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Let $X(t)$ be a stationary Gaussian process with mean $\mu = 0$, variance $\sigma^2 = 1$ and Gaussian correlation function:

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

Where $\tau = t_1 - t_2$. The parameter $\theta$ is known as the correlation length and satisfies:

$$\theta = 2\int_0^{\infty} \rho(\tau)\, d\tau$$

As $\theta \rightarrow \infty$ the correlation between points of the process becomes one and the realisations of $X(t)$ are homogeneous. The variance of any single realisation of $X(t)$ is zero. However I believe that the variance of this process $\sigma^2$ is still equal to 1 (considering multiple realisations).

However consider the case where $\theta \rightarrow 0$. The points of the process become completely uncorrelated and I have read that $X(t)$ becomes a white noise Gaussian process. I have also read that the variance of a continuous white noise process is infinite.

If $\theta \rightarrow 0$ does the correlation function $\rho(\tau)$ become the Dirac delta function? Does the variance of the process become infinite?

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  • $\begingroup$ The variance doesn't blow up as $\theta \to 0$ as you can check by just plugging in $\tau=0$. This is different from "standard white noise" (i.e. the derivative of the Wiener process) which indeed has infinite variance. $\endgroup$
    – Ian
    Aug 17 '16 at 15:15
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If $\theta \rightarrow 0$ does the correlation function $\rho(\tau)$ become the Dirac delta function? Does the variance of the process become infinite?

No, to both questions. For $\tau>0$, $\lim_{\theta\to 0}\rho_\theta(\tau) = 0$, but for $\tau=0$, $\lim_{\theta\to 0}\rho_\theta(0) =1$; hence, the correlation function becomes the Kronecker delta function:

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

For a Gaussian process, this implies that the process is a collection of i.i.d. Gaussian random variables, which is sometimes called "white noise" (not to be confused with so-called "delta-correlated" white noise, which has a correlation function proportional to the Dirac delta function).

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  • $\begingroup$ Thank you for your comment. I did not realise there was a distinction between white noise and delta-correlated white noise. $\endgroup$
    – egg
    Aug 18 '16 at 8:16
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The limit when $\theta\rightarrow \infty $ is zero, so the correlation function becomes the zero function, impying a total incorrelation between them. As there is no correlation between the time, then the process can be seen as a white Gaussian process.

The variance does not become infinite, but remains as $\sigma^2$.

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