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Let $\mathbf{Y}=\begin{pmatrix} \mathbf{Y_1}\\\mathbf{Y_2} \end{pmatrix}\sim N\left (\boldsymbol{\mu},\boldsymbol{\Sigma} \right ), $ $\boldsymbol{\mu}=\begin{pmatrix} \boldsymbol{\mu_1}\\\boldsymbol{\mu_2} \end{pmatrix}$ and $\boldsymbol{\Sigma}=\begin{pmatrix} \boldsymbol{\Sigma_{11}}& \boldsymbol{\Sigma}_{12} \\ \boldsymbol{\Sigma}_{12}& \boldsymbol{\Sigma}_{22} \\ \end{pmatrix}$ are are compatibly partitioned. Making a variable transform $$\left(\begin{array}{l} \boldsymbol{U}_{1} \\ \boldsymbol{U}_{2} \end{array}\right)=\left(\begin{array}{cc} \boldsymbol{I} & -\boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1} \\ \mathbf{0} & \boldsymbol{I} \end{array}\right)\left(\begin{array}{c} \boldsymbol{Y}_{1} \\ \boldsymbol{Y}_{2} \end{array}\right)-\begin{pmatrix} \boldsymbol{\mu}_{1}\\ \boldsymbol{\mu}_{2}\\ \end{pmatrix}.$$ The conditional distribution of $\boldsymbol{X_{1}}$ given $ \boldsymbol{X_{2}} $ denote as $\boldsymbol{X}_{1} \mid \boldsymbol{X}_{2}.$Symbol $\boldsymbol{X}_{1} \stackrel{\mathrm{d}}{=}\boldsymbol{X}_{2} $ means two random variables $\boldsymbol{X}_{1} $ and $\boldsymbol{X}_{2}$ have the same distribution.

Show that $$\boldsymbol{Y}_{1} \mid \boldsymbol{Y}_{2}\stackrel{\mathrm{d}}{=}\left[\boldsymbol{U}_{1}+\boldsymbol{\mu}_{1}+\boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1} \boldsymbol{U}_{2}\right] \mid \boldsymbol{U}_{2}.$$

From the inverse transform $$\left(\begin{array}{l} \boldsymbol{U}_{1} \\ \boldsymbol{U}_{2} \end{array}\right)=\left(\begin{array}{cc} \boldsymbol{I} & -\boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1} \\ \mathbf{0} & \boldsymbol{I} \end{array}\right)\left(\begin{array}{c} \boldsymbol{Y}_{1}-\boldsymbol{\mu}_{1} \\ \boldsymbol{Y}_{2}-\boldsymbol{\mu}_{2} \end{array}\right),$$ $\boldsymbol{U}_{1}$ and $\boldsymbol{U}_{2}$ are independent, $\boldsymbol{Y}_{1}=\boldsymbol{\mu}_{1}+\boldsymbol{U}_{1}+\boldsymbol{\Sigma}_{12} \boldsymbol{\Sigma}_{22}^{-1} \boldsymbol{U}_{2},\boldsymbol{Y}_{2}=\boldsymbol{\mu}_{2}+\boldsymbol{U}_{2}.$ I can not go further about this.I don't know what's the formal definition about $\boldsymbol{Y_{1}}|\boldsymbol{Y_{2}}\stackrel{\mathrm{d}}{=}\boldsymbol{X_{1}}|\boldsymbol{X_{2}}.$

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  • $\begingroup$ I don't see what the confusion is: $Y_1 \mid U_2 \stackrel{d}{=} Y_1 \mid Y_2$ since $\mu_2$ is a constant and so it doesn't affect the conditional distribution. $\endgroup$
    – Yashaswi Mohanty
    Commented Mar 8, 2023 at 22:06
  • $\begingroup$ @YashaswiMohanty I believe the problem is subtler than that. Generally, even when $(U_1,U_2)=f(Y_1,Y_2)$ is a measurable one-to-one transformation, it is not the case that the conditional distributions $U_1\mid U_2$ and $Y_1\mid Y_2$ must be the same. Out of this misconception arise many so-called "geometric paradoxes" in probability theory, such as the Bertran Paradox. $\endgroup$
    – whuber ♦
    Commented Mar 11, 2023 at 12:49
  • $\begingroup$ Hmm, I’m not sure why this would cause problems in this specific example; $\sigma(Y_2) = \sigma(U_2 + c) = \sigma(U_2)$ and so $P(Y_1 \in A \mid Y_2) = P(Y_1 \in A \mid U_2)$ for all $A \in \mathcal{B}(\mathbb{R})$? $\endgroup$
    – Yashaswi Mohanty
    Commented Mar 13, 2023 at 10:10
  • $\begingroup$ @YashaswiMohanty I did not state that would cause problems, but only wished to point out that there indeed is something to demonstrate here: the result is not automatic. $\endgroup$
    – whuber ♦
    Commented Mar 13, 2023 at 15:12

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