How to calculate the value of $E[X^4], E[X^6],E[X^8] $....?

Question

I learned that when X is a normal random variable , $X$~ $N(0,1)$ ,

$E[X^2]=1$

$E[X^4]=1.3=3$

$E[X^6]=1.3.5=15$

$E[X^8]=1.3.5.7=105$

For the general case , when variance is s , how do you do for $E[X^4], E[X^6],E[X^8] $....?

I know that $E[X^2]=Var(X)-(E[X])^2$ = s .

I got stuck for higher degree. Please help.

If you change only the variance but keep the mean at $0$ then multiply your $E(X^{2k})$ by $v^k$ where $v$ is the variance. — Did, Mar 17 '14 at 19:32

score 2 · Answer 1 · answered Mar 17 '14 at 19:14

2

Let $X$ have mean $\mu$ and variance $\sigma^2$. Then $X=\sigma Z+\mu$ where $Z$ is standard normal.

Thus $X^n=(\sigma Z+\mu)^n$. Expand using the Binomial Theorem, and use the linearity of expectation and your knowledge about the $E(Z^k)$.

answered Mar 17 '14 at 19:14

André Nicolas

1 Answers1