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.

  • $\begingroup$ 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. $\endgroup$ May 16, 2017 at 5:32

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$ 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

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