Probability for more general sumprocess

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 $$a<0, the sum process $$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

$$P(S_{T_{a,b}}=a)=\frac{1-(\frac{p}{q})^b}{1-(\frac{p}{q})^{b-a}},$$

$$E[T_{a,b}]=\frac{b}{p-q}-\frac{b-a}{p-q}\cdot\frac{1-\left(\frac{p}{q}\right)^b}{1-\left(\frac{p}{q}\right)^{b-a}}.$$

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? :)

• 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! :) – user408858 Nov 3 '18 at 1:03
• 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 $E[X_1]=p-q$. – user408858 Nov 3 '18 at 16:38