91Ó°ÊÓ

We have a random vector \( \mathbf{X} = (X_1, X_2, X_3) \) with a three-dimensional normal distribution, mean vector \( \boldsymbol{\mu} = \mathbf{0} \), and covariance matrix \( \mathbf{\Lambda} \). The task is to find the distribution of \( X_1 + X_3 \) given that \( X_2 = 0 \).

Identify submatrices of the covariance matrix \( \mathbf{\Lambda} \). Let \( \mathbf{\Lambda}_{11} \) be the covariance matrix of \( (X_1, X_3) \) and \( \mathbf{\Lambda}_{22} \) be the variance of \( X_2 \), and \( \mathbf{\Lambda}_{12} \) be the covariance vector between \( (X_1, X_3) \) and \( X_2 \).- \( \mathbf{\Lambda}_{11} = \begin{bmatrix} 3 & 1 \ 1 & 1 \end{bmatrix} \)- \( \mathbf{\Lambda}_{22} = [2] \)- \( \mathbf{\Lambda}_{12} = \begin{bmatrix} -2 \ 0 \end{bmatrix} \)

Using the conditional distribution formulas for a multivariate normal distribution, find the conditional mean and covariance of \( (X_1, X_3) \) given \( X_2 = 0 \):- The conditional mean \( \boldsymbol{\mu}_{1|2} = \mathbf{\Lambda}_{12} \mathbf{\Lambda}_{22}^{-1}(X_2 - \mu_2) = \begin{bmatrix} -1 \ 0 \end{bmatrix}(0) = \begin{bmatrix} 0 \ 0 \end{bmatrix} \).- The conditional covariance \( \mathbf{\Sigma}_{1|2} = \mathbf{\Lambda}_{11} - \mathbf{\Lambda}_{12} \mathbf{\Lambda}_{22}^{-1} \mathbf{\Lambda}_{12}^T = \begin{bmatrix} 3 & 1 \ 1 & 1 \end{bmatrix} - \begin{bmatrix} -2 \ 0 \end{bmatrix}[2]^{-1}\begin{bmatrix} -2 & 0 \end{bmatrix} = \begin{bmatrix} 3 - 2 & 1 \ 1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 1 \ 1 & 1 \end{bmatrix}. \)

Having the conditional distribution \( (X_1, X_3) \sim N(\begin{bmatrix} 0 \ 0 \end{bmatrix}, \begin{bmatrix} 2 & 1 \ 1 & 1 \end{bmatrix}) \), we find the distribution of their sum:- The mean of \( X_1 + X_3 \) is \( 0 + 0 = 0 \).- The variance is \( \operatorname{Var}(X_1) + \operatorname{Var}(X_3) + 2\operatorname{Cov}(X_1, X_3) = 2 + 1 + 2 \times 1 = 5 \).Thus, the distribution is \( X_1 + X_3 \sim N(0, 5) \).