Finding expectation through conditioning

Question

I want to find $E(X^2Y^2)$. Now $Var(XY) = E(X^2Y^2)-E(XY)^2.$ Since $E(X)$ and $E(Y)$ both are 0, $Cov(X,Y) = E(XY)$ and $Cov(X,Y) = \rho \sigma_{x}\sigma_{y}. $ Therefore, $Cov(X,Y)=\rho.$ Therefore from previous argument, we have $E(XY)=\rho$. How do I proceed further?

How do I calculate $E[Y^2|X]$ ? $\quad Y^2$ will be a chi square distribution with degree of freedom 1 right? How do I proceed further? — Equation_Charmer, Oct 28 '22 at 16:20
You could use the formulas at https://stats.stackexchange.com/a/71303/919, for instance. — whuber, Oct 28 '22 at 21:32

score 2 · Accepted Answer · answered Oct 29 '22 at 03:30

Following up the hint from Moderator whuber, consider that when $X$ is and $Y$ are jointly normal random variables, the conditional distribution of $Y$, conditioned on $X = x$, is a normal distribution with mean $\mu_y + \rho\left. \left.\frac{\sigma_y}{\sigma_x}\right(x-\mu_x\right)$ and variance $\sigma_y^2(1-\rho^2)$. Surely you can compute $E[Y^2\mid X=x]$ from this information without needing to evaluate any integrals and the like?

Finding expectation through conditioning

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