3
$\begingroup$

Given AR(1): $$X_t - \mu = \phi(X_{t-1}-\mu) + \epsilon_t$$

where $$ \mu = 0.85 \\ \phi=0.59 $$

and $$ W_t = X_t - X_{t-1} $$

Compute $$ Corr(W_t,W_{t-1})=-0.205 \\ Cov(W_t,W_{t-4})=-0.43 \\ Corr(W_t,W_{t-4})=-0.04 $$

Could anybody help please?

$\endgroup$
  • $\begingroup$ Add the self-study tag. $\endgroup$ – Michael Chernick Jun 6 at 23:52
  • $\begingroup$ Rather than giving your own answer as a hasty edit at the end of the question, it is perfectly acceptable for you to remove this part and then undelete your answer containing same. It is perfectly acceptable (even encouraged) for a questioner to give their own answer to a question. $\endgroup$ – Reinstate Monica Jun 7 at 0:40
  • $\begingroup$ If it's already answered elsewhere (as the OP suggests), that part of the question would be a duplicate, in which case it would be better not posted as an answer. I think the question could probably be edited down to simply asking about covariance and (perhaps) linking to the other post as a related question. $\endgroup$ – Glen_b Jun 7 at 1:43
  • $\begingroup$ Thank you for the advice, i will edit right now $\endgroup$ – Nico Mark Jun 7 at 7:56
0
$\begingroup$

Hint: use bilinearity $$ \text{Cov}\left(W_{t}, W_{t-1} \right) = \text{Cov}\left(X_t - X_{t-1}, X_{t-1} - X_{t-2} \right) = 2\gamma_X(1) - \gamma_X(2) - \gamma_X(0) $$ where $\gamma_X(h) = \text{Cov}(X_{t+h},X_t)$ is the autocovariance function of your original process.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.