Expectation of special sum and indicator function, i.i.d. random variables

Question

Let $X_1, X_2, X_3, \ldots \in \mathscr{L}^1(\mathbb{P})$ be independent, identically distributed random variables with values in $\left[0, \infty\left[\right. \right.$ and $N \in \mathscr{L}^1(\mathbb{P})$ a random variable independent of $X_1, X_2, X_3, \ldots$ with values in $\mathbb{N}$. Let the random variable $S_N$ be given by $$ S_N:=\sum_{n=1}^{\infty} S_n \mathbf{1}_{\{N=n\}} $$ where $S_n:=\sum_{k=1}^n X_k, n \in \mathbb{N}$. Show that $$ \mathbb{E}\left(S_N\right)=\mathbb{E}(N) \mathbb{E}\left(X_1\right) $$ holds.

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First of all, I think I have to show that $S_N$ is a random variable. Nevertheless, I got this: $$ \mathbb{E}\left(S_N\right)=\mathbb{E}\left(\sum_{n=1}^{\infty} S_n \mathbf{1}_{\{N=n\}}\right)=\sum_{n=1}^{\infty}\mathbb{E}\left(S_n \mathbf{1}_{\{N=n\}}\right)=\sum_{n=1}^{\infty}\mathbb{E}(S_n) \mathbb{E}(\mathbf{1}_{\{N=n\}}) $$ and here I dont know if it is right to say that this equals to $$ \sum_{n=1}^{\infty}\mathbb{E}(S_n) \mathbb{E}(N)=\sum_{n=1}^{\infty}\mathbb{E}\left(\sum_{k=1}^n X_k\right) \mathbb{E}(N)=\sum_{n=1}^{\infty}\sum_{k=1}^n\mathbb{E}\left(X_k\right) \mathbb{E}(N)=\mathbb{E}(X_1) \mathbb{E}(N) $$ I am not sure and I think I broke some math laws here. Any hints on this expectation and how to show that $S_N$ is a random variable? Thanks in advance!

Answer 1

That $S_N$ is a random variable simply follows from the fact that it is the pointwise limit of $\left(\sum_{n=1}^\ell S_n\mathbf{1}_{\{N=n\}}\right)_{\ell\geqslant 1}$ (the series is convergent since the sets $\{N=n\},n\in\mathbb N$ are pairwise disjoint.

The expectation you want is not equal to $\sum_{n=1}^{\infty}\mathbb{E}(S_n) \mathbb{E}(N)$, actually, unless $X_1$ is centered, $\sum_{n=1}^{\infty}\mathbb{E}(S_n)$ is divergent.

However, it was correct until this point and you can use that $$ \mathbb E(S_n) \mathbb{E}\left(\mathbf{1}_{\{N=n\}}\right)=n\mathbb E(X_1)\mathbb{E}(\mathbf{1}_{\{N=n\}}) $$ then $$n\mathbb{E}\left(\mathbf{1}_{\{N=n\}}\right)= \mathbb{E}\left(n\mathbf{1}_{\{N=n\}}\right)= \mathbb{E}\left(N\mathbf{1}_{\{N=n\}}\right),$$ which gives $$ \mathbb E(S_n) \mathbb{E}\left(\mathbf{1}_{\{N=n\}}\right)=\mathbb E(X_1)\mathbb{E}\left(N\mathbf{1}_{\{N=n\}}\right) $$ and summing over $n\in\mathbb N$ gives the wanted formula.