Markovian Gaussian stationary process with continuous paths

Question

Could you, please, help me figure out the following problem.

We call a stationary Gaussian process $\xi_t$ (with continuous paths) an Ornstein-Uhlenbeck process if its correlation function $\mathbb{E}\xi_{s+t}\bar \xi_s$ is $$ \mathbb{E}\xi_{s+t}\bar \xi_s=R(t)=R(0)e^{-\alpha|t|}, \alpha\ge 0 $$

The question is that Ornstein-Uhlenbeck process is the unique Markovian stationary Gaussian process.

By Markovian I mean that $$ \mathbb{E}\left(f(\xi_t|\xi_{t_n},\ldots, \xi_{t_1}\right)=\mathbb{E}\left(f(\xi_t|\xi_{t_n}\right). $$ for $t_1\le t_2\le\ldots\le t_n\le t$ and any Borel-measurable funciton $f$.

This problem amazed me at first glance.

I can't figure out how to deduce 'exponential decay' of correlation function $R(t)$ from Markovian property.

Answer 1

1. It turns out that a centered Gaussian process is Markov if and only if its covariance function $\Gamma: \mathbb{R}\times\mathbb{R} \to \mathbb{R}$ satisfies the equality:

$$\Gamma(s,u)\Gamma(t,t)=\Gamma(s,t)\Gamma(t,u)\ \ \ \ (1)$$

for all $s<t<u$.

It turns out to be difficult to find a clear proof of the above fact; one that I found but did not understand well was: Example 4.5, p.119 of Random Processes for Engineers by Bruce Hajek, or Example 4.7 on pp. 120-121.

2. Moreover, unlike most stochastic processes, Gaussian processes are uniquely and completely determined by their covariance functions (I believe up to the mean function).

I.e., given any two stochastic process, if it is known that they are Gaussian, and further that their covariance functions are exactly the same, then they must be the exact same process (again, I believe up to mean function).

Hence, if you can show that a Gaussian process is stationary and satisfies equation (1) if and only if it has covariance function equal to that of the Ornstein-Uhlenbeck process, then you are done.

3. With regards to:

"how to deduce 'exponential decay' of correlation function R(t)R(t) from Markovian property"

I believe this follows from equation (1) combined with the multiplicative form of the Cauchy functional equation: https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation.

See also here: Overview of basic facts about Cauchy functional equation

Basically, any measurable function such that $f(x+y)=f(x)+f(y)$ (additivity) must be linear, and any measurable function $g$ such that $g(x+y)=g(x)g(y)$ must be an exponential.

Answer 2

Thank you so much for the explanation. Your first tip about the form of correlation function $R(s,t)$ for Gaussian Markov processes was really helpful.

1. Assume that we know that Gaussian process $\xi_t$ with the correlation function $R(s,t)$ is Markovian, then it implies that $$R(s,u)R(t,t)=R(s,t)R(t,u)$$ as you told. I have no enough knowledge to prove it rigorously. I'll work on it and if I succeed, I'll surely post it here.

2. As $R(s,t)$ is stationary, it depends only on $s-t$, that is $R(s,t)=R(s-t)$. Then $R(x+y)R(0)=R(x)R(y)$. Denoting $f(x)=\frac{R(x)}{R(0)}$, we come to $$ f(x+y)=f(x)f(y).\quad (2) $$

3. The rest is not so difficult. We only need to prove that if $f(x)$ is continuous and $f\not\equiv 0$ , then the functional equation (2) implies $$ f(x)=e^{-\alpha|x|},\alpha>0 $$

3.1 For $x=m\in \mathbb{Z}$: $f(m)=f(1+\ldots +1)=f(1)^m.$

3.2 For $x=\frac{p}{q}\in \mathbb{Q}$: $f(\frac{p}{q})=f(\frac{1}{q}+\ldots +\frac 1q)=f(\frac 1q)^p.$

But $f(1)=f(\frac 1q+\ldots+\frac 1q)=f(\frac 1q)^q$, therefore $f(\frac 1q)=f(1)^{\frac 1q}$. So $f(\frac{p}{q})=f(1)^{\frac pq}$.

3.2 As $\mathbb{Q}$ is dense in $\mathbb{R}$ by continuity of function $f$ we conclude that $f(x)=f(1)^x$ for all $x>0$, or equivalently $f(x)=e^{x\ln f(1)}$.

After some tedious manipulation, we can extend it for $x<0$ as well and obtain $$ f(x)=e^{-\alpha |x|} $$ for $\alpha >0$, so $R(x)=R(0)e^{-\alpha |x|}$.

Thank you so much, guys, for help!