We say Xt with paramters (µ,σ) is brownian process if (Xt-s - X t) ~N (µs,σ2 s) AMONG other conditons .

Here we don't speak about any particular distribution for X t. We only say it is a brownian motion and its increments are normally distributed.

But when it comes to standard brownian motion ( Wt) , why do we say it has a normal distribution i.e Wt ~ N(0,t).

Does that mean I can say any brownian motion process Xt with parameters is µ,σ is also normally distributed N(µt,σ2 s)?

I am new to this topic and if the question does not have a logic, please enlighten with your inputs

edit 1: it would be helpful if the explanation is more intuitive than mathematical

  • $\begingroup$ standard just means that $\mu=0,\sigma^2=1,X_0=0$. All the normal properties of BM also apply to standard BM. "standard" is just a specific set of parameters for BM. $\endgroup$
    – Alex C
    Commented Aug 26, 2018 at 3:05

1 Answer 1


A standard Brownian motion $ \{W_t, t \in \mathbb{R} \}$ starts from 0 which means $W_{0} = 0$ with probability one, add to that $W_t - W_{s} \sim N(0,t-s)$. Replace s with 0 and the you get $W_t =W_t - W_{0} \sim N(0,t)$

From this standard Brownian motion you can construct another one with a "drift" $\mu$ : $X_t = \mu t + \sigma W_t$ and you'll get $X_t - X_{s} \sim N(\mu(t-s), \sigma^{2}(t-s))$ and then particularly:

$X_{t-s} - X_{t} \sim N(\mu s, \sigma^{2}s)$

$X_{t} \sim N(\mu t, \sigma^{2}t) $

  • $\begingroup$ And importantly the increments are independent $\endgroup$
    – Ivan
    Commented Aug 25, 2018 at 14:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.