3
$\begingroup$

My lecture notes give the following definition:

A stochastic process $(X_t)_{t\in\mathbb{Z}}$ is called autoregressive of order $p$ if it satisfies: $$X_t=\phi_1X_{t-1}+...+\phi_{t-p}X_{t-p}+W_t.$$ Where $W_t$ is independent from $(X_s)_{s<t}$, and $\phi_p \neq 0$.

Then it is stated that if the process is weak-sense stationary it follows that the $W_t$'s are identically distributed. How does this follow? Or, if incorrect, could you give me a counterexample?

EDIT: Previously I hadn't made clear that I mean weak-sense stationary.

Improve this question
$\endgroup$
1
  • $\begingroup$ I think it is incorrect. Assume $W_t$ is AR(1) or MA(1) and you should be able to find out that $X_t$ is still stationary. But then tecnically $X_t$ would not be AR(p). Thus perhaps trying an ARCH or GARCH model for $W_t$ could be an alternative. $\endgroup$
    – Richard Hardy
    May 16, 2017 at 5:32

1 Answer 1

Reset to default
2
$\begingroup$

I am not a full expert in time series, but I think this is not correct. It is true, that the mean of $W_t$ is restricted to be the same for all $t$ as by imposing weak stationarity, the process must satisfy, that $$E[X_t]=E[X_{t+\tau}]\text{ } \forall \tau \in \mathbb{N} $$ which would probably imply that $E[W_t]=E[W_{t+\tau}]\text{ } \forall \tau \in \mathbb{N}$. But secondly we have by weak stationarity, that

$$E(X_t−E[X_t])(X_{k}−E[X_t])=E(X_{t+\tau}−E[X_{t}])(X_{k+\tau}−E[X_t]) =\gamma(t-k)$$ which does to covariance function $\gamma$ restrict to depend on the the shift in time $t-k$. From this, the expectation of $W_t$ must indeed stay the same, but the covariance is not restricted to be zero on the off-diagonal. Therefore, my opinon is that the statment is false in general.

If a violation of of the $\phi_p \neq 0$ condition is allowed, we have the trival case $$X_t = \mu + W_t $$ with weak stationarity the autocovariance of $W_t$ is allowed to depend on the shift of time $t-k$ and is hence not iid in the sence that the realizations of the process $\{W_t\}_{t=1}^{T}$ are independent.

Improve this answer
$\endgroup$
3
  • $\begingroup$ There was a requirement in the OP that $\phi_p\neq 0$, so your last example is ruled out. But it conveys a certain point, so I am just noting this aspect. $\endgroup$
    – Richard Hardy
    May 16, 2017 at 9:16
  • $\begingroup$ Ahh you are totally right. I will add a change to my comment. THX $\endgroup$
    – Michael L.
    May 16, 2017 at 9:44
  • $\begingroup$ Even in the case $\phi=0$ the independence of the $W_s$'s is guaranteed by the condition that $W_t$ is independent from $(X_s)_{s<t}$ in the definition. Same for the case $\phi\neq0$. So, in a counterexample the identical distribution part would be violated and not the independence part. $\endgroup$
    – Winkelried
    May 16, 2017 at 12:11

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.