# Distribution of sum of $m$ independent random variables

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.

• @Pedro Thank you, but please don't forget. Because I really need the help here – Stan_Allen Apr 4 '15 at 15:53
• For the first part, you can have a look here: math.stackexchange.com/questions/250059/… or there. – Clement C. Apr 4 '15 at 15:57
• 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. – Pedro Apr 4 '15 at 15:58
• You just take the convolution of all the individual pdfs – texasflood Apr 4 '15 at 15:59
• See here www-sigproc.eng.cam.ac.uk/foswiki/pub/Main/NGK/… page 33 – texasflood Apr 4 '15 at 16:00

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).