Covariance of Gaussian Process

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  • #1
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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?
 

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  • #2
mathman
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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?

Naive(?) question. What is AC?
 
  • #3
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Sorry- AC is absolutely continuous.
 
  • #4
mathman
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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.
 
  • #5
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No no - by RV I mean random vector. I just mean that for any choice of sampling times, the resulting covariance matrix is invertible.
 
  • #6
mathman
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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.
 
  • #7
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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!
 
  • #8
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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!

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
 
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  • #9
mathman
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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!
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.
 
  • #10
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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.
 
  • #11
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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.

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]
 
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  • #12
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How is that related to anything?
 
  • #13
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How is that related to anything?

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.
 
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  • #14
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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”.
 
  • #15
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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”.

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?
 
  • #16
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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”.
 
  • #17
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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”.

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.
 
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