# Valuation of Game Options in Jump-Diffusion Model and with Applications to Convertible Bonds

Authors: Lei Wang; Zhiming Jin
Journal: Journal of Applied Mathematics and Decision Sciences
 (2009)

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## Abstract

Game option is an American-type option with added feature that the writer can exercise the option at any time before maturity. In this paper, we consider some type of game options and obtain explicit expressions through solving Stefan(free boundary) problems under condition that the stock price is driven by some jump-diffusion process. Finally, we give a simple application about convertible bonds.


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## Body

## 1. Introduction

Let(Ω,ℱ,P) be a probability space hosting a Brownian
motion W={Wt:t≥0} and an independent Poisson process N={Nt:t≥0} with the constant arrival rate λ,
both adapted to some filtration F={ℱt}t≥0 satisfying usual conditions. Consider the
Black-Scholes market. That is, there is only one riskless bond B and a risky asset S. They satisfy, respectively,(1.1)dBt=rBtdt,t≥0,dSt=St−[μdt+σdWt−y0(dNt−λdt)]for some constants μ∈R,r,σ>0 and y0∈(0,1). Note that the absolute value of relative jump sizes is equal to y0,
and jumps are downwards. It can be comprehended as a downward tendency of the
risky asset price brought by bad news or default and so on. From Itô formula we
can obtain (1.2)St=S0exp{(μ−12σ2+λy0)t+σWt}(1−y0)Nt.Suppose thatX={Xt:t≤T} and Y={Yt:t≤T} be two continuous stochastic processes defined
on (Ω,ℱ,F,P) such that for all 0≤t≤T,Xt≤Yt
a.s.. The game option is a contract between a
holder and writer at time t=0. It is a general American-type option with the added property that the writer
has the right to terminate the contract at any time before expiry time T. If the holder exercises first, then he/she may obtain the value of X at the exercise time and if the writer
exercise first, then he/she is obliged to pay to the holder the value of Y at the time of exercise. If neither has
exercised at time T and T<∞,
then the writer pays the holder the value XT. If both decide to claim at the same time then the lesser of the two claims is
paid. In short, if the holder will exercise with strategy τ and the writer with strategy γ,
we can conclude that at any moment during the life of the contract, the holder
can expect to receive Z(τ,γ)≜Xτ1(τ≤γ)+Yγ1(γ<τ). For a detailed description and the valuation of game options, we refer the
reader
to Kifer [1],
Kyprianou [2], Ekström [3], Baurdoux and Kyprianou [4], Kühn et al. [5], and so on.It is well known that in the no-arbitrage pricing
framework, the value of a contract contingent on the assetS is the maximum of the expectation of the total
discounted payoff of the contract under some equivalent martingale measure. Since the market is incomplete, there are more than one equivalent
martingale
measure. Following Dayanik and Egami [6], let the restriction to ℱt of every equivalent martingale measure Pα in a large class admit a Radon-Nikodym
derivative in the form of(1.3)dPαdP|ℱt≜ηt,dηt=ηt−[βdWt+(α−1)(dNt−λdt)],t≥0,η0=1for some constants β∈R and α>0. The constants β and α are known as the market price of the diffusion
risk and the market price of the jump risk, respectively, and satisfy the drift
condition(1.4)μ−r+σβ−λy0(α−1)=0.Then the discounted value
process {e−rtSt:t≥0} is a (Pα,F)-martingale. By the Girsanov theorem, the
process {Wtα≜Wt−βt:t≥0} is a Brownian motion under the measure Pα,
and {Nt:t≥0} is a homogeneous Poisson process with the
intensity λα≜λα independent of the Brownian motion Wα under the same measure. The infinitesimal
generator of the process S under the probability measure Pα is given by(1.5)𝒜αf(x)≜(r+λαy0)x∂f∂x+12σ2x2∂2f∂x2+λα[f(x(1−y0))−f(x)],on the collection of
twice-continuously differentiable functions f(⋅). It is easily checked that (𝒜α−r)f(x)=0 admits two solutions f(x)=xk1 and f(x)=xk2,
where k1<0<1=k2 satisfy(1.6)12σ2k(k−1)+(r+λαy0)k−(r+λα)+λα(1−y0)k=0.Suppose that Pxα is the equivalent martingale measure for S under the assumption that S0=x for a specified market price α(⋅) of the jump risk, and denote Exα to be expectation under Pxα. The following theorem is the Kifer pricing result.Theorem 1.1.
