Can someone explain how to solve the following problem?
If $\{E_n\}$ are iid exponentially distributed with parameter $\alpha$, $N$ is geometric with $P(N=n) = p(1-p)^{n-1}$, $n\geq 1$ and $N$ is independent of $\{E_n\}$, show that $\sum_{n=1}^N E_n$ has an exponential distribution by computing the Laplace transform. What is the parameter? What is the relation of this result to thinning?
I am not very familiar with Laplace transforms, so I am not sure what I am meant to see here. If I try to compute the probability distribution, I think I obtain the result that $\sum_{n=1}^N E_n \sim Exp(p\alpha)$ but I don't know how to get this result using Laplace transforms. This is what I have so far, is it right?
\begin{align*} E[e^{-t\sum_{n=1}^N E_n}] &= E[(e^{-tE_1})^N]\\ &= \sum_{n=1}^\infty E[(e^{-tE_1})^N| N=n] P(N=n) \\ &= \sum_{n=1}^\infty E[(e^{-tE_1})^n] p(1-p)^{n-1} \\ &= p \sum_{n=1}^\infty \Big(\int_0^\infty e^{-tx} \alpha e^{-\alpha x}dx\Big)^n (1-p)^{n-1} \\ &= p\sum_{n=1}^\infty \Big[-\frac{\alpha}{t+\alpha}\Big]^n (1-p)^{n-1} \end{align*}
$\int_0^\infty e^{-tx} \alpha e^{-\alpha x}dx =\frac {\alpha } {\alpha +t}$. The minus sign in your computation is a mistake. Now the final sum you get is just $c\sum\limits_{k=1}^{\infty} s^{k}$ where $c=\frac p {1-p}$ and $s=\frac {\alpha (1-p)} {\alpha +t}$. This is a geometric sum and so we get $\frac p {1-p} \frac s {1-s}$ which becomes $\frac {\alpha p} {t+\alpha p}$. This is the Laplace trasnform of exponential distribution with parameter $\alpha p$.