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For two stochastic process $x_t$, $y_t$, how can the cross-covariance function between $x_t$ and $y_t$ at lag k different from that at lag -k? For some reason, I can't scrape my head around this. Please explain.

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Consider the simple example $y_t=x_{t-1}$. Then element 21 of the auto cross-covariance function at lag $k=1$ is $$ \gamma_{21,1}=\operatorname{Cov}(x_{t-1},y_t)=\operatorname{Cov}(x_{t-t},x_{t-1})=\gamma_{11,0} $$ whereas at lag $k=-1$ it is $$ \gamma_{21,-1}=\operatorname{Cov}(x_{t+1},y_t)=\operatorname{Cov}(x_{t+1},x_{t-1})=\gamma_{11,2}. $$ But note that \begin{align} \gamma_{12,k} &=\operatorname{Cov}(x_{t-k},y_t) \\&=\operatorname{Cov}(y_t,x_{t-k}) \\&=\operatorname{Cov}(y_{t+k},x_t)=\gamma_{21,-k} \end{align} so $\Gamma_k=\Gamma_{-k}^T$.

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  • $\begingroup$ So in general, is there any scenario when the two autocovariance equal? $\endgroup$ – denizen of the north Jul 1 '17 at 14:20
  • $\begingroup$ One example would be any vector ARMA process if all matrices are symmetric and with shared eigenvectors. But that is essentially just a rotation of independent ARMA processes and so perhaps not so interesting. $\endgroup$ – Jarle Tufto Jul 1 '17 at 19:35

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