# Solved – 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?

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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 $$Xsim text{Bin}(S,frac{alpha}{alpha+beta})$$. 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(-tsum_{i=1}^S B_i)|S] = text{E}[prod_{i=1}^S e^{-tB_i}|S] = prod_{i=1}^Stext{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}_Sbig[e^{St}text{E}[e^{-Xt}|S]big] = text{E}_Sbig[e^{St}Big(frac{alpha e^{-t}+beta}{alpha+beta}Big)^Sbig]\ & = text{E}big[Big(frac{alpha +beta e^t}{alpha+beta}Big)^Sbig] 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 $$beta$$.