# The First Passage Time and the Dividend Value Function for One-Dimensional Diffusion Processes between Two Reflecting Barriers

Authors: Chuancun Yin; Huiqing Wang
Journal: International Journal of Stochastic Analysis
 (2012)

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## Abstract

We consider the general one-dimensional time-homogeneous regular diffusion process between two reflecting barriers. An approach based on the Itô formula with corresponding boundary conditions allows us to derive the differential equations with boundary conditions for the Laplace transform of the first passage time and the value function. As examples, the explicit solutions of them for several popular diffusions are obtained. In addition, some applications to risk theory are considered.


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## Body

## 1. Introduction and the Model

Diffusion processes with one or two barriers appear in many applications in economics, finance, queueing, mathematical biology, and electrical engineering. Among queueing system applications, reflected Ornstein-Uhlenbeck and reflected affine processes have been studied as approximations of queueing systems with reneging or balking [1, 2]. Motivated by Ward and Glynn’s one-sided problem, Bo et al. [3] considered a reflected Ornstein-Uhlenbeck process with two-sided barriers. In this paper, we consider the expectations of some random variables involving the first passage time and local times for the general one-dimensional diffusion processes between two reflecting barriers.LetX={Xt,t≥0} be a one-dimensional time-homogeneous reflected diffusion process with barriers a and b, which is defined by the following stochastic differential equation:
(1.1)dXt=μ(Xt)dt+σ(Xt)dBt+dLt-dUt,X0=x∈(a,b),
where B(t) is a Brownian motion in ℛ, L={Lt,t≥0} and U={Ut,t≥0} are the regulators of point a and b, respectively. Further, the processes L and U are uniquely determined by the following properties (see, e.g., [4]):(1)
botht→Lt and t→Ut are continuous nondecreasing processes with L0=U0=0,(2)
L and U increase only when X=a and X=b, respectively, that is, ∫0t‍I{Xs=a}dLs=Lt and ∫0t‍I{Xs=b}dUs=Ut, for t≥0.It is well known that under certain mild regularity conditions on the coefficientsμ(x) and σ(x), the SDE (1.1) has a unique strong solution for each starting point (see, e.g., [5]). The solution Xt is a time-homogeneous strong Markov process with infinitesimal generator
(1.2)𝒜f(x)=12σ2(x)f′′(x)+μ(x)f′(x),x∈(a,b),
acting on functions on [a,b] subject to boundary conditions: f'(a)=f'(b)=0.Define the first passage time(1.3)τy=inf{t≥0:Xt=y},
where τy=∞ if Xt never reaches y.Forλ>0, η>0, θ>0, we consider the Laplace transform φ, and the value functions ψ, ψ1, and ψ2 on x∈[a,b]:
(1.4)φ(x)=Ex[e-λτy],(1.5)ψ(x)=Ex[η∫0∞‍e-λtdUt-θ∫0∞‍e-λtdLt],(1.6)ψ1(x)=Ex[∫0∞‍e-λtdUt],(1.7)ψ2(x)=Ex[∫0∞‍e-λtdLt].The rest of the paper is organized as follows. Section2 studies the Laplace transform of the first passage time. Section 3 deals with the value function. Some applications in risk theory are considered in Section 4.


