Why are vectors $X_2$ and $X_3$ bivariate normally distributed?

Question

I have a stochastic vector $\mu = \begin{bmatrix}1 \\ 0 \\ 1\end{bmatrix}$ and $\Sigma= \begin{bmatrix}1 & 0 & -1\\0 & 2 & 0 \\ -1 & 0 & 3\end{bmatrix}$.

I have to proof that $X_2$ and $X_3$ are independent. I can tell from the covariance matrix that $\operatorname{Cov}(X_2,X_3) = 0$, but the solution manual also states that $X_2$ and $X_3$ have the property of being bivariate normally distributed.

How can I tell this is the case? The proof for this seems way more complicated than these simple matrices suggest.

Answer 1

What is the definition of the 'stochastic vector' $\mu$? It looks like the stochastic vector is supposed to have normally distributed components. If this is the case, then any linear combination of its components (eg a component of $\Sigma \mu$) is also normally distributed.

Answer 2

The definition of $X=(X_1,\ldots,X_n)$ following a multivariate normal distribution is that $$ \sum_{i=1}^n a_iX_i $$ follows a (one-dimensional) normal distribution for all choices of $a_1,\ldots,a_n\in\mathbb{R}$.

Now, assume that $(X_1,X_2,X_3)$ is multivariate normal distributed, which by definition means that $$ a_1X_1+a_2X_2+a_3X_3 $$ is (one-dimensional) normal distributed for all choices of $a_1,a_2,a_3\in\mathbb{R}$. Let's show that $(X_1,X_2)$ indeed follows a multivariate normal distribution as well. To that end, let $a_1,a_2\in\mathbb{R}$ be arbitrary. Then $$ a_1X_1+a_2X_2=a_1X_1+a_2X_2+a_3X_3 $$ with $a_3=0$ and since the right-hand side is normal distributed then so is the left-hand side. Since $a_1$ and $a_2$ were arbitrary, we are done.