Moment generating function of a conditional distribution

Question

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?

Answer 1

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 $\beta$.