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Let $X_1, X_2$, ... be i.i.d random variables all whose characteristic functions are $\phi_{X_i} = \phi(t), i = 1, 2, ...$

If $N$ is a random variable taking values in the positive integers, and if $N$ is independent of $X_1, X_2$, ..., determine $\phi_{S_N}$ where $S_N = X_1 + X_2 + ... + X_N.$

Attempt:

$$ \phi_{S_N}(t) = E(e^{itS_N}) = E(exp\{it(X_1 + X_2 + ... + X_N)\} $$

$$ = E(e^{itX_1}e^{itX_2}...e^{itX_N}) $$

$$ = E(e^{itX_1})...E(e^{itX_N})$$

$$ = \phi_{X_1}\phi_{X_2}...\phi_{X_N} $$

$$ = [\phi(t)]^N $$

Official Answer: $ E\Big([\phi(t)]^N\Big) $

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The problem in your answer is that $\phi(t)^N$ is a random variable which may not be constant.

First observe that $e^{itS_N}=\sum_{n\geqslant 1}e^{itS_n}\mathbf 1\{N=n\}$. Then take the expectation on both sides. In order to compute $\mathbb E\left[e^{itS_n}\mathbf 1\{N=n\}\right]$, first use the independence of $N$ with $S_n$ to write it as $\mathbb E\left[e^{itS_n}\right]\mathbb P\{N=n\}$. Then use what you did for a fixed $n$.

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  • $\begingroup$ I don't understand the problem you mention. The expectation of the product of random variables is the product of the expectations, when the random variables are independent, which happens to be the case. After that, I just used the definition of characteristic function. Also, I don't understand what you told me to do.. if you would please explain a little more or start the correct solution, please. $\endgroup$
    – Sigma
    Oct 22, 2019 at 21:46
  • $\begingroup$ You are in a special case of math.stackexchange.com/questions/3267837/… "The expectation of the product of random variables is the product of the expectations, when the random variables are independent, which happens to be the case." This is true when the number of elements of the product is fixed. When it is random, it is an other story. $\endgroup$ Oct 22, 2019 at 21:49
  • $\begingroup$ I didn’t know that, although I don’t understand why. I mean, N is still just a positive integer. $\endgroup$
    – Sigma
    Oct 22, 2019 at 21:53
  • $\begingroup$ But not constant: this is a random variable. $\endgroup$ Oct 22, 2019 at 21:58
  • $\begingroup$ Okay.. as I said, I was wondering if you would please start the correct solution, as i don’t understand what you told me to do $\endgroup$
    – Sigma
    Oct 22, 2019 at 22:00

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