# Re: Mgf of Exponential distribution

*From*: Paul Rubin <rubin@xxxxxxx>*Date*: Fri, 06 Jun 2008 11:25:17 -0400

thampw@xxxxxxxxxxx wrote:

Hi all,

I'm reading the proof for mgf of exponential distribution from

http://ocw.mit.edu/NR/rdonlyres/Mathematics/18-05Spring-2005/45FFA38A-2A87-4FCD-A1A9-46845A8F3D39/0/18_05_lec18.pdf

I'm basically struggling to understand one small part of the proof. On

page 53 & 54, it gives the following:

mgf = ... = int 0 to infinity [a*exp((t-a)*x)]dx = a*(exp((t-a)*x))/(t-

a)| infinity & 0 = 1 - a/(t-a) = ...

OK, how do we obtain one when we substitute infinity into a*(exp((t-

a)*x))/(t-a)? I thought exp(infinity) = infinity?

I think I'm missing something really easy...

Thanks.

Actually, the author of the page booted a grounder. Assume 0 < t < alpha (not just t < infinity as stated). Then exp((t-alpha)*x) -> 0 as x -> infinity, so the definite integral reduces to -alpha/(t-alpha) = alpha/(alpha-t). Note that the next bit right after your sticking point, 1 - alpha/(t-alpha) = alpha/(alpha-t), is also not generally true. So two wrongs made a right here.

/Paul

.

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