$\newcommand{\E}{\mathbb{E}}$
Theorem (Wald's identity): Suppose $\{X_i\}_{n \in \mathbb{N}}$ is a sequence of i.i.d random variables with $\E X_1 < \infty$. Let $\tau$ be a stopping time with respect to the usual filtration $\{\mathcal{F}_n = \sigma(X_1, X_2, \ldots, X_n)\}_{n \in \mathbb{N}}$, such that $\E\tau < \infty$. Define $S_t = \sum_{k = 1}^{t} X_k$. Then $\E S_\tau = \E\tau \E X_1$.
"Proof:" $\E S_\tau = \E[\E[S_\tau\mid\tau]] = \E[\tau\E X_1] = \E\tau\E X_1$.
The most controversial appears to be the second equality, i.e. $\E[S_\tau|\tau] = \tau\E X_1$. So, I have two questions:
- Why is this wrong, if it is.
- And how to justify the second equality rigorously.
$\newcommand{\E}{\mathbb{E}}$ For the first question, i think it is true. For the second one : We can write for every stopping time $\tau$, $S_\tau$ as $1_{\tau \geq 1} X_1 + ... + 1_{\tau \geq n} X_n$ etc. Now $ \E[S_{\tau}]=\E[\sum 1_{\tau \geq n} X_n] =\sum \E [(1-1_{\tau \leq n-1}) X_n]$. $1-1_{\tau \leq n-1}$ is $F_{n-1}$ measurable so we get $\sum \E[1_{\tau \geq n}\E[X_n|F_{n-1}]]$ But $X_n$ is independent from $F_{n-1}$, so we get $\sum \E1_{\tau \geq n}]\E[X_n]=\E\tau\E X_n =\E\tau \E X_1$. By Radon-Nikodym we can say that $\E[S_{\tau}|\tau]=\tau \E X_n$