I'm watching a lecture series on physical application of stochastic processes, and the professor is making some pretty wild jumps that I don't follow when deriving the moment generating function for the sum of Poissons.
Let $M \sim \text{Poisson}(\mu)$, $N \sim \text{Poisson}(\nu)$, $M \bot N$, and $S = M + N$.
$$P(s) = \sum_{m=0}^\infty\sum_{n=0}^\infty \frac{e^{-\mu}\mu^m}{m!}\frac{e^{-\nu}\nu^n}{n!} \text{ such that } m+n = s$$
Say we have a generating function of the form $\sum_{s=0}^\infty f(s) = P(s)z^s$. Expanding we have:
$$ f(s) = \sum_{s=0}^\infty\sum_{m=0}^\infty\sum_{n=0}^\infty \frac{e^{-\mu}\mu^m}{m!}\frac{e^{-\nu}\nu^n}{n!} z^s \text{ such that } m + n = s $$
(1) Now for the first handwavey part ... He just drops the summation over $s$ and substitutes $m+n$. This doesn't seem legal.
$$ f(s) = \sum_{m=0}^\infty\sum_{n=0}^\infty \frac{e^{-\mu}\mu^m}{m!}\frac{e^{-\nu}\nu^n}{n!} z^{m+n} $$
(2) And now for the jump that has scant explanation. The above reduces to:
$$f(s) = e^{\mu(z-1)+\nu(z-1)}$$
I can't see how the constraint drops at (1) and how the closed form is derived at (2).
What is going on here?
So I actually watched the video, and what the lecturer is doing is different than your interpretation. He does fumble a bit, but it is correct reasoning.
Let's review what is going on. The probability mass function (PMF) of the sum $S = X+Y$ is $$\Pr[S = s] = \sum_{m=0}^\infty \sum_{n=0}^\infty e^{-\mu} \frac{\mu^m}{m!} e^{-\nu} \frac{\nu^n}{n!} \delta_{m+n, s},$$ where he uses the notation $$\delta_{m+n,s} = \mathbb 1(m+n = s) = \begin{cases}1, & m+n = s \\ 0, & m+n \ne s. \end{cases}$$ That is to say, the sum is a double sum but only over those summands for which the indices of summation add to $s$. So for example, if I want $\Pr[S = 4]$, I would sum over the ordered pairs $$(m,n) \in \{(4,0), (3,1), (2,2), (1,3), (0,4)\}.$$
Now, the probability-generating function (PGF) of $S$ is given by $$\mathcal P(z) = \operatorname{E}[z^S] = \sum_{s=0}^\infty z^s \Pr[S = s].$$ This leads to the triple sum $$\mathcal P(z) = \sum_{s=0}^\infty \sum_{m=0}^\infty \sum_{n=0}^\infty \frac{e^{-(\mu+\nu)}}{m! n!} \mu^m \nu^n z^s \delta_{m+n,s}.$$ Now this is the part you say is "hand-waving." The thing to understand here is that the summand is $0$ if $m+n \ne s$. Therefore, the term $z^s$ can only ever equal $z^{m+n}$; it cannot take on any other value for any other pair $(m,n)$ in the summation. Moreover, because the sum is over all $s$, every combination of $(m,n)$ is represented by the double sum over $m$ and $n$ if you remove the $\delta_{m+n,s}$ term and the sum over $s$. This is how he goes from the above to $$\begin{align*}\mathcal P(s) &= \sum_{m=0}^\infty \sum_{n=0}^\infty \frac{e^{-(\mu + \nu)}}{m! n!} \mu^m \nu^n z^m z^n \\ &= \sum_{m=0}^\infty e^{-\mu} \frac{(\mu z)^m}{m!} \sum_{n=0}^\infty e^{-\nu} \frac{(\nu z)^n}{n!} \\ &= e^{-\mu} e^{\mu z} e^{-\nu} e^{\nu z} \\ &= e^{(\mu + \nu)(z-1)}. \end{align*}$$ Now if this doesn't convince you, we can do it formally step by step. Because the limits of summation do not depend on each other, we may interchange the sum order: $$\mathcal P(z) = \sum_{m=0}^\infty \sum_{n=0}^\infty \frac{e^{-(\mu + \nu)}}{m! n!} \mu^m \nu^n \sum_{s=0}^\infty z^s \delta_{m+n,s}.$$ Now instead of thinking the values of $m$ and $n$ are restricted depending on $s$, conceptualize it the other way around: the summand $z^s$ is nonzero only if $s = m+n$ for whatever choice of $m$ and $n$ exist in the outer double sum. Therefore, for each choice of $(m,n)$, the inner sum over $s$ contains only one term, the one that corresponds to $s = m+n$. Formally, $$\sum_{s=0}^\infty z^s \delta_{m+n,s} = z^{m+n}.$$ And the rest follows as before.