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: Recall the conditional variance formula for the multivariate normal.
If \((X_1,X_2,X_3)^T\) follows a trivariate normal distribution with mean \(\mu\) and covariance matrix \(\Sigma\), partition the vector into \(X_1\) and \((X_2,X_3)^T\). Then, conditionally on \(X_2,X_3\), \(X_1\) is again normal with
\[ \text{Var}(X_1\mid X_2,X_3)=\Sigma_{11}-\Sigma_{12}\,\Sigma_{22}^{-1}\,\Sigma_{21} \]
where \(\Sigma_{11}\) is the variance of \(X_1\), \(\Sigma_{12}\) is the row vector of covariances of \(X_1\) with \((X_2,X_3)\), \(\Sigma_{21}=\Sigma_{12}^T\), and \(\Sigma_{22}\) is the covariance matrix of \((X_2,X_3)\). This formula does not involve the conditioning values themselves, only the covariance structure, so the given values \(X_2=1,X_3=-1\) will not enter the final number.

Step 2: Identify the pieces from Sigma.
\[ \Sigma=\begin{pmatrix}2&1&1\\1&2&1\\1&1&2\end{pmatrix} \]
So
\[ \Sigma_{11}=2,\qquad \Sigma_{12}=\begin{pmatrix}1&1\end{pmatrix},\qquad \Sigma_{22}=\begin{pmatrix}2&1\\1&2\end{pmatrix} \]

Step 3: Invert Sigma22.
The determinant is \(\det(\Sigma_{22})=2(2)-1(1)=3\). So
\[ \Sigma_{22}^{-1}=\frac{1}{3}\begin{pmatrix}2&-1\\-1&2\end{pmatrix} \]

Step 4: Compute Sigma12 times Sigma22 inverse.
\[ \begin{pmatrix}1&1\end{pmatrix}\cdot\frac{1}{3}\begin{pmatrix}2&-1\\-1&2\end{pmatrix} =\frac{1}{3}\begin{pmatrix}(1)(2)+(1)(-1) & (1)(-1)+(1)(2)\end{pmatrix} =\frac{1}{3}\begin{pmatrix}1&1\end{pmatrix} \]

Step 5: Multiply by Sigma21.
\[ \frac{1}{3}\begin{pmatrix}1&1\end{pmatrix}\begin{pmatrix}1\\1\end{pmatrix}=\frac{1}{3}(1+1)=\frac{2}{3} \]

Step 6: Subtract from Sigma11.
\[ \text{Var}(X_1\mid X_2=1,X_3=-1)=2-\frac{2}{3}=\frac{4}{3} \]

Step 7: Convert to decimal.
\[ \frac{4}{3}\approx1.3333 \]

Final Answer:
Rounded to two decimal places, \[ \boxed{\text{Var}(X_1\mid X_2=1,X_3=-1)\approx1.33} \]