Is weakly stationary equivalent to $I(0)$?

I'm currently reading some time series lecture notes. It says that:

Weakly stationary (or wide-sense stationary) processes are said to be $I(0)$ (integrated of order $0$).

Let's call the above definition A.

On the other hand the Wikipedia page on order of integration says

A time series is integrated of order $0$ if it admits a moving average representation with $\sum _{k=0} ^\infty |b_k|^2 < \infty$, where $b$ is the vector of moving average weights. This is a necessary but not sufficient condition for a processes to be stationary.

Let's call the above definition B.

My question is: I am unsure if if definitions A and B for $I(0)$ are equivalent. Can someone please point me to a proof or a counter example? Many thanks!

• Perhaps adding in exactly what "integrated of order $0$" means to you as well as an exact reference to where "wiki" gives an alternative definition of $I(0)$ will be helpful.... – Dilip Sarwate Jun 15 '16 at 12:39
• @DilipSarwate, yes you are certainly correct that my post was badly organized. It is now edited. Thank you. – pippy Jun 15 '16 at 13:05

The definition I see there (and in the book referenced therein) is $(1-L)^dx_t=z_t$, and a process is of order 0 if $(1-L)^0x_t$ is stationary (when used without prefix it means strict stationarity). But when $d=0$ we simply get $z_t=x_t$, so a process is of order 0 if and only if it is stationary.

From this definition, $I(0)\Rightarrow$ weakly stationary but not vice versa, i.e. $I(0)$ is a sufficient condition for stationarity, but weak stationarity is only a necessary condition for $I(0)$.

• To amplify what yoki wrote, I(0) means stationary. Weakly stationary means that the first two moments of a process (mean and variance) are time independent, where as stationary means that ALL moments are time invariant. – aginensky Jun 15 '16 at 13:46
• @aginensky, I(0) and stationary are not equivalent. See the beginning of chapter 3 (e.g. p. 35) in Johansen's book "Likelihood-based inference in cointegrated vector autoregressive models". – Richard Hardy Jul 10 '16 at 19:53
• @Hardy- embarassing mis- statement on my part. I merely meant to point out the second sentence on. – aginensky Jul 12 '16 at 3:38