Simple symmetric random walk and limit of expected value at stopping time.

Question

Take $\{X_n\}$ a simple symmetric random walk on state space $\mathbb{Z}$. With $X_0 = 0$ and stopping time : T = $inf\{n \geq 0 : X_n = -a\}$ then, $lim_{M \rightarrow \infty} E[X_M | T > M] = \infty$

I have defined: $K = min(T , M)$
Take some arbitrary $M \in \mathbb{Z}$ so $M < \infty$ therefore $P(K \leq M) = 1 \implies E[X_K] = E[X_0] = 0$ by properties of martingales.
Given, $M$ is an upper bound for $K$ : $E[X_M] = E[X_0] = 0$.

I am hoping to establish the conditional expectation, any hints would be appreciated!

Answer 1

By the law of total probability we have $$\begin{align*}0 = E[X_K] &= E[X_K \mid T > M] P(T > M) + E[X_K \mid T \le M] P(T \le M)\end{align*}$$ Then observe that:

1. $E[X_K \mid T > M] = E[X_M \mid T > M]$

2. $E[X_K \mid T \le M] = -a$

Conclude that $$E[X_M \mid T > M] = a \frac{P(T \le M)}{P(T > M)}.$$

By recurrence of simple random walk, we have $P(T < \infty) = 1$, so as $M \to \infty$ we have $P(T \le M) \to 1$ and $P(T > M) \to 0$.