I have a task:
$X$ is random variable with normal distribution $N(0,1)$, $Y$ has $N(0,2)$ distribution. $X,Y$ are independent. Calculate: $$E(X^3\mid X+Y)$$
I know how to do it from definition. But I want to complicate $X^3$ and use of measurable and independence of $X+Y$.
Thanks in advance.
Yes, it is possible to proceed by using normal regression, as you suggest in the comment. Indeed, denoting $S = (X+Y)/3$, $Z = X-S$, and using the independence of $Z$ and $S$, $$ \mathrm E[X^3\mid X+Y] = \mathrm E[(Z+S)^3\mid X+Y] = \mathrm E[Z^3 + 3Z^2 S + 3Z S^2 + S^3\mid X+Y] \\ = \mathrm E[Z^3] + 3S\mathrm E[Z^2] + 3S^2 \mathrm E[Z] + S^3. $$ Now noting that $Z$ has a centered Gaussian distribution, $$ \mathrm E[X^3\mid X+Y] = 3S\mathrm{Var}(Z)+ S^3 = \frac{8(X+Y)}{9} + \frac{(X+Y)^3}{27}. $$ (You have to check the computations!)