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求显式过程(分 a=ba=baba\ne b

Determine the stochastic process

专题
Finance / 金融
难度
L4

题目详情

Determine the stochastic process XtX_{t} such that

dXt=(Xt+a)(Xt+b)(Xt+a+b2)dt+(Xt+a)(Xt+b)dWt\begin{array}{c}{{d X_{t}=\left(X_{t}+a\right)\left(X_{t}+b\right)\left(X_{t}+\frac{a+b}{2}\right)d t}}\\ {{+\left(X_{t}+a\right)\left(X_{t}+b\right)d W_{t}}}\end{array}

with X0=ξX_0 = \xi , where aa and bb are positive numbers. Consider the cases a=ba = b and aba \neq b separately.

解析

SDE:

dXt=(Xt+a)(Xt+b)(Xt+a+b2)dt+(Xt+a)(Xt+b)dWt.dX_t=(X_t+a)(X_t+b)\left(X_t+\frac{a+b}{2}\right)dt+(X_t+a)(X_t+b)dW_t.

(1) aba\ne b:设

Yt=lnXt+aXt+b.Y_t=\ln\frac{X_t+a}{X_t+b}.

Itô 可得漂移抵消,

dYt=(ba)dWtYt=lnξ+aξ+b+(ba)Wt.dY_t=(b-a)dW_t\Rightarrow Y_t=\ln\frac{\xi+a}{\xi+b}+(b-a)W_t.

Rt=ξ+aξ+be(ba)WtR_t=\frac{\xi+a}{\xi+b}e^{(b-a)W_t},则

Xt+aXt+b=RtXt=Rtba1Rt.\boxed{\frac{X_t+a}{X_t+b}=R_t\Rightarrow X_t=\frac{R_t b-a}{1-R_t}}.

(2) a=ba=b:令 Ut=Xt+aU_t=X_t+a,则 dUt=Ut3dt+Ut2dWtdU_t=U_t^3dt+U_t^2dW_t

Yt=1/UtY_t=-1/U_t,Itô 给出 dYt=dWtdY_t=dW_t,所以

Yt=Wt1ξ+a.Y_t=W_t-\frac{1}{\xi+a}.

从而

Xt=a1Wt1ξ+a.\boxed{X_t=-a-\frac{1}{W_t-\frac{1}{\xi+a}}}.