# Variance condition for (weak) stationarity

I read that one condition for weak stationarity is that a series needs to have a constant mean.

My (rather short) question: Does weak stationarity require the variance to be constant as well or is having a finite variance sufficient (or does one imply the other)?

This distinction not seem to be treated in a uniform manner in the literature I encountered. (However, maybe it is me who overlooks something obvious)

Yes, weak stationarity requires both constant variance and constant mean (over time). To quote from wikipedia: A wide-sense stationary random processes only require that 1st moment (i.e. the mean) and autocovariance do not vary with respect to time.

• Quick follow-up question: Does an autocovariance that does not vary with respect to time already imply a constant variance? Or is the "constant variance" only fulfilled by the two conditions (autocov. and mean)?
– Kuma
Jul 21, 2017 at 8:22
• The condition formally states that the autocovariance function must depend on the lag and not on time, this is equivalent to say that variance is constant over time. Jul 21, 2017 at 8:38
• Thanks for the explanation and the link, I was somewhat confused there.
– Kuma
Jul 21, 2017 at 9:05

To give another view than that of Digio, I have actually only encountered the requirement for a finite second moment¹, and not for a constant one; at least in books and academic papers, as opposed to online resources (presentations, blogposts, etc.).

I thus believe the formal definition for a weak (or wide-sense) stationary process is:

1. The first moment of $$x_i$$ is constant; i.e. $$∀t, E[x_i]=𝜇$$
2. The second moment of $$x_i$$ is finite for all $$t$$; i.e. $$∀t, E[x_i²]<∞$$ (which also implies of course $$E[(x_i-𝜇)²]<∞$$; i.e. that variance is finite for all $$t$$)
3. The cross moment ― i.e. the auto-covariance ― depends only on the difference $$u-v$$; i.e. $$∀u,v,\tau, cov(x_u, x_v)=cov(x_{u+\tau}, x_{v+\tau})$$

However, I believe that the apparent confusion between the two conditions, and the fact that in some places a requirement for constant variance is stated instead of a finite one, is due to the fact that this indeed follows directly from the three conditions above.

The third condition implies that every lag $$\tau \in \mathbb{N}$$ has a constant covariance value associated with it:

$$cov(X_{t_1}, X_{t_2}) = K_{XX}(t_1,t_2) = K_{XX}(t_2-t_1,0) = K_{XX}(\tau)$$

Note that this directly implies that the variance of the process is also constant, since we get that for all $$t \in \mathbb{N}$$

$$Var(X_t) = cov(X_t, X_t) = K_{XX}(t,t) = K_{XX}(0) = d$$

for some constant $$d$$.

1 When writing second moment I mean $$E[x_i^2]$$, and not variance, which is the second central moment.