Consider $(X_n)_{n\ge1}$ iid with

$$P(X_1=1)=p,\quad P(X_1=-1)=1-p$$

for some $p\in(0,1)$, $a,b\in\mathbb{Z}$ with $a<0<b$, the sum process $S_n:=\sum_{i=1}^nX_i$ and the stopping time

$$T_{a,b}:=\min\left\{n\ge 1: S_n\in\{a,b\}\right\}.$$

I want to show that



So, I was already able to show the formula for $E[T_{a,b}]$ by using the first result and Wald's identity. This is pretty straight forward, but I am struggling with $P(S_{T_{a,b}}=a)$. I was able to show that $$\left(\frac{q}{p}\right)^{S_n}\text{ and } S_n-n(p-q)$$ are martingales with respect to the filtration $(\mathcal{F}_n)_{n\ge 1}$, $\mathcal{F}_n:=\sigma(S_n)$, but I do not understand how to use this hint. Can someone give me another hint? :)

Thanks in advance!


Sure, here is a hint: Look at your first martingale. You know what the value of it is at 0. What values can it take at the stopping time? (Hint: it can only take two values) What is it's expectation equal to? (Technical note: You need to verify one of the hypotheses of optional stopping theorem here: https://en.wikipedia.org/wiki/Optional_stopping_theorem Your martingale satisfies one of these, and that is enough)

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    $\begingroup$ Oh, poor me, I was always trying to calculate $E[S_{T_{a,b}}]$ to find out $P(S_{T_{a,b}}=a)$... I have it now. Thank you! :) $\endgroup$ – user408858 Nov 3 '18 at 1:03
  • $\begingroup$ But still I wonder why I need the second martingale... $\endgroup$ – user408858 Nov 3 '18 at 1:03
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    $\begingroup$ Ok, once you know $P(T_{a,b}<\infty)=1$ you have again using the optional stopping theorem that $E[S_{T_{a,b}}-T_{a,b}(p-q)]=E[S_0-0(p-q)]=0$, which gives you exactly the same equation stated in Wald's identity, since $E[X_1]=p-q$. $\endgroup$ – user408858 Nov 3 '18 at 16:38

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