# Multivariate normal distribution with one constant term

Assume that we have a random vector $$X$$ taking values from $$\mathbb{R}^{k}$$ which follows a multivariate normal distribution, i.e. $$X \sim \mathcal{N}(\mu, \Sigma)$$, where $$\mu \in \mathbb{R}^{k}$$ is some constant and $$\Sigma$$ is $$k \times k$$ positive semi-definite matrix.

Next, let us consider a random vector $$X' \in \mathbb{R}^{k+1}$$ contracted from $$X$$ with one more constant element, say $$c$$, added to its end. From the definition of multivariate random vector, the vector $$X'$$ is also a normal random vector.

What is its covariance matrix in this case? Would it be correct to say (and write) that $$\Sigma'$$ is $$(k+1)\times (k+1)$$ matrix which is equal to $$\Sigma$$ with added the last raw and column of all zeros, i.e.

$$\begin{equation*} \Sigma' = \begin{pmatrix} \Sigma & \mathbf{o} \\ \mathbf{o}^{T} & 0 \end{pmatrix}, \end{equation*}$$ where $$\mathbf{o} \in \mathbb{R}^{k}$$ is a vector of zeros.

PS I have never seen a covariance matrix with a diagonal element equal to zero.

• Yes it is correct. Since $X'$ has a degenerate normal distribution its covariance matrix would be singular. May 5, 2020 at 14:25
• Dear StubbornAtom, the question is not about the singularity of covariance matrix, it is about a zero diagonal element.
– ABK
May 5, 2020 at 14:27
• Zero diagonal element because variance of the constant is zero. This is fine because distribution of $X'$ is singular (does not have a density). May 5, 2020 at 14:32
• ok.. It is clear that it does not have density... I have never seen this in the literature the cov matrix with zero diag element
– ABK
May 5, 2020 at 14:33

The last element in the diagonal of $$\Sigma'$$ would normally be equal to $$0$$ since the last element of the variable $$X_n'$$ is constant.
This comes from the covariance formula, where we have $$var(c,c) = E[(c-E(c))(c-E(c))]$$
Since $$c$$ is constant we also have: $$E(c) = c$$
So $$var(c,c)=0$$