I have given the AR(1) process as followed: $$ y_t = \phi y_{t-1} + e_t $$ where $$ e_t \sim WN(0,\sigma^2)$$ I need to prove that $$ cov(y_t,y_{t-j}) = \phi^j \sigma^2[1 + \sigma^2 + (\sigma^2)^2 + ...+ (\sigma^2)^{t-1-j}]$$ This is a nonstationary process.
I don't really know where to start at except for the expansion of the AR(1) process. Any help would be appreciated. Thank you guys.

  • $\begingroup$ The last equation looks suspicious. There is $j$ on the right hand side but not on the left hand side. Also, why is the process nonstationary? Is $\phi \geqslant 1$? $\endgroup$ – Richard Hardy Sep 6 '15 at 10:14
  • $\begingroup$ that was a typo the j is supposed to be on the otherside too (fixed) and there is not restriction on $$phi < 1 $$ $\endgroup$ – Tan Nguyen Sep 6 '15 at 11:36
  • $\begingroup$ Have you checked related questions on this site? What about this one or this one? $\endgroup$ – Richard Hardy Sep 6 '15 at 12:16
  • $\begingroup$ I cannot see how this is correct. Any innocent-looking side assumption? $\endgroup$ – Alecos Papadopoulos Sep 8 '15 at 1:17

The expression cannot be correct. The correct one is

$${\rm Cov}(y_t,y_{t-j}) = \phi^j \sigma^2[1 + \phi^2 + (\phi^2)^2 + ...+ (\phi^2)^{t-1-j}]$$

How can we prove it (show it actually, there is no "proof" involved)? Well,

$$y_t = \phi y_{t-1} + e_t \\ \implies y_t = \sum_{k=1}^t\phi^{t-k}e_k \\ \implies y_{t-j} = \sum_{k=1}^{t-j}\phi^{t-j-k}e_k \\$$


$${\rm Cov}(y_t,y_{t-j}) = E\left[\left(\sum_{k=1}^t\phi^{t-k}e_k\right)\left(\sum_{k=1}^{t-j}\phi^{t-j-k}e_k\right)\right]$$

Hmmm... the first sum contains errors that lie "in the future" of the second sum... some breaking up of the sums will do it, don't you think? Be careful with the indices, it's all that matters here.

Signal when you solved it.

| cite | improve this answer | |

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.