (93) we obtain
(112)lim⁡(t,x)→(t0,K)P[t,x][σs+h*<τs*]=1,
and the result follows.Next, we deal with functionsP(t,x)=F(t,ex), and so it is natural to consider the domain D0=(0,β)×(k,ln⁡K) for some positive k<ln⁡K (which is, essentially, the same domain after the space coordinate change) and let
(113)c=PA,t(β,ln⁡K)=lim⁡x→log⁡KPA,t(β,x).
Let v(t,x) be a function solving the equation ((∂/∂t)+A)v(t,x)=0 with A defined by (27) and satisfying the boundary conditions
(114)v(t,ln⁡K)=c,v(t,k)=Pt(t,k)for0≤t≤β,v(β,x)=Pt(β,x)fork<x<ln⁡K.
Since these boundary conditions are continuous then (see [14]) they are satisfied by a unique solution in C1,2[D] of the above equation. Let w(t,x) be a function on D-0 such that
(115)Pt(t,x)=w(t,x)+v(t,x)∀(t,x)∈D-0∖(β,ln⁡K).
Thus, w(t,x)∈C1,2[D′] and it satisfies the same parabolic equation in D0 as (∂P/∂t)(t,x) and v(t,x). Its boundary values are
(116)w(t,ln⁡K)=-c,w(t,k)=0for0≤t≤β,w(β,x)=0fork<x<ln⁡K.
From the continuity of v(t,x) on D-0 we see that it is bounded there and since ∂P/∂t is also bounded there we obtain the same result for the function w as for v. Hence,
(117)w(t,x),v(t,x)∈C1,2[D0]∩L∞[D0].


## 4.2. Integrability ofwt(t,x) and wx(t,x)

Now we will analyze the functionw(t,x). Let Zt[u,x]=(u+t,Xtx) be the diffusion process in the plane whose infinitesimal generator is equal to L1=(∂/∂t)+A on the space of C2 functions. For each ε>0 define Dε=(0,β-ε)×(k+ε,ln⁡K-ε). Let Γε be the parabolic boundary of Dε. For every ε>0 which is sufficiently small we can find a smooth function w-(t,x) with compact support on the plane such that in D-ε it is equal to w(t,x). By the Dynkin formula we obtain that for every (u,x)∈Dε,
(118)E[u,x][w-(Zτ(Γε))]=w-(u,x)+E[u,x][∫0τ(Γε)‍L1w-(Zs)ds],
where τ(Q) denotes the arrival time to Q by the process Zt[u,x]. Note that since w(t,x)=w-(t,x) for (t,x)∈D-ε we can replace w- by w in the above formula and since Zs[u,x]∈D0 for s≤τ we obtain that L1w-(Zs)=0. It follows that for every ε>0,
(119)w(u,x)=E[u,x][w(Zτ(Γε))].
Now fix (u,x)∈D0 and a continuous path ω0. Let ℰ={Zτ(Γ1/n)[u,x](ω0)}n0<n⊂D-0, where n0 is such that (u,x)∈D1/n0. The sequence of times {τ(Γ1/n)(ω)}n>n0 is nondecreasing with respect to n and so it has a limit ρ≤T. Let γ be an accumulation point in ℰ; that is, lim⁡k→∞Zτ(Γ1/nk)[u,x](ω0)=γ for some subsequence nk. Define d(y,Γ0)=inf⁡{|y-x|:x∈Γ0} and note that this function is continuous on D-0 and it is 0 if and only if y∈Γ0. Since d(Yτ(Γ1/nk)[u,x](ω0),Γ0)≤1/nk for each k we conclude that γ∈Γ0 and since lim⁡k→∞τ(Γ1/nk)=ρ it follows that Zρ[u,x](ω0)=lim⁡k→∞Zτ(Γ1/nk)[u,x](ω0)=γ. Hence, τ(Γ0)(ω0)=ρ. By the definition w(t,x) is continuous except at the point (β,ln⁡K), but because P[u,x][Zτ(Γ0)=(β,ln⁡K)]=0 for every (u,x)∈D0 we can ignore paths that reach the point (β,ln⁡K), and so
(120)lim⁡ε→0w(Zτ(Γε)[u,x])=w(Zτ(Γ0)[u,x])P[u,x]a.s.Corollary 20.
For every(t,x)∈D0,
(121)w(t,x)=E[t,x][w(Zτ(Γ0))]=-cE[t,x][𝕀{τ(Γ01)<τ(Γ02∪Γ03)}]=-cP[t,x][τ(Γ01)<τ(Γ02∪Γ03)],
where Γ01={(t,ln⁡K):0≤t≤β}, Γ02={(β,x):k≤x≤ln⁡K}, and Γ03={(t,k):0≤t≤β}.Proof.
From (117) we know that the function w(t,x) is bounded and so we can use the Lebesgue bounded convergence theorem and from the boundary conditions on w(t,x) it follows that
(122)w(t,x)=lim⁡ε→0E[t,x][w(Zτ(Γε))]=E[t,x][lim⁡ε→0w(Zτ(Γε))]=E[t,x][w(Zτ(Γ0))]
which gives the first equality of the corollary, while the second equality follows from (116) and the third equality is obvious.Let(t,x),(t′,x)∈D0 and assume that t≤t′. Then it is not difficult to understand that
(123)P[t,x][τ(Γ01)<τ(Γ02∪Γ03)]hhhh≥P[t′,x][τ(Γ01)<τ(Γ02∪Γ03)],
and so w(t,x) is nonincreasing in t for every x which implies that
(124)∂w∂t(t,x)≥0∀(t,x)∈D0.
It is also easy to see that for 0≤t<T and 0≤x≤x′≤ln⁡K,
(125)P[t,x][τ(Γ01)<τ(Γ02∪Γ03)]hhhh≤P[t,x′][τ(Γ01)<τ(Γ02∪Γ03)],
and so
(126)∂w∂x≤0∀(t,x)∈D0.Lemma 21.
The functionswt and wx are in L1[D0].Proof.
We will use (124) in order to prove the result for wt(t,x). The case of wx(t,x) can be proven similarly by using (126). Using (116), (124), and the continuity of w(0,x) on {0}×[k,ln⁡K] we obtain that
(127)∫D0‍|∂w∂t|dtdx=∫kln⁡K‍∫0β‍∂w∂tdtdx=lim⁡ε→0∫kln⁡K-ε‍∫0β‍∂w∂tdtdx=lim⁡ε→0∫kln⁡K-ε‍(w(β,x)-w(0,x))dx=-lim⁡ε→0∫kln⁡K-ε‍w(0,x)dx=-∫kln⁡K‍w(0,x)dx<∞.
Using (126) in place of (124) the proof of integrability of wx is similar.


## 4.3. Integrability ofvt(t,x) and vx(t,x)

We continue this section by analyzing the functionv(t,x) solving the equation L1v=0 with the boundary conditions given by (114). Let C1,2[D-0] be the set of all functions which have one derivative in t and two derivatives in x both uniformly continuous in D0.Lemma 22.
There exists a functionz(t,x)∈C1,2[D-0] such that
(128)z(t,x)=v(t,x)∀(t,x)∈Γ0.Proof.
Recall thatPA,t(T,x)=Pt(T,k) for k≤x<ln⁡K and note that the functions PA,t(T,x),PA,t(t,x), and Pt(t,k) as function of (t,x) belong to the space C1,2[D-0]. Set
(129)z~(t,x)=ln⁡K-xln⁡K-k(Pt(t,k)+PA,t(T,x)-PA,t(T,k))+x-kln⁡K-kPA,t(t,x).
Then z~(t,x)∈C1,2[D-] since it is a linear combination of functions from this space. We also have
(130)z~(t,k)=ln⁡K-kln⁡K-k(Pt(t,k)+PA,t(T,k)-PA,t(T,k))h=Pt(t,x)∀0≤t≤βz~(t,ln⁡K)=PA,t(t,ln⁡K)when0≤t≤β,∀k≤x≤ln⁡K,z~(T,x)=ln⁡K-xln⁡K-k(Pt(T,k)+PA,t(T,x)-PA,t(T,k))hh+x-kln⁡K-kPA,t(T,x)=PA,t(T,x).
Thus, we obtain
(131)z(t,x)=ln⁡K-xln⁡K-kz~(t,x)+x-kln⁡K-kz~(T,x)∈C1,2[D-].
Since
(132)z(t,k)=z~(t,k)=Pt(t,x),z(t,ln⁡K)=z~(T,ln⁡K)=PA,t(T,ln⁡K)=c,z(T,x)=z~(T,x)=PA,t(T,x),
it follows that
(133)z(t,x)=v(t,x)∀(t,x)∈Γ.Next, definef(t,x)=-Lz(t,x). From Lemma 22 we obtain that f(x,t) is bounded in D0 and so it belongs to Lp[D0] for every 1≤p≤∞. Set v~(t,x)=v(t,x)-z(x,t) and observe that
(134)Lv~(t,x)=f(t,x),v~(t,x)=0∀(t,x)∈Γ0.
We conclude that the function v~(t,x) is the unique solution of the following problem (see [16]).Theorem 23.
Let1≤p<∞, then for any f(t,x)∈Lp[D0] there exists a unique function v~ such that(i)
v~∈Lp[0,T;W2,p(0,1)]∩Lp[0,T;W01,p(0,1)],(ii)
∂v~/∂t∈Lp[D0],(iii)
Lv~(t,x)=f(t,x) for every (t,x)∈D0,(iv)
v~|Γ0=0.From assertions (i) and (ii) of Theorem23 we obtain that the functions v~x(t,x) and v~t(t,x) are both in Lp[D] for every 0≤p<∞ and since z(t,x)∈C1,2[D-0] we obtain the following.Corollary 24.
For every1≤p<∞ the functions vt(t,x) and vx(t,x) belong to the space Lp[D0].We can now summarize most of the results of this section as follows.Proposition 25.
Lets(β)<k<ln⁡K<k′ and define
(135)D0=(0,β)×(k,ln⁡K),D0′=(0,β)×(ln⁡K,k′).
Then the function Pt(t,x) is continuous at every point in the domain D-0∖{(β,ln⁡K)}, and there exist two functions w(t,x) and v(t,x) on D0 such that
(136)Pt(t,x)=w(t,x)+v(t,x)forevery(t,x)∈D-0∖{(β,ln⁡K)},(137)w(t,x),v(t,x)∈C1,2(D0)∪L∞[D0],
and both functions are solutions of the parabolic equation L1u=0. Furthermore, w(t,x) is continuous in D0 and it satisfies
(138)w(t,ln⁡K)=PA,t(β,ln⁡K),w(t,b)=0when0≤t≤β,w(β,x)=0whenk<x<ln⁡K,(139)wt(t,x),wx(t,x)∈L1[D].
Finally, v(t,x)∈C(D-) and for every 1≤p<∞,
(140)vt(t,x),vx(t,x)∈Lp[D].
The same decomposition of Pt(t,x) with the same properties holds true in the domain D0′.Proof.
Taking the same functionsv and w as in (115) we see that (137) is actually the same as (117) and the fact that both v and w are solution of L1u=0 is clear from their definitions. Next we see that (138) is the same as (116), that (139) is the same as Lemma 21, and that (140) is, in fact, Corollary 24. Observe that we did not use in this section the fact that k<ln⁡K, so all the proofs are also applicable to the case k′>ln⁡K and the domain D0′.From (136), (139), and (140), and estimating Pxx via other derivatives in view of (29) we obtain the following.Corollary 26.
LetD~={(t,x):0<t<β,k-μt<x<ln⁡K-μt} and
(141)u(t,x)=e-rtP(t,x+μt).
Then
(142)∂2u∂t2∈L1[D~].


## 4.4. Price Function When Initial Stock Price Is Large

LetF(t,x),P(t,x), and u(t,x) be as above. Recall that in the domain (0,T)×(ln⁡K,∞) the function P(t,x) satisfies the equation L1P=0 and it is continuous in the closure of [0,T]×[ln⁡K,∞) and P(T,x)=(K-ex)+=0 for x>ln⁡K. Define
(143)v(t,x)=u(T-t,κ2x+ln⁡K+|μ|T),
where u is given by (141) and set G=(0,T)×(0,∞). It follows from Proposition 25 that(1)
v(t,x)∈C1,2[G]∪C[G-],(2)
vxx(t,x)=vt(t,x) for every (t,x)∈G,(3)
v(t,0)=u(T-t,ln⁡K+|μ|T) is continuous,(4)
v(0,x)=0 for every x>0,(5)
v(t,x) is bounded (since P(t,x) is).Since a bounded solution of the heat equation inG is unique (see [13]) then for every (t,x)∈G,
(144)v(t,x)=-2∫0t‍∂K∂x(t-τ,x)v(τ,0)dτ,whereK(t,x)=14πte-(x2/4t),
and so
(145)v(t,x)=14π∫0t‍xe-(x2/4(t-τ))v(τ,0)dτ(t-τ)3/2.
Differentiating v we obtain polynomials Qk,n(s,x) such that for all k,n∈ℕ,
(146)∂k+nv∂nt∂kx(t,x)=∫0t‍Qn,k((t-τ)-1/2,x)e-x2/4(t-τ)v(τ,0)dτ.
If N is large enough and c>0 then ((t-τ)N/xN)Qk,n((t-τ)-1/2,x) is a polynomial in (t-τ)1/2 and 1/x and it is bounded on (0,T)×(c,∞). Since supy≥0yNe-y<∞ for any N∈ℕ we can set y=x2/4(t-τ) deriving that, for any N∈ℕ and (t,x)∈(0,T)×(c,∞),
(147)∂k+nv∂nt∂kx(t,x)=∫0t‍4N(t-τ)Nx2NQn,k((t-τ)-1/2,x)yNe-yv(τ,0)dτ≤(4x)N∫0t‍((t-τ)NxNQn,k(x,(t-τ)-1/2))cccccccccccccccc×yNe-yv(τ,0)dτ≤CxN.
For some C=C(N)>0. Hence, the following results hold true.Corollary 27.
For anyk,n positive integers k,n, and c>0,
(148)∂k+nv(t,x)∂kt∂nx∈L2[(0,T)×(c,∞)].Corollary 28.
Let(2/κ)(ln⁡K+|μ|T)<k′ and G~={(t,x):0<t<β,k′-μ<x<∞}. Then
(149)∂2u∂t2(t,x)∈L2[G~].


## 5. Proof of Main Theorem

We split the proof into two cases forx≤ln⁡K and for x>ln⁡K.


