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 $0<t<u<\infty$
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 $0<t_i<t_{i+1}<\infty$
Best Answer
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 $0<t_i<t_{i+1}<\infty$ 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 $0<t<u<\infty$
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.