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Show that for X ~ Gamma($\alpha$, $\beta$), for positive constant $\nu$,

$E[X^\nu] = \dfrac{\beta^\nu*\Gamma(\nu + \alpha)}{\Gamma(\alpha)}$.

I have the following solution:

Solution

However, I don't understand how we get that $\dfrac{1}{\Gamma(\alpha)*\beta^\alpha}*\int_{0}^{\infty}x^{(\nu+\alpha)-1}e^{-x/\beta}dx = \dfrac{\Gamma(\nu+\alpha)*\beta^{\nu+\alpha}}{\Gamma(\alpha)*\beta^{\alpha}}$

Would appreciate any help on how this step was completed. Basically, I don't understand how: $\int_{0}^{\infty}x^{(\nu+\alpha)-1}e^{-x/\beta}dx = \Gamma(\nu+\alpha)*\beta^{\nu+\alpha}$.

I see that the left side is close to the definition of the Gamma function, but can't see how exactly to turn it into the right side.

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  • $\begingroup$ If it helps, substitute $\frac{x}{\beta}=t$. Basically it is the definition of Gamma function. $\endgroup$ – StubbornAtom Nov 5 '18 at 6:36
  • $\begingroup$ If that substitution results in $(-\beta)\int_{0}^{\infty}(\beta*t)^{\nu+\alpha-1}e^{-t}dt$, then what would I do? $\endgroup$ – bob Nov 5 '18 at 6:57
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One way to understand the calculation is to recall that for a gamma distribution with shape $\alpha$ and scale $\beta$, $$f_X(x) = \frac{x^{\alpha-1} e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)}, \quad x > 0.$$ The denominator, being independent of $x$, suggests that $1/(\beta^\alpha \Gamma(\alpha))$ is the required multiplicative factor for the density such that $$\int_{x=0}^\infty f_X(x) \, dx = 1.$$ In other words, $$\beta^\alpha \Gamma(\alpha) = \int_{x=0}^\infty x^{\alpha - 1} e^{-x/\beta} \, dx.$$ This holds true for any $\alpha, \beta > 0$. Now, with this in mind, $$\operatorname{E}[X^\nu] = \int_{x=0}^\infty x^\nu f_X(x) \, dx = \int_{x=0}^\infty \frac{x^{\nu + \alpha - 1} e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)} \, dx = \frac{\beta^{\nu + \alpha} \Gamma(\nu + \alpha)}{\beta^\alpha \Gamma(\alpha)}\int_{x=0}^\infty \frac{x^{\nu + \alpha - 1} e^{-x/\beta}}{\beta^{\nu + \alpha}\Gamma(\nu + \alpha)} \, dx.$$ This is what the provided solution does, but the motivation for doing so should now be clear, because now the integrand represents a gamma density with shape parameter $\nu + \alpha$, and rate $\beta$. Therefore, its integral over its support is also $1$, provided $\nu + \alpha > 0$. It follows that $$\operatorname{E}[X^\nu] = \frac{\beta^\nu \Gamma(\nu + \alpha)}{\Gamma(\alpha)}, \quad \nu > -\alpha,$$ as claimed.

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