### 5.1. Casex≤lnK

We begin by proving the upper bound in (12). Since the option holder can exercise at time 0 it is clear from the definition of P(t,x) in (5) that P(t,x)≥ψ(x) for every x>0. Furthermore, by Proposition 3(iv) for each fixed t the function P(t,x) as a function of x is nonincreasing. Therefore, P(t,x)≥P(t,ln⁡K)=δ when x≤ln⁡K. From the definition (8) of the stopping time σ(n) it is not difficult to see that in the present case when σ(n)<T,
(150)x+μσ(n)+κWσ(n)<ln⁡K,
and so
(151)P(σ(n),x+μσ(n)+κWσ(n))≥δ.
Hence for every τ∈𝒯(n) we obtain
(152)E[e-rτ∧σ(n)(ψ(x+μt+κWτ(n))𝕀τ≤σ(n)+δ𝕀σ(n)<τ)]≤E[e-rτ∧σ(n)P(τ∧σ(n),x+μτ∧σ(n)+κWτ∧σ(n)(n))]=E[u(τ∧σ(n),Xτ∧σ(n)(n))].
By Proposition 4,
(153)E[u(τ∧σ(n),Xτ∧σ(n)(n))]=u(0,x)+E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))],
where, as before, u(t,x)=e-rtP(t,x+μt). Taking the sup with respect to all τ∈𝒯(n) in the inequality (152) and using the fact that u(0,x)=P(0,x) we obtain that
(154)P1(n)(x)-P(0,x)≤sup⁡τ∈𝒯(n)E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))].
Thus, in order to bound P1(n)(x)-P(0,x) from the above it suffices to find an upper bound of the right hand in (154).Next, we split the domain[0,T]×ℝ into three parts
(155)C={(t,x)∈[0,T-h]:μt+x>s(t+h)+|μ|h+κh},S={(t,x)∈[0,T-h]:μt+x≤s(t)-|μ|h-κh},B={(t,x)∈[0,T-h]×ℝ:s(t)-|μ|h-κh≤μt+x≤s(t+h)+|μ|h+κh}.
In order to estimate the right hand side of (154) we split it into three parts according to the domains C, S, and B; that is,
(156)E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))]=E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀((j-1)h,X(j-1)h)∈C]+E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀((j-1)h,X(j-1)h)∈S]+E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀((j-1)h,X(j-1)h)∈B].
By Proposition 3(ii) after the time β the prices of the American and game put options coincide which enables us to conclude that u(t,x)=e-rtPA(t,x+μt) for t≥β and that the sets Ct≥β={(t,x)∈C:t≥β}, St≥β={(t,x)∈S:t≥β}, and Bt≥β={(t,x)∈B:t≥β} are the same as the corresponding parts of the domains C-, S-, and B- introduced in [3] for the case of American put options. Therefore, we can use the following results from Sections 4.2 and 4.3 in [3].Proposition 29.
There exists a constantC>0 such that, for every τ∈𝒯(n),
(157)E[∑j=kβ(τ/h)∨kβ‍𝒟|u((j-1)h,X(j-1)h(n))|𝕀((j-1)h,X(j-1)h)∈C]≤C(ln⁡nn)4/5,
where kβ=min⁡{k:kh≥β} and
(158)E[∑j=kβ(τ/h)∨kβ‍𝒟u((j-1)h,X(j-1)h(n))𝕀((j-1)h,X(j-1)h)∈B]≤Cn3/4.Observe also thatP(t,x)=K-ex in the domain S, and so we can use there Lemma 2 from Section 4 of [3].Lemma 30.
For every(t,x)∈S one has 𝒟u(t,x)≤0, and so
(159)E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀((j-1)h,X(j-1)h)∈S]≤0.Thus, for an upper bound of the right side of (154) we can ignore the second term in the right hand side of (156) and estimate only two remaining terms starting with the first term in the right hand side of (156).Proposition 31.
There is a constantC>0 such that, for all n∈ℕ,
(160)E[∑j=1h-1(τ∧σ(n))‍|𝒟u((j-1)h,X(j-1)h(n))|𝕀((j-1)h,X(j-1)h)∈C]≤Cn-3/4.Proof.
We have(161)E[∑j=1h-1(τ∧σ(n))‍|𝒟u((j-1)h,X(j-1)h(n))|𝕀((j-1)h,X(j-1)h)∈C]=E[∑j=1(h-1(τ∧σ(n)))∧kβ‍|𝒟u((j-1)h,X(j-1)h(n))|×𝕀((j-1)h,X(j-1)h)∈C∑j=1(h-1(τ∧σ(n)))∧kβ‍]+E[∑j=kβ(h-1(τ∧σ(n)))∨kβ‍|𝒟u((j-1)h,X(j-1)h(n))|×𝕀((j-1)h,X(j-1)h)∈C∑j=kβ(h-1(τ∧σ(n)))∨kβ‍].
Proposition 29 provides a bound for the second term in the right hand side of (161), and so it remains to deal only with the first term there. Note that if jh<σ(n)∧β(n) and (jh,Xjh(n))∈C then
(162)c~1(j)=s(jh)-μjh+κh≤Xjh(n)≤ln⁡K-2κh-μjh=c~2(j),
where the equalities above are just definitions of c~1 and c~2. Observe also that since x<ln⁡K and jh<σ(n) then by the definition of the stopping times σ(n) the process Xjh(n)+μjh does not exceed ln⁡K-2Kh. By Proposition 5,
(163)𝒟u(t,x)=1κ∫0h‍dy∫-κyκy‍dz(z∂2u∂t∂x(t+y2,x+z)hhhhhhhhhhhhhhhh+δ(u)(t+y2,x+z)z∂2u∂t∂x).
Relying on the same computation as in Section 4 of [3] we see that for (t,x)∈C and x<ln⁡K-|μ|h-κh,
(164)|𝒟u(t,x)|≤hκ∫tt+h‍ds∫x-κhx+κh‍dz|∂2u∂t2(s,z)|.
Thus, for 0≤j<kβ,
(165)E(|𝒟u(jh,Xjh(n))|𝕀{(jh,Xjh(n))∈C}∩{jh<σ(n)})≤∫c~1(j)c~2(j)‍|Du(jh,y)|dPXjh(y)≤∫c~1(j)c~2(j)‍(h2κ∫jhjh+h‍ds∫y-κhy+κh‍|∂2u∂t2(s,z)|dz)dPXjh(y)=h2κ∫jh(j+1)h‍ds∫c~1(j)+κhc~2(j)+κh‍dz|∂2u∂t2(s,z)|×∫max⁡⁡(c~1(j),z-κh)min⁡⁡(c~2(j),z+κh)‍dPXjh(y)≤h2κ∫jh(j+1)h‍ds∫c~1(j)-κhc~2(j)+κh‍dz|∂2u∂t2(s,z)|×P[|Xjh(n)-z|≤κh].
From (24) we see that there is a constant C>0 independent of j and n such that
(166)P[|Xjh(n)-z|≤κh]≤Cj+1.
Hence, for jh<σ(n),
(167)E(|𝒟u(jh,Xjh(n))|𝕀(jh,Xjh(n))∈C𝕀jh<σ(n))≤h2κ∫jh(j+1)h‍ds∫c~1(j)-κhc~2(j)+κh‍dz|∂2u∂t2(s,z)|Cj+1=Ch2κ∫jh(j+1)h‍dsh(j+1)∫c~1(j)-κhc~2(j)+κh‍dz|∂2u∂t2(s,z)|≤C1n∫jh(j+1)h‍dss∫c~1(j)-κhc~2(j)+κh‍dz|∂2u∂t2(s,z)|.
Define
(168)c1(t)=s(t)-μt,c2(t)=ln⁡K-μt-κh,
where s(t)=ln⁡(b(t)) is the free boundary of the option holder and b(t) was introduced at the beginning of Section 3. Observe that for every j and any jh≤s≤(j+1)h,
(169)c~1(j)-κh≥c1(s),c~2(j)+κh≤c2(s).
Summing up the above estimates we obtain
(170)∑j=0kβ-1‍E(|𝒟u(jh,Xjh(n))|𝕀{(jh,Xjh(n))∈C}∩{jh<σ(n)})≤C2n+C1n∫hβ‍dss∫c1(s)c2(s)‍dz|∂2u∂t2(s,z)|=C2n+C1n(∫hh‍dss∫c1(s)c2(s)‍dz|∂2u∂t2(s,z)|ccccccccccccccccc+∫hβ‍dss∫c1(s)c2(s)‍dz|∂2u∂t2(s,z)|),
where the term C2/n comes from the first term E|𝒟u(0,x)| of the sum which can be estimated easily using the fact that ut(t,x) and uxx(t,x) are bounded for small t.
LetG={(t,x):0<t<β,c1(t)<x<ln⁡K-μt} and note that G⊂E∪D~, where E and D~ are defined in Corollaries 16 and 26 which imply that (∂2u/∂t2)(s,z)∈L1[F]. Hence,
(171)∫hβ‍dss∫c1(s)c2(s)‍dz|∂2u∂t2(s,z)|≤C1n1/4∫hβ‍ds∫c1(s)c2(s)‍dz|∂2u∂t2(s,z)|≤Cn1/4.
Next, we estimate the first integral in brackets in the right hand side of (170). Let s(β)<k<ln⁡K,k′=(ln⁡K-k)/2 and split the integral in question as follows:
(172)∫hh‍dss∫c1(s)c2(s)‍dz|∂2u∂t2(s,z)|=∫hh‍dss∫c1(s)k′-μs‍dz|∂2u∂t2(s,z)|+∫hh‍dss∫k′-μsc2(s)‍dz|∂2u∂t2(s,z)|.
From Corollary 16 we know that the function (∂2u/∂t2)(s,z) is in L2[E~], where
(173)E~={(s,z):0<s<T,c1(t)<z<k′-μt}⊂Eσ
(for an appropriate b<ln⁡K in the definition of Eσ). Therefore we can use the Cauchy-Schwarz inequality to obtain
(174)∫hh‍dss∫c1(s)k′-μs‍dz|∂2u∂t2(s,z)|≤(∫hh‍dss∫c1(s)k′-μs‍dz)1/2(∫hh‍∫c1(s)k′-μs‍|∂2u∂t2(s,z)|2dz)1/2≤Cln⁡n.
Now we are left with the second integral in the right hand side of (172). We will show that there is a constant C>0 such that
(175)ℐn=∫hh‍dss∫k′-μsc2(s)‍dz|∂2u∂t2(s,z)|≤Cn1/4.
Recall that u(t,x)=e-rtP(t,x+μt), and so
(176)∂2u∂t2(t,x)=e-rt(r2P(t,x+μt)-2rPt(t,x+μt)-2rμPx(t,x+μt))+e-rt(μ2Pxx(t,x+μt)+2μPxt(t,x+μt)+Ptt(t,x+μt)μ2).
Observe that the functions P(t,x),Px(t,x),Pt(t,x), and Pxx(t,x) are all bounded for small t. Indeed, P≤K+δ, while Pt is bounded in the domain of integration in (175) for small h in view of (115), (117), (136), and (137). Next, Px is bounded by Theorem 8.1 from [8]. Finally, Pxx is bounded since in the domain in question P and its first derivatives are bounded and P satisfies the equation ((∂/∂t)+A)P=0 (see (29)). Therefore, we can write
(177)ℐn≤∫hh‍dtt∫k′-μtc2(t)‍dx(|2μe-rtPtx(t,x+μt)|cccccccccccccccccccccc+|e-rtPtt(t,x+μt)|)+C1,
for some constant C1>0 independent of n. Recall that, for (x,t)∈D=(0,β)×(k,ln⁡K) by Proposition 25, Pt(t,x)=v(t,x)+w(t,x), where vt and vx belong to L2[D]. Hence, expressing Ptx and Ptt via vt,wt and vx,wx we can estimate the integral (177) containing vt and vx by means of the Cauchy-Schwarz inequality as it was done in (174). Replacing these integrals by C2ln⁡n we obtain
(178)ℐn≤∫hh‍dtt∫k′-μtc2(t)‍dx(|2μe-rtwx(t,x+μt)|cccccccccccccccccccccc+|e-rtwt(t,x+μt)|)+C2ln⁡n+C1.
By (124) and (126) the functions wt(t,x) and wx(t,x) do not change signs in D, and so it follows that
(179)∫hh‍dtt∫k′-μtc2(t)‍(|2μe-rtwx(t,x+μt)|+|e-rtwt(t,x+μt)|)dx=|∫hh‍dtt2|μ|e-rt∫k′ln⁡K-κh‍wx(t,x)dx|+|∫hh‍dtt∫k′-μtc2(t)‍e-rtwt(t,x+μt)dx|.
By Proposition 25, w(x,t) is bounded on D, and so the contribution of the first integral in the right hand side of (179) is bounded by a constant and it remains to estimate only the second integral there.
Next, we will need a more explicit representation of the functionw. Let
(180)z~(t,x)=e-rtw(t,x+μt).
Then in the domain E~={(t,x),0<t<β,k-μt<x<ln⁡K-μt},
(181)κ22z~xx(t,x)+z~t(t,x)=0.
Define
(182)z(t,x)=z~(T-t,κ2x)
and let
(183)E={(t,x):0<t<T,2(k-μt)κ<x<2(ln⁡K-μt)κ}.
In the domain E the function z(t,x) satisfies the heat equation
(184)zxx(t,x)=zt(t,x).
If we let
(185)d1(t)=2(k-μ(T-t))κ,d2(t)=2(ln⁡K-μ(T-t))κ,
then from the boundary values of w(t,x) we obtain
(186)z(0,x)=0ford1(0)<x<d2(0),z(t,d1(t))=0,z(t,d2(t))=e-r(T-t)for0<t≤T.
Note that z(t,x) is a bounded continuous function on the boundaries (t,di(t)),i=1,2,0<t≤T of E. Hence, by Chapter 14 of [13] we can represent z(t,x) in the form
(187)z(t,x)=∫0t‍∂H∂x(x-d1(τ),t-τ)ϕ1(τ)dτ+∫0t‍∂H∂x(x-d2(τ),t-τ)ϕ2(τ)dτ,
where H(t,x)=(1/4πt)e-x2/4t is the fundamental solution and the functions ϕi(t),i=1,2, are bounded continuous on the interval (0,T]. From the definition of z~ we see that
(188)z~t(t,x)=-re-rtw(t,x+μt)+e-rtwt(t,x+μt).
Since w(t,x) is bounded then for some constant C1>0 independent of n,
(189)|∫hh‍dtt∫k′-μtc2(t)‍e-rtwt(t,x+μt)dx|≤|∫hh‍dtt∫k′-μtc2(t)‍zt(t,x)dx|+C1.
From the representation (187) of z(t,x) we obtain that
(190)|∫hh‍dtt∫k′-μtc2(t)‍zt(t,x)dx|≤κ2|∫hh‍dtt∫(2/κ)(k′-μt)(2/κ)c2(t)‍ddt×∫0T-t‍∂H∂x(x-d1(τ),T-t-τ)ϕ1(τ)dτdx∫(2/κ)(k′-μt)(2/κ)c2(t)‍|+κ2|∫hh‍dtt∫(2/κ)(k′-μt)(2/κ)c2(t)‍ddt×∫0T-t‍∂H∂x(x-d2(τ),T-t-τ)ϕ2(τ)dτdx∫(2/κ)(k′-μt)(2/κ)c2(t)‍|.
Observe that as long as we keepx or t away from 0 the function H(x,t) is smooth and it has bounded derivatives with bounds depending on the range of t,x and their distance from zero. Next, if x satisfies
(191)2κ(k′-μt)<x<2κc2(t)=2κ(ln⁡K-μt-κh),
then
(192)k′-k≤2κ(k′-k+μ(T-t-τ))<x-d1(τ)for0<τ≤T-t.
Since k′>k we see that x-d1(τ) stays away from 0 on the entire interval (0,T-t]. It follows from the above that the function
(193)Φ1(t,x)=∫0T-t‍∂H∂x(x-d1(τ),T-t-τ)ϕ1(τ)dτ
has bounded derivatives with respect to t with bounds independent of n in the region {(t,x):h<t<h,(2/κ)(k′-μt)<x(2/κ)c2(t)}. We conclude that the first integral in the right hand side of (190) is bounded from above by a constant independent of n and it remains to estimate the second integral there.
Set(194)Φ2(t,x)=∫0T-t‍∂H∂x(x-d2(τ),T-t-τ)ϕ2(τ)dτ.
We see that if
(195)x<2κc2(t)=2κ(ln⁡K-μt-h),
then
(196)x-d2(τ)=x-2κ(ln⁡K-μ(T-τ))<2κ(μ(T-t-τ)-h).
In this case x-d2(τ) can be zero when τ∈[0,T-t], but this can happen only for a τ which is at least μ-1h apart from T-t. Thus, the function Φ2 is smooth with a bounded uniformly continuous derivative with respect to t though this bound may depend on n. Nevertheless, we still have the following:
(197)κ2|∫hh‍dtt∫(2/κ)(k′-μt)(2/κ)c2(t)‍ddt×(∫0T-t‍∂H∂x(x-d2(τ),T-t-τ)ϕ2(τ)dτ)dx∫(2/κ)(k′-μt)(2/κ)c2(t)‍|=κ2|∫hh‍dttddt(∫0T-t‍∫(2/κ)(k′-μt)(2/κ)c2(t)‍∂H∂xhhhhhhhhhhhhhh×(x-d2(τ),T-t-τ)dxϕ2(τ)dτ∫hh‍)|=κ2|∫hh‍dttddt(2κ∫0T-t‍(H(2κc2(t)-d2(τ),T-t-τ)hhhhhhhhhhhhhhhhhhhhhhhhh-H(2κ(k′-μt)ccccccccccccccccccccccccccccccccccccc-d2(τ),T-t-τ2κ))cccccccccccccccccccccc×ϕ2(τ)dτ∫hh‍)∫hh‍|≤κ2|∫hh‍dttddt×(∫0T-t‍H(2κc2(t)-d2(τ),T-t-τ)cccccccccccc×ϕ2(τ)dτ∫0T-t‍)∫hh‍|+κ2|∫hh‍dttddt(∫0T-t‍H(2κ(k′-ln⁡Khhhhhhhhhhhhhhhhhhhhhh+μ(T-t-τ)),T-t-τ∫0T-t‍)cccccccccccccccccccccccccccccccccc×ϕ2(τ)dτ2κ)∫hh‍|.
