二维布朗：离开椭圆的期望时间

Independent brownian 2

专题: Finance / 金融
难度: L4
来源: QuantQuestion

题目详情

Let $P_{t} = (Z_{t},W_{t})$ be a two- dimensional stochastic process that starts at $(0,0)$ . The processes $Z_{t}$ and $W_{t}$ are independent Brownian motions. What is the expected time for $P_{t}$ to exit the ellipse $\frac{y^{2}}{a^{2}} +\frac{y^{2}}{b^{2}} = 1$ , where $a$ and $b$ are fixed constants?

解析

令 $(Z_t,W_t)$ 为二维标准布朗运动，从 $(0,0)$ 出发，离开椭圆

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

的首次时间为 $\tau$ 。

对二维布朗运动， $u(x,y)=\mathbb{E}_{(x,y)}[\tau]$ 解 Poisson 方程

\frac12\Delta u=-1\ \text{于域内},\qquad u=0\ \text{于边界}.

取试探函数 $u(x,y)=A\Bigl(1-\frac{x^2}{a^2}-\frac{y^2}{b^2}\Bigr)$ ，则

\Delta u=A\left(-\frac{2}{a^2}-\frac{2}{b^2}\right).

代入得 $A(1/a^2+1/b^2)=1$ ，即

A=\frac{1}{\frac{1}{a^2}+\frac{1}{b^2}}=\frac{a^2b^2}{a^2+b^2}.

因此从原点出发

\boxed{\mathbb{E}[\tau]=u(0,0)=\frac{a^2b^2}{a^2+b^2}}.