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 $0<t_i<t_{i+1}<\infty$

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    Doesn't your computation of $\operatorname{Var}(B(t))$ immediately answer the question? – whuber Dec 6 at 19:12
  • Only half the question. It still remains to be confirmed if a Wiener process is weakly stationary. – Herman Autore Dec 6 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 at 23:16
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    I see what you're saying. Yes, you make sense. – Herman Autore Dec 6 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 $0<t_i<t_{i+1}<\infty$ 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.

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