Simple symmetric random walk on $\mathbb{Z}$

546 Views Asked by At

Let $S_n$ be the simple symmetric walk on $\mathbb{Z}$ (prob go forward = prob go backward = $1/2$) and let $N = \inf\{n \geq 0 : S_n = 0\}$ be the hitting time at $0$. Then I would like to verify that

i) Prove that $\mathbb{P}(\sup_{n\geq 1} S_{N \wedge n}\geq m)= 1/m $ for all $m > 0$

ii) Use i) to show that $N < \infty$ a.s. (This proves that $S_{N \wedge n} \rightarrow 0$ a.s.)

I am not sure how to begin. I m quite new to random walk. Please help.

1

There are 1 best solutions below

0
On

For i), take one step from the origin, without loss of generality to $1$, and then consider the walk to end if it reaches either $0$ or $m$. A recurrence relation for the probability of ending at $m$ shows that this probability increases linearly from $0$ at $0$ to $1$ at $m$ and hence is $\frac1m$ at $1$.

For ii), note that the walk almost surely does not remain confined in any finite interval, and thus is almost surely unbounded. And the probabilities of reaching points at increasing distance from the origin before returning to the origin converge to $0$.