Is there a closed form for $E(Y^n)$, where $Y$ is a random variable with a gamma distribution with parameters $\alpha$, $\beta$?
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$\begingroup$ Isn't this just a matter of reading it off the moment generating function? Gives alpha(alpha+1)...(alpha+n-1)beta^n. $\endgroup$– George LowtherNov 19, 2009 at 21:57
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$\begingroup$ mm yeah i forgot about those. I just figured it out by doing the integrals by hand. $\endgroup$– ClaudiuNov 19, 2009 at 22:51
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$\begingroup$ It's been a while since you posted this, but I will claim my answer is better than the one you accepted because it shows you how routine the integral is, and exactly how to evaluate it. $\endgroup$– Michael HardyJul 23, 2023 at 18:29
2 Answers
If the shape parameter is $\alpha$ and the scale parameter $\beta$, then $E(Y^r) = \beta^r \Gamma(\alpha + r)/\Gamma(\alpha)$ for real $r > 0$.
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$\begingroup$ Odd that you write $\Gamma(\alpha+r)/\Gamma(\alpha)$ without mentioning that this simplifies to $\alpha(\alpha+1)(\alpha+2) \cdots (\alpha+r-1). \qquad$ $\endgroup$ Jul 23, 2023 at 18:44
For now I will assume you mean $$ \frac1{\Gamma(\alpha)} \left( \frac y\beta\right)^{\alpha-1} e^{-y/\beta} \, \frac{dy}\beta \text{ for } y>0 $$ rather than $$ \frac1{\Gamma(\alpha)} \left( \beta y\right)^{\alpha-1} e^{-\beta y} (\beta\, dy) \text{ for } y>0. $$ The expected value is: \begin{align} \operatorname E\left( Y^n \right) & = \int_0^\infty y^n f_Y(y)\, dy \\[12pt] & = \int_0^\infty y^n \cdot \frac 1 {\Gamma(\alpha)} \left( \frac y\beta\right)^{\alpha-1} e^{-y/\beta} \, \frac{dy}\beta \\[12pt] & = \frac{\beta^n}{\Gamma(\alpha)} \int_0^\infty \left( \frac y \beta \right)^{n+\alpha-1} e^{-y/\beta} \, \frac{dy}\beta \\[12pt] & = \frac{\beta^n \Gamma(\alpha+n)}{\Gamma(\alpha)} \\[12pt] & = \beta^n \, \underbrace{\alpha(\alpha+1)(\alpha+2)\cdots(\alpha+n-1)}_\text{$n$ factors}. \end{align}
So the only integral is the one that usually defines the gamma function.