Suppose
that for allx>0(1.7)Exα(sup0≤t≤Te−rtYt)<∞and if  T=∞   that  Pxα(limt↑∞e−rtYt=0)=1. Let  𝒮t,T   be the class of  F-stopping times valued in  [t,T], and  𝒮≡𝒮0,∞, then the price of the game option is given by (1.8)V(x)=infγ∈𝒮0,Tsupτ∈𝒮0,TExα(e−r(τ⋀γ)Zτ,γ)=supτ∈𝒮0,Tinfγ∈𝒮0,TExα(e−r(τ⋀γ)Zτ,γ).Further the optimal stopping
strategies for the holder and writer, respectively, are (1.9)τ∗=inf{t≥0:V(St)=Xt}⋀T,γ∗=inf{t≥0:V(St)=Yt}⋀T.


## 2. A Game
Version of the American Put Option (Perpetual Israeliδ-Penalty Put Option)

In this case,
continuous stochastic processes are, respectively, given by(2.1)Xt=(K−St)+,Yt=(K−St)++δ,where K>0 is the strike-price of the option, δ>0 is a constant and can be considered as penalty
for terminating contract by the writer. For the computation of the following,
let us first consider the case of the perpetual American put option with the
same parameter K. From Jin [7] we know
that the price of the option is(2.2)VA(x)=supτ∈𝒮Exα(e−rτ(K−Sτ)+)with the superscript A representing American. Through martingale
method we have the following.Theorem 2.1.
The
price of the perpetual American option is given by(2.3)VA(x)={K−xx∈(0,x∗],(K−x∗)(xx∗)k1x∈(x∗,∞),where  x∗=k1K/(k1−1), the optimal stopping strategy is (2.4)τ∗=inf{t≥0:St≤x∗}.Proposition 2.2.
VA(x)   is decreasing and convex on  (0,∞), and under equivalent martingale measure  Pxα, one has that  {e−rtVA(St):t≥0}   and  {e−r(t⋀τx∗)VA(St⋀τx∗):t≥0}   are supermartingale and martingale,
respectively.Now, let us consider this game option. It is obvious
that for the holder, in order to obtain the most profit, he will exercise whenS becomes as small as possible. Meanwhile, he
must not wait too long for this to happen, otherwise he will be punished by the
exponential discounting. Then the compromise is to stop when S is smaller than a given constant. While for
the writer, a reasonable strategy is to terminate the contract when the value
of the asset S equals to K. Then only the burden of a payment of the form δe−rτ is left. For this case, if the initial value
of the risky asset is below K then it would seem rational to terminate the
contract as soon as S hits K. On the other hand, if the initial value of the risky asset is above K,
it is not optimal to exercise at once although the burden of the payment at
this time is only δ. A rational strategy is to wait until the last moment that St≥K in order to prolong the payment. However, it
should be noted that the value of the δ must not be too large, otherwise it will be
never optimal for the writer to terminate the contract in advance.Theorem 2.3.
Letδ∗≜VA(K)=(K−x∗)(K/x∗)k1, one has the following.
(1) Ifδ≥δ∗,
then the price of this game option is equal to the price of the perpetual
American put option, that is, it is not optimal for the writer to terminate the
contract in advance.
(2) Ifδ<δ∗,
then the price of the game option is(2.5)V(x)={K−xx∈(0,k∗],Ax+Bxk1x∈(k∗,K),δ(xK)k1x∈[K,∞)with (2.6)A=δk∗k1−(K−k∗)Kk1Kk∗k1−k∗Kk1,B=K(K−k∗)−δk∗Kk∗k1−k∗Kk1,and the optimal stopping
strategies for the holder and writer, respectively, are (2.7)τ∗=inf{t≥0:St≤k∗},γ∗=inf{t≥0:St=K},where k∗ is the (unique) solution in (0,K) to the equation (2.8)(δ+K)(1−k1)xk1+K2k1xk1−1−K1+k1=0.Before the proof, we will first give
two propositions.Proposition 2.4.
Equation (2.8) has and only has one root in (0,K).Remark 2.5.
If
we denote the root of (2.8) in (0,K) by k∗,
then from Proposition 2.4 we know that K(K−k∗)−δk∗>0,
thus B>0.Proposition 2.6.
V(x)
defined by the right-hand sides of (2.5) is
convex and decreasing on  (0,∞).Proof.
