A problem with stopping time about random walk

Question

Let $(Z_i)_{i \ge 1}$be a sequence of i.i.d. random variables that are uniformly distributed on $\{1,1\}$ i.e., $P(Z_i = 1) = P(Z_i = -1) = \frac{1}{2}$ and define the Markov chain, for $n \ge 1$, $X_0 = 0$ \begin{equation*} X_n = X_{n-1} + Z_n \end{equation*} Now, let $a \in \mathbb{Z}$ be a non-zero and define \begin{equation*} T = \inf\{n \ge 1: X_n = a\} \end{equation*} which is, $T$ is the first hitting time of $a$ and a stopping time

Then, $E[X_T] = 0$ or $E[X_T] = a$?

(1) First, I am not sure $X_T$ is a measurable function, i.e., a random variable.

(2) If so, for all $\omega \in \Omega$, $X_{T(\omega)} (\omega) = a$ and in that case, $E[X_T] = a$. Is this the wrong argument?

(3) However, I think, according to Optimal Sample Theorem and Wald Identities, $E[X_T] = 0$.

I am really confused about this. Could you tell me which part is wrong and what the right argument is? Thank you very much.

Answer 1

1) $X_T$ is $\mathcal F_T:=\{A\in \mathcal F\mid A\cap\{T=n\}\in \mathcal F_n\}-$measurable.

2) Obviously (and as you remark), $\mathbb E[X_T]=a$ ($X_T$ is deterministic).

3) Why do you think that you can use Optional stopping theorem ?

Answer 2

I was just got stuck at this question so I decided to complete this question for future people who come here.

The reason we can't use Doob's thm is because $\mathbb{E}[T]=\infty$. I will prove it for $a=1$.
$\mathbb{E}[T]=\sum_{k=1}^\infty k\mathbb P(T=k)$
$\mathbb{P}(T=k)=\mathbb P(S_k=1 \wedge \forall m<k S_m<1)=\frac{C_k}{2^k}$ when $C_k$ is Catalan number. Using Stirling's formula we get $\frac{C_k}{2^k}\geq\frac{2^k}{k\sqrt k}e^{-\frac{1}{8k}}>\frac{2^k}{50k\sqrt k}$
And so
$\mathbb{E}[T]=\sum_{k=1}^\infty k\mathbb P(T=k)>\sum_{k=1}^\infty\frac{2^k}{50k\sqrt k}=\infty$