A Gaussian process that is continuous in probability but not almost surely

Question

Let $(X_t)_{t\in [0,1]}$ be a Gaussian process. $(X_t)$ is continuous in probability if and only if its mean and autocovariance function are continuous on $[0,1]$. Continuity in probability does not imply continuity almost surely. But I cannot find an example of an $(X_t)$ that is continuous in probability but not almost surely. What is a classical counterexample?

Answer 1

There's cheap counter example using Brownian motion.

Consider a standard Brownian motion $\{B_{t}\}$. Take $U$ a uniform random variable in $[0,1]$. Define $$\tilde{B}_{t}=\begin{cases} B_{t} & t\neq U\\ 0 & t=U\\ \end{cases} $$

It's easy to see the finite distributions are the same as the standard Brownian motion and in particular $cov(\tilde{B}_t,\tilde{B}_s)= s \land t$. However, this process is a.s. discontious, as $$P(\{\tilde{B}_t\} \text{is continous} | U=u)= P(B_u=0)=0$$

Answer 2

In the following we consider a separable stationary Gaussian process $X=\{X(t),t\in\mathbb{R}\}$, which takes real values. For simplicity we assume that $\mathsf{E}[X(t)]=0$ and denote the covariance function of $X$ by $B(t)=\mathsf{E}[X(t)X(0)]$.

Now suppouse that the spectral density of $X$ is \begin{equation*} f(\lambda)=\frac{1_{\{[3,\infty)\}}(\lambda)}{\lambda(\log\lambda)^2},\qquad \lambda\in\mathbb{R}_+, \end{equation*} and the covariance function of $X$ is \begin{equation*} B(t)=\mathsf{E}[X(t)X(0)]=\int_{\mathbb{R}_+}\cos(t\lambda)f(\lambda)\, \mathrm{d}\lambda, \quad t\in\mathbb{R}. \end{equation*} Hence $B(t)$ is continuous in $t$, and $X$ is continuous in quadratic mean and continuous in probability.

On the other hand, using Belyaev's result in paper "Local properties of the sample functions of stationary Gaussian processes, Theor. Prob. Appl., 5, 117-120.(1960) " (Theorem 3 If the stationary Gaussian process $X(t)$ has a spectral density $f(\lambda)$ such that $f(\lambda)\ge C/(\lambda(\log \lambda )^2) $ for some C>0, $\lambda_0\ge0$ and all $\lambda \ge \lambda_0 $, then almost all sample function of $X$ are unbounded on any interval of finite length), $X$ is not continuous almost surely.

Remark Since FZan already provided an interesting example. Here we emphasize the separable process.