# Computing Distribution of Conditional Expectation of Gaussian RV

I am trying to compute distribution of the following random variable \begin{align*} E[(X-E[X|Y])^2|Y] \end{align*} where $X \sim \mathcal{N}(0,\sigma^2_x)$ and $Z \sim \mathcal{N}(0,\sigma^2_Z)$ where $Z$ is independent of $X$ and $Y=X+Z$.

I know that $E[(X-E[X|Y])^2|Y]$ should have Gaussian distribution but I get lost in what is conditional expectation and what is random variable.

Here is my reasoning: I know that $E[X|Y=y]= \frac{\sigma_{XY}}{\sigma_Y^2}y=\frac{\sigma^2_x}{\sigma^2_x+\sigma^2_z}y$. So, then

\begin{align*} E[(X-E[X|Y])^2|Y]&=E\left[\left(X-\frac{\sigma^2_x}{\sigma^2_x+\sigma^2_z}Y\right)^2 \Big|Y\right]=E\left[\left(Y-Z-\frac{\sigma^2_x}{\sigma^2_x+\sigma^2_z}Y\right)^2 \Big|Y\right]\\ &= \frac{\sigma^2_z}{\sigma^2_x+\sigma^2_z}E[Y^2|Y]-2\frac{\sigma^2_z}{\sigma^2_x+\sigma^2_z}E[ZY|Y]+E[Z|Y]\\ &= \frac{\sigma^2_z}{\sigma^2_x+\sigma^2_z}Y^2-2\frac{\sigma^2_z}{\sigma^2_x+\sigma^2_z}YE[Z|Y]+E[Z|Y]\\ &= \frac{\sigma^2_z}{\sigma^2_x+\sigma^2_z}Y^2-2\frac{\sigma^2_z}{\sigma^2_x+\sigma^2_z}YE[Y-X|Y]+E[Y-X|Y]\\ &= \frac{\sigma^2_z}{\sigma^2_x+\sigma^2_z}Y^2-2\frac{\sigma^2_z}{\sigma^2_x+\sigma^2_z}(Y^2-YE[X|Y])+Y-E[X|Y]\\ &= \frac{\sigma^2_z}{\sigma^2_x+\sigma^2_z}Y^2-2\frac{\sigma^2_z}{\sigma^2_x+\sigma^2_z}(Y^2-\frac{\sigma^2_x}{\sigma^2_x+\sigma^2_z}Y^2)+Y-\frac{\sigma^2_x}{\sigma^2_x+\sigma^2_z}Y\\ &= -\frac{\sigma^2_z}{\sigma^2_x+\sigma^2_z}Y^2+\frac{2\sigma^2_z\sigma^2_x}{(\sigma^2_x+\sigma^2_z)^2}Y^2+Y-\frac{\sigma^2_x}{\sigma^2_x+\sigma^2_z}Y \end{align*}

but $Y^2$ can not be Gaussian. So, I am making a mistake somewhere. Could you guys help. Thank you

Note that $Y \sim {\mathcal N}(0, \sigma_y^2)$ where $\sigma_y^2 = \sigma_x^2 + \sigma_z^2$ and $\text{Cov}(X,Y) = \sigma_x^2$. Thus $X = \dfrac{\sigma_x^2}{\sigma_y^2} Y + W$ where $W$ is independent of $Y$ and $W \sim {\mathcal N}(0, \sigma_x^2 \sigma_z^2/\sigma_y^2)$. You have $E[X|Y] = \dfrac{\sigma_x^2}{\sigma_y^2} Y$, so $X - E[X|Y] = W$. The rest should be easy, since $W$ is independent of $Y$.
• Thanks. So then $E[(X-E[X|Y])^2|Y]=E[W^2|Y]=E[W^2]$, right? But $E[W^2]$ is not a random variable it's just a number, right? – Boby Jun 26 '15 at 15:39