Artificial intelligent assistant

Wald's equation: Modified assumption 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?

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.

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