Question

Let $X=(X_1,X_2,X_3)^{T}\sim N_3(\mu,\Sigma)$ where \[ \mu=\begin{pmatrix}3\\2\\4\end{pmatrix},\qquad \Sigma=\begin{pmatrix} 4 & 2 & 1\\ 2 & 3 & 0\\ 1 & 0 & 2 \end{pmatrix}. \] Find $E[X_1\mid(X_2=4,X_3=7)]$ (in integer).

Hint:

For partitioned multivariate normals, use $E[X_1\mid X_2]=\mu_1+\Sigma_{12}\Sigma_{22}^{-1}(x_2-\mu_2)$. Ensure correct matrix partitioning and mean substitution.

Answer 1

Correct Answer: 6

We use the conditional expectation formula for the multivariate normal distribution: \[ E[X_1\mid X_2=x_2, X_3=x_3] =\mu_1+\Sigma_{12}\Sigma_{22}^{-1}\begin{pmatrix}x_2-\mu_2
x_3-\mu_3\end{pmatrix}. \]
Step 1: Identify components.
Partition $\Sigma$ as \[ \Sigma= \begin{pmatrix} \Sigma_{11} & \Sigma_{12}
\Sigma_{21} & \Sigma_{22} \end{pmatrix}, \] where $\Sigma_{11}=4$, $\Sigma_{12}=(2,1)$, and \[ \Sigma_{22}= \begin{pmatrix} 3 & 0
0 & 2 \end{pmatrix}. \] Then $\Sigma_{22}^{-1}=\begin{pmatrix}1/3 & 0
0 & 1/2\end{pmatrix}$.
Step 2: Compute the conditional mean.
\[ \Sigma_{12}\Sigma_{22}^{-1} =(2,1) \begin{pmatrix}1/3 & 0
0 & 1/2\end{pmatrix} =(2/3,\,1/2). \] Hence \[ E[X_1\mid X_2=x_2,X_3=x_3] =3+(2/3)(x_2-2)+(1/2)(x_3-4). \] Substituting $x_2=4, x_3=7$: \[ E[X_1\mid X_2=4,X_3=7] =3+(2/3)(2)+(1/2)(3) =3+\tfrac{4}{3}+\tfrac{3}{2} =\tfrac{6+4+4.5}{?} =3+\tfrac{17}{6} =5.83\approx6. \] Thus, the required expectation is $\boxed{6}$.