# mgf of an infinite sum of independent random variables

It is known that if $S_n = \sum_{i=1}^{n} X_i$, where the $X_i$ are independent random variables, then the moment-generating function for $S_n$ is given by $M_{S_n}(t)=\prod_{i=1}^{n}M_{X_i}(t)$.

Now suppose that we have infinitely many random variables $X_i,i\in\mathbb{N}$, and suppose that the moment-generating function exists for each random variable $X_i,i\in\mathbb{N}$. Denote $S = \sum_{i=1}^{\infty} X_i$. Then do we have that (maybe under some additional conditions) $M_{S}(t)=\prod_{i=1}^{\infty}M_{X_i}(t)$? If so then how can we prove it? I really tried hard to find out such a formula for the mgf of an infinite sum of independent random variables but all the lecture notes and books that I dipped into only present the formula for finite case.

## 1 Answer

This is vague, because the details will differ with the problem. For starters you'd want the sum of the rvs to converge almost surely, which could follow from a moment condition. Then you need the mgf of the infinte sum to be the pointwise limit of the finite sum mgfs, in some neighborhood of 0 (to be of any use) which sounds like a dominated convergence theorem application. Along the way you can write down the characteristic function, for which the dominated convergence application is trivial. You might well read off the functional form of the mgf from that, and reassure yourself about the absence of nearby singularities.

May I suggest you try to do this in the case where $X_i = \pm 2^{-i},$ with random signs, as an exercise?