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So I have been following through some notes on the gamma function to try and understand how this is done, and I am stuck at one step:

So I have that $M_{X}(t) = E\left(e^{tX}\right) = \int_{0}^{\infty} e^{tx}f_{X}(x)dx $,

where $f_{X}(x) = \frac{x^{\alpha - 1}e^{\frac{-x}{\beta}}}{\beta^{\alpha}\gamma(\alpha)} $

$ \gamma(\alpha)$, is the gamma function.

= $\int_{0}^{\infty}e^{tx} \frac{x^{\alpha - 1}e^{\frac{-x}{\beta}}}{\beta^{\alpha}\gamma(\alpha)} dx $

= $\frac{1}{\beta^{\alpha}\gamma(\alpha)} \int_{0}^{\infty}x^{\alpha -1}e^{-x\left(\frac{1}{\beta} - t\right)} dx $

So the integral is finite if $\frac{1}{\beta} - t > 0$, Let $\frac{1}{\beta} - t = \frac{1}{\phi}$

= $\frac{1}{\beta^{\alpha}\gamma(\alpha)} \int_{0}^{\infty}x^{\alpha -1}e^{\frac{-x}{\phi}} dx $

Here is where I am unsure of how this jump is made:

= $\frac{1}{\beta^{\alpha}\gamma(\alpha)} [\gamma(\alpha)\phi^{\alpha}] $

The rest of the solution I am ok with, but how is this jump made from the 2nd to last step to this last step? Could someone explain the steps for me?

Many thanks!

Willie Wong
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JackReacher
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    Your exposition contains errors: (i) $f(x) dx$ is not "the gamma function". If I plug in 2, I don't get $\Gamma(2)$ coming out, for example; (ii) you have some mathematics beginning with "=". What's the LHS? (iii) the last step is simply a matter of recognizing the density function. – Glen_b Sep 25 '13 at 05:10
  • You are correct, I made an error in posting the question, which I have now amended. I'm trying to learn how the jump from the 2nd last step to the last step is made - could you elaborate a little further? – JackReacher Sep 25 '13 at 08:51
  • What's the integral of a density function over the region where it's non-zero? – Glen_b Sep 25 '13 at 10:23
  • ... So, if you see something proportional to a density, mutiply and divide by the quantity required to make it a density, leaving a constant times something that integrates to a known (and very simple) value. – Glen_b Sep 25 '13 at 10:43

2 Answers2

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Once you reached the identity $$ M_X(t)=\frac{1}{\beta^{\alpha}\Gamma(\alpha)} \int_{0}^{\infty}x^{\alpha -1}\mathrm e^{-x/\phi(t)} \mathrm dx, $$ where $\phi(t)^{-1}=\beta^{-1}-t$, the change of variable $x=\phi(t) u$ yields $$ M_X(t)=\frac{1}{\beta^{\alpha}\Gamma(\alpha)}\phi(t)^{\alpha} \int_{0}^{\infty}u^{\alpha -1}\mathrm e^{-u} \mathrm du. $$ One can identify the last integral without computing it since $M_X(0)=1$, hence $$ M_X(t)=\phi(t)^{\alpha}\phi(0)^{-\alpha}, $$ and I am sure you can complete the proof.

Did
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Gamma density with parameters (\alpha,\beta) is given by (\frac{\beta^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}) X has density (\frac{1}{\theta}\textbf{1}_{{0\le X\le \theta}}) Hence posterior density is proportional to

(\frac{\beta^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}\frac{1}{\theta}\textbf{1}_{{0\le X\le \theta}})

seed
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