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Let $(X_t, Y_t)$ be a stationary 2D Gaussian process, therefore $\mathbb{E}\left(X_t\right) = \mathbb{E}(Y_t) = 0$.

I am looking for an explicit example of a valid auto-covariance matrix, i.e: $$ R(h) = \begin{pmatrix} \mathbb{E}\left(X_t X_{t+h}\right) & \mathbb{E}\left(X_t Y_{t+h}\right) \\ \mathbb{E}\left(Y_t X_{t+h}\right) & \mathbb{E}\left(Y_t Y_{t+h}\right) \end{pmatrix} $$ such that $R(h)$ is not symmetric for $h\not=0$. Thank you.

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I fail to see why this was down-voted... If I am being signaled about a flaw in the question, please speak up. I'll try to improve it. – Sasha Dec 21 '12 at 22:07
Somebody downvoted this question? Wow. – Did Dec 21 '12 at 22:17
up vote 2 down vote accepted

Consider your favorite stationary Gaussian process $(X_t)_t$ and define $(Y_t)_t$ by $Y_t=X_{t+\ell}$, for some nonzero $\ell$. Then $\mathbb E(X_tY_{t+h})=\mathbb E(X_0X_{\ell+h})$ is the covariance at distance $|\ell+h|$, while $\mathbb E(Y_tX_{t+h})=\mathbb E(X_0X_{h-\ell})$ is the covariance at distance $|\ell-h|$.

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Thanks a lot! Very useful for me. – Sasha Dec 21 '12 at 22:03
This construction leads to a Gaussian process with degenerate covariance of some distinct-positive-time slices of the process, that is covariance of $(X_{t_1}, Y_{t_1}, \ldots, X_{t_n}, Y_{t_n} )$. I guess I need to revise the question. – Sasha Dec 29 '12 at 6:07
I actually found that a vector valued ARMA process fits the bill. – Sasha Dec 29 '12 at 6:24

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