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This period I am learning Brownian motion and I am struggling with this question:

Suppose that $X=(X_t,t\geq 0)$ is a Brownian Motion BM($\mu$,$\sigma^2$) with $\mu=0.2$ and $\sigma=1$ $(X_0=0)$. We define $\tau=inf\{t\geq0: X_t=1.2 \quad or \quad -1\}$ as the random time when $X_t$ is equal to $1.2$ or $-1$ for the first time.Find the probability density function of $\tau$.

I have read the answers of this question but I still can't find the answer. Can anyone help me?

Thanks in advance!

Billy
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1 Answers1

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The linked question only handles the case of the exit time for the interval $(-\infty,a]$, namely, $\tau=\inf\{t\ge 0: X_t=a\}$. In Karatzas & Shreve, Brownian Motion and Stochastic Calculus pp. 98 the interval $[0,a]$ is treated for a standard BM ($\mu=0,\sigma^2=1$) starting at $x\in[0,a]$: their equation (8.24) can be read as \begin{align} P^x[\tau\in dt]&=\frac{1}{\sqrt{2\pi t^3}}\sum_{n=-\infty}^{+\infty}\Bigg[(2na+x)\exp\Bigg\{-\frac{(2na+x)^2}{2t}\Bigg\}\\ &\quad+(2na+a-x)\exp\Bigg\{-\frac{(2na+a-x)^2}{2t}\Bigg\}\Bigg]\,dt. \end{align} where $\tau=T_0\wedge T_a$ and $T_z=\inf\{t\ge 0:X_t=b\}$ so that $\tau=\inf\{t\ge 0:X_t=0\text{ or }X_t=a\}\,.$

Kurt G.
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