# 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?

• Just $\beta^{\alpha+1}$ has been factored out. Clear now? – callculus Nov 1 '19 at 20:12
• Please see if my self-posted answer is correct: – Five9 Nov 1 '19 at 20:29
• Which answer, Five9? – callculus Nov 1 '19 at 20:30
• Lol took longer than I expected to type out sorry. Here it is now below. – Five9 Nov 1 '19 at 20:43

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: $$=\frac{\beta^{c}\Gamma(a+c)}{\Gamma(\alpha)}$$
• I was struggling to get my last equation to look like a gamma function $\Gamma(\alpha) = \int_0^\infty{x^{\alpha-1}e^{-x} dx}$ but my goal was to find $E(Y^c)$ so I just failed to see that $\alpha-1+c$ was literally what I needed (if $\alpha$ was replaced with $\alpha-1$) it just looked weird since the $-1$ was before the $c$. I'm going to clean up this thread in a bit so it's more clear. – Five9 Nov 1 '19 at 21:20