This question already has an answer here:

enter image description here

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.


marked as duplicate by whuber probability Feb 24 at 16:43

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


\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}

I will leave the simplification as an exercise.


Not the answer you're looking for? Browse other questions tagged or ask your own question.