Covariance matrix does not always exist?

AI Thread Summary
A covariance matrix for a random vector does not always exist, contradicting the notion that it is guaranteed. The discussion highlights that if a covariance matrix does exist, it is typically only positive semi-definite rather than strictly positive definite or non-singular. An example provided involves a probability density function that diverges, illustrating a scenario where the covariance cannot be calculated. The key takeaway is that the existence of a covariance matrix is conditional and not absolute. Understanding these nuances is crucial for accurate risk management analysis.
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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