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

Question

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!

Answer 1

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 $I(0)$.