# Moment generating function questions with two distributions

Suppose $X_1,X_2,\dots,X_n$ are independent each with exponential distribution with parameter $\lambda$ and $N$ has a Poisson distribution ($\lambda$) and is independent of $X_is$

a) Determine the moment generating function $S_N= X_1 + X_2+\dots+X_N$.

b) Use your answer in (a) to find mean of $S_N$.

c) Find the distribution $Y= \sum_{i=1}^n 2\lambda X_i$ identify the distribution of $Y$ by name and specify all parameter values.

I can only suspect this question has a lot to do with Gamma distribution's moment generating function. Need serious help on this kind of question, any help is really appreciated

• en.wikipedia.org/wiki/Erlang_distribution – Alex Aug 13 '13 at 19:53
• thanks! now i know what it looks like, but I still don't know how to get $S_N$. so you add $x_is$ and multiply it by poisson? that's the closest thing i can get.. – Roy Wang Aug 13 '13 at 20:56
• For (a), do you mean $S_N=X_1+\ldots+X_N$? If so, I think you need to compute the conditional expectation $\mathbb{E}[e^{tS_N}|N=n]$. – eeeeeeeeee Aug 13 '13 at 23:23

As pedrosuavo suggests, the key here is conditional expectation; and, more specifically, the formula $$\tag{1} \DeclareMathOperator{\E}{\mathbb{E}}\E\bigl[\E[e^{tS_N}\mid N]\bigr]=\E[e^{tS_N}].$$ Conitioned on $N=n$, $S_N$ is distributed as $S_n$, so that $$\E[e^{tS_N}\mid N=n]=\E[e^{tS_n}\mid N=n]=\E[e^{tS_n}],$$ since $S_n$ and $N$ are independent. But, since $X_1,\ldots,X_n$ are independent and identically distributed, we have $$\E[e^{tS_n}]=\E[e^{tX_1}\cdots e^{tX_n}]=\prod_{i=1}^{n}\E[e^{tX_i}]=(\E[e^{tX_1}])^n$$ provided $n\geq 1$. This being the case, we know that $$\E[e^{tX_1}]=\int_0^{\infty}\lambda e^{-\lambda x}e^{tx}\,dx.$$ Thus the function $t\mapsto \E[e^{tX_1}]$ has domain $t\in(-\infty,\lambda)$, and for such $t$ we have $\E[e^{tX_1}]=\frac{\lambda}{\lambda-t}$.
So, for $n\geq 1$, and $t<\lambda$, we have $$\E[e^{tS_N}\mid N=n]=\left(\frac{\lambda}{\lambda-t}\right)^n.$$ What about when $n=0$? We have $S_0=0$, so that $\E[e^{tS_0}]=1=\left(\frac{\lambda}{\lambda-t}\right)^0$.
So, going back to the formula (1), we have that for $t<\lambda$, \begin{align} \E[e^{tS_N}]&=\sum_{n=0}^{\infty}\E[e^{tS_N}\mid N=n]\cdot P(N=n)\\ &=\sum_{n=0}^{\infty}\left(\frac{\lambda}{\lambda-t}\right)^n\frac{e^{-\lambda}\lambda^n}{n!}\\ &=\exp\left[\frac{t\lambda}{\lambda-t}\right]. \end{align} Based on this, we can compute $\E[S_N]$; if $\phi(x)$ is the MGF for $S_N$, then we know that $\E[S_N]=\phi'(0)=1$.
(Conveniently, this matches our heuristic idea that $\E[S_N]$ should be something like $\E[N]\cdot\E[X_i]=\lambda\cdot\lambda^{-1}$.)