Why a Stationary Time Series Has Symmetric Autocovariance

Summary

The document asks why the autocovariance function of a stationary time series is symmetric in its lag: the covariance at lag k equals the covariance at lag −k. The explanation uses two basic properties of covariance and stationarity rather than a model-specific result.

Reversing the order of the two variables in a covariance does not change its value. For a stationary series, covariance depends only on the separation between observations, not on their absolute time indices. Thus, shifting the indices in the covariance at lag k and reversing the variables expresses it as the covariance at the opposite lag, while stationarity leaves the value unchanged. This establishes the stated symmetry.

The argument is concise and assumes the relevant covariances exist. In practice, the standard autocovariance function is defined for a weakly stationary series with finite second moments; strict stationarity alone does not guarantee finite covariance. The note establishes a property of the autocovariance, not a claim that observations at opposite lags are independent or that the series itself is time-reversible.

Key ideas

• Covariance is unchanged when its two arguments are reversed.
• For a stationary series, covariance depends on the lag between observations rather than their calendar time.
• Together, these properties imply that autocovariance has the same value at positive and negative lags.
• The usual result presumes that the series has finite second moments so its autocovariance exists.
• Symmetry of autocovariance does not imply independence or time reversibility.

Tags

# Auto-covariance function of station time series


# Auto-covariance function of station time series












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, $\gamma_k=cov(z_t,z_{t-k})$

## Answer by mark leeds (score 3)

https://quant.stackexchange.com/a/42958

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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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.