Can a sample continuous Gaussian process $(X_t)$ have $Cov(X_t,X_s)=0$ while $Var(X_u)\neq 0$ for all $u \in [s,t]$?

Question

Let $(X_t)$ be a sample continuous Gaussian process on $[0,1]$. We suppose that $(X_t)$ is non-trivial, i.e it is not constant on $[0,1]$.

If $Cov(X_s,X_t)=0$ for some $s,t$ in $[0,1]$, does it imply that $Var(X_u)=0$ for some $u$ in $[s,t]$ ?

Here is where this came to my mind:

Suppose that $(X_t)$ is furthermore Markovian, then $Cov(X_s,X_t)=0$ can happen if at some point $u$ between $s$ and $t$, we have $Var(X_u)=0$.

Now suppose that $(X_t)$ is not Markovian then $Cov(X_s,X_t)=0$ can happen if either $Var(X_s)=0$ or $Var(X_t)=0$.

I was wondering if these were the only cases where $Cov(X_s,X_t)=0$ can happen for a sample continuous Gaussian process.

Answer 1

Let $(B_t)_{0\le t\le 1}$ be a standard Brownian motion. Fix $0<s<1/2<t<1$ and take $c=2s$. Define $X_u=B_u-cB_{1/2}$ for $s\le u\le t$.

Then the covariance of $X_s$ and $X_t$ is $E(X_sX_t)= s-c/2-cs+(1/2)=s(1-c)-c(1-c)/2=0$. But for $u\in [s,t]$, the variance of $X_u$ is $$ E(X_u^2)= \cases{2u(1/2-u)+2(s-u)^2,&$s\le u\le 1/2$,\cr u^2-1/2+2(s-1/2)^2,&$1/2\le u\le t$.\cr}, $$ which doesn't vanish provided $s$ is small enough, say $0<s<0.1$.