We see that in the second term in the right hand side k′-ln⁡K+μ(T-t-τ) can take on the value 0 for τ∈(0,T-t], but then τ is at least c=μ-1|k′-ln⁡K| apart from T-t and now the separation constant c does not depend on n. Thus, we can bound the second term there from above by a constant and it remains to estimate the first term which we do as follows:
(198)I=κ2|∫hh‍dttcccccccc×ddt∫0T-t‍H(2κc2(t)-d2(τ),T-t-τ)ccccccccccccccc×ϕ2(τ)dτ∫hh‍|≤C2∫hh‍dttcccccccc×∫0T-t‍|1(T-t-τ)3/2ccccccccccccccccc×exp⁡(-(((2/κ)(μ(T-t-τ)cccccccccccccccccccccccccccccc-κh))2)×(4(T-t-τ))-1)cccccccccccccccccccc×ϕ2(τ)1(T-t-τ)3/2|dτ+C2∫hh‍dttcccccccccc×∫0T-t‍|1(T-t-τ)1/2ccccccccccccccccccc×exp⁡(-(((2/κ)(μ(T-t-τ)ccccccccccccccccccccccccccccccc-κh))2)×(4(T-t-τ))-1)cccccccccccccccccccc×ϕ2(τ)1(T-t-τ)1/2|dτ+C2∫hh‍dtt∫0T-t‍|h(T-t-τ)5/2exp⁡(-κ2h2(T-t-τ))cccccccccccccccccccccccc×exp⁡(μh2-μ22κ2(T-t-τ))cccccccccccccccccccccccc×ϕ2(τ)h(T-t-τ)5/2|dτ,
where C2>0 is a constant independent of n. Analyzing the integral with respect to τ in the second term in the right-hand side above by considering different possible values of T-t-τ we conclude that this integral is bounded by a constant independent of n. Next we observe that |exp⁡⁡((μh/2)-(μ2/2κ2)(T-t-τ))ϕ2(τ)| is also bounded by a constant independent of n too. Hence, we obtain
(199)I≤C3+C3∫hh‍dtt∫0T-t‍1(T-t-τ)3/2hhhhhhhhhhhhhh×exp⁡⁡(-h2(T-t-τ))dτ+C3∫hh‍dtt∫0T-t‍h(T-t-τ)5/2hhhhhhhhhhh×exp⁡⁡(-h2(T-t-τ))dτ
for a constant C3>0 independent of n. Set ρ=h/2(T-t-τ) and note that dρ/dτ=-h/22(T-t-τ)3/2 and dρ2/dτ=-h/4(T-t-τ)2. We proceed by changing variables arriving at
(200)I≤C4+C4∫hh‍dtt∫h/2(T-t)∞‍1he-ρ2dρ+C4∫hh‍dtt∫h/2(T-t)∞‍1hρe-ρ2dρ2≤C4+C51h∫hh‍dtt≤C4+C52(1+1h1/4)≤C6n1/4,
for some constants C4,C5,C6>0 independent of n and (175) follows. Combining (175) and (174) we obtain from (172) that
(201)∫hh‍dss∫c2(s)c1(s)‍dz|∂2u∂t2(s,z)|≤Cn1/4.
Finally, Proposition 29 follows from (201), (170), and (171).Next, we turn our attention to the domainB. First, we will prove the following result.Lemma 32.
There exists a constantC>0 such that for all n∈ℕ,
(202)E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀((j-1)h,X(j-1)h)∈B]≤Cn-3/4.Proof.
LetBt<β(n) and Bt≥β(n) be the set of all points (t,x)∈B such that t<β(n) and t≥β(n), respectively. We split (202) according to these two regions, namely,
(203)E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀((j-1)h,X(j-1)h)∈B]=E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))cccccccccccc×𝕀((j-1)h,X(j-1)h)∈Bt<β(n)∑j=1h-1(τ∧σ(n))‍]+E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))ccccccccccccc×𝕀((j-1)h,X(j-1)h)∈Bt≥β(n)∑j=1h-1(τ∧σ(n))‍].
By Proposition 29 we have that for a constant C>0 independent of n,
(204)E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀((j-1)h,X(j-1)h)∈Bt≥β(n)]≤Cn-3/4.
Thus, it remains to estimate only the first term in the right-hand side. Let E={(t,x):0<t<β(n),a-μt<x<b-μt}, where a<s(0) and s(β(n)+h)+|μ|h+2κh<b<ln⁡K. For n large enough we can find such a b because s(t) is continuous and s(β(n))<ln⁡K. We know from Corollary 16 that u(t,x)∈H2[E]. Since C2[E] is dense in this space we can approximate u(t,x) by C2 functions to get equality (22) of Proposition 5 for u(t,x), as well. Since ut(t,x)+(κ2/2)uxx(t,x)≤0 in the domain E we obtain
(205)𝒟u(t,x)≤1κ∫0h‍dy∫-κyκy‍dz(z∂2u∂t∂x(t+y2,x+z))≤∫0h‍ydy∫-κyκy‍dz|∂2u∂t∂x(t+y2,x+z)|=12∫0h‍ds∫-κyκy‍dz|∂2u∂t∂x(t+s,x+z)|.
It follows that
(206)𝒟u(t,y)𝕀(t,y)∈B-≤12∫tt+h‍ds∫s(t)-λh-μts(t+h)+λh-μt‍𝕀|z-y|≤κh|∂2u∂t∂x(s,z)|dz,
where λ=|μ|+κ. Hence,
(207)E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀{((j-1)h,X(j-1)h)∈Bt<β(n)}]≤12(∑j=1kβ‍∫jh(j+1)h‍dτ∫s(jh)-λh-μjhs(jh+h)+λh-μjh‍P(|Xjh(n)-z|≤κh)cccccccccccccccccccccccccccccccc×|∂2u∂t∂x(s,z)|dz∑j=1kβ‍)+Cn.
Here kβ=⌈β/h⌉, and the term C/n is the contribution of 𝒟u(0,X0n)=𝒟u(0,x)≤C/n which holds true from by the definition of the operator 𝒟 and boundedness of ut and uxx for small t. From Corollary 16 we see that there exists a constant C1>0 such that
(208)∫ab‍|∂2u∂t∂x(t,z)|2dz≤C1when0≤t≤β(n).
This together with (24), the Cauchy-Schwarz inequality, and the inequality 1/τ≥1/2jh, which is satisfied when j≥1 and jh≤τ≤2jh, yields that
(209)E[∑j=1h-1(τ∧β(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀{((j-1)h,X(j-1)h)∈Bt<β(n)}]≤hC2∑j=1kβ‍∫jh(j+1)h‍dττ((s(j+1)h)-s(jh)+2λh)1/2+C2n.
From Proposition6 and Lipschitz continuity of the function P(t,x) in t≤β(n) uniformly in x≤ln⁡K (see Theorem 8.1 in [8]) we obtain that for some constant C3>0,
(210)|s(t1)-s(t2)|≤|t1-t2|C3whenever0≤t1,t2≤β(n).
Hence,
(211)E[∑j=0h-1(τ∧β(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀{((j-1)h,X(j-1)h)∈Bt<β(n)}]≤C4n3/4,
for some constant C4>0 independent of n.By combining the results of Lemma30, Proposition 31, and Lemma 32 together with (156) we obtain that the upper bound P1(n)(x)-P(0,x)<C/n3/4 for some constant C>0 independent of n and of x≤ln⁡K.Next, we will obtain a lower bound for the approximation errorP1(n)(x)-P(0,x) when x≤ln⁡K. Set
(212)τ(n)=inf⁡⁡{t:μ[th]h+Xt(n)<s([th]h+h)+|μ|h+κh}.
By Proposition 31,
(213)E[u(τ(n)∧σ(n),Xτ(n)∧σ(n)(n))]=u(0,x)+E[∑j=1τ(n)∧σ(n)/h‍𝒟u((j-1)h,X(j-1)h(n))]=u(0,x)+E[∑j=1h-1(τ∧σ(n))‍∣𝒟u((j-1)h,X(j-1)h(n))ccccccccccccccc×𝕀((j-1)h,X(j-1)h)∈C∑j=1h-1(τ∧σ(n))‍]≥P(0,x)-Cn-3/4.Setα=αn=T-(1/n2/3) and let τA(n) be defined by (212) with s there replaced by the free boundary sA for the American put option (see Section 2.2 in [3]). Define also τα(n)=τ(n)𝕀{τ(n)+h<α}+T𝕀{τ(n)+h≥α} and τA,α(n)=τA(n)𝕀{τA(n)+h<α}+T𝕀{τA(n)+h≥α}. We will rely on the following estimate from Section 4.5 in [3].Lemma 33.
There exists a constantC>0 independent of nℕ such that
(214)|E[uA(τA(n),XτA(n)(n))-erτA,α(n)ψ(μτA,α(n)+XτA,α(n))]|≤Cn2/3,
where uA(t,x)=e-trPA(t,x+μt) with PA given by (3).Remark 34.
Note thatsA(t)=s(t) for β≤t<T, and so τA(n)∨β=τ(n)∨β and τA,α(n)∨β=τα(n)∨β.From now on we assume thatn is large enough so that β(n)<α. From the definition of P1(n)(x) we have
(215)P1(n)(x)≥E[e-rτα(n)∧σ(n)(ψ(μτα(n)+Xτα(n)(n))𝕀{τα(n)≤σ(n)}+δ𝕀{σ(n)<τα(n)})].
Hence, if we prove that for some constant C>0 independent of n(216)J=|e-rτα(n)∧σ(n)E[u(τ(n)∧σ(n),Xτ(n)∧σ(n)(n))]-E[e-rτα(n)∧σ(n)(ψ(μτα(n)+Xτα(n)(n))𝕀{τα(n)≤σ(n)}+δ𝕀σ(n)<τα(n))]|≤Cn,
then by (215) and (213) we could conclude that
(217)-Cn≤P1(n)(x)-P(0,x).
We split the left hand side of (216) into three parts
(218)J=E[{(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))u(τ(n)∧β(n),Xτ(n)∧β(n)(n))-e-rτ(n)∧β(n)(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))}×𝕀{τα(n)≤σ(n)∧β(n)}{(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))u(τ(n)∧β(n),Xτ(n)∧β(n)(n))]+E[{u(σ(n),Xσ(n)(n))-e-rσ(n)δ}𝕀{σ(n)<τα(n)}]+E[𝕀{β(n)<τα(n)≤σ(n)}{u(τ(n),Xτ(n)(n))-e-rτα(n)(ψ(μτα(n)+Xτα(n)(n)))}×𝕀{β(n)<τα(n)≤σ(n)}].
This equality is true since τ(n)=τα(n)=τ(n)∧β on the set τα(n)≤σ(n)∧β(n)<α. We begin with the last term. First note that on the set β(n)≤τα(n)≤σ(n) we have, in particular, β(n)≤σ(n) and so σ(n)=T by Remark 34. In the case τα(n)>β(n) we have τα(n)=τA,α(n) and τ(n)=τA(n) and so from Lemma 33 we derive that
(219)|E[{u(τ(n),Xτ(n)(n))-e-rτα(n)(ψ(μτα(n)+Xτα(n)(n)))}×𝕀{β(n)<τα(n)≤σ(n)}](ψ(μτα(n)+Xτα(n)(n)))|≤|E[{u(τA(n),XτA(n)(n))-e-rτA,α(n)(ψ(μτA,α(n)+XτA,α(n)(n)))}]|≤Cn2/3.Next, we deal with the first term in the right hand side of (218) where τα(n)=τ(n)≤σ(n)∧β(n). This means that before time β(n) the process X(n) is stopped near the boundary s(t) and
(220)μτ(n)+Xτ(n)(n)<s(τ(n)+h)+|μ|h+σh.
By the definition, u(τ(n),Xτ(n)(n))=e-rτ(n)P(τ(n),X(n)+μτ(n)). Thus, we have
(221)E[{(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))u(τ(n)∧β(n),Xτ(n)(n))-e-rτ(n)∧β(n)(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))}×𝕀{τα(n)≤σ(n)∧β(n)}{(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))u(τ(n)∧β(n),Xτ(n)(n))]=E[{e-rτ(n)(P(τ(n),X(n)+μτ(n))-ψ(μτ(n)+Xτ(n)(n)))}×𝕀{τα(n)≤σ(n)∧β(n)}(μτ(n)+Xτ(n)(n))].
If μτ(n)+Xτ(n)(n)≤s(τ(n)), then P(τ(n),X(n)+μτ(n))-ψ(μτ(n)+Xτ(n)(n))=0, so we can assume that
(222)s(τ(n))<μτ(n)+Xτ(n)(n)<s(τ(n)+h)+|μ|h+σh.
To continue we need the following lemma.Lemma 35.
There is a constantC>0 independent of n such that for every point (t,x) satisfying s(t)≤μt+x≤s(t+h)+|μ|h+σh and 0≤t≤β(n),
(223)|P(t,μt+x)-ψ(μt+x)|≤Cn.Proof.
The functionP(t,x) is Lipschitz continuous when t≤β and 0≤x≤μβ+ln⁡K (see [8]), and so
(224)|P(t,μt+x)-P(t+h,μt+x)|≤Cn,
for some C>0 independent of n. If μt+x≤s(t+h) then P(t+h,μt+x)=ψ(μt+x) and we are done. Now assume that s(t+h)<μt+x<s(t+h)+λ/n, where λ=n(|μ|h+σh).
From Corollary15 it follows that for every t<T and a<ln⁡K the function Pxx(t,x) is continuous in x on the closed interval [s(t),a], so we can write
(225)P(t+h,μt+x)=P(t+h,s(t+h))+Px(t+h,s(t+h))λn+Pxx(t+h,s(t+h))λ22n+α,
where α=α(h) satisfies lim⁡h→0(α/h)=0. From the property of smooth fit (see [8]) it follows that Px(t+h,s(t+h))=ψx(s(t+h)), and so for some C>0 independent of n,
(226)|P(t+h,μt+x)-ψ(μt+x)|≤Cn.Using (222) and the above lemma we obtain
(227)|E[{(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))u((μτ(n)∧β(n)+Xτ(n)∧β(n)(n))τ(n)∧β(n),Xτ(n)∧β(n)(n))-e-rτ(n)∧β(n)(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))}×𝕀{τα(n)≤σ(n)∧β(n)}(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))]|≤Cn.
Hence, we are done with the first term in the right hand side of (218) and it remains to estimate the second one. Since σ(n)<τα(n)≤T the process X(n) is stopped near the writer’s boundary. Namely, we have
(228)ln⁡K-|μ|h-σh<μσ(n)+Xσ(n)(n)≤ln⁡K.
Since P(t,ln⁡K)=δ when t≤β, β(n)-β<h, and P is Lipschitz continuous (see Theorem 8.1 of [8]) we obtain that
(229)|P(σ(n),μσ(n)+Xσ(n)(n))-δ|≤Cn,
for some C>0 independent of n. Hence,
(230)E[(u(σ(n),Xσ(n)(n))-e-rσ(n)δ)𝕀{σ(n)<τα(n)}]≤Cn.
It follows that there exists C>0 independent of n such that, for every x≤ln⁡K,
(231)-Cn<P1(n)(x)-P(0,x).Next, we will derive a lower bound for the second approximation functionP2(n)(x) defined by (11), still assuming that x≤ln⁡K. According to (213) in order to obtain
(232)P2(n)(x)-P(0,x)≥-Cn2/3,
it suffices to show that
(233)E[u(τ(n)∧σ(n),Xτ(n)∧σ(n)(n))]-P2(n)(x)≤Cn2/3.
We have
(234)E[u(τ(n)∧σ(n),Xτ(n)∧σ(n)(n))]-P2(n)(x)≤E[u(τ(n)∧σ(n),Xτ(n)∧σ(n)(n))-e-rτα(n)∧σ(n)×(ψ(μτα(n)+Xτα(n)(n))𝕀{τα(n)≤σ(n)}+(ψ(μσ(n)+Xσ(n)(n))+δ)𝕀{σ(n)<τα(n)})e-rτα(n)∧σ(n)]=E[{u(τ(n)∧β(n),Xτ(n)∧β(n)(n))-e-rτ(n)∧β(n)(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))}ccccccc×𝕀{τα(n)≤σ(n)∧β(n)}Xτ(n)∧β(n)(n)]+E[𝕀{σ(n)<τα(n)}{u(σ(n),Xσ(n)(n))-e-rσ(n)(ψ(μσ(n)+Xσ(n)(n))+δ)}ccccccc×𝕀{σ(n)<τα(n)}Xτα(n)(n)]+E[{u(τ(n),Xτ(n)(n))-e-rτα(n)(ψ(μτα(n)+Xτα(n)(n)))}ccccccc×𝕀{β(n)<τα(n)≤σ(n)}Xτα(n)(n)].
Indeed, the first inequality is true since P2(n)(x) is defined as the sup on τ∈𝒯(n) and we chose a specific one, that is, τα(n). The equality is true due to the same reason that (218) holds true. We see that the first term in the right hand side of (234) is the same as the first term in (218) and by (227) it is less than C/n for some constant C. The second term is nonpositive because, for every (t,x) we have P(t,x)≤ψ(x)+δ and u(t,x)=e-rtP(t,μt+x), so we can just remove it from the equation. The last term is the same as the last term of (218) and from Lemma 33 we obtain that this term is less than or equal C/n2/3 for an appropriate C. These arguments yield (233) and hence (232), as well. For the upper bound we already know that P1(n)(x)-P(0,x)≤C/n3/4 and from the definition of P1(n) and P2(n) it is not hard to see that |P2(n)-P1(n)|≤C/n. It follows from above that there exist C>0 such that, for every x≤ln⁡K,
(235)-Cn3/2≤P2(n)(x)-P(0,x)≤Cn.


