Reference request: Where is this exercise concerning the exponential distribution found in textbooks?

Question

\begin{align} & T_1,T_2,T_3,\ldots\sim\text{i.i.d.} \\ & \text{with distribution } e^{-t/\mu} (dt/\mu)\text{ for } t>0. \\[8pt] & \Pr(N=n) = p^{n-1}q \text{ for } n = 1,2,3,\ldots \\ & \text{where $p+q=1,$ $p,q>0$} \\ & \text{and $N$ is independent of } T_1,T_2,T_3,\ldots \\ & \text{so that } \operatorname E(N) = 1/q. \\[8pt] & T= \sum_{n\,=\,1}^N T_n \\ & \text{so } \operatorname E(T) = \mu/q. \end{align} An exercise is to show that the distribution of $T$ is exponential: $e^{-qt/\mu} (q\,dt/\mu) \text{ for } t>0.$

MY QUESTION IS whether this is somehow a standard exercise known to everyone except me, or more literally: Did I just somehow miss that this one is in various textbooks? And while I'm at it, do others solve this differently from the way I did (I'll post something below)?

Answer 1

This is fairly standard. One easy way to prove it is to use Laplace transforms. The Laplace transform of $\exp(\lambda)$ is $\frac {\lambda} {\lambda+t}$. You can compute the Laplace transform of $T$ by conditioning on $N$ and then compute the geometric sum you get. You will get $\frac {q/\mu} {q/m+t}$ which is the Laplace transform of $\exp(q/\mu)$.

Answer 2

This will answer the exercise but not the main question.

A standard exercise posits that $M\sim\operatorname{Poisson}(\lambda)$ and $X\mid M \sim \operatorname{Binomial}(M,q),$ and asks for proof of the following:

• $X\sim\operatorname{Poisson}(q\lambda),$
• $M-X\sim\operatorname{Poisson}((1-q)\lambda),$
• $X$ and $M-X$ are independent.

Then one has the "thinned Poisson process": The number of arrivals in any time interval is $\lambda$ times the length of the interval, and the numbers of arrivals in disjoint intervals are independent, and one "thins" the process by accepting each arrival with probability $q$ independently of the Poisson process. The exercise above shows that the "thinned" process must also be a Poisson process, so its interarrival times are exponentially distributed. Finally, one observes that those exponentially distributed interarrival times are just what is described in the question above.

So the next question is: What is a more direct way to do it? Without mentioning the Poisson distribution. I expect this will be easy, but I'm about to crash and I won't think about that further tonight.

This does not actually answer the main question above.

Answer 3

Here's a more direct argument than what either I or Kavi Rama Murthy posted earlier, but this still does not answer the reference request: \begin{align} & \Pr(T>t) = \operatorname E(\Pr(T>t\mid N)) \\[8pt] = {} & \sum_{n\,=\,1}^\infty \Pr(N=n) \Pr(T>t\mid N= n) \\[8pt] = {} & \sum_{n\,=\,1}^\infty p^{n-1} q \int_t^\infty \frac 1 {(n-1)!} \left( \frac s \mu \right)^{n-1} e^{-s/\mu} \, \left( \frac{ds}\mu \right) \\[8pt] = {} & \int_t^\infty q \sum_{n\,=\,1}^\infty \frac{(ps/\mu)^{n-1}}{(n-1)!} e^{-s/\mu} \left( \frac{ds}\mu \right) \\[8pt] = {} &\int_t^\infty qe^{ps/\mu} e^{-s/\mu} \left( \frac{ds}\mu \right) = \int_t^\infty e^{-qs/\mu} \left( \frac{q\,ds}\mu \right) = e^{-qt/\mu}. \end{align}