Moment generating function of a conditional distribution Let $S$ ~ Poisson$(\alpha + \beta)$, and $X|_{S = s}$ ~ Binomial$(s, \alpha/(\alpha + \beta))$, $\alpha > 0, \beta > 0$
Suppose Z = S - X is independent from X. What is the distribution of Z?
I've tried the following:
\begin{align}
M_Z(t) &= E[e^{(S-X)t}] \\
 & = E[e^{St}]E[e^{-Xt}] \\ 
 & = E[e^{St}]E[E[e^{-Xt}|S]]\\
 & = E[e^{St}]E[(1-p+pe^{-t})^s], \text{where} \ p = \alpha/(\alpha + \beta) \\
 & = E[e^{St}]E[(1-(sp/s)+(sp/s)e^{-t})^s] \\
 & = E[e^{St}]E[(1+(1/s)(-sp+spe^{-t}))^s] \\
 & = E[e^{St}]E[e^{(-sp+spe^{-t})}], \ \text{as s → ∞}  \\
 & = E[e^{St}]E[e^{sp(e^{-t}-1)}] \\
 & = \ ? \\
 & = M_S(t){E[M_U(-t)]}, \ U \sim Poisson(sp)\\
 & = \ ? \\
 & \sim Skellam(\alpha + \beta, sp)
\end{align}
I feel like I'm missing a step at the question mark. $e^{sp(e^{-t}-1)}$ is the MGF of a -U but I'm missing some steps (I'm ignoring the expectation for now)? Are the steps correct enough to salvage the solution?
 A: I will ignore your assumption that $Z=S-X$ is independent from $X$ because I don't think that is true. Now, first, if $S$ is known then $X\sim \text{Bin}(S,\frac{\alpha}{\alpha+\beta})$. That means $X=\sum_{i=1}^S B_i$, conditional on $S$, where the Bernoulli variables $B_i$ are independent and 1 with probability $\frac{\alpha}{\alpha+\beta}$ and 0 otherwise. So,
\begin{align*}
  \text{E}[e^{-Xt}|S] & = \text{E}[\exp(-t\sum_{i=1}^S B_i)|S] = \text{E}[\prod_{i=1}^S e^{-tB_i}|S] = \prod_{i=1}^S\text{E}[e^{-tB_i}]\\
& =\prod_{i=1}^S \frac{\alpha e^{-t}+\beta}{\alpha+\beta} = \Big(\frac{\alpha e^{-t}+\beta}{\alpha+\beta}\Big)^S
\end{align*}
Now,
\begin{align*}
\text{E}[e^{tZ}] &= \text{E}[e^{(S-X)t}] = \text{E}_S\big[e^{St}\text{E}[e^{-Xt}|S]\big]
 = \text{E}_S\big[e^{St}\Big(\frac{\alpha e^{-t}+\beta}{\alpha+\beta}\Big)^S\big]\\
& = \text{E}\big[\Big(\frac{\alpha +\beta e^t}{\alpha+\beta}\Big)^S\big] 
\end{align*}
and using the fact that the generating function of a Poisson random variable $S$ with mean $\alpha+\beta$ is $\text{E}[z^S]=e^{(\alpha+\beta)(z-1)}$,
\begin{align*}
\text{E}[e^{tZ}] &= \exp \Big((\alpha+\beta)(\frac{\alpha +\beta e^t}{\alpha+\beta}-1)\Big) = e^{\beta (e^t -1)}
\end{align*}
which is the MGF of a Poisson distribution with mean $\beta$. 
