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/1805Spring2005/45FFA38A2A874FCDA1A946845A8F3D39/0/18_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((ta)*x)]dx = a*(exp((ta)*x))/(t
a) infinity & 0 = 1  a/(ta) = ...
OK, how do we obtain one when we substitute infinity into a*(exp((t
a)*x))/(ta)? 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((talpha)*x) > 0 as x > infinity, so the definite integral reduces to alpha/(talpha) = alpha/(alphat). Note that the next bit right after your sticking point, 1  alpha/(talpha) = alpha/(alphat), is also not generally true. So two wrongs made a right here.
/Paul
.
 FollowUps:
 Re: Mgf of Exponential distribution
 From: thampw
 Re: Mgf of Exponential distribution
 References:
 Mgf of Exponential distribution
 From: thampw
 Mgf of Exponential distribution
 Prev by Date: Mgf of Exponential distribution
 Next by Date: Re: Histograms using data with errors
 Previous by thread: Mgf of Exponential distribution
 Next by thread: Re: Mgf of Exponential distribution
 Index(es):
Relevant Pages
