Find conditional expectation from a 3-dimensional random vector?

Question

I have a three-dimensional random vector $Y$ ~ $N(\mu , \Sigma)$ whose moment-generating function is:

$M_Y(t_1,t_2,t_3)=exp(2t_1+3t_2-3t_3+4t_1^2+5t_2^2+3t_3^2-t_1t_2+2t_1t_3)$.

It was easy to calculate $\mu$

because $M_Y(t)=exp(\mu^Tt+\frac12 t^T\Sigma t), \mu_1 = 2, \mu_2 = 3, \mu_1 = -3.$

Also, $\Sigma=\left[\matrix{8&-1&2 \\ -1&10&0 \\ 2&0&6}\right]$. (If there was any wrong points in this calculation, please let me know).

This is what I have done now....

However, I know only how to calculate the conditional mean when the vector is two-dimentional: $E[Y_1|Y_2=y_2]=E[Y_1]+\Sigma_{12}\Sigma_{22}^{-1}(y_2-\mu_2)$

$VAR[Y_1|Y_2=y_2]=\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}$ (* May anyone let me know the process of having the inverse of covariance in the $\Sigma$?)

For 3D, I still did not get good idea how to calculate $E[Y_1|Y_2=y_2, Y_3=y_3] $ and $VAR[Y_1|Y_2=y_2, Y_3=y_3]$.

For example: $E[Y_1|Y_2=\frac72, Y_3=-2] $ and $VAR[Y_1|Y_2=\frac72, Y_3=-2]$. How is the process changing compared to a 2D one, and where should I start?

Answer 1

Partition the $\Sigma$ matrix as required. If you are interested in the conditional distribution of $Y_{1}$ given $\left(\begin{array}{c} Y_{2}\\ Y_{3}\\ \end{array}\right)$, then the partition the covariance matrix as

\begin{equation*} \Sigma = \left( \begin{array}{c|cc} 8 &-1& 2\\ \hline -1& 10& 0\\ 2& 0& 6\\ \end{array}\right) \end{equation*} Identify this partition as \begin{equation*} \Sigma = \left( \begin{array}{c|c} \Sigma_{11} & \Sigma_{12}\\ \hline \Sigma_{21}& \Sigma_{22}\\ \end{array}\right) \end{equation*} where $\Sigma_{11}=8$, $\Sigma_{12}=(-1\;\; 2)$, $\Sigma_{21}=\left(\begin{array}{c} -1\\ 2 \end{array}\right)$ and $\Sigma_{22}=\left(\begin{array}{cc} 10 & 0\\ 0& 6\\ \end{array}\right)$ Find the inverse of $\Sigma_{22}$ and plug-in the values into your expressions.