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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.
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$\newcommand{\E}{\mathbb{E}}$ 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 $\E[S_{\tau}|\tau]=\tau \E X_n$

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  • $\begingroup$ Maybe a silly question, but why can you interchange the sum and the expectation in your third sentence? $\endgroup$ – Mriganka Basu Roy Chowdhury May 26 at 6:36
  • $\begingroup$ @MrigankaBasuRoyChowdhury if you change $X_i$'s with their absolute value, then the thing in the inner part of the expectation will be positive hence by monotone convergence theorem you can take the sum out. But this also implies that without absolute value it also hold, by dominated converge theorem $\endgroup$ – dankmemer May 26 at 11:20
  • $\begingroup$ But then, we will need that $\mathbb{E}(\sum_{k = 1}^{\tau}|X_k|) < \infty$. This seems to be a stronger condition than $\mathbb{E}(|S_\tau|) < \infty$ which is necessary for defining $\mathbb{E}(S_\tau)$. Anyway, thanks for the help! $\endgroup$ – Mriganka Basu Roy Chowdhury May 26 at 11:40
  • $\begingroup$ And, can't it be inferred from the conditions of the theorem that $\mathbb{E}(|S_\tau|) < \infty$? Otherwise, if we're required to verify this statement before applying the result, then a lot of the power of this theorem is lost. $\endgroup$ – Mriganka Basu Roy Chowdhury May 26 at 11:42
  • $\begingroup$ But sure enough, the statement would hold for nonnegative $X_t$, so we can indeed write $\mathbb{E}(\sum_{k = 1}^\tau |X_k|) = \mathbb{E}(\tau)\mathbb{E}(|X_1|)$. If we can show that this is $< \infty$ in a particular setting, we would be able to use the said theorem as $|S_\tau| \leq \sum_{k = 1}^\tau |X_k|$, but still, maybe the said theorem also holds for general RVs. $\endgroup$ – Mriganka Basu Roy Chowdhury May 26 at 11:44

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