Why do the following facts implies that the cov matrix is PSD. Is anyone willing to help?

Asked Jan 18 '20 at 20:20

Active Jan 20 '20 at 17:51

Viewed 163 times

Let $V=(v_{ij})$ be a covariance matrix associated to the random vector r.

My professor writes that the covariance matrix is PSD. Indeed for all $i,j = 1,...,n$ we have $v_{ij} = E((r_i −E(r_i))(r_j −E(r_j))$ so, calling $ρ_{ij} := \frac{ v_{ij}} {\sqrt{v_{ii}v_{jj}}}$, ($ρ_{ij}$ is called correlation of $r_i$ and $r_j$) we get $ρ_{ij}≤ |1|$. The two above facts imply V is PSD.

But why do the above two facts implies that the matrix is PSD?? Is anyone willing to help??

edited Jan 20 '20 at 17:51

asked Jan 18 '20 at 20:20

Alchemy

3

Does this answer your question? What is the proof that covariance matrices are always semi-definite? – NCh Jan 19 '20 at 07:51
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I do not think so. I mean, it does not use the facts mentioned – Alchemy Jan 19 '20 at 12:58
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Isn't the first fact alone enough? – cangrejo Jan 19 '20 at 16:39
It is true the that the second fact (correlation coefficient) in effect gives you PSD for a 2x2 matrix (i.e. in effect via Sylvester Determinant Criterion noting that top left component is positive). The n x n case comes from the first fact alone. Consider centered random vector $A := r - E[r]$. For $\mathbf x \in \mathbb R^n$, the covariance matrix obeys $\mathbf x^T \operatorname E(A A^\top) \mathbf x = \sum_{i=1}^n\sum_{j=1}^n x_i x_j E[A_i A_j] = E[\sum_{i=1}^n\sum_{j=1}^n x_i x_j A_i A_j] = E[(\sum_{i=1}^n x_i A_i)^2]\geq 0$ and hence is PSD – user8675309 Jan 19 '20 at 22:17
What he might mean is: Covariance $\Rightarrow$ PSD $\Rightarrow$ $|\rho_{ij}|\le 1$. The implications are not reversable. – A.Γ. Jan 19 '20 at 22:45
Well symmetry is pretty easy. Did you figure that out? – ViktorStein Jan 19 '20 at 23:12
Yes, sure. I figured out symmetry – Alchemy Jan 19 '20 at 23:15

Why do the following facts implies that the cov matrix is PSD. Is anyone willing to help?

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