## 2. Laplace Transform

Bo et al. [3] consider the Laplace transform Ex[e-λτy] for a reflected Ornstein-Uhlenbeck process with two-sided barriers. In this section we consider the Laplace transform of the first passage time for the general reflected diffusion process X defined by (1.1).Theorem 2.1.
Letx∈[a,b], λ>0, and assume that f1(y), f2(y) satisfy the following equations, respectively:
(2.1)𝒜f1(y)=λf1(y),y∈(a,b),f1′(a)=0,𝒜f2(y)=λf2(y),y∈(a,b),f2′(b)=0.
Iff1(y)≠0 for y∈[x,b] and f2(y)≠0 for y∈[a,x], then
(2.2)Ex[e-λτy]=f1(x)f1(y),y∈[x,b],Ex[e-λτy]=f2(x)f2(y),y∈[a,x].Proof.
Applying the Itô formula for semimartingales toh(t,Xt)=e-λtf(Xt) with f∈Cb2([a,b]) we obtain
(2.3)h(t,Xt)=h(0,X0)+∫0t‍∂h∂sh(s,Xs)ds+∫0t‍∂h∂xh(s,Xs)dXs+12∫0t‍∂2h∂x2h(s,Xs)d[Xs,Xs]=f(X0)+∫0t‍-λe-λsf(Xs)ds+∫0t‍e-λsf′(Xs)dXs+12∫0t‍e-λsf′′(Xs)σ2(Xs)ds=f(x)+∫0t‍e-λs(𝒜-λ)f(Xs)ds+∫0t‍e-λsf′(Xs)dBs+f′(a)∫0t‍e-λsdLs-f′(b)∫0t‍e-λsdUs.
Since τy<∞ is a stopping time and x∈[a,b], it follows from the optional sampling theorem that
(2.4)Ex(e-λτyf(Xτy))=f(x)+Ex(∫0τy‍e-λs(𝒜-λ)f(Xs)ds)+f′(a)Ex(∫0τy‍e-λsdLs)-f′(b)Ex(∫0τy‍e-λsdUs).
By the definitions of τy, U, and L we have
(2.5)Ex[∫0τy‍e-λsdLs]=0,y∈[a,x];Ex[∫0τy‍e-λsdUs]=0,y∈[x,b];X(τy)=y.
Substituting them into (2.4) one gets
(2.6)Ex[e-λτyf(y)]=f(x)+Ex[∫0τy‍e-λs(𝒜-λ)f(Xs)ds]+f′(a)Ex[∫0τy‍e-λsdLs],y∈[x,b].Ex[e-λτyf(y)]=f(x)+Ex[∫0τy‍e-λs(𝒜-λ)f(Xs)ds]-f′(b)Ex[∫0τy‍e-λsdUs],y∈[a,x].
The result follows.Remark 2.2.
Although neitherf1(·) nor f2(·) in the Theorem 2.1 is unique, but each of their ratios is unique.As  illustrations of Theorem2.1, we consider some examples.Example 2.3.
Bessel process:dXt=((d-1)/2Xt)dt+dBt, where d>1 is a real number.
We consider the differential equation(1/2)ψ′′(x)+((d-1)/2x)ψ′(x)=λψ(x), λ>0. It is well known that the increasing and decreasing solutions are, respectively:
(2.7)ψ+(x)=x-νIν(2λx),ψ-(x)=x-νKν(2λx),
where ν=(d-2)/2 and Iν and Kν are the usual modified Bessel functions.
Then, we can givef1(x),f2(x) as follows:
(2.8)f1(x)=C1ψ+(x)+C2ψ-(x),f2(x)=C3ψ+(x)+C4ψ-(x),
where the constants C1, C2 and C3, C4 can be derived from f1′(a)=0 and f2′(b)=0, respectively. We can obtain their ratios, respectively:
(2.9)M=C1C2=-2λaKν′(2λa)-νKν(2λa)2λaIν′(2λa)-νIν(2λa),N=C3C4=-2λbKν′(2λb)-νKν(2λb)2λbIν′(2λb)-νIν(2λb).
