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.
Best 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 $$ \gamma_j=E(Y_{t+j}Y_{t})=E(Y_{t}Y_{t+j})=E(Y_{t}Y_{t-(-j)})=\gamma_{-j} $$