Covariance of Gaussian Process

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Discussion Overview

The discussion revolves around the non-degeneracy of centered, stationary Gaussian processes, specifically focusing on how to determine if the covariance matrix of sampled random vectors is invertible. Participants explore the implications of different covariance functions and their relationship to absolute continuity and stochastic analysis.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants inquire about determining non-degeneracy based on the covariance function of a Gaussian process, questioning how to ensure that the resulting random vector is absolutely continuous.
  • One participant suggests that sample functions of a stationary Gaussian process are generally not absolutely continuous, using Brownian motion as an example.
  • Another participant clarifies that by "random vector," they mean that the covariance matrix must be invertible for any choice of sampling times.
  • It is noted that the covariance matrix is positive semi-definite, which raises questions about its invertibility.
  • Some participants provide counterexamples, such as a constant stochastic process, to illustrate that non-degeneracy is not a general property and depends on the specific process.
  • There is discussion about the relationship between the variance-covariance matrix and the sampling of random vectors, with some arguing that singular matrices can arise from standardized Gaussian stationary distributions.
  • One participant emphasizes the need for clarity regarding the connection between their question and the responses given, reiterating the focus on basic non-degeneracy and the invertibility of the covariance matrix.
  • Examples of covariance functions are provided, with one participant stating that certain functions indicate degeneracy while others do not.

Areas of Agreement / Disagreement

Participants express differing views on the conditions under which the covariance matrix is invertible, with no consensus reached on the general properties of non-degeneracy in Gaussian processes. Multiple competing perspectives on the implications of specific covariance functions are presented.

Contextual Notes

Participants mention that the invertibility of the covariance matrix may depend on the specific covariance function used and the sampling times chosen. There are unresolved questions regarding the implications of absolute continuity and the conditions under which singular matrices may arise.

Palindrom
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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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Palindrom said:
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?
 
Sorry- AC is absolutely continuous.
 
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.
 
No no - by RV I mean random vector. I just mean that for any choice of sampling times, the resulting covariance matrix is invertible.
 
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.
 
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!
 
Palindrom said:
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 X_{i},X_{j} 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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Palindrom said:
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
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
Palindrom said:
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
 
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  • #12
How is that related to anything?
 
  • #13
Palindrom said:
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
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
Palindrom said:
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
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
Palindrom said:
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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