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For independent gaussians with following the normal distribution with expectation zero and variance one, how do I compute:

$E(X^2|X-2Y), E(X^3|X-2Y)$

I know that $X-2Y$,$X+2Y$ are independent. However, this does not seem to be enough to deduce the result, without a restriction such as:

$E(X^2|X-2Y)=-2E(Y^2|X-2Y)$

(I am not sure if this holds).

Dole
  • 2,653

1 Answers1

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$X=\frac {(X+2Y)+(X-2Y)} 2$. Compute $X^{2}$ and $X^{3}$ in terms of $X+2Y$ and $X-2Y$ from this. [ $$E(X^{2}|X-2Y)=\frac {E((X+2Y)^{2}|X-2Y)+E((X-2Y)^{2}|X-2Y)+2E((X+2Y)E((X-2Y)|X-2Y)} 4$$ $$=[E((X+2Y)^{2} +(X-2Y)^{2}+2(X-2Y)E(X+2Y)] /4=[E((X+2Y)^{2} +(X-2Y)^{2}] /4$$ $$=\frac 5 4+\frac {(X-2Y)^{2}} 4.$$

This answer was written assuming that the OP was right is saying that $X+2Y$ and $X-2Y$ are independent. They are not, so a slight modification is required. See my comment below for the modification.

Kavi Rama Murthy
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