Autocorrelation function of Itô process

Question

I'm working with a time independent (vector) Itô SDE such as:

$$ dX = a(X) dt + b(X) dW. $$

I've looked (numerically) at several examples and it seems that the autocovariance function $r_{xx}(\Delta t)$ is continuous and goes to zero exponentially fast at large $\Delta t$.

Is that true for sufficiently nice $a$, $b$?

Answer 1

$\newcommand\Cov{\operatorname{Cov}}\newcommand\De\Delta$No. Even in the nicest case when $a=0$ and $b=1$, so that $X=W$, we have $$\Cov(X_t,X_{t+\De t})=t\underset{\De t\to\infty}{\not\longrightarrow}0 \tag{1}\label{1}$$ for any real $t\ge0$.

On the other hand, for any smooth enough $a$ and $b$ and any real $t\ge0$ and $\De_t\ge0$, the random variable $X_{t+\De t}-X_t$ is independent of $X_t$ (and even of the entire pre-history $(X_s\colon s\in[0,t])$), so that $$\Cov(X_t,X_{t+\De t}-X_t)=0 \tag{2}\label{2}$$ and, moreover, $$\Cov(X_s-X_r,X_{t+\De t}-X_t)=0 \tag{3}\label{3}$$ for all real $r,s$ in $[0,t]$. Perhaps, this is the reason for your numerical observations. (The equalities in \eqref{2} and \eqref{3} also follow from the equality in \eqref{1}.)

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N.B.: The correct way to write the SDE is this: $dX_t=a(X_t)\,dt+b(X_t)\,dW_t$.