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Covariance matrix does not always exist?

  1. Dec 13, 2013 #1
    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).
  2. jcsd
  3. Dec 13, 2013 #2


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    As a simple example, imagine the random variable in one dimension whose pdf is
    [tex] p(x) = 1/x^2 [/tex]
    for x > 1, and 0 otherwise.

    The covariance matrix in this case is simply the variance of the pdf, which is
    [tex] \int_{1}^{\infty} x^2 \frac{1}{x^2} dx [/tex]
    which doesn't exist as the integral diverges.
  4. Dec 13, 2013 #3
    Consider 1d case, e.g. Pareto.
  5. Dec 13, 2013 #4
    Thanks that helps!
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