Show that $M_{x}(t) = \left(\frac{1}{1-{\beta} t}\right)^{\alpha}$

Question

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!

Answer 1

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.

Answer 2

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}})