Linear transformation of a Gamma Distribution with shape $\alpha$ and rate $1$ I am a bit confused as to why a particular theorem I found was true. It claims that if $Z \sim \text{Gam}(\alpha,1),$ then $X = Z / \beta \sim \text{Gam}(\alpha, \beta).$ 
I am unsure as to how to prove it. Wouldn't a Gam function with $\beta = 1$ yield gamma function, how does a gamma function over $\beta$ yield a Gamma distribution? Also, what would be the practical application of this?
 A: I'm betting there is a section in your probability text that handles this sort of thing. Maybe it's called something like Transformations of random variables.
There are several methods. I will use what is sometimes called the PDF method or the method of density functions.
In one book the relevant theorem states: Let $X$ have PDF $f_X(x).$ If $h(x)$ is either an increasing or decreasing function for all $x$ such that $f_X(x) > 0,$ then $Y = h(X)$ has the density function
$$f_Y(y) = f_X(h^{-1}(y))\times|dh^{-1}/dy|,$$
where $dh^{-1}/dy = d(h^{-1}(y))/dy.$
In your case, $y = h(x) = x/\beta,$ so $x = h^{-1}(y) = \beta y,$ and
$|dx/dy| = |dh^{-1}/dy| = |\beta| = \beta.$ (Because $h$ is an increasing
function, it's derivative is positive, and the absolute value signs are not
really necessary.) 
Moreover, the PDF of $\mathsf{Gamma}(\alpha, 1)$ is 
$f_X(x) = \frac{1}{\Gamma(\alpha)}x^{\alpha-1}e^{-x},$ for $x > 0.$
So 
$$f_Y(y) = \frac{1}{\Gamma(\alpha)}(\beta y)^{\alpha-1}e^{-\beta y}\times\beta
= \frac{\beta^\alpha}{\Gamma(\alpha)}y^{\alpha - 1}e^{-\beta y},$$
for $y > 0.$ Which we recognize as the density function of 
$\mathsf{Gamma}(\text{shape} = \alpha, \text{rate} = \beta).$
Notice that $E(X) = \alpha$ and $Var(X) = \alpha.$
Accordingly, $E(Y) = E(X/\beta) = \alpha/\beta$
and $Var(Y) = Var(X/\beta) = \alpha/\beta^2,$ which match the
usual formulas for the mean and variance of  $\mathsf{Gamma}(\alpha, \beta).$ 

I will illustrate this transformation using a simulation in R statistical software, for 
$\alpha = 5$ and $\beta = 2.$
m = 10^5; x = rgamma(m, 5, 1);  y = x/2
mean(x);  sd(x)
## 5.002914  # aprx E(X) = 5
## 2.240974  # aprx SD(X) = 2.236
mean(y);  sd(y)
## 2.501457  # aprx E(Y) = 5/2
## 1.120487  # aprx SD(Y) =  1.118

Below are histograms of the simulated distributions, along with the
respective density functions. The effect of the transformation has
been to make $Y$-values half as large as the corresponding $X$-values.
Thus the histogram of $Y$ is 'half as wide' as the histogram for $X$.
But the area under each density histogram
and PDF must be unity. So the histogram for $Y$-values must be 'twice
as tall'. The role of $|dh^{-1}/dy| = |dx/dy| = \beta = 2$ has been to
inflate the height of $f_Y$ as required.

A: Let $X\sim gamma(\alpha,1)$ and $Y\sim gamma(\alpha, \beta)$. Notice that $$P[Y\leq y]=P\left[\frac{X}{\beta}\le y\right]=P\left[X\le y\beta\right]$$  differentiate by chain rule
$$f_Y(y)=F'(Y)=\beta f_X(\beta x)=\beta \frac{(\beta x)^{\alpha-1}}{\Gamma(\alpha)}e^{-\beta x}\sim gamma(\alpha, \beta)$$
