A counterexample on the continuity of Gaussian processes in a rather complex setting

Question

Let $(X_t)_{t\in[0,1]}$ be a Gaussian process with mean $0$ and variance $\sigma^2(t)$.

We know that:

• $(X_t)$ is right-continuous a.s.
• The autocovariance of the process is continuous on $[0,1]$, i.e., $(X_t)$ is continuous in probability on $[0,1]$.
• $X_{1/2}=0$ a.s. and $\sigma(t)\neq 0$ for all $t\neq 1/2$.
• $X_s$ is independent from $X_u$ for any $s<1/2$ and $u\geq1/2$.

Is it possible for $X_t$ to jump at $t=1/2$? That is, the left limit at $1/2$ is not equal to $0$ a.s.?

I would argue that it is not possible, that is, $X_t$ is almost surely continuous at $1/2$, because the continuity of $\sigma$ implies that $X_t$ will converge a.s. to its mean when approaching $1/2$ from the left. But in some papers, authors argue that is possible for $X_t$ to jump at $1/2$, so I guess there is a counterexample to my claim that I am not seeing.

Answer 1

At first, according your assumption, $\{X(t),t\in [0,1]\}$ is a Gaussian process and continuous in probability on $[0,1]$, then $\{X(t),t\in [0,1]\}$ is continuous in mean square on $[0,1]$ and $\sigma^2(t)=\mathsf{E}[X^2(t)]$ is also continuous on $[0,1]$.

In the following we consider a separable real Gaussian process $Y=\{Y(t),t\in[0,1]\}$ with $\mathsf{E}[Y(t)]=0$ and covariance function \begin{align*} B(s,t)&=\mathsf{E}[Y(s)Y(t)]\\ &=\begin{cases} \displaystyle{\int_{0}^{\infty}\frac{\cos (\lambda(t-s))}{2(e+\lambda) [\log(e+\lambda)]^{3/2}}}\mathrm{d}\lambda & 0\le s,t< \tfrac12,\\ 0, & 0\le s< \tfrac12, \tfrac12\le t\le 1, \\ 0, & 0\le t< \tfrac12, \tfrac12\le s\le 1, \\ \displaystyle{\int_{0}^{\infty}\frac{3\cos (\lambda(t-s))}{2(e+\lambda) [\log(e+\lambda)]^{5/2}}}\mathrm{d}\lambda & \tfrac12\le s,t\le 1 \end{cases} \end{align*} Then \begin{equation*} B(t,t)=1,\quad t\in [0,1]. \end{equation*} Using Belyaev's result in paper "Local properties of the sample functions of stationary Gaussian processes, Theor. Prob. Appl., 5, 117-120.(1960) ", almost all sample fonctions of $Y$ are continuous in $[\frac12,1]$ and unbounbed on any interval $I\subset [0,\frac12)$.

Now let \begin{gather*} X(t)=\sigma(t)Y(t),\qquad 0\le t\le 1, \tag{1}\\ \mathsf{E}[X(s)X(t)]=\sigma(s)\sigma(t)B(s,t),\qquad 0\le s,t \le 1. \end{gather*}

It is easy to verify that $\mathsf{E}[X(s)X(t)]$ is continuous on $[0,1]\times[0,1]$ and $\{X(t),t\in [0,1]\}$ is continuous in mean. Furthermore, $\{X(t),t\in [0,1]\}$ satisfy all conditions required above but the No.1: $\{X(t),t\in [0,\tfrac12)\}$ is not right-continuous a.s. and $\{X(t),t\in [\tfrac12,1]\}$ is right-continuous a.s.. Meanwhile, almost all sample functions of $X$ are unbounded on any interval $I\subset [0,\frac12)$, hence there is a discontinuous of second kind at $t=1/2$.

Sorry, I couldn't provide an example satisfying all conditions required above.

Remark: Regarding the discontinuity of stationary Gaussian process, please refer the paper: R. L. Dobrushin, Properties of sample functions of a stationary Gaussian process, Theory Prob. Applications, 5 1960, pp. 120-122.