# Two Gaussian processes with same variances and means but different covariances.

Let $$(X_t)$$ and $$(Y_t)$$ be two Gaussian stochastic processes. I am trying to find expressions for these processes such that $$\mathrm E[X_t]=\mathrm E[Y_t]$$, $$\mathrm {Var}(X_t)= \mathrm {Var}(Y_t)$$ for all $$t\geq 0$$ but $$\mathrm {Cov}(X_s,X_t) \neq \mathrm {Cov} (Y_s,Y_t)$$ for some $$s,t \in \mathbb R^+$$.

It is possible because mean and variance alone do not determine a Gaussian process in law, we need the covariance. That is the whole motivation behind trying to construct the above example.

In the discrete time case, this is simply saying that we want two different Gaussian couplings with the same marginals. I know how to do that.

For the continuous time case above, I can use the discrete time case for the finite dimensional law of $$(X_t)$$ and $$(Y_t)$$, which solves the question "in law". What I am struggling with is finding a actual expressions (omega by omega) $$X_t=..., Y_t=...$$, which would yield laws that satisfy the contraints in the question. Something built using Brownian motion for example ?

• Could we consider $X_t = W_t, Y_t = W_{t^2} / \sqrt{t}$ such that $E[X_t] = E[Y_t] = 0$, $Var[X_t] = Var[Y_t] = t$ and $Cov[X_s, X_t] = \min\{s, t\}, Cov[Y_s, Y_t] = \min\{s^2, t^2\} / \sqrt{st}$
– BGM
Nov 23, 2021 at 0:28
• Brownian motion is an overkill. Take e.g. $X_t$ being iid $\mathcal N(0,1)$, $Y_t \equiv X_0$. Nov 23, 2021 at 10:09
• @BGM You should add your answer so I can accept it, thank you! Nov 24, 2021 at 12:01

Let $$W_t$$ be standard Wiener process. Define $$X_t = W_t$$ and

$$Y_t = \begin{cases} 0 & \text{when} & t = 0 \\ \displaystyle \frac {W_{t^2}} {\sqrt{t}} & \text{when} & t > 0 \end{cases}$$

Then both $$X_t$$ and $$Y_t$$ are Gaussian process, with common mean $$E[X_t] = E[Y_t] = 0$$

and common variance

$$Var[X_t] = t = \frac {t^2} {(\sqrt{t})^2} = Var\left[\frac {W_{t^2}} {\sqrt{t}} \right] = Var[Y_t]$$

However the autocovariance is different:

$$Cov[X_s, X_t] = \min\{s, t\}$$

$$Cov[Y_s, Y_t] = \frac {\min\{s^2, t^2\}} {\sqrt{st}} = \begin{cases} \displaystyle \frac {s^2} {\sqrt{st}} = s \sqrt{\frac {s} {t}} \leq s & \text{when} & s \leq t \\ \displaystyle \frac {t^2} {\sqrt{st}} = t \sqrt{\frac {t} {s}} < t & \text{when} & s > t \end{cases}$$

So we have $$Cov[Y_s, Y_t] \leq Cov[X_s, X_t]$$ with equality holds only when $$s = t$$.