# Why is E[Cov(Y|X)]=Cov(Y|X) true in this case?

I'm given this equality when $X=Y+Z$ and $Y,Z$ are independent standard Gaussians. Why is this equality true?

So let's say $X$ and $Y$ are any Gaussian vectors of any size where $$\pmatrix{X \\Y}\sim \mathcal{N}(\pmatrix{\mu_X\\\mu_Y},\pmatrix{\Sigma_{XX} & \Sigma_{XY} \\ \Sigma_{YX} & \Sigma_{YY}})$$ It is well known that $$X|Y \sim \mathcal{N}(\mu_{X}+\Sigma_{XY}\Sigma_{YY}^{-1}(Y-\mu_{Y}),\Sigma_{XX}-\Sigma_{XY}\Sigma_{YY}^{-1}\Sigma_{YX})$$ You can see the conditional covariance matrix $Var(X|Y)=\Sigma_{X}-\Sigma_{XY}\Sigma_{YY}^{-1}\Sigma_{YX}$ is constant and does not depend on $Y$. Thus, we have $E[Var(X|Y)]=Var(X|Y)$