2
$\begingroup$

Let $S_n=\sum_{i=1}^n X_i$ be a standard Gaussian random walk with i.i.d. increments $X_i \sim \mathcal{N}(0,1)$. Define the first exit time $\tau$ as $\tau=\inf_n\{|S_n|\geq a \}$. I'm trying to work out the moments of $\tau$ and also the distribution of $\tau$.

I find this question extremely difficult for me. I know that for simple random walks or Brownian motions, $E[\tau]=a^2$ by either solving the differential equation or using the reflection principle. However I simply do not know how to start with this question. Really appreciate any help you can provide.

$\endgroup$
1
$\begingroup$

Let's pretend for convenience that $\tau=\inf_n\{|S_n|> a \}$ (it won't change the answer). Let $X_0\in [-a,a]$ be the starting point of the random walk, and let $f(x)=\frac1{\sqrt{2\pi}}e^{-x^2/2}$ be the pdf of $\mathcal N(0,1)$.

If you let $t(x)=E[\tau|X_0=x]$ for $x\in[-a,a]$, then $$t(x)=\int_{-a}^af(u-x)(1+t(u))du=\int_{-a-x}^{a-x}f(z)(1+t(z+x))dz$$ so $$t'(x)=(t(a)+1)(f(x+a)-f(x-a))$$ And since $t(-a)=t(a)$, $$t(x)=(t(a)+1)\int_{-a}^x(f(u+a)-f(u-a))du+t(a)$$ $$t(x)=\frac12(t(a)+1)(erf(\frac{a-x}{\sqrt2})+erf(\frac{a+x}{\sqrt2})-erf(\sqrt2 a))+t(a)$$ $$t(0)=\frac12(t(a)+1)(2erf(\frac{a}{\sqrt2})-erf(\sqrt2 a))+t(a)$$

Unfortunately I don't know how to figure out what $t(a)$ is. Maybe I'm missing some initial condition or clever symmetry.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.