Let ${X_t}$ be a stationary time seres, and define ${Y_t}$ as follows;
${Y_t}$ = $X_t$ for t odd
${Y_t}$ = $X_t$+3 for t even
a) Show that Cov(${Y_t}$,${Y_{t-k}}$) is free of t for all lags k
I'm really unsure how to get the expected value here as E[$X_t$,$X_t+3$...] will just repeat forever. Any help would be great.
The value of $\operatorname{Cov}(Y_t, Y_{t-k}) $ would always be one of the 4 following cases (depending on the parity of $t$ an d $k$): $$\operatorname{Cov}(Y_t, Y_{t-k}) = \operatorname{Cov}(X_t, X_{t-k}) $$ $$\operatorname{Cov}(Y_t, Y_{t-k}) = \operatorname{Cov}(X_t + 3, X_{t-k}) $$ $$\operatorname{Cov}(Y_t, Y_{t-k}) = \operatorname{Cov}(X_t, X_{t-k} + 3) $$ $$\operatorname{Cov}(Y_t, Y_{t-k}) = \operatorname{Cov}(X_t + 3, X_{t-k} + 3) $$ Since $X$ is stationary, $\mathbb{E}(X_t)$ and $\operatorname{Cov}(X_t, X_{t-k}) $ do not depend on $t$ thus in every case $\operatorname{Cov}(Y_t, Y_{t-k}) $ takes a value that does not depend on $t$.