Covariance matrix does not always exist?

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Discussion Overview

The discussion revolves around the existence of a covariance matrix for random vectors, particularly in the context of a sample question from risk management material. Participants explore theoretical aspects of covariance matrices, including conditions under which they may or may not exist.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant questions the assertion that a covariance matrix always exists for a random vector, citing a sample question that suggests otherwise.
  • Another participant provides an example of a probability density function (pdf) where the covariance matrix does not exist due to the divergence of the integral used to calculate variance.
  • A third participant references the Pareto distribution as a specific case to consider in the one-dimensional scenario.

Areas of Agreement / Disagreement

Participants express differing views on the existence of covariance matrices, with some arguing that they may not always exist, while others provide examples that support this claim. The discussion remains unresolved regarding the generality of the existence of covariance matrices.

Contextual Notes

The discussion highlights limitations in the assumptions regarding the existence of covariance matrices, particularly in relation to specific probability distributions and their properties.

Phoeniyx
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Hey guys. I am going through the PRM (risk manager) material and there is a sample question that is bugging me. The PRM forum is relatively dead, and they don't usually go that deep into the theory anyway. So wanted to ask you guys.

Shouldn't a random vector always have a covariance matrix? Why is the "answer" below saying that it doesn't always have to exist? i.e. why is (c) wrong?

Q: A covariance matrix for a random vector:
a) Is strictly positive definite, if it exist
b) Is non-singular, if it exist
c) Always exists
d) None of the above

A: This question is full of red herrings. A covariance matrix may not exist, which contradicts c). If it does exist, it is in general only positive semi-definite, which contradicts both a) and b) hence d).
 
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As a simple example, imagine the random variable in one dimension whose pdf is
p(x) = 1/x^2
for x > 1, and 0 otherwise.

The covariance matrix in this case is simply the variance of the pdf, which is
\int_{1}^{\infty} x^2 \frac{1}{x^2} dx
which doesn't exist as the integral diverges.
 
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Consider 1d case, e.g. Pareto.
 
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Thanks that helps!
 

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