# $\mathbb E[X^4]$ 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 :)

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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 at 0:25
Indeed. Thank you. –  André Mar 18 at 0:26
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.