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

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.

• It is also possible that the path of $X$ have disconinuity of second kind at $t=1/2$. Commented Jul 17, 2023 at 9:00
• @JGWang What is that? Commented Jul 17, 2023 at 13:47
• I didn't agree following judgment: because the continuity of $\sigma$ implies that $X_t$ will converge a.s. to its mean when approaching 1/2 from the left. Commented Jul 18, 2023 at 0:51
• @JGWang If you have a counterexample, that would be great. Commented Jul 23, 2023 at 16:09

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.

• Thanks, I will check that paper asap. Do you think if I impose that $(X_t)$ must also have left-limits, there won't be counterexamples anymore? Commented Jul 26, 2023 at 0:49
• 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. Commented Jul 26, 2023 at 1:19
• Randomly revisiting this question, I noticed that the autocovariance of your process $X$ is not continuous, hence does not satisfy requirement number 2. Commented Jan 17 at 20:20
• @W.Volante Thank you for your replication. I add "At first ..." at beginning and a remark at last. Commented Jan 18 at 8:07
• The process defined by $X_t= \sqrt{\sigma^2(t)} Y_t$ is not continuous in probability because the covariance function of $(Y_t)$ is not a continuous function. It is supposed to be continuous in probability everywhere on $[0,1]$. Commented Jan 18 at 19:26