Prove/show the expected value of a transformed gamma distribution UNDERGRADUATE PROBABILITY ~
In the answer to this I'm having trouble understanding what is going on between the first and second line of the answer. I was studying the answer to show $ E(Y^c) = \frac{\beta^c \Gamma(\alpha+c)}{\Gamma(\alpha)}$ so I can't copy the solution because I'm not sure how $ u = x/\beta $ came about in the linked answer. My calculus is a bit rusty that might be why?
 A: Answering my own question after I figured it out:
To show that $E(Y^c) = \frac{\beta^{c}\Gamma(a+c)}{\Gamma(a)}$ use the E(X) formula
$$
E(Y^c)=\int{y^c P(y)=\int{y^{c}\frac{1}{\beta^a \Gamma(a)}y^{a-1}e^{-y/\beta}dy}}
$$
$$
=\frac{1}{\beta^{\alpha} \Gamma(\alpha)}\int{y^{\alpha+c-1} e^{-y/\beta} dy}
$$
Here's where I made a mistake: I also gave the exponential's $y$ a $c$, making it $e^{-y/\beta}$ when it doesn't need it. I was trying to factor this in the next steps which was impossible...Let $ u = \frac{y}{\beta}$ and thus $du/dy=\frac{y}{\beta}$
$$
=\frac{\beta^{a+c-1+1}}{\beta^{\alpha} \Gamma(\alpha)}
\int{(\frac{y}{\beta})^{\alpha+c-1} e^{-y/\beta}\frac{1}{\beta}dy}
$$
Substitute u's and simplify the coefficient:
$$
=\frac{\beta^{c}}{\Gamma(\alpha)}
\int{(u)^{\alpha+c-1} e^{-u}du}
$$
And following the definition of a gamma function, the integral is just a gamma function with parameter a+c:
$$
=\frac{\beta^{c}\Gamma(a+c)}{\Gamma(\alpha)}
$$
-Sorry not sure how to left-align and adjust those brackets to full size lmao
