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

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.

• The well known white noise process seems to satisfy most - if not all - your requirements. Commented Apr 1, 2022 at 17:50
• @KurtG. to my knowledge, white noise is not a stochastic process but a stochastic tempered distribution. It would be the "derivative" of Brownian motion if it existed. I am looking for a sample continuous Gaussian process. Commented Apr 1, 2022 at 19:46

Let $$(B_t)_{0\le t\le 1}$$ be a standard Brownian motion. Fix $$0 and take $$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.