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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?

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    Statement unclear. What are distributions of $Z$ and $X?$ What is 'Beta' ? What topics studied recently? – 2017-01-19
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    Sorry, Z is of Gamma distribution with shape a parameter alpha and scale parameter of 1.X is of Gamma distribution with shape parameter alpha and scale parameter beta. The topic of discussion is Gamma distribution. https://en.wikipedia.org/wiki/Gamma_distribution – 2017-01-19
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    Here it is written out, my apologies for being unclear: https://i.gyazo.com/34017a3622277722c80efc1bbe0ec3f4.png – 2017-01-19
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    Thanks. I've used your image to edit your Question. Please re-edit if I introduced errors. // This seems to be the distribution of a linear transformation (multiplying by constant $1/\beta$). The proof should be straightforward. (Maybe you are confusing 'gamma distribution' and 'gamma function'. Maybe try using $a$ for $\alpha$ and $b$ for $\beta.$) Or start by finding dist'n of $2Z$ and see how that goes.// Practical application might be in reliability or queueing theory. Sum of two exponential random variables (same rate) has gamma dist'n with shape parameter 2. – 2017-01-19
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    Hmm, I tried to give it a bit more thought and I am having trouble identifying the proof. https://i.gyazo.com/6db34b98b5dd33ae2e2d0e1d9f5ba402.png is the result of X. Divinding by a constant 1/B does not yield the pdf of a gamma function. – 2017-01-19
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    I hope my Answer clears up your difficulty. My guess is you are ignoring the $dx/dy$ factor. – 2017-01-19
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    Yep! That was exactly it. Thank you very much! – 2017-01-19

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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.

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