# $\mathbb E[X^4]$ with $X \sim \mathcal N(0,1)$

I have to calculate $(Y = X^2)$ $$\rho_{X,Y} = \frac{\mathbb Cov(X,Y)}{\sigma_X \cdot \sigma_Y}$$ But for this I have to calculate $\mathbb Var(Y)$ and thus $\mathbb E[Y^2] = \mathbb E[X^4]$. I dont think that integration helps. I would appreciate some litte hint :)

-
Integration does help. The basic idea is to integrate by parts using $u\,dv = x^3\cdot xe^{-x^2/2}$ since $v = -e^{-x^2/2}$ is a known quantity. Hopefully, then you can recognize the integral $\int v\,du$ as something whose value you can deduce without the formality of integration. – Dilip Sarwate Mar 18 '13 at 0:25
Indeed. Thank you. – Andre Mar 18 '13 at 0:26

## 1 Answer

$\newcommand{\cov}{\operatorname{cov}}$ $\newcommand{\E}{\mathbb E}$ $$\cov(Y,X)=\E(YX) - \E(Y)\E(X) = \E(X^3)-\E(Y)\E(X).$$ Notice that $\E(X^3)=0$ because the distribution of $X^3$ is symmetric about $0$ (and you can show the integral converges by a comparison test), and $\E(X)=0$. Therefore $\cov(Y,X)=0$.

You don't need $\sigma_Y$ because the numerator is $0$. (But one can evaluate the integral.)

Also, it wouldn't hurt to notice what the graph of $y=x^2$ looks like and consider the implications of the symmetry of the distibution of $X$ and the symmetry of that curve about the $y$-axis. If $\rho>0$, that would mean that on average, $Y$ increases as $X$ increases, and if $\rho<0$, that would mean that on average, $Y$ decreases as $X$ increases. But the symmetries tell you that neither of those happens.

-
Thanks. I like answers which give some more intuition behind the pure math:) – Andre Mar 18 '13 at 17:35