This is a question from my university list:
Let $\{x_1, x_2, \ldots, x_n\}$ be statistically independent and identically distributed random variables, each with an exponential probability density function of parameter $1$, that is: $$ p_{x_i} (X)=a \cdot e^{-aX} u(X) \quad ,\,i=1,2,\ldots $$
Let $n$ be a discrete random variable statistically independent of each $x_i$, with a probability density function given by $$ p_n (N)=\sum_{k=0}^\infty\frac{e^{-1}}{k!} δ(N-k). $$
Set the random variable $$ y = \sum_{i=1}^n x_i $$ where, by definition, $y = 0$ if $n = 0$. Determine:
a) the characteristic function $M_y(v)$;
However, I can't solve this problem problem, I found this solution:
$$M_y = \frac{1}{(a-j \cdot v)^{n}}$$
I don't know how to eliminate the random variable "$n$" from the equation and get a expression that only depends on from v(My(v)). Could someone help me how to complete the solving of this problem???
Use the definition of the characteristic function and separate the instances of expectation values for $n$ and $x_i$ to obtain:
$$M_y(v)=\mathbb{E}[e^{ivy}]=\mathbb{E}[e^{iv\sum_{i=1}^{n}{x_i}}]=\mathbb{E_n[E_{x_i}}[\prod_{i=1}^ne^{iv{x_i}}]]=\mathbb{E}_n[(\mathbb{E_x}[e^{ivx}])^n]$$
However
$$M_x(v)=\mathbb{E}[e^{ivx}]=\int_{0}^{\infty}e^{ivx}ae^{-ax}dx=\frac{a}{a-iv}$$
and now all that's left is to compute the expectation over the state space of the $n$ variable:
$$\begin{align}\mathbb{E}_n[(M_x(v))^n]&=\sum_{N=0}^{\infty}(M_x(v))^N\sum_{k=0}^{\infty}\frac{1}{ek!}\delta(N-k)\\&=\sum_{k=0}^{\infty}\frac{1}{ek!}\sum_{k=0}^{\infty}(M_x(v))^N\delta(N-k)\\ &=\sum_{k=0}^{\infty}\frac{1}{e}\frac{(M_x(v))^k}{k!}\\&=e^{M_x(v)-1}\end{align}$$
and hence we find that
$$M_y(v)=\exp\Big(\frac{iv}{a-iv}\Big)$$
EDIT: Explanation of the first line
Note that the variables $x_1,...,x_n,n$ are independent. This allows us to construct the joint probability distribution
$$f(X_1,..., X_N,N)=p_n(N)\prod_{i=1}^N p_{x_i}(X_i)$$
We can easily see that this is a distribution since trivially
$$\sum_{N=0}^{\infty}\int~\prod_{i=1}^N ~dX_i~f(X_1,..,X_N,N)=1$$
It is indeed a funny looking distribution, because it's state space varies with $N$, and hence that variable has to be chosen first! Nevertheless, with this expression we can compute expectation values like the characteristic function:
$$M_y(v)=\sum_{N=0}^{\infty}p_n(N)\prod_{i=1}^N dX_i \exp(iv\sum_{k=1}^N X_k)f(\{X_i\},N)=\mathbb{E}_N[\mathbb{E}_{\{x_i\}}[e^{iv\sum x_i}]]\\=\sum_{N=0}^{\infty}p_n(N)\Big(\int_{0}^{\infty} dXe^{itX}p_x(X)\Big)^N=\mathbb{E}_N[(M_x(v))^N]$$
The reason why the first line in the equation above is true is essentially because we can pull out $p_n(N)$ because the variables are independent, and view what's left as an expectation value taken over the rest of the variables.