# Prove that $\text{Var} \tau = \frac{1 − (p − q)^2}{(p-q)^3}$ where $\tau$-stopping time

Let $S_n = \xi_1 + \dots + \xi_n$ be asimetric random walk such that $P(\xi_i = 1) = p > \frac{1}{2}$ and $P(\xi_i = -1) = q$. Let $\sigma^2 =1-(p-q)^2$ and let $X_n=(S_n-n-(p-q)n)^2 - \sigma^2n$

($X_n$ is martingale). Prove that $\text{Var} \tau = \frac{1 − (p − q)^2}{(p-q)^3}$

where $\tau = \min\{n:S_n=1\}$

I would use fact that $\text{Var}S_\tau = \sigma^2 E \tau + \mu^2 \text{Var} \tau$ where $\mu = E \xi_i$ and $\sigma^2 = \text{Var} \xi_i$. But I don't have $\text{Var}S_\tau$ and $E\tau$ to find $\text{Var} \tau$. Maybe I should use fact that $X_n$ is martingale but how?

I will grateful for yours help.

• Is there a more precise description of $\tau$ other than being just a stopping time? – gmath Nov 10 '14 at 14:49
• Sorry missed the definition! Anyway, $S_\tau =1$ almost surely. And you can find $E(\tau)$ using the martingale $S_n-n(p-q)$ and optional stopping. – gmath Nov 10 '14 at 14:50
• If I use Wald's formula I have: $ES_\tau = E \tau E \xi_1$ but from Doob's theorem $ES_\tau=ES_1 = E \xi_1$ so $E \tau = 1$. Am I right? – Thomas Nov 10 '14 at 15:34
• I dont understand how you are using Doob's Theorem. – gmath Nov 10 '14 at 17:20