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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.

ki3i
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Stan_Allen
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  • @Pedro Thank you, but please don't forget. Because I really need the help here – Stan_Allen Apr 04 '15 at 15:53
  • For the first part, you can have a look here: http://math.stackexchange.com/questions/250059/sum-of-independent-gamma-distributions-is-a-gamma-distribution or there. – Clement C. Apr 04 '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 04 '15 at 15:58
  • You just take the convolution of all the individual pdfs – texasflood Apr 04 '15 at 15:59
  • See here http://www-sigproc.eng.cam.ac.uk/foswiki/pub/Main/NGK/3F1RandProcSV.pdf page 33 – texasflood Apr 04 '15 at 16:00
  • The distribution of the sum of N iid RV has pdf which is the convolution of their individual pdfs, hence the mgf is the product of their mgf and as then are all the same the mgf of the sum is the N-th power of the pdf of any one or the RVs. Now compare this N-th power with the mgf of the distribution you wish to show it equal to. This also works with the charateristic functions rather than mgf's. – Conrad Turner Apr 04 '15 at 18:07
  • For the first part, take a look at this document at pages $8$ and $9$. They show how to find the convolution (thus, the sum) of $2$ exponentially distributed random variables. I am not sure but maybe if you have found the sum of the first $2$ and the first $3$ random variables, you can maybe proof the sum of the others by induction. – Pedro Apr 06 '15 at 06:26

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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.

Batman
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