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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?

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

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  • $\begingroup$ Wait jokes that was stupid I got it. $\endgroup$ – Five9 Nov 1 '19 at 21:15
  • $\begingroup$ Really? What was the mistake? $\endgroup$ – callculus Nov 1 '19 at 21:16
  • $\begingroup$ 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. $\endgroup$ – Five9 Nov 1 '19 at 21:20
  • $\begingroup$ I´m looking forward to see what you found out. $\endgroup$ – callculus Nov 1 '19 at 21:24
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    $\begingroup$ Sorry was rushing to finish the assignment :) Here it is, lmk how it looks $\endgroup$ – Five9 Nov 3 '19 at 3:20

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