0
$\begingroup$

Let $c, d \in \mathbb{Z}$ be such that $c < 0 < d$. Let $\tau_k = \inf_{n \geq 0} \left\{n: S_n = k\right\}$ be the first hitting time of state $k$, where $S_n$ denotes the position of a symmetric random walk on $\mathbb{Z}$ after $n$ steps. The question is to show that $\mathbb{P}(\tau_D < \tau_C) = \frac{-c}{d-c}$. The random walk starts at position $0$ at time $0$.

I have been beating my head on this for a little bit, and I suspect it is a lot simpler than I am thinking. I played around with finding the probability of $\tau_A = n$ but wasn't able to use that to find the answer to the question. Any suggestions?

$\endgroup$
2
$\begingroup$

Hint: Let $\tau=\min(\tau_C,\tau_D)$. By Wald's equation, $E[S_\tau]=0$. On the other hand, $S_\tau$ is equal to either $d$ or $c$, according to whether $\tau_C$ is less than or more than $\tau_D$.

There is the hairy detail that in order to use Wald's equation, you need to show that $E[\tau]<\infty$. To prove this, note that if $X_i=1$ for $d-c$ steps in a row at any point, then the process will leave the interval $[c,d]$ if it has not already. This shows that $P(\tau>(d-c)k)\le (1-\frac1{2^{d-c}})^k$. This bound on the tail probabilities of $\tau$ can be used to show $E[\tau]$ is finite.

$\endgroup$

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.