View Full Version : gamma moments integral short question

leo nidas
05-25-2010, 05:06 PM
Hi there,

About the moment generating function of the gamma distr.

f(x)=λ^α/(Γ(α))x^(α-1)exp(-λx). is the pdf.

MGF=E(exp(tX))=int(exp(tx)f(x),0 to Inf)

=λ^α/(Γ(α)) int(x^(α-1)*(exp(-x(λ-t)),0,Inf)

the last integral it is said that it exists if x(λ-t)>0.

Why? Don't we care about the term x^(α-1)?

Apart from the above:

If we have int(x^(α-1)*(exp(-x(λ-t)),0,Inf) for -Inf<a<Inf
then can we say again that the integral exists if x(λ-t)>0?

Thans in advance for any answers!

05-26-2010, 02:35 AM
You need to ensure that \lambda - t > 0
such that when x \rightarrow +\infty, e^{-(\lambda - t)x} \rightarrow 0
Otherwise, the integral will diverge to infinity.

In your question, we already prespecified \alpha > 0
to be the parameter of the gamma distribution.
And note that the gamma function \Gamma(\alpha-1)
is well defined \forall \alpha > 0
So for the moment generating function, the range of \alpha
does not change and thus does not specify again in the calculation.
To be more specific, you can write
M_T(t) = (\frac {\lambda} {\lambda - t})^\alpha, t < \lambda,
\alpha > 0, \lambda > 0

For the last question, the integral will goes to infinity \forall
\lambda - t .
Same reason as the first question. In the integrand,
you have a polynomial and an exponential function. The exponential
function will dominate (either decay to 0 or grow up to infinity)
So it depends on the variable -(\lambda - t)x \rightarrow +\infty or -(\lambda - t)x \rightarrow -\infty
But since in your case x will go to infinity at both end,
the exponential function will grow up anyway at one end.

leo nidas
05-26-2010, 03:28 AM
I understand why the superscripy of e should be negative. But even if it is then as x tends to Inf x^(a-1) might also go to infinity. I am sure I am missing something here.. Is it because that exp(-..) goes faster to 0? Is this what you mean by saying that the exponential function will dominate the polynomial?


And for the second case you say that the integral will go always to inf? How can that be, since for some a (i.e. for all a>0) we get the same integral as in the first case..

Thanx for you answer, I really appreciate your time!

05-26-2010, 04:46 AM
Yes. exponetial function will goes faster to 0 so the integral converge

Now if \lambda - t > 0 , then
e^{-(\lambda - t)x} \rightarrow 0 as x \rightarrow +\infty
e^{-(\lambda - t)x} \rightarrow +\infty as x \rightarrow -\infty

The otherway round.
Now if \lambda - t < 0 , then
e^{-(\lambda - t)x} \rightarrow +\infty as x \rightarrow +\infty
e^{-(\lambda - t)x} \rightarrow 0 as x \rightarrow -\infty

So in any case the integral will go to infinity

* The arguments here are not rigorous and just for illustration.