# Stationary Gaussian process whose correlation parameter approaches zero.

Consider a mean-zero ($\mu = 0$), unit-variance ($\sigma^2$) Gaussian random process $X(t)$. This process is strictly stationary (the joint-probability distribution does not vary with $t$). The covariance function is isotropic (covariance structure does not depend on $t$) and is given by:

$$C(\tau) = \exp\left(-\frac{\tau^2}{\theta}\right)$$

Where $\tau = t_1 - t_2$.

I am interested in the case where $\theta \rightarrow 0$. In my previous post it was explained that the covariance function $C(\tau)$ becomes the Kronecker delta function:

$$C(\tau) = \delta_\tau = \begin{cases} 0 &\text{if } \tau \neq 0, \\ 1 &\text{if } \tau = 0. \end{cases}$$

However according to Stochastic Filtering Theory by Kallianpur (1980, p. 10) a process with this covariance function is not measurable. AFAIK this means that the integral of such a process is not defined. Consider the following stochastic integral:

$$I = \int_0^LX(t)\, dt$$

Is it correct to say that the integral $I$ is only defined when $\theta > 0$ since $X(t)$ is only measurable if $\theta > 0$?

This post suggests that it may be better to take the correlation function as the Dirac delta function since Dirac delta correlated white noise is measurable and consequently the integral $I$ is defined. However a Gaussian process that is mean-zero with the Dirac delta function is known as Gaussian white noise which is said to have infinite variance which would imply:

$$\text{Var}[X(t)] = \sigma^2 = \begin{cases} 1 &\text{if } \theta \neq 0, \\ \infty &\text{if } \theta = 0. \end{cases}$$

This does not seem correct since the variance of $X(t)$ should be independent of $\theta$. How is the process $X(t)$ defined as $\theta \rightarrow 0$?

Note: sorry if this counts as a duplicate question but I have really been struggling with this topic and could use some guidance specific to my exact situation rather than the general case.

• Even if you ignore the obvious question of what exactly a Gaussian random variable with infinite variance is, standard white noise still does not have measurable paths. This problem always occurs when the values of the process at any two distinct times are uncorrelated; intuitively measurable functions have a tiny amount of regularity, which makes them "usually" not change by too much on short time intervals, in some very weak sense). – Ian Aug 25 '16 at 15:31
• However, the time integral of standard white noise in the sense of distribution theory, i.e. the Wiener process or standard Brownian motion, does have measurable paths. This integral is not a Lebesgue integral; essentially it just has the right power spectrum to be the integral of standard white noise (where here we define standard white noise in terms of its power spectrum being uniform). My suggestion would be that you study this rather than white noise. Then after you have understood SDEs, you can ask about the specifics of approximating weakly correlated noise with white noise. – Ian Aug 25 '16 at 15:32
• Part of the rationale for my suggestion is that approximating weakly correlated noise with white noise is actually quite subtle. For example, in an equation like Newton's laws with weakly correlated random forces exerted on a particle of small mass, the rate at which the correlation time goes to zero vs. the rate at which the mass goes to zero has a drastic impact on the asymptotic behavior. – Ian Aug 25 '16 at 15:35
• Thanks for your comment Ian. It is probably better to look at the spectral density function of $X(t)$ rather than the covariance function. I will look into this and update my question. – Fundamental Engineer Aug 26 '16 at 7:57
• I am not sure this makes sense but here are my 5 cents. If you look instead on $\int_0^L \frac{d}{dt}Y(t) dt$ where you can carry the integral out.., while you choose the process $Y(t)$ so that its derivative gives you your process $X(t)$. Can this be done? I am not sure, just suggesting... the link between the covariance of the derivative and the covariance of the original process has a very nice and known relationship. See page 52 formula 2.4.8 here. – them Aug 26 '16 at 9:50