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How to show that for any stationary time series its auto-covariance function is symmetric about the origin, that is $\gamma_{k}=\gamma_{-k}$ where, $\gamma_k=cov(z_t,z_{t-k})$

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Hi: Subtract $k$ from $z_t$ and add $k$ to $z_{t-k}$. Then you have $cov(z_{t-k,} z_{t})$ which by definition is $\gamma_{-k}$. But, by stationarity, this has to be equal to $cov(z_{t}, z_{t-k})= \gamma_{k}$ because the covariance is only a function of the lag difference.

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