Substituting them into (2.2), we get
(2.10)Ex[e-λτy]=Mx-νIν(2λx)+x-νKν(2λx)My-νIν(2λy)+y-νKν(2λy),y∈[x,b],Ex[e-λτy]=Nx-νIν(2λx)+x-νKν(2λx)Ny-νIν(2λy)+y-νKν(2λy),y∈[a,x].Example 2.4.
The Ornstein-Uhlenbeck process [6] is as follows:
(2.11)dXt=v(k-Xt)dt+σdBt,v,σ>0,k∈[a,b].
In mathematical finance, the Ornstein-Uhlenbeck process above is known as Vasicek model for the short-term interest rate process [7]. We consider the differential equation (1/2)σ2ψ′′(x)+v(k-x)ψ′(x)=λψ(x).
In the casek=0, σ=1, the two independent solutions to (1/2)ψ′′(x)-vxψ′(x)=λψ(x) are
(2.12)ψ+(x)=H-λ/v(-vx)=2-λ/2ve(1/2)vx2D-λ/v(-2vx),ψ-(x)=H-λ/v(vx)=2-λ/2ve(1/2)vx2Dλ/v(2vx),
where Hν(·) and Dν(·) are, respectively, the Hermite and parabolic cylinder functions [8]. Then, as the way used in Example 2.3, we obtain the ratios of the constants C1, C2 and C3, C4, respectively:
(2.13)M=C1C2=-vaDλ/v(2va)+2vDλ/v′(2va)vaD-λ/v(2va)-2vD-λ/v′(2va),N=C3C4=-vbDλ/v(2vb)+2vDλ/v′(2vb)vbD-λ/v(2vb)-2vD-λ/v′(2vb).
Substituting them into (2.2), we get
(2.14)Ex[e-λτy]=Me(1/2)vx2D-λ/v(-2vx)+e(1/2)vx2Dλ/v(2vx)Me(1/2)vy2D-λ/v(-2vy)+e(1/2)vy2Dλ/v(2vy),y∈[x,b],Ex[e-λτy]=Ne(1/2)vx2D-λ/v(-2vx)+e(1/2)vx2Dλ/v(2vx)Ne(1/2)vy2D-λ/v(-2vy)+e(1/2)vy2Dλ/v(2vy),y∈[a,x].
For the general k and σ, the two independent solutions are, respectively
(2.15)ψ+(x)=H-λ/v(-vσ(x-k))=2-λ/2ve(1/2)(v/σ2)(x-k)2D-λ/v(-2vσ(x-k)),ψ-(x)=H-λ/v(vσ(x-k))=2-λ/2ve(1/2)(v/σ2)(x-k)2D-λ/v(2vσ(x-k)).
Then, as the way used in Example 2.3, we obtain the ratios of the constants C1, C2 and C3, C4, respectively
(2.16)M=C1C2=-v(a-k)D-λ/v((2v/σ)(a-k))+2σD-λ/v′((2v/σ)(a-k))v(a-k)D-λ/v(-(2v/σ)(a-k))-2σD-λ/v′(-(2v/σ)(a-k)),N=C3C4=-v(b-k)D-λ/v((2v/σ)(b-k))+2σD-λ/v′((2v/σ)(b-k))v(b-k)D-λ/v(-(2v/σ)(b-k))-2σD-λ/v′(-(2v/σ)(b-k)).
Substituting them into (2.2), we get
(2.17)Ex[e-λτy]=h(x)(MD-λ/v(-(2v/σ)(x-k))+D-λ/v((2v/σ)(x-k)))h(y)(MD-λ/v(-(2v/σ)(y-k))+D-λ/v((2v/σ)(y-k))),y∈[x,b],Ex[e-λτy]=h(x)(ND-λ/v(-(2v/σ)(x-k))+D-λ/v((2v/σ)(x-k)))h(y)(ND-λ/v(-(2v/σ)(y-k))+D-λ/v((2v/σ)(y-k))),y∈[a,x],
where h(x)=e(1/2)(v/σ2)(x-k)2.Remark 2.5.
If we takek=0, σ=1, a=0 and substitute the series forms of Hν(·) and Dν(·) into the above result, then it is the same as Bo et al. [3].Example 2.6.
The square root process of  Cox et al. [9]:
(2.18)dXt=v(k-Xt)dt+σXtdBt,v,σ>0,k∈[a,b].
