1
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .