Determine the support of a gaussian vector

Question

We consider $(X,Y,Z)$ a gaussian with mean $(1,2,1)$ and with covariance matrix:

$$ \begin{pmatrix} 1 & -2 & 1\\ -2 & 5 & -1\\ 1 & -1 & 2 \end{pmatrix} $$

I am asked to determine the support of $(X,Y,Z)$.

The solution says it is $3X + Y − Z = 4$. Where does this come from please?

Answer 1

If $\ W\ $ is a random vector with a multivariate normal distribution, mean $\ \mu\ $ and covariance matrix $\ \Sigma\ $, then the support of $\ W\ $ is \begin{align} \text{supp}(W)&=\mu+\mathcal{N}(\Sigma)^\perp\\ &=\mu+\mathcal{R}(\Sigma)\ , \end{align} where $\ \mathcal{N}(\Sigma)\ $ is the nullspace of $\ \Sigma\ $, and $\ \mathcal{R}(\Sigma)\ $ is its range.

Answer 2

I wrote it properly eventually:

• Compute Ker(K):

$K \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = 0 \Leftrightarrow \begin{pmatrix} 1 & -2 & 1 \\ -2 & 5 & -1 \\ 1 & -1 & 2 \\ \end{pmatrix} \begin{pmatrix} x \\ y \\ z \\ \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \end{pmatrix} \Leftrightarrow \begin{cases} x = -3z \\ y = -z \end{cases} $

Therefore, Ker(K) = span$\begin{pmatrix} 3 \\ 1 \\ -1 \\ \end{pmatrix}$

• The image of K is Im(K) = Ker(K)$^{\perp}$ and its equation is:

$3x+y-z+d=0$

• $\mu = \begin{pmatrix} 1 \\ 2 \\ 1 \\ \end{pmatrix} \in $ Support($W$) $\Rightarrow d = -4$

• Thus, Support($W$) has equation $\boxed{3x+y-z=4}$