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If $(Y_1, Y_2)$ are bivariate Normal with correlation $\rho$ and with $Y_1$ and $Y_2$ marginally $N(0,1)$, how can we find the mean and variance of $Y_2^3|Y_1^2$?

I am told that this can be done Without calculus and by representation. Anyone have any ideas?

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    think one of those should be a $Y_2$ Also, the mean's gotta be 0, right? – 2017-01-10
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    I fixed it, thanks! – 2017-01-10

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Can write $Y_2 = \sqrt{1-\rho^2}Z + \rho Y_1$ where $Z$ is an independent $N(0,1).$ Then $$Y_2^3 = (1-\rho^2)^{3/2}Z^3 + 3 (1-\rho^2)\rho Z^2Y_1 +3\sqrt{1-\rho^2}\rho^2 ZY_1^2 + \rho^3Y_1^3.$$ Taking the expected value conditional on $Y_1^2$ gives zero. This is because $E(Y_1^n|Y_1^2) = 0$ for any odd $n$, and the terms with even powers of Y's all have odd powers of Z.

Then we square $Y_2^3$ and take the conditional EV to get the variance. Here I won't write all the terms, only the ones that contribute to the expectation: $$ Y_2^6 = (1-\rho^2)^3Z^6 + 9(1-\rho^2)^2\rho^2Z^4Y_1^2 + 9(1-\rho^2)\rho^4Z^2Y_1^4+\rho^6Y_1^6+ 6(1-\rho^2)^2\rho^2Z^4Y_1^2 +6(1-\rho^2)\rho^4Z^2Y_1^4 $$

where the last two terms are is the only indirect terms from the square to contribute. Since $E(Y_1^n|Y_1^2) = Y_1^n$ for Y even, we just have

$$ E(Y_2^6|Y_1^2) = (1-\rho^2)^3E(Z^6) + 15(1-\rho^2)^2\rho^2E(Z^4)Y_1^2 + 15(1-\rho^2)\rho^4E(Z^2)Y_1^4+\rho^6Y_1^6 $$ Since $E(Z^6)=15,$ $E(Z^4)=3,$ and $E(Z^2)=1,$ $$ Var(Y^3_2|Y_1^2)= E(Y_2^6|Y_1^2) = 15(1-\rho^2)^3 + 45(1-\rho^2)^2\rho^2Y_1^2 + 15(1-\rho^2)\rho^4Y_1^4+\rho^6Y_1^6 $$

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    Could you explain why $E(Y_1^n|Y_1^2) = 0$ for odd $n$? What's the formal way of thinking about it? Thanks! – 2017-01-10
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    @user321627 Given $Y_1^2,$ you know that $Y_1 = \pm \sqrt{Y_1^2}.$ Symmetry of the original distribution says it has to be 50% either way. So e.g. $Y_1^3$ will be $\pm(Y_1^2)^{3/2}$ with 50 percent prob each sign, so the EV is zero. I know that's not formal, but formal would presumably require finding the joint distribution, which won't have a pdf cause it's degenerate... I would just use the same intuition to get through that mess. – 2017-01-10