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leo nidas
05-25-2010, 06:06 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)?

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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?

BGM
05-26-2010, 03:35 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.

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
http://en.wikipedia.org/wiki/Gamma_function

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, 04:28 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?

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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..

BGM
05-26-2010, 05:46 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.