# Gamma Distribution Moments

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.

• If it helps, substitute $\frac{x}{\beta}=t$. Basically it is the definition of Gamma function. – StubbornAtom Nov 5 '18 at 6:36
• If that substitution results in $(-\beta)\int_{0}^{\infty}(\beta*t)^{\nu+\alpha-1}e^{-t}dt$, then what would I do? – bob Nov 5 '18 at 6:57

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.