### 5.2. Casex>lnK

We begin with the upper bound onP1(n). We will show first that
(236)P1(n)(x)-P(0,x)≤sup⁡τ∈𝒯(n)E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))].
The proof is similar to the proof of (154), we just have to show that for every τ∈𝒯(n),
(237)P(τ∧σ(n),μτ∧σ(n)+Xτ∧σ(n)(n))≥ψ(μτ+Xτ(n))𝕀{τ≤σ(n)}+(δ-Ke(|μ|h+2κh))𝕀{σ(n)<τ}.
On the set τ≤σ(n) this inequality is clear since P(t,x)≥ψ(x). For the case σ(n)<τ observe that because x>ln⁡K we must have
(238)ln⁡K<μσ(n)+Xσ(n)(n)<ln⁡K+|μ|h+2κh.
By Theorem 8.1 in [8] the right derivative Fx(t,K+) at K satisfies 0>Fx(t,K+)>-1 for any t, and so 0≤F(t,K)-F(t,K+Cλ)≤Cλ for each C>0 provided 0≤λ≤λ(C) is small enough. Assume 0<λ<1; then eλ-1≤λeλ≤λe. Hence, taking C=Ke we have
(239)P(t,ln⁡K)-P(t,ln⁡K+λ)=F(t,K)-F(t,Keλ)≤F(t,K)-F(t,K+Keλ)≤Keλ.
Put λ=|μ|h+κh; then for σ(n)<τ and sufficiently large n,
(240)P(σ(n),μσ(n)+Xσ(n)(n))≥P(σ(n),ln⁡K+λ)≥δ-Keλ.
Hence, we obtain (237) which yields also (234). To bound the right hand side of (236) we split it similarly into the case x≤ln⁡K (see (156)) according to the three different regions C,B, and S. Since our process starts at x>ln⁡K, if ((j-1)h,X(j-1)h(n))∈B for some j then this must happen after the time β, and so we can use (158). The part that belongs to the region S is nonpositive, so we can ignore it, and so we will be left only with the region C.Lemma 36.
For the discrete processXt(n) such that X0(n)=x>ln⁡K one has
(241)E[∑j=1h-1(τ∧σ(n))‍𝒟|u((j-1)h,X(j-1)h(n))|𝕀{((j-1)h,X(j-1)h)∈C}]≤Cn-3/4.Proof.
It suffices to show that(242)E[∑j=1kβ∧(σ(n)/h)‍𝒟|u((j-1)h,X(j-1)h(n))|𝕀{((j-1)h,X(j-1)h)∈C}]≤Cn-3/4,
for some C>0 independent of n since after time β(n) we come back to the American option case. This is done in the same way as in Proposition 31, and so we provide only a sketch of the proof. Let c(s)=ln⁡K-μs+κh; then similarly to the proof of Proposition 31 we obtain
(243)∑j=0kβ-1‍E(|𝒟u(jh,Xjh(n))|𝕀{(jh,Xjh(n))∈C}∩{jh<σ(n)})≤C2n+C1n∫hβ‍dss∫c(s)∞‍dz|∂2u(∂t2)(s,z)|.
Let (2/κ)(ln⁡K+|μ|T)<k′ and split the integral in (243) into two parts
(244)∫hβ‍dss∫c(s)∞‍dz|∂2u∂t2(s,z)|=∫hβ‍dss∫c(s)k′-μs‍dz|∂2u(∂t2)(s,z)|+∫hβ‍dss∫k′-μs∞‍dz|∂2u(∂t2)(s,z)|.
Let E={(s,z):0<s<β,k′-μs<z<∞}; then by Corollary 28 we see that (∂2u/∂t2)(s,z)∈L2[E], and so we obtain similarly to (174) that, for some constant C>0,
(245)∫hβ‍dss∫k′-μs∞‍dz|∂2u(∂t2)(s,z)|<Cln⁡n.
In the first integral in the right hand side of (244) we do the same procedure as in (170)–(175) relying on Proposition 25 and deriving that for some constant C>0,
(246)∫hβ‍dss∫c(s)k′-μs‍dz|∂2u(∂t2)(s,z)|<Cn1/4.
Combining (243)–(246) we obtain (242) and complete the proof of the lemma.The lower bound forP1(n)(x)-P(0,x) when x>ln⁡K is obtained similarly to the case x≤ln⁡K. As in that case we use the stopping time τ(n) from (212) and from the above we see that (213) is true also for the case under consideration. We consider again τα(n) defined before Lemma 33 and similarly to (214) obtain that
(247)|E[u(τα(n)∧σ(n),Xτα(n)∧σ(n))]-E[e-rτα(n)∧σ(n)(ψ(μτα(n)+Xτα(n)(n))𝕀{τα(n)≤σ(n)}cccccccccccccccccccccc+(δ-Ke(|μ|h+σh))cccccccccccccccccccccc×𝕀{σ(n)<τα(n)})](μτα(n)+Xτα(n)(n))|.
In order to estimate (247) for x>ln⁡K we only need to split it into two parts, one for τα(n)≤σ(n) and the other one for σ(n)<τα(n). This is true in view of the fact that if we begin with x>ln⁡K and τα(n)≤σ(n) then we must have β(n)≤τα(n), and so we are back to the American option case and can use Lemma 33 for this case. If σ(n)<τα(n) then the process X(n) is stopped near the seller’s boundary and similar to (230) we can use the Lipschitz property of P to obtain
(248)E[(𝕀{σ(n)<τα(n)}u(σ(n),Xσ(n)(n))-e-rσ(n)(δ-Ke(|μ|h+2κh))cccccc×𝕀{σ(n)<τα(n)})(σ(n),Xσ(n)(n))]≤Cn.
From here we can proceed similar to the case of x≤ln⁡K and obtain the lower bound for P1(n) proving (12) for P1(n).Next, we turn to the second approximation functionP2(n), still in the case of x>ln⁡K. For the upper bound we use Lemma 35 as in the case x≤ln⁡K and proceed similarly to the proof of the upper bound for the first approximation function P1(n). The proof of the lower bound is similar to the case x≤ln⁡K and we obtain the result observing that if x>ln⁡K then P(t,x)<ψ(x)+δ=δ for any t∈[0,T].


