Question on moment generating function

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Guest

Posted: Sun Apr 13, 2008 2:24 am

Hi,

I came across this example in one of the books that I'm reading.

Given X~Exp(a), then for u<a

mgf x(a) = a * integral from 0 to infinity (exp(ux)exp(-ax)dx = a/(a-
u)

I know that mgf x(a) = integral from -infinity to infinity
(exp(ux)dFx(x)), however, how I could substitute the dFx(x) term with
the given X term to obtain the results given above? Sorry, my stats is
a bit scratchy.

Thanks.

Paul Rubin

Posted: Sun Apr 13, 2008 10:11 am

Guest

thampw@hotmail.com wrote:

Quote:

Since X ~ Exp(a), for x >= 0 F(x) = 1 - exp(-ax) and dF(x) = f'(x)dx =
a*exp(-ax)*dx. For x < 0, F(x) = 0 and dF(x) = 0; so the lower limit of
integration changes from -infinity to 0.

Note the restriction u < a on the formula for the MGF.

/Paul

Guest

Posted: Tue Apr 15, 2008 10:46 pm

Thanks for the explanation, Paul.

Cheers!

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