# Showing the covariance and autocorrelation functions of a stationary time series are symmetric around 0

I need to show that the covariance and autocorrelation functions of a stationary time series are symmetric around zero. From my understanding, this entails $$\gamma(h) = \gamma(-h)$$ $$\rho(h) = \rho(-h)$$ I also know that $$\gamma(s, t) = cov(Y_s, Y_t) = E\{(Y_s-\mu_s)(Y_t-\mu_t)\}$$ $$\rho(s,t) =\frac{\gamma(s,t)}{\{\gamma(s,s)\gamma(t,t)\}^{1/2}}$$ and that for a stationary process specifically the mean $$\mu_t$$ is constant and $$\gamma(s,t)$$ depends only on $$t - s$$. However, I am unsure of where to go with said information aside from $$\gamma(t, t+h) = \gamma(0,h) = \gamma(h,0) = \gamma(0, -h)$$ but I'm pretty sure that doesn't really constitute as showing something.

I also need to consider a stationary bivariate time series $$\{(X_t,Y_t)\}_{t\in\mathbb{Z}}$$, defining the cross-covariance function as $$\gamma_{XY}(s,t) = cov(X_s, Y_t)$$ , and state whether this is also symmetric around zero, but I figure I'd have a better grasp on that if I understood the logic of the first part.

Update

Based on @ChristophHanck's comment, I've determined the covariance calculation as follows: $$\gamma(h)=E\{(Y_t-\mu_t)(Y_{t+h}-\mu_t)\}=E[Y_tY_{t+h}]+E[-Y_t\mu_t]+E[-Y_{t+h}\mu_t]+E[\mu_t^2]$$ $$=E[Y_tY_{t-(-h)}]+E[-Y_t\mu_t]+E[-Y_{t-(-h)}\mu_t]+E[\mu_t^2]=E\{(Y_t-\mu_t)(Y_{t-(-h)}-\mu_t)\} = \gamma(-h)$$ However, I am unsure of how to proceed with this knowledge into the correlation function unless I can assume that $$\gamma(s,s)$$ and $$\gamma(t,t)$$ are constants $$\sigma_s^2$$ and $$\sigma_t^2$$ respectively and just use this variance solution as the primary driving force in the matter.

• Hint in furtherance of Christoph Hanck’s answer: Stationarity means, among many other things, that all the random variables in the time series have the same mean. – Dilip Sarwate Sep 29 '18 at 11:40

## 1 Answer

Take a zero mean process for simplicity.

Then, $$\gamma_j=E(Y_tY_{t-j})$$. Under stationarity, the point in time at which we compute the expectation does not matter. Hence, we may add $$j$$ to each time index to get $$\gamma_j=E(Y_{t+j}Y_{t})=E(Y_{t}Y_{t+j})=E(Y_{t}Y_{t-(-j)})=\gamma_{-j}$$

• Thanks for the response! Using that logic with a mean I guess would lead to $$\gamma(h) = E\{(Y_t-\mu_t)(Y_{t+h}-\mu_t)\} = E(Y_t Y_{t+h}) + E(-Y_t \mu_t) + E(-Y_{t+h} \mu_t) + E(\mu_{t}^2) = E(Y_t Y_{t-(-h)}) + E(-Y_t \mu_t) + E(-Y_{t-(-h)} \mu_t) + E(\mu_{t}^2) = E\{(Y_t-\mu_t)(Y_{t-(-h)}-\mu_t)\} = \gamma(-h)$$ – strwars Sep 28 '18 at 21:41
• I'm still a bit lost on the correlation and bivariate to be honest though. Do I just assume the standard deviations of $Y_s$ and $Y_t$ are constants as $\sigma_s$ and $\sigma_t$ respectively and use the fact that $\gamma(h) = \gamma(-h)$ to show $\rho(h) = \rho(-h)$ ? I greatly appreciate your assistance on this question @ChristophHanck ! – strwars Sep 29 '18 at 7:26