From the expression ofV(x) and Remark 2.5 we know that V(x) is convex on (0,K) and (K,∞). Thus, we only need to prove the convexity of V(x) at the point K,
that is, V′(K+)≥V′(K−). Through elementary calculations we obtain(2.9)V′(K−)=1Kk∗k1−k∗Kk1[δk∗k1−(K−k∗)Kk1+(K(K−k∗)−δk∗)k1Kk1−1],V′(K+)=δk1K.Then if we can prove
that(2.10)δk∗k1−(K−k∗)Kk1≤0,V′(K+)≥V′(K−) will hold. From (2.8) we can easily find that
when δ=δ∗, k∗=x∗. Further, as δ decreases the solution k∗ increases. Especially, when δ=0, k∗=K. So if 0<δ<δ∗,
we have x∗<k∗<K.
Now let us
verify the correctness of (2.10). If not, that is, δ>(K−k∗)(K/k∗)k1,
then from (2.8) we obtain(2.11)K1+k1−K2k1k∗k1−1−K(1−k1)k∗k1=δ(1−k1)k∗k1>(K−k∗)(1−k1)Kk1,rearranging it we
have(2.12)(k∗Kk1−Kk∗k1)(1−k1+k1Kk∗)>0.Since k∗>x∗,
so 1−k1+k1K/k∗>0,
whereas k∗Kk1−Kk∗k1<0,
which contradicts with (2.12). So the hypothesis is not true, that is, (2.10)
holds, which also implies that A≤0. So V(x) is decreasing on (0,∞).Proof of Theorem2.3.
(1) Suppose thatδ≥δ∗. From the expression of VA(x) we can easily find that(2.13)(K−x)+≤VA(x)≤(K−x)++δ.By means of Proposition 2.2 and the
Doob Optional Stopping Theorem, we have(2.14)VA(x)=infγ∈𝒮Exα[e−r(τ∗⋀γ)VA(Sτ∗⋀γ)]≤infγ∈𝒮Exα{e−rτ∗(K−Sτ∗)+1(τ∗≤γ)+e−rγ[(K−Sγ)++δ]1(γ<τ∗)}≤infγ∈𝒮supτ∈𝒮Exα{e−rτ(K−Sτ)+1(τ≤γ)+e−rγ[(K−Sγ)++δ]1(γ<τ)}=supτ∈𝒮infγ∈𝒮Exα{e−rτ(K−Sτ)+1(τ≤γ)+e−rγ[(K−Sγ)++δ]1(γ<τ)}≤supτ∈𝒮Exα[e−rτ(K−Sτ)+]=VA(x).That is, the price of the game
option is equal to the price of the perpetual American put option.
(2) Ifδ<δ∗,
according to the foregoing discussion and Theorem 1.1, there exists a number k such that the continuation region
is(2.15)C={x:g1(x)<V(x)<g2(x)}={x:k<x<∞,x≠K}with g1(x)=(K−x)+,g2(x)=(K−x)++δ,k∈(0,K) a constant to be confirmed, while the stopping
area is(2.16)D=D1∪D2,where D1={x:V(x)=g1(x)}={x:x≤k} is the stopping area of the holder, D2={x:V(x)=g2(x)}={x:x=K} is the stopping area of the writer. For search
of the optimal k∗ and the value of V(x),
we consider the following Stefan(free boundary) problem with unknown number k and V=V(x):  (2.17)V(x)=K−x,x∈(0,k],(𝒜α−r)V(x)=0,x∈(k,K)∪(K,∞),and additional conditions on the
boundary k and K are given by(2.18)limx↓kV(x)=K−k,limx→KV(x)=δ,limx↓k∂V(x)∂x=−1,limx↑∞V(x)=0.
By computing
Stefan problem we can easily obtain the expression ofV(x) (denote it by V˜(x)) defined by the right-hand sides of (2.5),
while from (2.18) we can obtain (2.8). Proposition 2.4 implies
that this equation has and only has one root in (0,K),
denote it by k∗. Accordingly, we can obtain the expression (2.6) of A and B and optimal stopping strategy τ∗ for the holder. Now we must prove that the
solution of the Stefan problem gives, in fact, the solution to the optimal
stopping problem, that is, V(x)=V˜(x). For that it is sufficient to prove that
(a)
∀τ∈𝒮,Exαe−r(τ⋀γ∗)Zτ,γ∗≤V˜(x);(b)
∀γ∈𝒮,Exαe−r(τ∗⋀γ)Zτ∗,γ≥V˜(x);(c)
Exαe−r(τ∗⋀γ∗)Zτ∗,γ∗=V˜(x).
First, from
Proposition2.6 we know that V˜(x) is a convex function on (0,∞) such that(2.19)(K−x)+≤V˜(x)≤(K−x)++δ.Since V˜(x)∈C1(0,K)∩C2(0,K)∖{k∗},
for x∈(0,K),
we can apply Itô formula to the process {e−r(t⋀γ∗)V˜(St⋀γ∗):t≥0} and have(2.20)e−r(t⋀γ∗)V˜(St⋀γ∗)=V˜(x)+∫0t⋀γ∗e−ru(𝒜α−r)V˜(Su)du+∫0t⋀γ∗e−ruσSuV˜'(Su)dWuα+∫0t⋀γ∗e−ru[V˜(Su−(1−y0))−V˜(Su−)](dNu−λαdu).Note that in (0,K), 𝒜αV˜(x)−rV˜(x)≤0,
while the last two integrals of (2.20) are local martingales, then by choosing
localizing sequence and apply the Fatou lemma, we obtain(2.21)Exαe−r(τ⋀γ∗)V˜(Sτ⋀γ∗)≤V˜(x),whereas(2.22)Zτ,γ∗=(K−Sτ)+1τ≤γ∗+[(K−Sγ∗)++δ]1γ∗<τ=(K−Sτ)+1τ≤γ∗+δ1γ∗<τ≤V˜(Sτ⋀γ∗).For the inequality we have used
(2.19), hence from (2.21) we have(2.23)Exαe−r(τ⋀γ∗)Zτ,γ∗≤V˜(x).It is simple for the case that x∈(K,∞) and the method is the same as before. Thus, we
obtain (a).
The proof of (b): apply Itô formula to the process{e−r(τ∗⋀t)V˜(Sτ∗⋀t):t≥0} and note that V˜ is only continuous at K,
we have(2.24)e−r(τ∗⋀t)V˜(Sτ∗⋀t)=V˜(x)+∫0τ∗⋀te−ru(𝒜α−r)V˜(Su)du+∫0τ∗⋀te−ruσSuV˜'(Su)dWuα+∫0τ∗⋀te−ru[V˜(Su−(1−y0))−V˜(Su−)](dNu−λαdu)+e−r(τ∗⋀t)[V˜'(K+)−V˜'(K−)]Lτ∗⋀tK,where LK is the local time at K of S. Since V˜(x) is convex on (0,∞),
hence V˜'(K+)−V˜'(K−)≥0. While in (k∗,∞)∖{K}, (𝒜α−r)V˜(x)=0,
then using the same method as before we have(2.25)Exαe−r(τ∗⋀γ)V˜(Sτ∗⋀γ)≥V˜(x).Moreover, since(2.26)V˜(Sτ∗⋀γ)=V˜(Sτ∗)1(τ∗≤γ)+V˜(Sγ)1(γ<τ∗)≤(K−Sτ∗)+1(τ∗≤γ)+[(K−Sγ)++δ]1(γ<τ∗)=Zτ∗,γ,we can obtain(2.27)Exαe−r(τ∗⋀γ)Zτ∗,γ≥V˜(x),∀γ∈ℳ.
The proof of (c): takingτ=τ∗,γ=γ∗,
it is sufficient to note that in (k∗,K),
we have 𝒜αV(x)−rV(x)=0 and(2.28)V˜(Sτ∗⋀γ∗)=V˜(Sτ∗)1(τ∗≤γ∗)+V˜(Sγ∗)1(γ∗<τ∗)=(K−Sτ∗)+1(τ∗≤γ∗)+δ1(γ∗<τ∗)=Zτ∗,γ∗.The same result is true for the
case that x∈(K,∞).


## 3. Game Option with Barrier

Karatzas and
Wang [8] obtain
closed-form expressions for the prices and optimal hedging strategies of
American put options in the presence of an up-and-out  barrier by
reducing this problem to a variational inequality. Now we will consider the
game option connected with this barrier option. Following Karatzas and Wang,
the holder may exercise to take the claim of this barrier
option(3.1)Xt=(K−St)+1(t<τh),0≤t<∞.Here h>0 is the barrier, whereas(3.2)τh=inf{t≥0:St>h}is the time when the option
becomes “knocked-out”. The writer is punished by an amount δ for terminating the contract
early(3.3)Yt=[(K−St)++δ]1(t<τh).First, let us consider this type of barrier option. The price is given by(3.4)VB(x)=supτ∈𝒮Exe−rτ(K−Sτ)+1(τ<τh)with the superscript B representing barrier. Similarly to Karatzas
and Wang we can obtain the following.Theorem 3.1.
The
price of American put-option in the presence of an up-and-out barrier is(3.5)VB(x)={K−xx∈(0,p∗],Ax+Bxk1x∈(p∗,h),0x∈[h,∞),where  A=(p∗−K)hk1/(hp∗k1−p∗hk1),B=(K−p∗)h/(hp∗k1−p∗hk1), and the optimal stopping strategy is (3.6)τ∗=inf{t≥0:St≤p∗},where  p∗   is the (unique)   solution in  (0,K)   to the equation (3.7)h(1−k1)xk1+Khk1xk1−1−Khk1=0.The proof of the theorem mainly depends on the
following propositions and the process will be omitted.Proposition 3.2.
The expression ofVB(x)
defined by (3.5) is convex and decreasing
on  (0,∞), and under risk-neutral measure  Pxα, one has that  {e−rtVB(St):t≥0} and {e−r(t⋀τ∗)VB(St⋀τ∗):t≥0}   are supermartingale and martingale,
respectively.Proposition 3.3.
Equation (3.7) has and only has one root in  (0,K).Now let us consider the game option with barrierh. The price is given by(3.8)V(x)=supτ∈𝒮infγ∈𝒮Ex(e−rτ(K−Sτ)+1(τ≤γ)⋅1(τ<τh)+e−rγ[(K−Sγ)++δ]1(γ<τ)⋅1(γ<τh)).For this game option, the logic
of its solution is similar to the former, and based on this consideration, we
have the following theorem.Theorem 3.4.
Letδ∗≜VB(K)=(K−p∗)(hKk1−Khk1)/(hp∗k1−p∗hk1),
one has the following.
(1) Ifδ≥δ∗,
then the price of this game option is equal to the price of American
put options in the presence of an  up-and-out barrier, that
is, it is not optimal for the writer to exercise early.
(2) Ifδ<δ∗,
then the price of the game option is given by(3.9)V(x)={K−xx∈(0,b∗],C1x+C2xk1x∈(b∗,K),D1x+D2xk1x∈[K,h),0x∈[h,∞),where(3.10)C1=δb∗k1−(K−b∗)Kk1Kb∗k1−b∗Kk1,C2=K(K−b∗)−δb∗Kb∗k1−b∗Kk1,D1=−δhk1hKk1−Khk1,D2=δhhKk1−Khk1,and b∗ is the (unique) solution in (0,K) to the equation(3.11)(δ+K)(1−k1)xk1+K2k1xk1−1−K1+k1=0,and the optimal stopping
strategies for the holder and writer, respectively, are(3.12)τ∗=inf{t≥0:St≤b∗},γ∗=inf{t≥0:St=K}.Proposition 3.5.
The functionV(x) defined by (3.9) is convex and decreasing on (0,∞).Proof.
Similar to Proposition2.6, we only need to prove the convexity of V(x) at the point K,
that is,(3.13)V′(K+)−V′(K−)=(D2−C2)k1Kk1−1+(D1−C1)≥0.Through lengthy calculations we
know that it is sufficient to show that(3.14)δ(hb∗k1−b∗hk1)≤(K−b∗)(hKk1−Khk1).Suppose that (3.14) does not
hold, that is, δ>(K−b∗)(hKk1−Khk1)/(hb∗k1−b∗hk1),
then from (3.11) we find that(3.15)K1+k1−K2k1xk1−1−K(1−k1)b∗k1=δ(1−k1)b∗k1>(1−k1)b∗k1(K−b∗)(hKk1−Khk1)(hb∗k1−b∗hk1),rearranging it we
have(3.16)h(1−k1)b∗k1+Khk1b∗k1−1−Khk1<0.From (3.11), through complex
verification we get that when δ=δ∗, b∗=p∗. Furthermore, as δ decreases the solution b∗ increases, especially when δ=0, b∗=K. So if 0<δ<δ∗,
we have p∗<b∗<K. Thus from the property of (3.7) we know that h(1−k1)b∗k1+Khk1b∗k1−1−Khk1>0,
which contradicts with (3.16). So the hypothesis is not true, that is, (3.14)
holds. It is evident that V(x) is decreasing.Remark 3.6.
It
is obvious that  (2.8) is the same as (3.11), however, their roots not
always be equal to each other. Because of these two cases, the scope of δ is different. Penalty with barrier is usually
smaller than the other, that is, VB(K)<VA(K).Proof of Theorem3.4.
(1) Suppose thatδ≥δ∗. From Proposition 3.2 we know that(3.17)(K−x)+≤VB(x)≤(K−x)++δ.By means of the Doob optional
stopping theorem and (3.17), we have(3.18)VB(x)=infγ∈𝒮Exα[e−r(τ∗⋀γ⋀τh)VB(Sτ∗⋀γ⋀τh)]≤infγ∈𝒮Exα{e−rτ∗(K−Sτ∗)+1(τ∗≤γ)1(τ∗<τh)+e−rγ[(K−Sγ)++δ]1(γ<τ∗)1(γ<τh)}≤infγ∈𝒮supτ∈𝒮Exα{e−rτ(K−Sτ)+1(τ≤γ)1(τ<τh)+e−rγ[(K−Sγ)++δ]1(γ<τ)1(γ<τh)}=supτ∈𝒮infγ∈𝒮Exα{e−rτ(K−Sτ)+1(τ≤γ)1(τ<τh)+e−rγ[(K−Sγ)++δ]1(γ<τ)1(γ<τh)}≤supτ∈𝒮Exα[e−rτ(K−Sτ)+1(τ<τh)]=VB(x).That is, the price of the game
option is equal to the price of American put-options in the presence of an
up-and-out barrier.
(2) Suppose thatδ<δ∗. Then we may conclude that the holder should search optimal stopping strategy in
the class of the stopping times of the form τb=inf{t≥0:St≤b} with b∈(0,K) to be confirmed. While the optimal stopping
strategy for the writer is γ∗=inf{t≥0:St=K}. Considering the following Stefan problem:(3.19)V(x)=K−x,x∈(0,b],(3.20)(𝒜−r)V(x)=0,x∈(b,K)∪(K,h),(3.21)V(x)=0,x∈[h,∞),(3.22)limx↓bV(x)=K−b,limx→KV(x)=δ,limx↑hV(x)=0,limx↓b∂V(x)∂x=−1.Through straightforward
calculations we can obtain the expression of V(x) (denote it by V˜(x)) defined by the right-hand sides of (3.9). From condition (3.22) we can obtain (3.11). Proposition 2.4 implies
that the root of this equation is unique in (0,K),
denote it by b∗ and consequently τb∗ by τ∗. Now we only need to prove that V(x)=V˜(x). For that it is sufficient to prove that(3.23)(a)∀τ∈𝒮,Exαe−r(τ⋀γ∗)Zτ,γ∗1(τ⋀γ∗<τh)≤V˜(x);(3.24)(b)∀γ∈𝒮,Exαe−r(τ∗⋀γ)Zτ∗,γ1(τ∗⋀γ<τh)≥V˜(x).
(c) Taking stopping timeτ=τ∗,γ=γ∗, we have(3.25)Exαe−r(τ∗⋀γ∗)Zτ∗,γ∗1(τ∗⋀γ∗<τh)=V˜(x).
First, from Proposition3.5 we know that V˜(x) is convex in (0,∞) and further(3.26)(K−x)+≤V˜(x)≤(K−x)++δ.Applying Itô formula to the process {e−r(t⋀γ∗⋀τh)V˜(St⋀γ∗⋀τh):t≥0},
we have(3.27)e−r(t⋀γ∗⋀τh)V˜(St⋀γ∗⋀τh)=V˜(x)+∫0t⋀γ∗⋀τhe−ru(𝒜α−r)V˜(Su)du+∫0t⋀γ∗⋀τhe−ruσSuV˜'(Su)dWuα+∫0t⋀γ∗⋀τhe−ru[V˜(Su−(1−y0))−V˜(Su−)](dNu−λαdu).It is obvious that when x∈(0,K)∪(K,h),
we have (𝒜−r)V˜(x)≤0. Since the second and the third integrals of the right-hand sides of (3.27) are
local martingales, so(3.28)Exαe−r(τ⋀γ∗⋀τh)V˜(Sτ⋀γ∗⋀τh)≤V˜(x),while(3.29)Exαe−r(τ⋀γ∗⋀τh)V˜(Sτ⋀γ∗⋀τh)=Exαe−r(τ⋀γ∗)V˜(Sτ⋀γ∗)1(τ⋀γ∗<τh)≥Exαe−r(τ⋀γ∗)Zτ,γ∗1(τ⋀γ∗<τh).The inequality is obtained from
(3.26), and combining (3.28) we obtain (3.23), that is, (a) holds.
Applying Itô formula to the process{e−r(τ∗⋀t⋀τh)V˜(Sτ∗⋀t⋀τh):t≥0},
we have(3.30)e−r(τ∗⋀t⋀τh)V˜(Sτ∗⋀t⋀τh)=V˜(x)+∫0τ∗⋀t⋀τhe−ru(𝒜−r)V˜(Su)du+∫0τ∗⋀t⋀τhe−ruσ(Su)V˜'(Su)dWuα+e−r(τ∗⋀t⋀τh)[V˜'(K+)−V˜'(K−)]Lτ∗⋀t⋀τhK+∫0τ∗⋀t⋀τhe−ru[V˜(Su−(1−y0))−V˜(Su−)](dNu−λαdu).The definition of LK is the same as Theorem 2.3. From the convexity
of V˜(x) we know that V˜'(K+)−V˜'(K−)≥0. Since when x∈(b∗,K)∪(K,h),(𝒜−r)V˜(x)=0,
so from above expression we have(3.31)Exαe−r(τ∗⋀γ⋀τh)V˜(Sτ∗⋀γ⋀τh)≥V˜(x).Similarly we
have(3.32)Exαe−r(τ∗⋀γ⋀τh)V˜(Sτ∗⋀γ⋀τh)=Exαe−r(τ∗⋀γ)V˜(Sτ∗⋀γ)1(τ∗⋀γ<τh)≤Exαe−r(τ∗⋀γ)Zτ∗,γ1(τ∗⋀γ<τh).From (3.31) and (3.32) we know
that (b) holds. Combining (a) and (b) we can easily obtain (c).


## 4. A Simple Example: Application to Convertible Bonds

To raise
capital on financial markets, companies may choose among three major asset
classes: equity, bonds, and hybrid instruments, such as convertible bonds. As
hybrid instruments, convertible bonds has been investigated rather extensively
during the recent years. It entitles its owner to receive coupons plus the
return of the principle at maturity. However, the holder can convert it into a
preset number of shares of stock prior to maturity. Then the price of the bond
is dependent on the price of the firm stock. Finally, prior to maturity, the
firm may call the bond, forcing the bondholder to either surrender it to the
firm for a previously agreed price or else convert it for stock as above. Therefore, the pricing problem has also a game-theoretic aspect. For more
detailed information and research about convertible bonds, one is
referred to
Gapeev and Kühn [9], Sîrbu et al.
[10, 11], and so on.Now, we will
give a simple example of pricing convertible bonds, as the application of
pricing game options. Consider the stock process which pays dividends at a
certain fixed rated∈(0,r),
that is,(4.1)dSt=St−[(μ−d)dt+σdWt−y0(dNt−λdt)].Then the infinitesimal generator
of S becomes(4.2)𝒜αf(x)≜(r−d+λαy0)x∂f∂x+12σ2x2∂2f∂x2+λα[f(x(1−y0))−f(x)],and (𝒜α−r)f(x)=0 admits two solutions f(x)=xk1 and f(x)=xk2 with k1<0<1<k2 satisfying(4.3)12σ2k(k−1)+(r−d+λαy0)k−(r+λα)+λα(1−y0)k=0.At any time,
the bondholder can convert it into a predetermined numberη>0 of stocks, or continue to hold the bond and
collecting coupons at the fixed rate c>0. On the other hand, at any time the firm can call the bond, which requires the
bondholder to either immediately surrender it for the fixed conversion price K>0 or else immediately convert it as described
above. In short, the firm can terminate the contract by paying the amount max{K,ηS} to the holder. Then, if the holder terminates
the contract first by converting the bond into η stocks, he/she can expect to (discounted)
receive(4.4)Lt=∫0tc⋅e−rudu+e−rtηSt,while if the firm terminate the
contract first, he/she will pay the holder(4.5)Ut=∫0tc⋅e−rudu+e−rt(K∨ηSt).Then, according to Theorem 1.1,
the price of the convertible bonds is given by(4.6)VCB(x)=infγ∈𝒮supτ∈𝒮Exα(Lτ1(τ≤γ)+Uγ1(γ<τ))=supτ∈𝒮infγ∈𝒮Exα(Lτ1(τ≤γ)+Uγ1(γ<τ)).Note that when c≥rK,
the solution of (4.6) is trivial and the firm should call the bond immediately. This implies that the bigger the coupon rate c,