Now consider the differential equation(2.19)12σ2xψ′′(x)+(vk-vx)ψ′(x)=λψ(x),λ>0.
If (2v/σ2)k is not an integer, the two linear independent solutions are
(2.20)ψ+(x)=M(λv,2vσ2k,2vσ2x),ψ-(x)=U(λv,2vσ2k,2vσ2x),
where M and U are the confluent hypergeometric functions of the first and second kinds, respectively. Then, as the way used in Example 2.3, we obtain the ratios of the constants C1, C2 and C3, C4, respectively:
(2.21)M1=C1C2=-U'(λ/v,(2v/σ2)k,(2v/σ2)a)M'(λ/v,(2v/σ2)k,(2v/σ2)a),N=C3C4=-U'(λ/v,(2v/σ2)k,(2v/σ2)b)M'(λ/v,(2v/σ2)k,(2v/σ2)b).
Substituting them into (2.2), we get
(2.22)Ex[e-λτy]=M1M(λ/v,(2v/σ2)k,(2v/σ2)x)+U(λ/v,(2v/σ2)k,(2v/σ2)x)M1M(λ/v,(2v/σ2)k,(2v/σ2)y)+U(λ/v,(2v/σ2)k,(2v/σ2)y),y∈[x,b].Ex[e-λτy]=NM(λ/v,(2v/σ2)k,(2v/σ2)x)+U(λ/v,(2v/σ2)k,(2v/σ2)x)NM(λ/v,(2v/σ2)k,(2v/σ2)y)+U(λ/v,(2v/σ2)k,(2v/σ2)y),y∈[a,x].Example 2.7.
The Gompertz Brownian motion process [10] is as follows:
(2.23)dXt=vX(t)(ln(k)-ln(Xt))dt+σXtdBt,v,σ>0,k∈[a,b].
Now consider the differential equation(2.24)12σ2x2ψ′′(x)+vx(lnk-lnx)ψ′(x)=λψ(x),λ>0.
The increasing and decreasing solutions are, respectively:
(2.25)ψ+(x)=M(λ2v,12,vσ2(lnxk+σ22v)2),ψ-(x)=U(λ2v,12,vσ2(lnxk+σ22v)2),
where M and U, as in Example 2.6, are the first and second Kummer’s functions, respectively. Then, as the way used in Example 2.3, we obtain the ratios of the constants C1, C2 and C3, C4, respectively:
(2.26)M1=C1C2=-U'(λ/2v,1/2,(v/σ2)(ln(a/k)+σ2/2v)2)M'(λ/2v,1/2,(v/σ2)(ln(a/k)+σ2/2v)2),N=C3C4=-U'(λ/2v,1/2,(v/σ2)(ln(b/k)+σ2/2v)2)M'(λ/2v,1/2,(v/σ2)(ln(b/k)+σ2/2v)2).
Substituting them into (2.2), we get
(2.27)Ex[e-λτy]=M1M(λ/2v,1/2,(v/σ2)(ln(x/k)+σ2/2v)2)+U(𝔸)M1M(λ/2v,1/2,(v/σ2)(ln(x/k)+σ2/2v)2)+U(𝔸),y∈[x,b],Ex[e-λτy]=NM(λ/2v,1/2,(v/σ2)(ln(x/k)+σ2/2v)2)+U(𝔸)NM(λ/2v,1/2,(v/σ2)(ln(x/k)+σ2/2v)2)+U(𝔸),y∈[a,x],
where (𝔸) denotes (λ/2v,1/2,(v/σ2)(ln(x/k)+σ2/2v)2).Remark 2.8.
For a certain choice of parameters fora and b in Examples 2.3–2.7, we get the Laplace transform of the first passage time of one-dimensional diffusion with one-sided barrier. For example, letting a→-∞ or b→+∞ in Example 2.4, one gets the Laplace transform of the first passage time of the Ornstein-Uhlenbeck process with one-sided barrier; see Nobile et al. [11], Ricciardi and Sato [12], Alili et al. [13], and Ditlevsen [6].


