0
$\begingroup$

Let's say say I have a standard AR(1) process, apart from the fact that is multiplied by the $ε_i$~$iid (0,1)$.

Would this affect the independence of the the AR(1) series? My intuition is no because this would just have a multiplicative effect and $ε_i$ is iid.

I'm very new to time series and getting increasingly confused the more I think about it!

$\endgroup$
  • $\begingroup$ To clarify, you mean each point in the process is multiplied by an independent Gaussian? And the "independence" you mean is the conditional independence of any two points in the process given a point between them? If so I believe the answer is yes, but it only requires independence of the multiplicative errors and not identical distribution. $\endgroup$ – Sheridan Grant Aug 17 '19 at 1:38
0
$\begingroup$

If you write $e_{t}y_{t}$ where y follows an AR process, since e is a iid white noise independent from y and its conditional mean is equal to the unconditional mean of 0 for every t, then you have:

  • a conditional mean of $E(ey_{t} | I_{t-1})= E(e | I_{t-1})E(y_{t} | I_{t-1}) =0$ as $E(e_{t} | I_{t-1})= E(e_{t} )=0$

  • a condition variance of $E((y_{t}e_{t})^{2} | I_{t-1})-E(y_{t}e_{t} | I_{t-1})^{2} = E((y_{t}e_{t})^{2} | I_{t-1})= E(y_{t}^{2} | I_{t-1})$

  • an unconditional mean of $E(ey_{t})= 0$

  • and an uncondition variance of $E((y_{t}e_{t})^{2})-E(y_{t}e_{t})^{2} = E(y_{t}^2)$

Notice however that if you fix a certain assumption on the distribution of y and e (which is not your case because you did not report a distribution for the shocks on y and assumed that e is just iid without specifying a pdf function), then notice that the new process y*e may have a different distribution compared to the ones of y and e. In this case for example, if you assume that the shocks in y are conditionally normally distributed and you make the assumption that e is a normal white noise, then consider that the product of two normally distributed variables follows a chi-squared distribution.

| cite | improve this answer | |
$\endgroup$

Your Answer

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