## 5.1. Casex≤lnK

We begin by proving the upper bound in (12). Since the option holder can exercise at time 0 it is clear from the definition of P(t,x) in (5) that P(t,x)≥ψ(x) for every x>0. Furthermore, by Proposition 3(iv) for each fixed t the function P(t,x) as a function of x is nonincreasing. Therefore, P(t,x)≥P(t,ln⁡K)=δ when x≤ln⁡K. From the definition (8) of the stopping time σ(n) it is not difficult to see that in the present case when σ(n)<T,
(150)x+μσ(n)+κWσ(n)<ln⁡K,
and so
(151)P(σ(n),x+μσ(n)+κWσ(n))≥δ.
Hence for every τ∈𝒯(n) we obtain
(152)E[e-rτ∧σ(n)(ψ(x+μt+κWτ(n))𝕀τ≤σ(n)+δ𝕀σ(n)<τ)]≤E[e-rτ∧σ(n)P(τ∧σ(n),x+μτ∧σ(n)+κWτ∧σ(n)(n))]=E[u(τ∧σ(n),Xτ∧σ(n)(n))].
By Proposition 4,
(153)E[u(τ∧σ(n),Xτ∧σ(n)(n))]=u(0,x)+E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))],
where, as before, u(t,x)=e-rtP(t,x+μt). Taking the sup with respect to all τ∈𝒯(n) in the inequality (152) and using the fact that u(0,x)=P(0,x) we obtain that
(154)P1(n)(x)-P(0,x)≤sup⁡τ∈𝒯(n)E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))].
Thus, in order to bound P1(n)(x)-P(0,x) from the above it suffices to find an upper bound of the right hand in (154).Next, we split the domain[0,T]×ℝ into three parts
(155)C={(t,x)∈[0,T-h]:μt+x>s(t+h)+|μ|h+κh},S={(t,x)∈[0,T-h]:μt+x≤s(t)-|μ|h-κh},B={(t,x)∈[0,T-h]×ℝ:s(t)-|μ|h-κh≤μt+x≤s(t+h)+|μ|h+κh}.
In order to estimate the right hand side of (154) we split it into three parts according to the domains C, S, and B; that is,
(156)E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))]=E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀((j-1)h,X(j-1)h)∈C]+E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀((j-1)h,X(j-1)h)∈S]+E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀((j-1)h,X(j-1)h)∈B].
By Proposition 3(ii) after the time β the prices of the American and game put options coincide which enables us to conclude that u(t,x)=e-rtPA(t,x+μt) for t≥β and that the sets Ct≥β={(t,x)∈C:t≥β}, St≥β={(t,x)∈S:t≥β}, and Bt≥β={(t,x)∈B:t≥β} are the same as the corresponding parts of the domains C-, S-, and B- introduced in [3] for the case of American put options. Therefore, we can use the following results from Sections 4.2 and 4.3 in [3].Proposition 29.
There exists a constantC>0 such that, for every τ∈𝒯(n),
(157)E[∑j=kβ(τ/h)∨kβ‍𝒟|u((j-1)h,X(j-1)h(n))|𝕀((j-1)h,X(j-1)h)∈C]≤C(ln⁡nn)4/5,
where kβ=min⁡{k:kh≥β} and
(158)E[∑j=kβ(τ/h)∨kβ‍𝒟u((j-1)h,X(j-1)h(n))𝕀((j-1)h,X(j-1)h)∈B]≤Cn3/4.Observe also thatP(t,x)=K-ex in the domain S, and so we can use there Lemma 2 from Section 4 of [3].Lemma 30.
For every(t,x)∈S one has 𝒟u(t,x)≤0, and so
(159)E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀((j-1)h,X(j-1)h)∈S]≤0.Thus, for an upper bound of the right side of (154) we can ignore the second term in the right hand side of (156) and estimate only two remaining terms starting with the first term in the right hand side of (156).Proposition 31.
There is a constantC>0 such that, for all n∈ℕ,
(160)E[∑j=1h-1(τ∧σ(n))‍|𝒟u((j-1)h,X(j-1)h(n))|𝕀((j-1)h,X(j-1)h)∈C]≤Cn-3/4.Proof.
We have(161)E[∑j=1h-1(τ∧σ(n))‍|𝒟u((j-1)h,X(j-1)h(n))|𝕀((j-1)h,X(j-1)h)∈C]=E[∑j=1(h-1(τ∧σ(n)))∧kβ‍|𝒟u((j-1)h,X(j-1)h(n))|×𝕀((j-1)h,X(j-1)h)∈C∑j=1(h-1(τ∧σ(n)))∧kβ‍]+E[∑j=kβ(h-1(τ∧σ(n)))∨kβ‍|𝒟u((j-1)h,X(j-1)h(n))|×𝕀((j-1)h,X(j-1)h)∈C∑j=kβ(h-1(τ∧σ(n)))∨kβ‍].
Proposition 29 provides a bound for the second term in the right hand side of (161), and so it remains to deal only with the first term there. Note that if jh<σ(n)∧β(n) and (jh,Xjh(n))∈C then
(162)c~1(j)=s(jh)-μjh+κh≤Xjh(n)≤ln⁡K-2κh-μjh=c~2(j),
where the equalities above are just definitions of c~1 and c~2. Observe also that since x<ln⁡K and jh<σ(n) then by the definition of the stopping times σ(n) the process Xjh(n)+μjh does not exceed ln⁡K-2Kh. By Proposition 5,
(163)𝒟u(t,x)=1κ∫0h‍dy∫-κyκy‍dz(z∂2u∂t∂x(t+y2,x+z)hhhhhhhhhhhhhhhh+δ(u)(t+y2,x+z)z∂2u∂t∂x).
Relying on the same computation as in Section 4 of [3] we see that for (t,x)∈C and x<ln⁡K-|μ|h-κh,
(164)|𝒟u(t,x)|≤hκ∫tt+h‍ds∫x-κhx+κh‍dz|∂2u∂t2(s,z)|.
Thus, for 0≤j<kβ,
(165)E(|𝒟u(jh,Xjh(n))|𝕀{(jh,Xjh(n))∈C}∩{jh<σ(n)})≤∫c~1(j)c~2(j)‍|Du(jh,y)|dPXjh(y)≤∫c~1(j)c~2(j)‍(h2κ∫jhjh+h‍ds∫y-κhy+κh‍|∂2u∂t2(s,z)|dz)dPXjh(y)=h2κ∫jh(j+1)h‍ds∫c~1(j)+κhc~2(j)+κh‍dz|∂2u∂t2(s,z)|×∫max⁡⁡(c~1(j),z-κh)min⁡⁡(c~2(j),z+κh)‍dPXjh(y)≤h2κ∫jh(j+1)h‍ds∫c~1(j)-κhc~2(j)+κh‍dz|∂2u∂t2(s,z)|×P[|Xjh(n)-z|≤κh].
From (24) we see that there is a constant C>0 independent of j and n such that
(166)P[|Xjh(n)-z|≤κh]≤Cj+1.
Hence, for jh<σ(n),
(167)E(|𝒟u(jh,Xjh(n))|𝕀(jh,Xjh(n))∈C𝕀jh<σ(n))≤h2κ∫jh(j+1)h‍ds∫c~1(j)-κhc~2(j)+κh‍dz|∂2u∂t2(s,z)|Cj+1=Ch2κ∫jh(j+1)h‍dsh(j+1)∫c~1(j)-κhc~2(j)+κh‍dz|∂2u∂t2(s,z)|≤C1n∫jh(j+1)h‍dss∫c~1(j)-κhc~2(j)+κh‍dz|∂2u∂t2(s,z)|.
Define
(168)c1(t)=s(t)-μt,c2(t)=ln⁡K-μt-κh,
where s(t)=ln⁡(b(t)) is the free boundary of the option holder and b(t) was introduced at the beginning of Section 3. Observe that for every j and any jh≤s≤(j+1)h,
(169)c~1(j)-κh≥c1(s),c~2(j)+κh≤c2(s).
Summing up the above estimates we obtain
(170)∑j=0kβ-1‍E(|𝒟u(jh,Xjh(n))|𝕀{(jh,Xjh(n))∈C}∩{jh<σ(n)})≤C2n+C1n∫hβ‍dss∫c1(s)c2(s)‍dz|∂2u∂t2(s,z)|=C2n+C1n(∫hh‍dss∫c1(s)c2(s)‍dz|∂2u∂t2(s,z)|ccccccccccccccccc+∫hβ‍dss∫c1(s)c2(s)‍dz|∂2u∂t2(s,z)|),
where the term C2/n comes from the first term E|𝒟u(0,x)| of the sum which can be estimated easily using the fact that ut(t,x) and uxx(t,x) are bounded for small t.
LetG={(t,x):0<t<β,c1(t)<x<ln⁡K-μt} and note that G⊂E∪D~, where E and D~ are defined in Corollaries 16 and 26 which imply that (∂2u/∂t2)(s,z)∈L1[F]. Hence,
(171)∫hβ‍dss∫c1(s)c2(s)‍dz|∂2u∂t2(s,z)|≤C1n1/4∫hβ‍ds∫c1(s)c2(s)‍dz|∂2u∂t2(s,z)|≤Cn1/4.
Next, we estimate the first integral in brackets in the right hand side of (170). Let s(β)<k<ln⁡K,k′=(ln⁡K-k)/2 and split the integral in question as follows:
(172)∫hh‍dss∫c1(s)c2(s)‍dz|∂2u∂t2(s,z)|=∫hh‍dss∫c1(s)k′-μs‍dz|∂2u∂t2(s,z)|+∫hh‍dss∫k′-μsc2(s)‍dz|∂2u∂t2(s,z)|.
From Corollary 16 we know that the function (∂2u/∂t2)(s,z) is in L2[E~], where
(173)E~={(s,z):0<s<T,c1(t)<z<k′-μt}⊂Eσ
(for an appropriate b<ln⁡K in the definition of Eσ). Therefore we can use the Cauchy-Schwarz inequality to obtain
(174)∫hh‍dss∫c1(s)k′-μs‍dz|∂2u∂t2(s,z)|≤(∫hh‍dss∫c1(s)k′-μs‍dz)1/2(∫hh‍∫c1(s)k′-μs‍|∂2u∂t2(s,z)|2dz)1/2≤Cln⁡n.
Now we are left with the second integral in the right hand side of (172). We will show that there is a constant C>0 such that
(175)ℐn=∫hh‍dss∫k′-μsc2(s)‍dz|∂2u∂t2(s,z)|≤Cn1/4.
Recall that u(t,x)=e-rtP(t,x+μt), and so
(176)∂2u∂t2(t,x)=e-rt(r2P(t,x+μt)-2rPt(t,x+μt)-2rμPx(t,x+μt))+e-rt(μ2Pxx(t,x+μt)+2μPxt(t,x+μt)+Ptt(t,x+μt)μ2).
Observe that the functions P(t,x),Px(t,x),Pt(t,x), and Pxx(t,x) are all bounded for small t. Indeed, P≤K+δ, while Pt is bounded in the domain of integration in (175) for small h in view of (115), (117), (136), and (137). Next, Px is bounded by Theorem 8.1 from [8]. Finally, Pxx is bounded since in the domain in question P and its first derivatives are bounded and P satisfies the equation ((∂/∂t)+A)P=0 (see (29)). Therefore, we can write
(177)ℐn≤∫hh‍dtt∫k′-μtc2(t)‍dx(|2μe-rtPtx(t,x+μt)|cccccccccccccccccccccc+|e-rtPtt(t,x+μt)|)+C1,
for some constant C1>0 independent of n. Recall that, for (x,t)∈D=(0,β)×(k,ln⁡K) by Proposition 25, Pt(t,x)=v(t,x)+w(t,x), where vt and vx belong to L2[D]. Hence, expressing Ptx and Ptt via vt,wt and vx,wx we can estimate the integral (177) containing vt and vx by means of the Cauchy-Schwarz inequality as it was done in (174). Replacing these integrals by C2ln⁡n we obtain
(178)ℐn≤∫hh‍dtt∫k′-μtc2(t)‍dx(|2μe-rtwx(t,x+μt)|cccccccccccccccccccccc+|e-rtwt(t,x+μt)|)+C2ln⁡n+C1.
By (124) and (126) the functions wt(t,x) and wx(t,x) do not change signs in D, and so it follows that
(179)∫hh‍dtt∫k′-μtc2(t)‍(|2μe-rtwx(t,x+μt)|+|e-rtwt(t,x+μt)|)dx=|∫hh‍dtt2|μ|e-rt∫k′ln⁡K-κh‍wx(t,x)dx|+|∫hh‍dtt∫k′-μtc2(t)‍e-rtwt(t,x+μt)dx|.
By Proposition 25, w(x,t) is bounded on D, and so the contribution of the first integral in the right hand side of (179) is bounded by a constant and it remains to estimate only the second integral there.
Next, we will need a more explicit representation of the functionw. Let
(180)z~(t,x)=e-rtw(t,x+μt).
Then in the domain E~={(t,x),0<t<β,k-μt<x<ln⁡K-μt},
(181)κ22z~xx(t,x)+z~t(t,x)=0.
Define
(182)z(t,x)=z~(T-t,κ2x)
and let
(183)E={(t,x):0<t<T,2(k-μt)κ<x<2(ln⁡K-μt)κ}.
In the domain E the function z(t,x) satisfies the heat equation
(184)zxx(t,x)=zt(t,x).
If we let
(185)d1(t)=2(k-μ(T-t))κ,d2(t)=2(ln⁡K-μ(T-t))κ,
then from the boundary values of w(t,x) we obtain
(186)z(0,x)=0ford1(0)<x<d2(0),z(t,d1(t))=0,z(t,d2(t))=e-r(T-t)for0<t≤T.
Note that z(t,x) is a bounded continuous function on the boundaries (t,di(t)),i=1,2,0<t≤T of E. Hence, by Chapter 14 of [13] we can represent z(t,x) in the form
(187)z(t,x)=∫0t‍∂H∂x(x-d1(τ),t-τ)ϕ1(τ)dτ+∫0t‍∂H∂x(x-d2(τ),t-τ)ϕ2(τ)dτ,
where H(t,x)=(1/4πt)e-x2/4t is the fundamental solution and the functions ϕi(t),i=1,2, are bounded continuous on the interval (0,T]. From the definition of z~ we see that
(188)z~t(t,x)=-re-rtw(t,x+μt)+e-rtwt(t,x+μt).
Since w(t,x) is bounded then for some constant C1>0 independent of n,
(189)|∫hh‍dtt∫k′-μtc2(t)‍e-rtwt(t,x+μt)dx|≤|∫hh‍dtt∫k′-μtc2(t)‍zt(t,x)dx|+C1.
From the representation (187) of z(t,x) we obtain that
(190)|∫hh‍dtt∫k′-μtc2(t)‍zt(t,x)dx|≤κ2|∫hh‍dtt∫(2/κ)(k′-μt)(2/κ)c2(t)‍ddt×∫0T-t‍∂H∂x(x-d1(τ),T-t-τ)ϕ1(τ)dτdx∫(2/κ)(k′-μt)(2/κ)c2(t)‍|+κ2|∫hh‍dtt∫(2/κ)(k′-μt)(2/κ)c2(t)‍ddt×∫0T-t‍∂H∂x(x-d2(τ),T-t-τ)ϕ2(τ)dτdx∫(2/κ)(k′-μt)(2/κ)c2(t)‍|.
Observe that as long as we keepx or t away from 0 the function H(x,t) is smooth and it has bounded derivatives with bounds depending on the range of t,x and their distance from zero. Next, if x satisfies
(191)2κ(k′-μt)<x<2κc2(t)=2κ(ln⁡K-μt-κh),
then
(192)k′-k≤2κ(k′-k+μ(T-t-τ))<x-d1(τ)for0<τ≤T-t.
Since k′>k we see that x-d1(τ) stays away from 0 on the entire interval (0,T-t]. It follows from the above that the function
(193)Φ1(t,x)=∫0T-t‍∂H∂x(x-d1(τ),T-t-τ)ϕ1(τ)dτ
has bounded derivatives with respect to t with bounds independent of n in the region {(t,x):h<t<h,(2/κ)(k′-μt)<x(2/κ)c2(t)}. We conclude that the first integral in the right hand side of (190) is bounded from above by a constant independent of n and it remains to estimate the second integral there.
Set(194)Φ2(t,x)=∫0T-t‍∂H∂x(x-d2(τ),T-t-τ)ϕ2(τ)dτ.
We see that if
(195)x<2κc2(t)=2κ(ln⁡K-μt-h),
then
(196)x-d2(τ)=x-2κ(ln⁡K-μ(T-τ))<2κ(μ(T-t-τ)-h).
In this case x-d2(τ) can be zero when τ∈[0,T-t], but this can happen only for a τ which is at least μ-1h apart from T-t. Thus, the function Φ2 is smooth with a bounded uniformly continuous derivative with respect to t though this bound may depend on n. Nevertheless, we still have the following:
(197)κ2|∫hh‍dtt∫(2/κ)(k′-μt)(2/κ)c2(t)‍ddt×(∫0T-t‍∂H∂x(x-d2(τ),T-t-τ)ϕ2(τ)dτ)dx∫(2/κ)(k′-μt)(2/κ)c2(t)‍|=κ2|∫hh‍dttddt(∫0T-t‍∫(2/κ)(k′-μt)(2/κ)c2(t)‍∂H∂xhhhhhhhhhhhhhh×(x-d2(τ),T-t-τ)dxϕ2(τ)dτ∫hh‍)|=κ2|∫hh‍dttddt(2κ∫0T-t‍(H(2κc2(t)-d2(τ),T-t-τ)hhhhhhhhhhhhhhhhhhhhhhhhh-H(2κ(k′-μt)ccccccccccccccccccccccccccccccccccccc-d2(τ),T-t-τ2κ))cccccccccccccccccccccc×ϕ2(τ)dτ∫hh‍)∫hh‍|≤κ2|∫hh‍dttddt×(∫0T-t‍H(2κc2(t)-d2(τ),T-t-τ)cccccccccccc×ϕ2(τ)dτ∫0T-t‍)∫hh‍|+κ2|∫hh‍dttddt(∫0T-t‍H(2κ(k′-ln⁡Khhhhhhhhhhhhhhhhhhhhhh+μ(T-t-τ)),T-t-τ∫0T-t‍)cccccccccccccccccccccccccccccccccc×ϕ2(τ)dτ2κ)∫hh‍|.
