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

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The equality is true not because of the 'independent' or 'standard' part, but because of the 'Gaussian' part.

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 $E[Var(X|Y)]=Var(X|Y)$

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  • $\begingroup$ I think this answer would be better if you didn't appeal to results for random vectors, used consistent notation, and said more clearly how this answers the question $\endgroup$ – Taylor Feb 11 '17 at 5:28
  • $\begingroup$ What part of my notations inconsistent? I get that sigma_XX/sigma_YY part. Also I think I answered OP's question directly by stating his statement as truth at the end. $\endgroup$ – Julius Feb 11 '17 at 5:31
  • $\begingroup$ Yep. Also I think you reversed the order of the x and ys. And there's no mention of the Z. But the conditional variance being non random point has been made $\endgroup$ – Taylor Feb 11 '17 at 6:22

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