From reading Showing that there do not exist uncountably many independent, non-constant random variables on $ ([0,1],\mathcal{B},\lambda) $. and the answer to What is meant by a continuous-time white noise process?, I believe we cannot construct a stochastic process with covariance given by the dirac delta $\delta$ without resorting to generalized functions.

My understanding is that this implies that we cannot construct a stochastic process $X(t)$ where for any collection of unique $t_1,t_2,...$ the random variables $X(t_1), X(t_2), ...$ are mutually orthogonal.

Does this imply that is it is impossible to construct a Gaussian Process that has a diagonal covariance matrix? If not, why?


1 Answer 1


This is not true. Limiting to $\Big([0,1], \mathcal{B}, \lambda \Big)$ is crucial in cited problem.

On the other hand, from Kolmogorov existence theorem it is straight forward to check that:

If $\ (\mu_{t})_{t \in T} \ $ is any family of distributions on $\ \mathbb{R}, \ $ then there exists an independent family $\ (X_{t})_{t\in T} \ $ of such random variables, that $\ X_{t} \ $ is distributed according to $\ \mu_{t} \ $ for all $\ t\in T. \ $

In the above statement $\ T \ $ is an interval contained in $\ \mathbb{R}_{+}. \ $

By taking $\ \mu_{t}=\mathcal{N}(0,1) \ $ for all $ \ t\in \mathbb{R}_{+} \ $ we see that:

  • $(X_{t})_{t\in \mathbb{R}_{+}}$ is gaussian.
  • $\mathrm{Cov}(X_{t},X_{s})=0 \ $ for all $\ s\not= t.$

This approach has nothing to do with generalized functions.

  • $\begingroup$ Thanks for the answer. However, I am still a bit confused. How exactly does this allow us to get around the problem? Is the idea that the uncountable family of independent random variables that the Kolmogorov extension theorem guarantees exist are defined over some probability space $(\Omega, \Sigma, \mu)$ such that $L^2(\Omega, \mu)$ is not separable? $\endgroup$
    – gigalord
    Jan 17, 2020 at 13:02
  • $\begingroup$ I think that You are right. $L^{2}([a,b], \lambda)$ is separable because of Lebesgue measure, but not necesarly for other possible measures and $\sigma$-fields. $\endgroup$
    – user617199
    Jan 17, 2020 at 18:52
  • $\begingroup$ I dug a bit deeper into this, and exercise 1.3 of these course notes galton.uchicago.edu/~lalley/Courses/386/GaussianProcesses.pdf seems to suggest that there is no probability space where we can define such a gaussian process $\endgroup$
    – gigalord
    Jan 20, 2020 at 13:34
  • $\begingroup$ Is the challenge in this case the continuity condition? So Kolmogorov guarantees the existence of the random variable-valued function from $\mathbb{R}$, but this function is not guaranteed to be continuous? $\endgroup$
    – gigalord
    Jan 20, 2020 at 14:24
  • $\begingroup$ Ofcourse Kolmogorov theorem does not guarantee continuity of paths. Even in case of Brownian motion we need to prove that there exists modification with continues paths in some other, additional way. But You haven't asked about this property in Your question. $\endgroup$
    – user617199
    Jan 20, 2020 at 15:58

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