We see that in the second term in the right hand side k′-ln⁡K+μ(T-t-τ) can take on the value 0 for τ∈(0,T-t], but then τ is at least c=μ-1|k′-ln⁡K| apart from T-t and now the separation constant c does not depend on n. Thus, we can bound the second term there from above by a constant and it remains to estimate the first term which we do as follows:
(198)I=κ2|∫hh‍dttcccccccc×ddt∫0T-t‍H(2κc2(t)-d2(τ),T-t-τ)ccccccccccccccc×ϕ2(τ)dτ∫hh‍|≤C2∫hh‍dttcccccccc×∫0T-t‍|1(T-t-τ)3/2ccccccccccccccccc×exp⁡(-(((2/κ)(μ(T-t-τ)cccccccccccccccccccccccccccccc-κh))2)×(4(T-t-τ))-1)cccccccccccccccccccc×ϕ2(τ)1(T-t-τ)3/2|dτ+C2∫hh‍dttcccccccccc×∫0T-t‍|1(T-t-τ)1/2ccccccccccccccccccc×exp⁡(-(((2/κ)(μ(T-t-τ)ccccccccccccccccccccccccccccccc-κh))2)×(4(T-t-τ))-1)cccccccccccccccccccc×ϕ2(τ)1(T-t-τ)1/2|dτ+C2∫hh‍dtt∫0T-t‍|h(T-t-τ)5/2exp⁡(-κ2h2(T-t-τ))cccccccccccccccccccccccc×exp⁡(μh2-μ22κ2(T-t-τ))cccccccccccccccccccccccc×ϕ2(τ)h(T-t-τ)5/2|dτ,
where C2>0 is a constant independent of n. Analyzing the integral with respect to τ in the second term in the right-hand side above by considering different possible values of T-t-τ we conclude that this integral is bounded by a constant independent of n. Next we observe that |exp⁡⁡((μh/2)-(μ2/2κ2)(T-t-τ))ϕ2(τ)| is also bounded by a constant independent of n too. Hence, we obtain
(199)I≤C3+C3∫hh‍dtt∫0T-t‍1(T-t-τ)3/2hhhhhhhhhhhhhh×exp⁡⁡(-h2(T-t-τ))dτ+C3∫hh‍dtt∫0T-t‍h(T-t-τ)5/2hhhhhhhhhhh×exp⁡⁡(-h2(T-t-τ))dτ
for a constant C3>0 independent of n. Set ρ=h/2(T-t-τ) and note that dρ/dτ=-h/22(T-t-τ)3/2 and dρ2/dτ=-h/4(T-t-τ)2. We proceed by changing variables arriving at
(200)I≤C4+C4∫hh‍dtt∫h/2(T-t)∞‍1he-ρ2dρ+C4∫hh‍dtt∫h/2(T-t)∞‍1hρe-ρ2dρ2≤C4+C51h∫hh‍dtt≤C4+C52(1+1h1/4)≤C6n1/4,
for some constants C4,C5,C6>0 independent of n and (175) follows. Combining (175) and (174) we obtain from (172) that
(201)∫hh‍dss∫c2(s)c1(s)‍dz|∂2u∂t2(s,z)|≤Cn1/4.
Finally, Proposition 29 follows from (201), (170), and (171).Next, we turn our attention to the domainB. First, we will prove the following result.Lemma 32.
There exists a constantC>0 such that for all n∈ℕ,
(202)E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀((j-1)h,X(j-1)h)∈B]≤Cn-3/4.Proof.
LetBt<β(n) and Bt≥β(n) be the set of all points (t,x)∈B such that t<β(n) and t≥β(n), respectively. We split (202) according to these two regions, namely,
(203)E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀((j-1)h,X(j-1)h)∈B]=E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))cccccccccccc×𝕀((j-1)h,X(j-1)h)∈Bt<β(n)∑j=1h-1(τ∧σ(n))‍]+E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))ccccccccccccc×𝕀((j-1)h,X(j-1)h)∈Bt≥β(n)∑j=1h-1(τ∧σ(n))‍].
By Proposition 29 we have that for a constant C>0 independent of n,
(204)E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀((j-1)h,X(j-1)h)∈Bt≥β(n)]≤Cn-3/4.
Thus, it remains to estimate only the first term in the right-hand side. Let E={(t,x):0<t<β(n),a-μt<x<b-μt}, where a<s(0) and s(β(n)+h)+|μ|h+2κh<b<ln⁡K. For n large enough we can find such a b because s(t) is continuous and s(β(n))<ln⁡K. We know from Corollary 16 that u(t,x)∈H2[E]. Since C2[E] is dense in this space we can approximate u(t,x) by C2 functions to get equality (22) of Proposition 5 for u(t,x), as well. Since ut(t,x)+(κ2/2)uxx(t,x)≤0 in the domain E we obtain
(205)𝒟u(t,x)≤1κ∫0h‍dy∫-κyκy‍dz(z∂2u∂t∂x(t+y2,x+z))≤∫0h‍ydy∫-κyκy‍dz|∂2u∂t∂x(t+y2,x+z)|=12∫0h‍ds∫-κyκy‍dz|∂2u∂t∂x(t+s,x+z)|.
It follows that
(206)𝒟u(t,y)𝕀(t,y)∈B-≤12∫tt+h‍ds∫s(t)-λh-μts(t+h)+λh-μt‍𝕀|z-y|≤κh|∂2u∂t∂x(s,z)|dz,
where λ=|μ|+κ. Hence,
(207)E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀{((j-1)h,X(j-1)h)∈Bt<β(n)}]≤12(∑j=1kβ‍∫jh(j+1)h‍dτ∫s(jh)-λh-μjhs(jh+h)+λh-μjh‍P(|Xjh(n)-z|≤κh)cccccccccccccccccccccccccccccccc×|∂2u∂t∂x(s,z)|dz∑j=1kβ‍)+Cn.
Here kβ=⌈β/h⌉, and the term C/n is the contribution of 𝒟u(0,X0n)=𝒟u(0,x)≤C/n which holds true from by the definition of the operator 𝒟 and boundedness of ut and uxx for small t. From Corollary 16 we see that there exists a constant C1>0 such that
(208)∫ab‍|∂2u∂t∂x(t,z)|2dz≤C1when0≤t≤β(n).
This together with (24), the Cauchy-Schwarz inequality, and the inequality 1/τ≥1/2jh, which is satisfied when j≥1 and jh≤τ≤2jh, yields that
(209)E[∑j=1h-1(τ∧β(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀{((j-1)h,X(j-1)h)∈Bt<β(n)}]≤hC2∑j=1kβ‍∫jh(j+1)h‍dττ((s(j+1)h)-s(jh)+2λh)1/2+C2n.
From Proposition6 and Lipschitz continuity of the function P(t,x) in t≤β(n) uniformly in x≤ln⁡K (see Theorem 8.1 in [8]) we obtain that for some constant C3>0,
(210)|s(t1)-s(t2)|≤|t1-t2|C3whenever0≤t1,t2≤β(n).
Hence,
(211)E[∑j=0h-1(τ∧β(n))‍𝒟u((j-1)h,X(j-1)h(n))𝕀{((j-1)h,X(j-1)h)∈Bt<β(n)}]≤C4n3/4,
for some constant C4>0 independent of n.By combining the results of Lemma30, Proposition 31, and Lemma 32 together with (156) we obtain that the upper bound P1(n)(x)-P(0,x)<C/n3/4 for some constant C>0 independent of n and of x≤ln⁡K.Next, we will obtain a lower bound for the approximation errorP1(n)(x)-P(0,x) when x≤ln⁡K. Set
(212)τ(n)=inf⁡⁡{t:μ[th]h+Xt(n)<s([th]h+h)+|μ|h+κh}.
By Proposition 31,
(213)E[u(τ(n)∧σ(n),Xτ(n)∧σ(n)(n))]=u(0,x)+E[∑j=1τ(n)∧σ(n)/h‍𝒟u((j-1)h,X(j-1)h(n))]=u(0,x)+E[∑j=1h-1(τ∧σ(n))‍∣𝒟u((j-1)h,X(j-1)h(n))ccccccccccccccc×𝕀((j-1)h,X(j-1)h)∈C∑j=1h-1(τ∧σ(n))‍]≥P(0,x)-Cn-3/4.Setα=αn=T-(1/n2/3) and let τA(n) be defined by (212) with s there replaced by the free boundary sA for the American put option (see Section 2.2 in [3]). Define also τα(n)=τ(n)𝕀{τ(n)+h<α}+T𝕀{τ(n)+h≥α} and τA,α(n)=τA(n)𝕀{τA(n)+h<α}+T𝕀{τA(n)+h≥α}. We will rely on the following estimate from Section 4.5 in [3].Lemma 33.
There exists a constantC>0 independent of nℕ such that
(214)|E[uA(τA(n),XτA(n)(n))-erτA,α(n)ψ(μτA,α(n)+XτA,α(n))]|≤Cn2/3,
where uA(t,x)=e-trPA(t,x+μt) with PA given by (3).Remark 34.
Note thatsA(t)=s(t) for β≤t<T, and so τA(n)∨β=τ(n)∨β and τA,α(n)∨β=τα(n)∨β.From now on we assume thatn is large enough so that β(n)<α. From the definition of P1(n)(x) we have
(215)P1(n)(x)≥E[e-rτα(n)∧σ(n)(ψ(μτα(n)+Xτα(n)(n))𝕀{τα(n)≤σ(n)}+δ𝕀{σ(n)<τα(n)})].
Hence, if we prove that for some constant C>0 independent of n(216)J=|e-rτα(n)∧σ(n)E[u(τ(n)∧σ(n),Xτ(n)∧σ(n)(n))]-E[e-rτα(n)∧σ(n)(ψ(μτα(n)+Xτα(n)(n))𝕀{τα(n)≤σ(n)}+δ𝕀σ(n)<τα(n))]|≤Cn,
then by (215) and (213) we could conclude that
(217)-Cn≤P1(n)(x)-P(0,x).
We split the left hand side of (216) into three parts
(218)J=E[{(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))u(τ(n)∧β(n),Xτ(n)∧β(n)(n))-e-rτ(n)∧β(n)(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))}×𝕀{τα(n)≤σ(n)∧β(n)}{(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))u(τ(n)∧β(n),Xτ(n)∧β(n)(n))]+E[{u(σ(n),Xσ(n)(n))-e-rσ(n)δ}𝕀{σ(n)<τα(n)}]+E[𝕀{β(n)<τα(n)≤σ(n)}{u(τ(n),Xτ(n)(n))-e-rτα(n)(ψ(μτα(n)+Xτα(n)(n)))}×𝕀{β(n)<τα(n)≤σ(n)}].
This equality is true since τ(n)=τα(n)=τ(n)∧β on the set τα(n)≤σ(n)∧β(n)<α. We begin with the last term. First note that on the set β(n)≤τα(n)≤σ(n) we have, in particular, β(n)≤σ(n) and so σ(n)=T by Remark 34. In the case τα(n)>β(n) we have τα(n)=τA,α(n) and τ(n)=τA(n) and so from Lemma 33 we derive that
(219)|E[{u(τ(n),Xτ(n)(n))-e-rτα(n)(ψ(μτα(n)+Xτα(n)(n)))}×𝕀{β(n)<τα(n)≤σ(n)}](ψ(μτα(n)+Xτα(n)(n)))|≤|E[{u(τA(n),XτA(n)(n))-e-rτA,α(n)(ψ(μτA,α(n)+XτA,α(n)(n)))}]|≤Cn2/3.Next, we deal with the first term in the right hand side of (218) where τα(n)=τ(n)≤σ(n)∧β(n). This means that before time β(n) the process X(n) is stopped near the boundary s(t) and
(220)μτ(n)+Xτ(n)(n)<s(τ(n)+h)+|μ|h+σh.
By the definition, u(τ(n),Xτ(n)(n))=e-rτ(n)P(τ(n),X(n)+μτ(n)). Thus, we have
(221)E[{(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))u(τ(n)∧β(n),Xτ(n)(n))-e-rτ(n)∧β(n)(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))}×𝕀{τα(n)≤σ(n)∧β(n)}{(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))u(τ(n)∧β(n),Xτ(n)(n))]=E[{e-rτ(n)(P(τ(n),X(n)+μτ(n))-ψ(μτ(n)+Xτ(n)(n)))}×𝕀{τα(n)≤σ(n)∧β(n)}(μτ(n)+Xτ(n)(n))].
If μτ(n)+Xτ(n)(n)≤s(τ(n)), then P(τ(n),X(n)+μτ(n))-ψ(μτ(n)+Xτ(n)(n))=0, so we can assume that
(222)s(τ(n))<μτ(n)+Xτ(n)(n)<s(τ(n)+h)+|μ|h+σh.
To continue we need the following lemma.Lemma 35.
There is a constantC>0 independent of n such that for every point (t,x) satisfying s(t)≤μt+x≤s(t+h)+|μ|h+σh and 0≤t≤β(n),
(223)|P(t,μt+x)-ψ(μt+x)|≤Cn.Proof.
The functionP(t,x) is Lipschitz continuous when t≤β and 0≤x≤μβ+ln⁡K (see [8]), and so
(224)|P(t,μt+x)-P(t+h,μt+x)|≤Cn,
for some C>0 independent of n. If μt+x≤s(t+h) then P(t+h,μt+x)=ψ(μt+x) and we are done. Now assume that s(t+h)<μt+x<s(t+h)+λ/n, where λ=n(|μ|h+σh).
From Corollary15 it follows that for every t<T and a<ln⁡K the function Pxx(t,x) is continuous in x on the closed interval [s(t),a], so we can write
(225)P(t+h,μt+x)=P(t+h,s(t+h))+Px(t+h,s(t+h))λn+Pxx(t+h,s(t+h))λ22n+α,
where α=α(h) satisfies lim⁡h→0(α/h)=0. From the property of smooth fit (see [8]) it follows that Px(t+h,s(t+h))=ψx(s(t+h)), and so for some C>0 independent of n,
(226)|P(t+h,μt+x)-ψ(μt+x)|≤Cn.Using (222) and the above lemma we obtain
(227)|E[{(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))u((μτ(n)∧β(n)+Xτ(n)∧β(n)(n))τ(n)∧β(n),Xτ(n)∧β(n)(n))-e-rτ(n)∧β(n)(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))}×𝕀{τα(n)≤σ(n)∧β(n)}(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))]|≤Cn.
Hence, we are done with the first term in the right hand side of (218) and it remains to estimate the second one. Since σ(n)<τα(n)≤T the process X(n) is stopped near the writer’s boundary. Namely, we have
(228)ln⁡K-|μ|h-σh<μσ(n)+Xσ(n)(n)≤ln⁡K.
Since P(t,ln⁡K)=δ when t≤β, β(n)-β<h, and P is Lipschitz continuous (see Theorem 8.1 of [8]) we obtain that
(229)|P(σ(n),μσ(n)+Xσ(n)(n))-δ|≤Cn,
for some C>0 independent of n. Hence,
(230)E[(u(σ(n),Xσ(n)(n))-e-rσ(n)δ)𝕀{σ(n)<τα(n)}]≤Cn.
It follows that there exists C>0 independent of n such that, for every x≤ln⁡K,
(231)-Cn<P1(n)(x)-P(0,x).Next, we will derive a lower bound for the second approximation functionP2(n)(x) defined by (11), still assuming that x≤ln⁡K. According to (213) in order to obtain
(232)P2(n)(x)-P(0,x)≥-Cn2/3,
it suffices to show that
(233)E[u(τ(n)∧σ(n),Xτ(n)∧σ(n)(n))]-P2(n)(x)≤Cn2/3.
We have
(234)E[u(τ(n)∧σ(n),Xτ(n)∧σ(n)(n))]-P2(n)(x)≤E[u(τ(n)∧σ(n),Xτ(n)∧σ(n)(n))-e-rτα(n)∧σ(n)×(ψ(μτα(n)+Xτα(n)(n))𝕀{τα(n)≤σ(n)}+(ψ(μσ(n)+Xσ(n)(n))+δ)𝕀{σ(n)<τα(n)})e-rτα(n)∧σ(n)]=E[{u(τ(n)∧β(n),Xτ(n)∧β(n)(n))-e-rτ(n)∧β(n)(ψ(μτ(n)∧β(n)+Xτ(n)∧β(n)(n)))}ccccccc×𝕀{τα(n)≤σ(n)∧β(n)}Xτ(n)∧β(n)(n)]+E[𝕀{σ(n)<τα(n)}{u(σ(n),Xσ(n)(n))-e-rσ(n)(ψ(μσ(n)+Xσ(n)(n))+δ)}ccccccc×𝕀{σ(n)<τα(n)}Xτα(n)(n)]+E[{u(τ(n),Xτ(n)(n))-e-rτα(n)(ψ(μτα(n)+Xτα(n)(n)))}ccccccc×𝕀{β(n)<τα(n)≤σ(n)}Xτα(n)(n)].
Indeed, the first inequality is true since P2(n)(x) is defined as the sup on τ∈𝒯(n) and we chose a specific one, that is, τα(n). The equality is true due to the same reason that (218) holds true. We see that the first term in the right hand side of (234) is the same as the first term in (218) and by (227) it is less than C/n for some constant C. The second term is nonpositive because, for every (t,x) we have P(t,x)≤ψ(x)+δ and u(t,x)=e-rtP(t,μt+x), so we can just remove it from the equation. The last term is the same as the last term of (218) and from Lemma 33 we obtain that this term is less than or equal C/n2/3 for an appropriate C. These arguments yield (233) and hence (232), as well. For the upper bound we already know that P1(n)(x)-P(0,x)≤C/n3/4 and from the definition of P1(n) and P2(n) it is not hard to see that |P2(n)-P1(n)|≤C/n. It follows from above that there exist C>0 such that, for every x≤ln⁡K,
(235)-Cn3/2≤P2(n)(x)-P(0,x)≤Cn.


