# Autocovariance of the process $Y_t = X_t - X_{t-1}$

Let $$\{X_t, t=1,2,...\}$$ be a stationary process. Obtain the autocovariance function of $$Y_t = X_t - X_{t-1}$$.\

Solution. Since $$X_t$$ is stationary, then $$E(X_t)$$ is constant and $$Cov(X_t, X_s)$$ depends on $$s$$ and $$t$$ only through $$|s-t|$$. Anyway, the autocovariance is

\begin{align} \gamma_Y(s,t) &= Cov(Y_t, Y_s) \\ &= Cov(X_t - X_{t-1}, X_s - X_{s-1}) \\ &= Cov(X_t, X_s) - Cov(X_t, X_{s-1}) - Cov(X_{t-1}, X_s) + Cov(X_{t-1}, X_{s-1}) \\ &= \gamma(t,s) - \gamma(t, s-1) - \gamma(t-1, s) + \gamma(t-1, s-1). \end{align}

I know the derivation is incomplete because I didn't use the hypothesis that $$X_t$$ is stationary. I contacted my professor and he said "since the $$X_t$$ is stationary, you can write the autocovariance function as a function of the lag, and combine the covariances that appear in the last equation". I'm not sure how to do that, though.

Thanks in advance for any help.

Because $$X_t$$ is stationary by hypothesis, then $$\gamma(t,s-1)=\gamma(t-1,s) = \gamma(1)$$. For the same reason, $$\gamma(t-1,s-1) = \gamma(t,s)$$, so we can simplify the last expression as
$$\gamma_Y(t,s) = 2\gamma(t,s) - 2\gamma(1)$$
which depends on $$t$$ and $$s$$ only through $$|s-t|$$, also proving that $$Y_t$$ is stationary.