## 3. The Value Function

In this section we study the value functions (1.5)–(1.7). Using Itô’s formula, we derive differential equation with boundary conditions for ψ.Theorem 3.1.
The functionψ defined by (1.5) satisfies the differential equation
(3.1)12σ2(x)ψ′′(x)+μ(x)ψ′(x)=λψ(x),a<x<b,
with the boundary conditions ψ'(a)=θ, ψ'(b)=η.Proof.
Applying the Itô’s formula for semimartingales toh(t,Xt)=e-λtψ(Xt) with ψ∈Cb2([a,b]) we obtain
(3.2)h(t,Xt)=h(0,X0)+∫0t‍∂h∂sh(s,Xs)ds+∫0t‍∂h∂xh(s,Xs)dXs+12∫0t‍∂2h∂x2h(s,Xs)d[Xs,Xs]=ψ(X0)+∫0t‍-λe-λsψ(Xs)ds+∫0t‍e-λsψ′(Xs)dXs+12∫0t‍e-λsψ′′(Xs)σ2(Xs)ds=ψ(x)+∫0t‍e-λs(𝒜-λ)ψ(Xs)ds+∫0t‍e-λsψ′(Xs)dBs+ψ′(a)∫0t‍e-λsdLs-ψ′(b)∫0t‍e-λsdUs,
where we have used that ψ'(Xt)dLt=ψ'(a)dLt and ψ'(Xt)dUt=ψ'(b)dUt. From (3.2) we have
(3.3)Ex(e-λtψ(Xt))-ψ(x)=Ex∫0t‍e-λs(𝒜-λ)ψ(Xs)ds+Ex(ψ′(a)∫0t‍e-λsdLs-ψ′(b)∫0t‍e-λsdUs).
Letf be a solution of
(3.4)𝒜f(x)=λf(x),x∈(a,b),f′(a)=0,f′(b)=1.
In place of (3.3), we have
(3.5)Ex(e-λtf(Xt))-f(x)=-∫0t‍e-λsdUs.
Letting t→∞, we get
(3.6)∫0∞‍e-λsdUs=f(x)<∞.
Likewise
(3.7)∫0∞‍e-λsdLs<∞.
Letting t→∞ in (3.3) and noting (3.6) and (3.7), we get the desired result.Corollary 3.2.
The functionψ1(x)=Ex[∫0∞‍e-λtdUt] is solution to the differential equation
(3.8)12σ2(x)ψ1′′(x)+μ(x)ψ1′(x)=λψ1(x),a<x<b,
with the boundary conditions ψ1′(a)=0,ψ1′(b)=1.Corollary 3.3.
The functionψ2(x)=Ex[∫0∞‍e-λtdLt] is solution to the differential equation
(3.9)12σ2(x)ψ2′′(x)+μ(x)ψ2′(x)=λψ2(x),a<x<b,
with the boundary conditions ψ2′(a)=-1, ψ2′(b)=0.For diffusions in Examples2.3–2.7 we can obtain the explicit expressions for ψ, ψ1, and ψ2. Now we consider the Ornstein-Uhlenbeck process only.Example 3.4.
The Ornstein-Uhlenbeck process is as follows:(3.10)dXt=v(k-Xt)dt+σdBt,v,σ>0,k∈[a,b].
From Example 2.4, the two independent solutions of differential equation
(3.11)12σ2ψ′′(x)+v(k-x)ψ′(x)=λψ(x)
are, respectively,
(3.12)ψ+(x)=H-λ/v(-vσ(x-k))=2-λ/2ve(1/2)(v/σ2)(x-k)2D-λ/v(-2vσ(x-k)),ψ-(x)=H-λ/v(vσ(x-k))=2-λ/2ve(1/2)(v/σ2)(x-k)2D-λ/v(2vσ(x-k)).
The general solution of (3.11) is of the form
(3.13)ψ(x)=C1ψ+(x)+C2ψ-(x),
where the constants C1 and C2 are determined by the boundary conditions ψ'(a)=θ, ψ'(b)=η. They are
(3.14)C1=θψ+′(b)-ηψ+′(a)ψ-′(a)ψ+′(b)-ψ+′(a)ψ-′(b),C2=ηψ-′(a)-θψ-′(b)ψ-′(a)ψ+′(b)-ψ+′(a)ψ-′(b).


