# Determine expected value for continuous random variable [duplicate]

I know that $$E[X^n]$$ is found by $$\displaystyle\int_{0}^\infty{x^nf_x(x)dx}$$
I simplified this to $$\displaystyle\int_{0}^\infty{ \frac{x^{\frac{v}{2}-1+n}e^{\frac{-x}{2}}}{\displaystyle\int_{0}^\infty{{x^{\frac{v}{2}-1}e^{\frac{-x}{2}}}}dx} dx }$$

But I don't know how to proceed, since i shouldn't solve the denominator.

• Check what is gamma ditribution. Feb 24 '19 at 8:55
• This question is a special case of stats.stackexchange.com/questions/104241/… (found by searching our site for "Gamma moment"). One doesn't need to integrate by parts at all, as explained at stats.stackexchange.com/questions/198595/… (found with the same search): the answer is a simple matter of dividing one constant by another.
– whuber
Feb 24 '19 at 16:42

\begin{align}E[X^n]& = \frac1{c_V}\int_0^\infty x^{n+v/2-1}\exp(-\frac{x}2) \, dx\\ &=\frac1{c_V}\left[\left.-2x^{n+v/2-1}\exp\left(-\frac{x}2 \right)\right|_0^\infty\right.\\&\left.+2(n+v/2-1)\int_0^\infty x^{(n-1)+v/2-1}\exp(-\frac{x}2) \, dx\right]\\ &=\frac{2(n+\frac{v}{2}-1)}{c_v}\int_0^\infty x^{(n-1)+v/2-1}\exp(-\frac{x}2) \, dx\\ &=2\left(n+\frac{v}2-1\right)E[X^{n-1}]\end{align}