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The question is: Let $\mathrm {X} $, $\mathrm Y$ $\sim$ $\mathcal N(0,1)$. Knowing that they're independent variables find $\mathbb E(X^4+Y^3|X+Y)$

I did: $\mathbb E|X|=\mathbb E|Y|<\infty$, therefore $\mathbb E(X|X+Y)=\mathbb E(Y|X+Y)=\frac{X+Y}{2}$, and hence I came up with an idea that: $$\mathbb E(X^4+Y^3|X+Y)=\mathbb E(X^4|X+Y)+\mathbb E(Y^3|X+Y)=(\frac{X+Y}{2})^4+(\frac{X+Y}{2})^3$$ However this seems quite trivial and I am afraid I am making some obvious mistakes. I'd be really grateful if somebody could check it (and maybe show some hints how to do it proberly given that my reasoning is wrong)

user201226
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2 Answers2

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The mistake you're making is that $E[X^4|X+Y] \neq E[X|X+Y]^4$. So you're right, it's not that trivial.

I'm not too sure on how to solve the problem entirely. One approach that may work is that $E[f(Z) | Z] = f(Z)$, and you may be able to use $E[(X+Y)^3|X+Y]$ in a clever way to simplify some terms out.

Titandrake
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You can do it in a hard way :) Let $Z=X+Y$

$$\begin{bmatrix} X \\ Y \\ Z \end{bmatrix}\sim \mathcal N\Bigg(\begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix},\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{bmatrix}\Bigg)$$

So,

$$X|Z\sim \mathcal N(\mu_X+\frac{\sigma_X}{\sigma_Z}\rho_{XZ}(z-\mu_Z),(1-\rho_{XZ}^2)\sigma_X^2)=\mathcal N\bigg (\frac{1}{2}z,\frac{1}{2}\bigg)$$ $$Y|Z\sim \mathcal N\bigg(\frac{1}{2}z,\frac{1}{2}\bigg)$$

Then,

$$E[X^4|Z]=\bigg(\frac{z^2+6}{4}\bigg)^2-\frac{3}{2}$$ $$E[Y^3|Z]=\frac{z}{2}\bigg(\frac{z^2+6}{4}\bigg)$$

Finally, $$E[X^4+Y^3|Z]=\bigg(\frac{z^2+z+6}{4}\bigg)^2-\bigg(\frac{z}{4}\bigg)^2-\frac{3}{2}$$