Proof of @@\mu_{A|B} = E[X_A | X_B = x_B]@@.

We assume a linear model relating @@X_A@@ and @@X_B@@:

$$X_A = \alpha + \beta X_B + \epsilon$$

where @@\alpha@@ is the intercept, @@\beta@@ is the matrix of regression coefficients, and @@\epsilon@@ is the error term.

We assume @@E[\epsilon] = 0@@ and that @@\epsilon@@ is independent of @@X_B@@.

The matrix of regression coefficients @@\beta@@ that minimizes the variance of the error (in the sense of ordinary least squares for multivariate variables) is given by:

$$\beta = \Sigma_{AB} \Sigma_B^{-1}$$

Here, @@\Sigma_{AB} = \text{Cov}(X_A, X_B)@@ and @@\Sigma_B = \text{Cov}(X_B, X_B)@@.

We take the expectation of the linear model:

$$E[X_A] = E[\alpha + \beta X_B + \epsilon]$$

Using the linearity of expectation:

$$E[X_A] = E[\alpha] + E[\beta X_B] + E[\epsilon]$$$$E[X_A] = \alpha + \beta E[X_B] + E[\epsilon]$$

Substituting the known means (@@E[X_A] = \mu_A@@, @@E[X_B] = \mu_B@@) and our assumption (@@E[\epsilon] = 0@@):

$$\mu_A = \alpha + \beta \mu_B$$

Solving for @@\alpha@@, we get:

$$\alpha = \mu_A - \beta \mu_B$$

Now, we compute the conditional expectation @@\mu_{A|B} = E[X_A | X_B = x_B]@@.

$$\mu_{A|B} = E[\alpha + \beta X_B + \epsilon | X_B = x_B]$$

By linearity of conditional expectation:

$$\mu_{A|B} = E[\alpha | X_B = x_B] + E[\beta X_B | X_B = x_B] + E[\epsilon | X_B = x_B]$$

Since @@\alpha@@ and @@\beta@@ are constants, @@E[\alpha | X_B = x_B] = \alpha@@.

When @@X_B@@ is given as @@x_B@@, @@E[\beta X_B | X_B = x_B] = \beta x_B@@.

Due to the independence of @@\epsilon@@ and @@X_B@@, @@E[\epsilon | X_B = x_B] = E[\epsilon] = 0@@.

Combining these terms, we have:

$$\mu_{A|B} = \alpha + \beta x_B$$

Substitute the expression for @@\alpha@@ from Step 2:

$$\mu_{A|B} = (\mu_A - \beta \mu_B) + \beta x_B$$

Rearranging the terms:

$$\mu_{A|B} = \mu_A + \beta(x_B - \mu_B)$$

Finally, substitute the expression for @@\beta@@ from Step 1:

$$\mu_{A|B} = \mu_A + \Sigma_{AB} \Sigma_B^{-1} (x_B - \mu_B)$$

This completes the derivation.

Block Matrix Inverse:Derivation of $\tilde{C}$ and $\tilde{B}$ and @@\tilde{A}@@ and @@\tilde{B}^T@@

We can derive the components by solving the system $MM^{-1} = I$, where $I = \begin{pmatrix} I_p & 0 \\ 0 & I_q \end{pmatrix}$.

1. Derivation of $\tilde{C}$ and $\tilde{B}$ (Right-hand column of $M^{-1}$)

From @@MM^{-1} = I@@, we use the top-right and bottom-right blocks of the identity matrix:

\[ \begin{aligned} (1) \quad A\tilde{B} + B\tilde{C} &= 0 \\ (2) \quad B^T\tilde{B} + C\tilde{C} &= I_q \end{aligned} \]

From (1), left-multiply by $A^{-1}$:

\[ \tilde{B} = -A^{-1}B\tilde{C} \]

Substitute this into (2):

\[ \begin{aligned} B^T(-A^{-1}B\tilde{C}) + C\tilde{C} &= I_q \\\\ (-B^T A^{-1} B + C)\tilde{C} &= I_q \\\\ (C - B^T A^{-1} B)\tilde{C} &= I_q \\\\ \end{aligned} \]

Thus, @@\tilde{C}@@ is the inverse of the Schur complement $M/A$:

\[ \tilde{C} = (C - B^T A^{-1} B)^{-1} = (M/A)^{-1} \]

2. Derivation of @@\tilde{A}@@ and @@\tilde{B}^T@@ (Left-hand column of @@M^{-1}@@)

By analogy, we use the top-left and bottom-left blocks of the identity matrix:

\[ \begin{aligned} (3) \quad A\tilde{A} + B\tilde{B}^T &= I_p \\ (4) \quad B^T \tilde{A} + C \tilde{B}^T &= 0 \end{aligned} \]

From (4), left-multiply by $C^{-1}$:

\[ \tilde{B}^T = -C^{-1}B^T \tilde{A} \]

Substitute this into (3):

\[ \begin{aligned} A\tilde{A} + B(-C^{-1}B^T \tilde{A}) &= I_p \\\\ (A - BC^{-1}B^T)\tilde{A} &= I_p \end{aligned} \]

Thus, $\tilde{A}$ is the inverse of the Schur complement $M/C$:

\[ \tilde{A} = (A - BC^{-1}B^T)^{-1} = (M/C)^{-1} \]