0
$\begingroup$

Let $A_m$ be the sum of $m$ identically distributed random variables that are independent and that have an exponential distribution with parameter $\mu$. How do I prove that $A_m$ has a gamma distribution with parameters $m$ and $\mu$? And how do I prove that $M_s=\max\{m:A_m\leq s\}$ has a Poisson distribution?

I don't really know how to tackle such a problem.

$\endgroup$
  • $\begingroup$ @Pedro Thank you, but please don't forget. Because I really need the help here $\endgroup$ – Stan_Allen Apr 4 '15 at 15:53
  • $\begingroup$ For the first part, you can have a look here: math.stackexchange.com/questions/250059/… or there. $\endgroup$ – Clement C. Apr 4 '15 at 15:57
  • $\begingroup$ I can already tell you that the sum of $m$ exponentially distributed random variables is in fact an Erlang distribution. A Gamma distribution with $m$ an integer is called an Erlang distribution. This erlang distribution is often used in queueing theory. $\endgroup$ – Pedro Apr 4 '15 at 15:58
  • $\begingroup$ You just take the convolution of all the individual pdfs $\endgroup$ – texasflood Apr 4 '15 at 15:59
  • $\begingroup$ See here www-sigproc.eng.cam.ac.uk/foswiki/pub/Main/NGK/… page 33 $\endgroup$ – texasflood Apr 4 '15 at 16:00
0
$\begingroup$

The characteristic function of an exponential($\lambda$) distribution is $\frac{ \lambda}{\lambda-i \omega}$, so the sum of $n$ i.i.d. Exp($\lambda$) distributions has characteristic function $(\frac{ \lambda}{\lambda-i \omega})^n$ since sums of independent rv's have their characteristic functions multiply. Then, you can recognize this as a gamma distribution with appropriate parameters by comparing to the form of a Gamma distribution (write $\frac{ \lambda}{\lambda-i \omega} = 1+\frac{ i \omega}{\lambda-i \omega}$ will make the comparison easier).

As for the latter, think of a Poisson process.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.