I found out that I simply can't rigorously prove that Wiener process is a stationary process, i.e. its finite-dimensional distributions don't change under shift in time. Let $W_t$ be a Wiener process:

  1. $W_0=0$ a.s.

  2. for $0\le t_0\le \ldots \le t_n$ random variables $W_{t_n}-W_{t_{n-1}}, \ldots, W_{t_2}-W_{t_{1}}, W_{t_1}-W_{t_{0}}$ are independent,

  3. $W_{t}-W_{s}\sim N(0, t-s)$.

Then why $W_t$ is a stationary process, i.e. for any measurable set $A\in \mathbb{R}^n$ $$ \mathbb{P}((W_{t_1},\ldots, W_{t_n})\in A)=\mathbb{P}((W_{t_1+h},\ldots, W_{t_n+h})\in A)? $$

For instance, if I take $n=1$, then of course $\mathbb{P}(W_{t}\in A)\neq \mathbb{P}(W_{t+h}\in A)$ because $W_{t}$ and $W_{t+h}$ dont have same distribution.

Sorry for probably simple question, but I seemed to get stuck.


Brownian motion isn't a stationary process in this sense.

What is true is that it has stationary increments: that for any $s,t$ and any $h$, $W_t - W_s$ has the same distribution as $W_{t+h}-W_{s+h}$. This is essentially what is expressed by your property 3. This is usually what people mean when they speak of Brownian motion being "stationary".

  • $\begingroup$ Thank you so much! Great answer! $\endgroup$ – Mushtandoid Oct 10 '15 at 21:57
  • $\begingroup$ This answer explains why Brownian motion can't be stationary. $\endgroup$ – Mars Aug 7 at 17:24

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