Originally Posted by

**akbar** Agree. I am asking this as it is "obviously" a requirement for the proof of Wald's equation, as it is shown on wikipedia:

Wald's equation - Wikipedia, the free encyclopedia
On the second proof, there is the requirement of $\displaystyle \mathbb{E}(X_i|T=t)=E(X_i)$ for $\displaystyle i\leq t$. So it makes sense in that case.

Note this condition of independence is not mentioned in the PlanetMath version of the result where the wiki article is taken from.

So I hope you understand my need of an expert's view on the matter.

Cordialement.

1. The wikipedia page states a very simple result (proof: $\displaystyle E[X_1+\cdots+X_T]=E[E[X_1+\cdots+X_T|T]]=E[T E[X_1]]=E[T]E[X_1]$), this is not the "full" Wald's equation; the version you need here is when $\displaystyle T$ is a stopping time, that's the one I quoted.

2. Of course the PlanetMath page is wrong, and doesn't even make sense: what does it mean that $\displaystyle X_1,\ldots,X_N$ are i.i.d. when $\displaystyle N$ is a random variable itself?

3. Your first answer suggested you knew the optional stopping theorem: then you have one proof with no independence needed between $\displaystyle T$ and the $\displaystyle X_i$'s. This proof works in your case since $\displaystyle T$ is bounded, and the steps $\displaystyle X_i$ as well.

4. In order to get the general result I mentioned in my first post, you would need to avoid the optional stopping theorem (which requires for instance a uniform integrability of $\displaystyle (S_{n\wedge T})_{n\geq 0}$) and write a direct (very short) proof:

$\displaystyle E[X_1+\cdots+X_T]=\sum_{n=1}^\infty E[X_n {\bf 1}_{\{T\geq n\}}]$

and, since T is a stopping time, $\displaystyle \{T\geq n\}(=\{T\leq n-1\}^c)$ depends on $\displaystyle X_1,\ldots,X_{n-1}$ and is independent of $\displaystyle X_n$, hence:

$\displaystyle E[X_1+\cdots+X_T]=\sum_{n=1}^\infty E[X_n]P(T\geq n)=E[X_1]\sum_{n=1}^\infty P(T\geq n)$ $\displaystyle =E[X_1]E[T]$.

...as simple as that. (For full rigour, I should first prove that $\displaystyle E[|X_1+\cdots+X_T|]<\infty$, which is obtained via the exact same lines plus the triangular inequality)