1
$\begingroup$

Let $(X_n)_{n\in \mathbb{N}}$ be an infinite sequence of real-valued random variables and let $N$ be a nonnegative integer-valued random variable, where all $X_n$ are integrable. Now consider Wald's equation:

$E[\sum_{n=1}^NX_n]=E[N]E[X_1]$

which is valid, if additionally $N$ is independent from the $X_n$. I've wondered, if Wald's equation would still be valid, if this assumption is not true, but if it holds that $\mathbb{1}_{N=n}, X_{n+1},X_{n+2},\dots$ are independent for all $n\in\mathbb{N}$.

Additionally, what if still $\mathbb{1}_{N=n}, X_{n+1},X_{n+2},\dots$ are independent for all $n\in\mathbb{N}$, but the $X_n$ are not iid, but only have the same expectation? Would Wald's equation then still be valid?

$\endgroup$
0
$\begingroup$

Think about how Wald's identity is proved and see what modifications can be made:

$$E\left[\sum_{n=1}^NX_n\right]=E\left[\sum_{n=1}^\infty X_n1_{N\ge n}\right]{\stackrel{*}{=}}\sum_{n=1}^\infty E[X_n1_{N\ge n}].$$

So Wald's identity follows provided (a) we can verify that $*$ is justified, and (b) for all $n\ge1$, we have $E[X_n1_{N\ge n}]=E[X_1]P(N\ge n)$. Certainly (b) holds if each $X_n$ is integrable with common mean and $X_n,1_{N\ge n}$ are independent for all $n\ge1$. If in addition either each $X_n$ is nonnegative or $N$ is integrable, then (a) holds as well. Your conditions seems not to be enough, but clearly we do not require $\{X_n\}$ iid and independent of $N$ either.

$\endgroup$
  • $\begingroup$ "Your conditions seems not to be enough" Actually they are since it suffices to use that $X_n$ and $1_{N\leqslant n-1}$ are independent. $\endgroup$ – Did Jun 17 '17 at 18:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.