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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

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  • $\begingroup$ 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

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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) $

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

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