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Let $(X1, X2, X3) $ be multivariate normally distributed. I want to prove that $$E(X1|X2,X3)=E(X_1|X_2)+\Sigma_{X_1X_3}\Sigma_{X_3 X_3}^{-1}X_3$$ I think one way of getting there is using with $Y=(X_2, X_3)$ that $$E(X_1|Y)=\mu_x+\Sigma_{X_1Y}\Sigma_{YY}^{-1}(Y-\mu_Y)$$ and then using the formula for block matrix inversion for $\Sigma_{YY}^{-1}$ but this seems very tedious. Is there a more elegant way to prove this?

Edit We need $E(X_3)=0$ and $\Sigma_{X_2 X_3}=0$, then the "block inversion" is trivial and the result follow immediately.

Joogs
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  • As far as I can see inverting a $2\times 2$ matrix seems necessary to prove this.
    • – Kavi Rama Murthy Apr 24 '19 at 08:22
  • In fact, I was missing multiple conditions, very silly. We need $E(X_3)=0$ and $\Sigma_{X_2 X_3}=0$. Then the inversion becomes very simple. – Joogs Apr 24 '19 at 08:29