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If $X$ has a $\Gamma(\alpha, 1)$ distribution would $\beta X$ not have a $\Gamma(\alpha \beta, 1)$ distribution by the summation property of Gamma distributions? I must be missing something.

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    $\begingroup$ you're not summing. You're scaling. You can prove both with moment generating functions, though. $\endgroup$ – Taylor Nov 30 '17 at 3:14
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Taylor's suggestion to use mgf is good. I show you how to find the distribution of $\beta X$ by transformation.

$X\sim \Gamma(\alpha,1) \Rightarrow f_{X}(x)=\frac{1}{\Gamma(\alpha)}x^{\alpha-1}e^{-x}$

Let $Y=\beta X\Rightarrow x=\frac{1}{\beta}y$

$\therefore |J|=\frac{\partial x}{\partial y}=\frac{1}{\beta}$

$ f_{Y}(y)=f_{X}(x)|J|=\frac{1}{\Gamma(\alpha)}(\frac{1}{\beta}y)^{\alpha-1}e^{-\frac{1}{\beta}y}\frac{1}{\beta}=\frac{1}{\Gamma(\alpha)\beta^{\alpha}}y^{\alpha-1}e^{-\frac{y}{\beta}}$

This is the pdf of $\Gamma(\alpha,\beta)$

$\therefore Y\sim \Gamma(\alpha, \beta)$

It is not $\Gamma(\alpha\beta,1)$ as you have thought

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