the more the payoff of the issuer, then they will choose to terminate the
contract immediately. So we will assume that c<rK in the following.Now, let us
first consider the logic of solving this problem. It is obvious thatηx≤VCB(x)≤K∨ηx for all x>0 (choose τ=0 and γ=0, resp.). Note when St≥K/η,Lt=Ut,
then VCB(x)=ηx for all x≥K/η. Hence the issuer and the holder should search optimal stopping in the class of
stopping times of the form(4.7)γa=inf{t≥0:St≥a},τb=inf{t≥0:St≥b},respectively, with numbers 0<a,b≤K/η to be determined. Note when the process S fluctuates in the interval (0,K/η),
it is not optimal to terminate the contract simultaneously by both issuer and
holder. For example, if the issuer chooses to terminate the contract at the
first time that S exceeds some point a∈(0,K/η),
then ηa<K,
and the holder will choose the payoff of coupon rather than converting the bond
into the stock, which is a contradiction. Similarly, one can explain another
case. Then only the following situation can occur: either a<b=K/η,
b<a=K/η,
or b=a=K/η.For search of the optimala∗,b∗ and the value of VCB(x),
we consider an auxiliary Stefan problem with unknown numbers a,b, and V(x)(4.8)(𝒜α−r)V(x)=−c,0<x<a⋀b,ηx<V(x)<ηx∨K,0<x<a⋀bwith continuous fit boundary
conditions(4.9)V(b−)=ηb,V(x)=ηxfor all x>b,b≤a=K/η,
and(4.10)V(a−)=K,V(x)=ηx∨Kfor all x>a,a≤b=K/η,
and smooth fit boundary conditions(4.11)V′(b−)=ηifb<a=Kη,V′(a−)=0ifa<b=Kη.By computing the Stefan problem
we can obtain that if(4.12)K>k2k2−1cr,then b∗<a∗=K/η,
and the expression of V(x) is given by(4.13)V(x)=ηb∗k2(xb∗)k2+crfor all 0<x<b∗,
with(4.14)b∗=k2η(k2−1)cr,and if(4.15)cr<K≤k2k2−1cr,then a∗=b∗=K/η,
and the value of V(x) is(4.16)V(x)=(K−cr)(ηxK)k2+crfor all 0<x<K/η.From the result we can observe that there are only two
regions forK,
and the situation a∗<b∗=K/η fails to hold. This implies that in this case,
when S fluctuates in the interval (0,K/η),
the issuer will never recall the bond. Now, we only need to prove that V(x)=VCB(x),
and the stopping times γ∗ and τ∗ defined by (4.7) with boundaries a∗ and b∗ are optimal.Applying Itô formula to the process{e−rtV(St):t≥0},
we have(4.17)e−rtV(St)=V(x)+∫0te−ru(𝒜α−r)V(Su)1(Su≠a∗,Su≠b∗,Su≠K/η)du+∫0te−ruσSuV′(Su)1(Su≠a∗,Su≠b∗,Su≠K/η)dWuα+∫0te−ru[V(Su−(1−y0))−V(Su−)](dNu−λαdu)+e−rt[V′(Kη+)−V′(Kη−)]LtK/η.Let(4.18)Mt=∫0te−ruσSuV′(Su)1(Su≠a∗,Su≠b∗,Su≠K/η)dWuα+∫0te−ru[V(Su−(1−y0))−V(Su−)](dNu−λαdu).Note for all 0<x<a∗,(𝒜−r)V(x)≤−c,
while for all 0<x<b∗,(𝒜−r)V(x)=−c. Since ηx≤V(x)≤ηx∨K,
so for 0<a∗≤K/η,0<b∗≤K/η,
we have(4.19)Lτ⋀γ∗=∫0τ⋀γ∗c⋅e−rudu+e−r(τ⋀γ∗)ηSτ⋀γ∗≤∫0τ⋀γ∗c⋅e−rudu+e−r(τ⋀γ∗)V(Sτ⋀γ∗)≤V(x)+Mτ⋀γ∗,Uτ∗⋀γ=∫0τ∗⋀γc⋅e−rudu+e−r(τ∗⋀γ)(ηSτ∗⋀γ∨K)≥∫0τ∗⋀γc⋅e−rudu+e−r(τ∗⋀γ)V(Sτ∗⋀γ)=V(x)+Mτ∗⋀γ.Because V(Sγ∗)=K∨ηSγ∗,V(Sτ∗)=ηSτ∗,
then by choosing localizing sequence and apply the Fatou lemma, we
obtain(4.20)Exα[Lτ1(τ≤γ∗)+Uγ∗1(γ∗<τ)]≤V(x)≤Exα[Lτ∗1(τ∗≤γ)+Uγ1(γ<τ∗)].Taking supremum and infimum for τ and γ of both sides, respectively, we can obtain the
result. While for (4.20), taking τ=τ∗,γ=γ∗,
we have(4.21)V(x)=Exα[Lτ∗1(τ∗≤γ∗)+Uγ∗1(γ∗<τ∗)].



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*Source: 945923-2009-05-06.xml*