Suppose I have a stationary process $\phi(t)$ with a known autocorrelation function

$$ A(\tau) \equiv \langle \phi(0) \phi(\tau) \rangle$$

and suppose I also know that $\phi(t)$ is Gaussian distributed.

If I have a particular realization of the process in which $\phi(0)=\phi_0$, what is the conditional distribution of $\phi(t)$ for later times in that realization?


1 Answer 1


"I know that $\phi(t)$ is Gaussian distributed" is not the same as saying "$\{\phi(t)\}$ is a Gaussian process" but I will assume that the latter is meant. With this assumption, and the additional assumption that the process is wide-sense-stationary (the autocorrelation function is listed as having only one argument), the process is strictly stationary and thus $\phi(t)$ and $\phi(0)$ are jointly Gaussian random variables. They have the same mean $\mu$ and variance $\sigma^2$, and their correlation coefficient is $\rho$ where $$\mu = \sqrt{\lim_{t\to \infty} A(t)}, \qquad \sigma^2 = A(0)-\mu^2, \qquad \rho(t) = \frac{A(t)-\mu^2}{\sigma^2}.$$ Consequently the conditional distribution of $\phi(t)$ given that $\phi(0)$ has taken on value $\phi_0$ is Gaussian with mean $\mu + \rho(t)(\phi_0 - \mu) = \rho(t) \phi_0 + (1-\rho(t))\mu$ and variance $\sigma^2(1-\rho(t)^2)$. Note that as $t\to\infty$, the conditional distribution of $\phi(t)$ approaches the unconditional distribution of $\phi(t)$: the distant past affects the present state less and less.

For more information about Gaussian processes, see, for example, this answer to a different question.

  • $\begingroup$ Could either explain why $\rho = (A(t) - \mu^2) / \sigma^2$ or provide a reference? $\endgroup$
    – DanielSank
    Commented Jun 14, 2015 at 18:48
  • 1
    $\begingroup$ $$A(t) = E[\phi(0)\phi(t)] = \operatorname{cov}(\phi(0), \phi(t)) + E[\phi(0)]E[\phi(t)] = \operatorname{cov}(\phi(0), \phi(t))+\mu^2.$$ $$\rho= \frac{\operatorname{cov}(\phi(0), \phi(t))}{\sqrt{\operatorname{var}(\phi(0))\operatorname{var}(\phi(t))}}$$ $\endgroup$ Commented Jun 15, 2015 at 2:09
  • $\begingroup$ This seems related to Doob's theorem. $\endgroup$
    – DanielSank
    Commented Jun 17, 2015 at 1:08

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