# Question

Let $$B(t)$$ be a standard Brownian motion (AKA a Wiener process).

Is $$B(t)$$ weakly or strictly stationary, particularly as defined here?

# My Thoughts

We know, by definition, that its increments are stationary, which implies it is weakly stationary—that its covariance only depends on the length of the intervals:

$$Cov(B(t),B(t+h)) = Cov(B(u),B(u+h))$$ for some $$h > 0$$, and $$0

This makes sense because the Wiener process is the integral of a Gaussian process. However, clearly by inspection of a graph we can see that as $$t$$ increases the variance increases, too. Moreover: $$var(B(t_i)) = t_i < var(B(t_{i+1}) = t_{i+1}$$ For $$0

• Doesn't your computation of $\operatorname{Var}(B(t))$ immediately answer the question? – whuber Dec 6 '18 at 19:12
• Only half the question. It still remains to be confirmed if a Wiener process is weakly stationary. – Herman Autore Dec 6 '18 at 23:14
• I don't think so: since the variance depends on time, it follows immediately that the process cannot be stationary or weakly second-order stationary. It is only weakly first-order stationary (because the mean is constantly zero). – whuber Dec 6 '18 at 23:16
• I see what you're saying. Yes, you make sense. – Herman Autore Dec 6 '18 at 23:18

A standard Brownian motion (AKA a Wiener process), $$B(t)$$ is neither strictly stationary nor weakly stationary.

# Strict stationarity

Strict stationarity requires the distribution not be a function of time $$t$$. For example, if some stochastic process $$X(t)$$ were strictly stationary, then for $$0 all $$X(t)$$ would be equal in distribution, or $$X(t_i) \overset{d}{=} X(t_{i+1})\ \forall\ i$$ But note that for $$B(t)$$: $$var(B(t_i)) = t_i < var(B(t_{i+1})) = t_{i+1}$$ so $$B(t_i) \overset{d}{\neq} B(t_{i+1})$$

# Weak stationarity

Weak stationarity requires that the covariance only be a function of the length of the interval, and not the interval's location in time. $$Cov(B(t),B(t+h)) = Cov(B(u),B(u+h))$$ for some $$h > 0$$, and $$0

But for $$B(t)$$:

\begin{align} Cov(B(t),B(t+h)) & < Cov(B(u),B(u+h)) \\ min(t,t+h) & < min(u,u+h) \\ t & < u \end{align}

Therefore B(t) is neither strictly nor weakly stationary. This is not to be confused with stationary intervals, however.