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[S]uppose that $X_1$ (sales), $X_2$ (price), $X_3$ (advertisement), and $X_4$ (sales assistants) are normally distributed with:

$$ \mu = \begin{pmatrix} 172.7 \\ 104.6 \\ 104.0 \\ 93.8 \end{pmatrix} \text{ and } \Sigma = \begin{pmatrix} 1037.21 & & & \\ -80.02 & 219.84 & & \\ 1430.70 & 92.10 & 2624.00 & \\ 271.44 & -91.58 & 210.30 & 177.36 \end{pmatrix} $$

(These are in fact the sample mean and the sample covariance matrix but in this case we pretend that they are the true parameter values.)

The conditional distribution of $X_1$ given $(X_2,X_3,X_4)$ is thus an univariate normal with mean

$$ \mu_1 + \sigma_{12} \Sigma_{22}^{-1} \begin{pmatrix} X_2 - \mu_2 \\ X_3 - \mu_3 \\ X_4 - \mu_4 \end{pmatrix} = 65.670 - 0.216 X_2 + 0.485 X_3 + 0.844 X_4 $$

and variance

$$ \sigma_{11.2} = \sigma_{11} - \sigma_{12} \Sigma_{22}^{-1} \sigma_{21} = 96.671 $$

I'm trying to understand this example and can't figure out how it partition the covariance matrix. In this case, what is $\Sigma_{22}^{-1}$ ? Thank you for your help.

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You need to complete the top-right of the covariance matrix by reflecting the bottom-left to give

$$ \Sigma = \left( \begin{array}{cccc} 1037.21 & -80.02 & 1430.7 & 271.44 \\ -80.02 & 219.84 & 92.1 & -91.58 \\ 1430.7 & 92.1 & 2624 & 210.3 \\ 271.44 & -91.58 & 210.3 & 177.36\end{array} \right)$$

$\sigma_{12}$ then apparently means the $1$st row of the matrix excluding the $1$st cell, so $$ \sigma_{12} = \left( \begin{array}{ccc} -80.02 & 1430.7 & 271.44 \end{array} \right)$$ and $\sigma_{21}$ is apparently the transpose of this (the $1$st column of the matrix excluding the $1$st cell)

while $\Sigma_{22}$ apparently means the covariance matrix without the $1$st row or column so

$$ \Sigma_{22} = \left( \begin{array}{ccc} 219.84 & 92.1 & -91.58 \\ 92.1 & 2624 & 210.3 \\ -91.58 & 210.3 & 177.36\end{array} \right)$$

with $\Sigma_{22}^{-1}$ being its inverse

$$ \Sigma_{22}^{-1} = \left( \begin{array}{ccc} 0.006427076 & -0.000543173 & 0.00396268 \\ -0.000543173 & 0.000467021 & -0.000834226 \\ 0.00396268 & -0.000834226 & 0.008673545 \end{array} \right)$$

and you can check that

$$\sigma_{12} \Sigma_{22}^{-1} \left( \begin{array}{c} X_2 \\ X_3 \\ X_4 \end{array} \right) = -0.215782095 X_2 + 0.485189757 X_3 +0.843726149 X_4$$ which gives the coefficients in your example. I will leave the arithmetic of the intercept and the variance to you, but it is similar matrix multiplication.

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