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I find the two very confusing as some seem to use them interchangeably and some don't seem to. Wiki says they're both the same "...is often called the standard Brownian motion" it says in the "Wiener Process" page.

I understand $B_t$ a Brownian motion is normally distributed $N(0,t)$. But is the "standard" Brownian motion distributed as $N(0,1)$? The name analogous to the standard normal distribution?

John Trail
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  • I think the standard Brownian is that $B_0=0$. – Nikita Evseev Dec 05 '15 at 19:18
  • Hi there, I'm not sure about that; in my notes, the Brownian motion = Wiener Process (which doesn't say "standard") $W_t$, one of the conditions it must satisfy(as a property) is $W_0=0$ with probability $1$. So I guess a Brownian motion must also have that property... – John Trail Dec 05 '15 at 19:22
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    Brownian motion has variance t. There are lots of other processes which are brownian motion but which maybe are not obviously brownian motion e.g. brownian motion shifted by a stop time. – Paul Dec 05 '15 at 19:39
  • "The name analogous to the standard normal distribution?" << No, it is not. Wiener process = Brownian motion = standard Brownian motion... there is no difference between these notions. – saz Dec 05 '15 at 19:58

2 Answers2

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Standard Brownian motion is the process you describe: a continuous Gaussian process $B_t$ whose distribution at time $t$ is normal with mean zero and variance $t$ (or in higher dimensions, mean zero and covariance $tI$).

Some authors also use the term "Brownian motion" to refer to translations or scalings of this process; e.g. $B_t + x_0$, which at time $t$ has distribution $N(x_0,t)$, or $c B_t$, which has distribution $N(0, c^2 t)$. So adding the word "standard" serves to clarify when we are not doing that; when we really do mean the process with $N(0,t)$ distribution at time $t$ (i.e. $x_0 = 0$ and $c=1$).

There is no process analogous to Brownian motion that has distribution $N(0,1)$ at time $t$. For one thing, it would have to have either a random starting point, or a jump immediately after time 0, which are typically things we don't want to have in our definition of Brownian motion. (However, you might like to look up the stationary Ornstein-Uhlenbeck process, which is a nontrivial continuous process with random starting point whose distribution at every time $t$ is indeed $N(0,1)$.)

In other contexts, the term "Brownian motion" can refer to other processes that are somehow analogous to standard Euclidean Brownian motion. For instance, on a Riemannian manifold $M$, one can define a stochastic process whose generator is the Laplace-Beltrami operator; this process is often called "Brownian motion on $M$" because it plays that role.

Nate Eldredge
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I had this question myself, and found the answer in page 7 of this link: https://www.ethz.ch/content/dam/ethz/special-interest/mavt/dynamic-systems-n-control/idsc-dam/Lectures/Stochastic-Systems/Brownian_Motion_and_Poisson_Process.pdf

Usually, if Brownian motion is mentioned without specification, it can be assumed to be standard.

It needs to be normally distributed with mean 0 and variance 1, and start in 0.

WilliamGram
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