## 5.2. Casex>lnK

We begin with the upper bound onP1(n). We will show first that
(236)P1(n)(x)-P(0,x)≤sup⁡τ∈𝒯(n)E[∑j=1h-1(τ∧σ(n))‍𝒟u((j-1)h,X(j-1)h(n))].
The proof is similar to the proof of (154), we just have to show that for every τ∈𝒯(n),
(237)P(τ∧σ(n),μτ∧σ(n)+Xτ∧σ(n)(n))≥ψ(μτ+Xτ(n))𝕀{τ≤σ(n)}+(δ-Ke(|μ|h+2κh))𝕀{σ(n)<τ}.
On the set τ≤σ(n) this inequality is clear since P(t,x)≥ψ(x). For the case σ(n)<τ observe that because x>ln⁡K we must have
(238)ln⁡K<μσ(n)+Xσ(n)(n)<ln⁡K+|μ|h+2κh.
By Theorem 8.1 in [8] the right derivative Fx(t,K+) at K satisfies 0>Fx(t,K+)>-1 for any t, and so 0≤F(t,K)-F(t,K+Cλ)≤Cλ for each C>0 provided 0≤λ≤λ(C) is small enough. Assume 0<λ<1; then eλ-1≤λeλ≤λe. Hence, taking C=Ke we have
(239)P(t,ln⁡K)-P(t,ln⁡K+λ)=F(t,K)-F(t,Keλ)≤F(t,K)-F(t,K+Keλ)≤Keλ.
Put λ=|μ|h+κh; then for σ(n)<τ and sufficiently large n,
(240)P(σ(n),μσ(n)+Xσ(n)(n))≥P(σ(n),ln⁡K+λ)≥δ-Keλ.
Hence, we obtain (237) which yields also (234). To bound the right hand side of (236) we split it similarly into the case x≤ln⁡K (see (156)) according to the three different regions C,B, and S. Since our process starts at x>ln⁡K, if ((j-1)h,X(j-1)h(n))∈B for some j then this must happen after the time β, and so we can use (158). The part that belongs to the region S is nonpositive, so we can ignore it, and so we will be left only with the region C.Lemma 36.
For the discrete processXt(n) such that X0(n)=x>ln⁡K one has
(241)E[∑j=1h-1(τ∧σ(n))‍𝒟|u((j-1)h,X(j-1)h(n))|𝕀{((j-1)h,X(j-1)h)∈C}]≤Cn-3/4.Proof.
It suffices to show that(242)E[∑j=1kβ∧(σ(n)/h)‍𝒟|u((j-1)h,X(j-1)h(n))|𝕀{((j-1)h,X(j-1)h)∈C}]≤Cn-3/4,
for some C>0 independent of n since after time β(n) we come back to the American option case. This is done in the same way as in Proposition 31, and so we provide only a sketch of the proof. Let c(s)=ln⁡K-μs+κh; then similarly to the proof of Proposition 31 we obtain
(243)∑j=0kβ-1‍E(|𝒟u(jh,Xjh(n))|𝕀{(jh,Xjh(n))∈C}∩{jh<σ(n)})≤C2n+C1n∫hβ‍dss∫c(s)∞‍dz|∂2u(∂t2)(s,z)|.
Let (2/κ)(ln⁡K+|μ|T)<k′ and split the integral in (243) into two parts
(244)∫hβ‍dss∫c(s)∞‍dz|∂2u∂t2(s,z)|=∫hβ‍dss∫c(s)k′-μs‍dz|∂2u(∂t2)(s,z)|+∫hβ‍dss∫k′-μs∞‍dz|∂2u(∂t2)(s,z)|.
Let E={(s,z):0<s<β,k′-μs<z<∞}; then by Corollary 28 we see that (∂2u/∂t2)(s,z)∈L2[E], and so we obtain similarly to (174) that, for some constant C>0,
(245)∫hβ‍dss∫k′-μs∞‍dz|∂2u(∂t2)(s,z)|<Cln⁡n.
In the first integral in the right hand side of (244) we do the same procedure as in (170)–(175) relying on Proposition 25 and deriving that for some constant C>0,
(246)∫hβ‍dss∫c(s)k′-μs‍dz|∂2u(∂t2)(s,z)|<Cn1/4.
Combining (243)–(246) we obtain (242) and complete the proof of the lemma.The lower bound forP1(n)(x)-P(0,x) when x>ln⁡K is obtained similarly to the case x≤ln⁡K. As in that case we use the stopping time τ(n) from (212) and from the above we see that (213) is true also for the case under consideration. We consider again τα(n) defined before Lemma 33 and similarly to (214) obtain that
(247)|E[u(τα(n)∧σ(n),Xτα(n)∧σ(n))]-E[e-rτα(n)∧σ(n)(ψ(μτα(n)+Xτα(n)(n))𝕀{τα(n)≤σ(n)}cccccccccccccccccccccc+(δ-Ke(|μ|h+σh))cccccccccccccccccccccc×𝕀{σ(n)<τα(n)})](μτα(n)+Xτα(n)(n))|.
In order to estimate (247) for x>ln⁡K we only need to split it into two parts, one for τα(n)≤σ(n) and the other one for σ(n)<τα(n). This is true in view of the fact that if we begin with x>ln⁡K and τα(n)≤σ(n) then we must have β(n)≤τα(n), and so we are back to the American option case and can use Lemma 33 for this case. If σ(n)<τα(n) then the process X(n) is stopped near the seller’s boundary and similar to (230) we can use the Lipschitz property of P to obtain
(248)E[(𝕀{σ(n)<τα(n)}u(σ(n),Xσ(n)(n))-e-rσ(n)(δ-Ke(|μ|h+2κh))cccccc×𝕀{σ(n)<τα(n)})(σ(n),Xσ(n)(n))]≤Cn.
From here we can proceed similar to the case of x≤ln⁡K and obtain the lower bound for P1(n) proving (12) for P1(n).Next, we turn to the second approximation functionP2(n), still in the case of x>ln⁡K. For the upper bound we use Lemma 35 as in the case x≤ln⁡K and proceed similarly to the proof of the upper bound for the first approximation function P1(n). The proof of the lower bound is similar to the case x≤ln⁡K and we obtain the result observing that if x>ln⁡K then P(t,x)<ψ(x)+δ=δ for any t∈[0,T].


## 6. Computations

In this section we exhibit computations of price functions and free boundaries of game and American put options. All graphs of functions related to game put options were plotted using the approximation functionP2(2000) (see (11)). The graphs for the American put options were computed using the approximation function PA(2000) from [3].Figure1 shows both free boundaries of the holder and of the writer of a game put option and also the free boundary of the holder of an American put option corresponding to the option parameters K=20,r=0.02,κ=0.15,T=0.5,  and  δ=0.15. Here K is the strike of the option, r is the interest rate, κ is the volatility, T is the time to maturity, and δ is the writer’s cancelation penalty in the case of game option. The horizontal axis represents here the time coordinate. The free boundary of the holder runs lower for American than for the game put option.Figure 1
Free boundaries of American and game put options.In Figure2 we plot the graphs of an American put option price function and of a game put option price functions as functions of the initial value of the stock. Here δ=1.0 and δ=1.5, while other parameters are K=20,r=0.02,κ=0.15,andT=10. To see what is what here we recall that prices of game options do not exceed prices of corresponding American options and higher penalties increase prices.Figure 2
The price functions of American and game put options.Figure3 shows the holder’s free boundary for American and game put options where we use the same setup and parameters as in Figure 1 adding also plots of free boundaries for the game put options with penalty values δ=0.3 and δ=0.5. To see which is which we observe that lower penalties lead to higher running holder’s free boundaries in Figure 3, while the free boundary of the American put option remains the same here as in Figure 1.Figure 3
Holder free boundaries of American and game put options.



---

*Source: 743030-2014-01-30.xml*