Undergrad On the expected value of a sum of a random number of r.v.s.

[psie]

TL;DR

I am confused about a proof concerning the expectation of a sum of a random number of random variables

There's a theorem in An Intermediate Course in Probability by Gut that says if $E|X|<\infty\implies EX=g_X'(1)$, where $g_X$ is the probability generating function. Now, consider the r.v. $S_N$, which is the sum of a random number $N$ of terms of i.i.d. r.v.s. $X_1,X_2,\ldots$ (everything's nonnegative integer-valued, and $N$ is independent of $X_1,X_2,\ldots$). One can derive the probability generating function for $S_N$, namely $g_{S_N}(t)=g_N(g_X(t))$. I am now reading a theorem that states;

Theorem If $EN<\infty$ and $E|X|<\infty$, then $ES_N=EN\cdot EX$.

The author proves this using the theorem I stated in the beginning, namely that $E|X|<\infty\implies EX=g_X'(1)$. What I don't understand is why we require $EN<\infty$ and $E|X|<\infty$. For $ES_N$ to exist via generating functions, we require $E|S_N|<\infty$, but I don't see how this means that we should require $EN<\infty$ and $E|X|<\infty$.

One idea that comes to mind is the following, but I'm not sure if this is correct: $$E|S_N|=E(|X_1+\ldots +X_N|)\leq E(|X_1|+\ldots +|X_N|)=E (N|X_1|)=EN E|X_1|,$$and so we see that $E|S_N|$ is finite if $EN$ and $E|X_1|$ are finite, as required by theorem. But I'm doubting if $E(|X_1|+\ldots +|X_N|)=E (N|X_1|)$ is correct. Grateful for any confirmation or help.

 

[Office_Shredder]

You can start with
$E|S_N|=\sum_{k=1^\infty} P(N=k) E|X_1+..+X_k|$

And now you are doing triangle inequalities on fixed number of terms

 

[psie]

You can start with
$E|S_N|=\sum_{k=1^\infty} P(N=k) E|X_1+..+X_k|$

And now you are doing triangle inequalities on fixed number of terms

Silly question maybe, but which variables is $S_N$ and consequently $|S_N|$ a function of? Certainly $N$, but is it correct to say it is also a function of $X_1,\ldots,X_N$?

 

[psie]

I think we should be able to write $$S_N = \sum_{j = 1}^{\infty}X_j \mathbf1_{j \leq N},$$ so $S_N$ is $\sigma((Y_n)_{n\in\mathbb N})$-measurable, where $Y_1=N, Y_2=X_1, Y_3=X_2, \ldots$.