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:


(I am not sure if this holds).


$X=\frac {(X+2Y)+(X-2Y)} 2$. 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.

  • $\begingroup$ Wait now I am thinking I made a mistake... $X-2Y$ may not actually be independent from $X+2Y$. It should probably be $2X+Y$ that is independent from $X-2Y$. $\endgroup$ – Dole Nov 27 '18 at 11:02
  • 1
    $\begingroup$ Use $X=\frac {(X-2Y)+2(2X+Y)} 5$. $\endgroup$ – Kavi Rama Murthy Nov 27 '18 at 11:49

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