Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Covariance of Gaussian Process

  1. Feb 15, 2010 #1
    This is probably a stupid question, but here goes: based on the covariance function of some (centered, stationary) Gaussian process - how can one determine non-degeneracy (here I mean for any choice of a finite number of sampling times, the resulting RV is AC).

    Ideas?
     
  2. jcsd
  3. Feb 15, 2010 #2

    mathman

    User Avatar
    Science Advisor

    Naive(?) question. What is AC?
     
  4. Feb 16, 2010 #3
    Sorry- AC is absolutely continuous.
     
  5. Feb 16, 2010 #4

    mathman

    User Avatar
    Science Advisor

    Sample functions (which is what I think you mean by RV) of a stationary Gaussian process are in general not absolutely continuous. I can't give a reference, but if you think of Brownian motion as a model, it is very jerky.
     
  6. Feb 17, 2010 #5
    No no - by RV I mean random vector. I just mean that for any choice of sampling times, the resulting covariance matrix is invertible.
     
  7. Feb 17, 2010 #6

    mathman

    User Avatar
    Science Advisor

    I have never had to address this question, so I can't tell you. However if I remember correctly the matrix is positive semi-definite. That could help - it makes me suspect that it is invertible.
     
  8. Feb 17, 2010 #7
    Not necessarily. Think of the random vector (X,X) for a standard normal X. The question of non-degeneracy for a Gaussian process is precisely this, and it's a pretty (basic and) important one when it comes to stochastic analysis. My problem is, as basic as it should
    be, I have a specific covariance function and I'm struggling to “take off”...

    Thanks anyway though!
     
  9. Feb 17, 2010 #8
    I'm not sure I understand your question regarding sampling times for stationary processes. In any case, for the variance-covariance matrix of two independent random sample vectors [tex]X_{i},X_{j}[/tex] having an equal number of comparable components, the matrix is symmetric, non-negative definite and, unless it is singular, invertible.r
     
    Last edited: Feb 18, 2010
  10. Feb 18, 2010 #9

    mathman

    User Avatar
    Science Advisor

    Is this example supposed to be a 2-vector with both components sampled at the same time? This would be the exception to the general statement. I suspect that as long as the vector components are samples at different times you will have an invertible matrix.
     
  11. Feb 19, 2010 #10
    Of course not. A counter example would be a “constant” stochastic process, equal at all times to the same standard normal RV. And that's not the only counter example. It's definitely not a general property, it depends on the given process.
     
  12. Feb 19, 2010 #11
    Covariance is defined for two random variables at the same mean times. A stationary process generally presumes a constant mean.
    http://math.lbl.gov/~kourkina/math220/chapter4.pdf [Broken]
     
    Last edited by a moderator: May 4, 2017
  13. Feb 19, 2010 #12
    How is that related to anything?
     
  14. Feb 20, 2010 #13
    I'll ignore your attitude for the benefit of others reading this thread. Singular v-c matrices may arise with sampling from standardized Gaussian stationary distributions. They can be dealt with analytically.

    http://www.riskglossary.com/link/positive_definite_matrix.htm

    I don't see how random sample timing or frequency comes in here, in terms of individual samples.

    EDIT: I'm assuming you know that singular matrices can be identified by simply obtaining the determinant.
     
    Last edited by a moderator: Apr 24, 2017
  15. Feb 20, 2010 #14
    I actually didn't mean to come off as having a bad attitude, but reading my post back I see how you could have gotten that feeling. Please accept my apology.

    I really meant I don't see the connection between my question and what you wrote. I'm not talking about any random times, just about simple, basic non-degeneracy; for each n times t_1,...,t_n, the random vector X=(X_{t_1},...,X_{t_n}) has a density function. This can be otherwise stated as “the covariance matrix of X is invertible”.
     
  16. Feb 20, 2010 #15
    Yes, this is almost always true for the variance-covariance matrix of any sample. However, given random sampling, it's possible that a singular (degenerate) matrix could be obtained. My second link shows how this can be handled analytically. If this is not your question, could you please rephrase it?
     
  17. Feb 20, 2010 #16
    Based on a given covariance function for some centered and stationary Gaussian process (i.e. R(t,s)=EX_tX_s), is there an technique for determining whether or not the process X possesses the non-degeneracy described above?

    As an example, if R(t,s)=1 for all s and t then the answer is “degenerate”, while for R(s,t)=min{s,t} the answer is “non-degenerate”.
     
  18. Feb 20, 2010 #17
    Well, it seems too simple, so I must be missing something. The determinant of the v-c matrix (or any matrix) of the components of a random vector will be zero if it is degenerate and non-zero if it's non-degenerate. If you're dealing with multiple matrices, the product of the determinants will be zero if at least one determinant is zero.
     
    Last edited: Feb 21, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook