2
$\begingroup$

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

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

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

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? :)

Thanks in advance!

$\endgroup$
1
$\begingroup$

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)

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.