Question

Let \((X_1,X_2,X_3)^T\) follow a trivariate normal distribution with mean vector \(\mu\) and covariance matrix \(\Sigma\) given by
\[\mu=\begin{pmatrix}1\\1\\1\end{pmatrix}\quad\text{and}\quad \Sigma=\begin{pmatrix}2&1&1\\1&2&1\\1&1&2\end{pmatrix}.\]
Then \(\text{Var}(X_1\mid X_2=1,X_3=-1)\) equals ______ (rounded off to two decimal places).

Hint:

Hint:
Use \(\text{Var}(X_1\mid X_2,X_3)=\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}\); the conditioning values themselves do not affect the conditional variance.

Answer 1

Correct Answer: 1.33

Step 1: Set up the best linear predictor of X1.
For jointly normal variables, the conditional mean of $X_1$ given $X_2,X_3$ is a linear function $a+b_2X_2+b_3X_3$, and the conditional variance is the leftover variance of $X_1$ after removing what this line explains. The coefficients $b_2,b_3$ solve the normal equations
\[ \text{Cov}(X_1,X_2)=b_2\text{Var}(X_2)+b_3\text{Cov}(X_2,X_3) \]
\[ \text{Cov}(X_1,X_3)=b_2\text{Cov}(X_2,X_3)+b_3\text{Var}(X_3) \]

Step 2: Plug in the numbers from Sigma.
From the matrix, $\text{Var}(X_2)=\text{Var}(X_3)=2$, $\text{Cov}(X_2,X_3)=1$, and $\text{Cov}(X_1,X_2)=\text{Cov}(X_1,X_3)=1$. So
\[ 2b_2+b_3=1 \]
\[ b_2+2b_3=1 \]

Step 3: Solve the pair by elimination.
From the first equation, $b_3=1-2b_2$. Put this in the second equation:
\[ b_2+2(1-2b_2)=1 \]
\[ b_2+2-4b_2=1 \]
\[ -3b_2=-1 \quad\Rightarrow\quad b_2=\frac{1}{3} \]
Then $b_3=1-\frac{2}{3}=\frac{1}{3}$. By the symmetry of the covariance matrix this equal split makes sense.

Step 4: Find the leftover, unexplained variance.
The conditional variance equals the total variance of $X_1$ minus the part explained by the best linear fit:
\[ \text{Var}(X_1\mid X_2,X_3)=\text{Var}(X_1)-\left[b_2\text{Cov}(X_1,X_2)+b_3\text{Cov}(X_1,X_3)\right] \]
\[ =2-\left[\frac{1}{3}(1)+\frac{1}{3}(1)\right]=2-\frac{2}{3}=\frac{4}{3} \]

Step 5: Note that the given values do not matter.
The numbers $X_2=1$ and $X_3=-1$ only shift the conditional mean, not this leftover variance, so they play no role in the final answer.
\[ \frac{4}{3}\approx1.33 \]
\[ \boxed{\text{Var}(X_1\mid X_2=1,X_3=-1)\approx1.33} \]