## 4. Applications to Risk Theory

LetXt denote the surplus of the company. If no dividends were paid, the surplus process follows the stochastic differential equation
(4.1)dXt=μ(Xt)dt+σ(Xt)dBt,t≥0,X0=x,
where B is a Brownian motion and μ and σ are Lipschitz-continuous functions.The company will pay dividends to its shareholders according to barrier strategy with parameterb>0. Whenever the surplus is about to go above the level b, the excess will be paid as dividends, and when the surplus is below b nothing is paid out. Let D(t) denote the aggregate dividends by time t. Thus the resulting surplus process Y is given by
(4.2)dYt=μ(Yt)dt+σ(Yt)dBt-dD(t),t≥0.
Let T=inf{t≥0:Yt=0} be the time of ruin. Note that when b<∞ ruin is certain, that is, P(T<∞)=1. We are interested in the Laplace transform of T. This model can be found in Paulsen [14], and some important special cases can be found in Gerber and Shiu [15], Cai et al. [16]. It follows from Theorem 2.1 that, for 0<x<b and λ>0, Ex(e-λT)=h(x), where h is the solution of
(4.3)𝒜h(x)=λh(x),x∈(0,b),h′(b)=0,h(0)=1.Assume that an insurance company is not allowed to go bankrupt and the beneficiary of the dividends is required to inject capital into the insurance company to keep its risk process stays nonnegative. Under such a dividend policy the controlled risk process with initial reservex>0 satisfies
(4.4)dX~t=μ(X~t)dt+σ(X~t)dBt+dLt-dDt,t≥0,
where Lt and Dt are local times at 0 and b, respectively. Lt ensures the insurance company will not ruin and Dt is the aggregate amount of paid dividends by time t. We consider the total expected discounted dividends minus the total expected discounted costs of injected capital:
(4.5)V(x;b)=Ex[∫0∞‍e-λtdDt-k∫0∞‍e-λtdLt],
where λ>0 is discounted factor and k is the cost per unit injected capital. Avram et al. [17] consider the problem in a Levy processes setting.According to Theorem3.1, we obtain the formula V(x,b), as long as we solve the equation
(4.6)12σ2(x)V′′(x)+μ(x)V′(x)=λV(x),0<x<b,
with the boundary conditions V'(0)=k, V'(b)=1.For0<x<b, let Tx=inf{t∣Xt~=x} be the first time when the surplus reaches the level x. It follows from Theorem 2.1 that, for 0<x<b and λ>0, Ex(e-λTx)=g(x), where g is the solution of
(4.7)𝒜g(x)=λg(x),x∈(0,b),g′(0)=0,g(b)=1.We now give two examples.Example 4.1.
In this example we consider the uncontrolled surplus of insurance company satisfyingXt=x+μt+σWt, where Wt is Brownian motion. The controlled surplus process X~ at time t follows the equation
(4.8)X~t=x+μt+σWt+Lt-Dt.
Now we consider the differential equation(4.9)12σ2V′′(x)+μV′(x)=λV(x),0<x<b,
with the boundary conditions V'(0)=k, V'(b)=1. Then
(4.10)V(x,b)=kerb-1s(erb-esb)esx-kesb-1r(erb-esb)erx,0<x<b,
where r and s are the positive root and negative root of the equation (σ2/2)ξ2+μξ-λ=0, respectively, that is,
(4.11)r=-μ+μ2+2σ2λσ2,s=-μ-μ2+2σ2λσ2.
For0<x<b and λ>0, Ex(e-λTx):=g(x) is the solution of
(4.12)12σ2g′′(x)+μg′(x)=λg(x),0<x<b,
with the boundary conditions g'(0)=0, g(b)=1. Solving it gives
(4.13)Ex(e-λTx)=resx-serxresb-serb,0<x<b.Example 4.2.
In this example we consider the Ornstein-Uhlenbeck-type model. The company’s surplus evolves according to(4.14)Xt~=x+∫0t‍(μ+ρXs~)ds+σBt+Lt-Dt.
The model is considered in Cai et al. [16] for the special case where Lt=0.
The diffusion and drift coefficients areσ(x)≡σ, μ(x)=μ+ρx. We consider the differential equation
(4.15)12σ2V′′(x,b)+(μ+ρx)V′(x,b)=λV(x,b),0<x<b,
with the boundary conditions V'(0)=k,V'(b)=1. In Cai et al. [16], they pointed out that the solution is given by
(4.16)V(x,b)=H(t)=C1(-t)1-cetM(1-a,2-c;-t)+C2etU(c-a,c;-t)
for certain coefficients C1 and C2, with c=1/2, a=-λ/2ρ, t=-(1/ρσ2)(μ+ρx)2. Here M and U are called the confluent hypergeometric functions of the first and second kinds, respectively. For more details on confluent hypergeometric functions, see Abramowitz and Stegun [8]. It follows from (3.7) in Cai et al. [16] that
(4.17)H′(t)=H(t)-C1(1-c)(-t)-cetM(1-a,2-c;-t)-C11-a2-c(-t)1-cetM(2-a,3-c;-t)+C2(c-a)etU(c-a+1,c+1;-t).
The conditions C1 and C2 can be determined by V'(0)=k, V'(b)=1, where V'(x)=-(2(μ+ρx)/σ2)H'(t). Solving it gives
(4.18)C1=σ22(C/(μ+ρx))e(μ+ρb)2/ρσ2-(kD/μ)eμ2/ρσ2AD-BC,C2=σ22(kB/μ)eμ2/ρσ2-(A/(μ+ρb))e(μ+ρb)2/ρσ2AD-BC,
where
(4.19)A=((μ2ρσ2)1/2-12(μ2ρσ2)-1/2)M(1-a,2-c;μ2ρσ2)-1-a2-c(μ2ρσ2)1/2M(2-a,3-c;μ2ρσ2),B=(((μ+ρb)2ρσ2)1/2-12((μ+ρb)2ρσ2)-1/2)M(1-a,2-c;(μ+ρb)2ρσ2)-1-a2-c((μ+ρb)2ρσ2)1/2M(2-a,3-c;(μ+ρb)2ρσ2),C=U(c-a,c;μ2ρσ2)+(c-a)U(c-a+1,c+1;μ2ρσ2),D=U(c-a,c;(μ+ρb)2ρσ2)+(c-a)U(c-a+1,c+1;(μ+ρb)2ρσ2).



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*Source: 971212-2012-10-23.xml*