1

Let $T_i$ for $i=1,2,...$ be a sequence of i.i.d exponential random variables with common parameter $\lambda$.

Let $N$ be a geometric random variable with parameter $(1/(p+1))$ that is independent of the sequence $T_i$.

Let $X$ be the sum of the $T_i$ from 1 to $N$ Show that the distribution of X is exponential.

I would like to use MGFs. I'm not sure how to incorporate the MGF of N in this case.

Bob
  • 1,157
  • One thing that would help you get started: the mean of $X$ will have to be $E[T] E[N]$, by Wald's identity. So that will at least tell you what exactly you want to get in the end. – Ian Oct 07 '15 at 23:24
  • https://math.stackexchange.com/questions/634158/pdf-of-a-sum-of-exponential-random-variables – StubbornAtom May 11 '21 at 18:59

2 Answers2

4

Consider a Poisson process of rate $\lambda$. Independently, each occurrence of the Poisson process is "special" with probability $q = 1/(p+1)$. Your $T$ is the waiting time until the first special occurrence.

You can also consider this from a different point of view: the special and the non-special occurrences form independent Poisson processes with rates $q\lambda$ and $(1-q)\lambda$. The waiting time until the first special occurrence is then exponential with rate $q\lambda$.

Robert Israel
  • 448,999
1

You can use MGFs to obtain the distribution of the conditional sum $X \mid N$ and show that this is gamma distributed with rate $\lambda$ and shape $N$. Then we would compute the unconditional/marginal distribution $X$ by summing over all $N = 0, 1, 2, \ldots$, weighted by the probability $\Pr[N = n]$.

heropup
  • 135,869
  • 1
    This works pretty well, with the condition that we choose the definition of Geometric distribution that forbids $N=0$ – John Fernley May 24 '19 at 12:06