I am trying to understand the proof of Theorem 4.8.7 in Durrett's Probability: Theory and Examples. The theorem is about symmetric simple random walk: let $\xi_1, \xi_2, \cdots$ be $i.i.d.$, $S_n = S_0 + \xi_1, \cdots + \xi_n$ where $S_0$ is a constant, and let $\mathcal{F}_n = \sigma(\xi_1, \cdots, \xi_n)$.
The theorem is: Suppose $P(\xi_i = 1) = P(\xi_i = -1) = 1/2$, $S_0 = x$, and let $N = \min\{n: S_n \notin (a, b)\}$. Then
(a) $P_x(S_N = a) = \frac{b-x}{b-a}$, $P_x(S_N = b) = \frac{x-a}{b-a}$, where $x$ is the starting point.
(b) $E_0 N = -ab$ and hence $E_x N = (b – x)(x – a)$.
My question is about the proof of part (b). Durrett first showed that $E_0 S^2_{N \wedge n} – E(N \wedge n) = 0$ and then used the limit of $E_0 S^2_{N \wedge n}$ to show that $E_0(N) = \lim E_0(N \wedge n) = \lim E_0 S^2_{N \wedge n} = -ab$.
The question is: Durrett used the stopping theorem for the bounded stopping time to show that $E_0 S^2_{N \wedge n} – E(N \wedge n) = 0$. The bounded stopping time theorem says that, if $X_n$ is a martingale and $N$ is a stopping time with $P(N \leq k) = 1$, then $EX_0 = EX_N = EX_k$. However, I don't see how he arrived at this conclusion using this theorem – where does the $E_0 S^2_{N\wedge n}$ term come from?
Thank you very much in advance.
For completeness I have attached the screenshots of the theorem and the proof:
Best Answer
The trick for answering questions by appealing to the optional stopping theorem often require choosing the right martingale. You have already observed that the random walk process $S_n$ is a martingale. There is another process associated to $S_n$ which is also a martingale:
$$R_n = S_n^2 - n$$
Now, since $n \land N$ is a bounded stopping time, the optional stopping theorem applied to the martingale $R_{n \land N}$ gives us that for each $n$: $$0 = \mathbb{E}(R_0) = \mathbb{E}(R_{n\land N}) = \mathbb{E}(S_{n \land N}^2) - \mathbb{E}(n \land N)$$ By sending $n \to \infty$, an application of MCT and LDCT gives us that $\mathbb{E}(N) = -ab$