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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.

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    $\begingroup$ Yes it is correct. Since $X'$ has a degenerate normal distribution its covariance matrix would be singular. $\endgroup$ May 5, 2020 at 14:25
  • $\begingroup$ Dear StubbornAtom, the question is not about the singularity of covariance matrix, it is about a zero diagonal element. $\endgroup$
    – ABK
    May 5, 2020 at 14:27
  • $\begingroup$ Zero diagonal element because variance of the constant is zero. This is fine because distribution of $X'$ is singular (does not have a density). $\endgroup$ May 5, 2020 at 14:32
  • $\begingroup$ 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 $\endgroup$
    – ABK
    May 5, 2020 at 14:33

1 Answer 1

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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 $$var(c,c)=0$$

I hope this answers your question.

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