We say X<sub>t</sub> with paramters (µ,σ) is brownian process if (X<sub>t-s</sub> - X <sub>t</sub>) ~N (µs,σ<sup>2</sup> s) AMONG other conditons . 

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

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

Does that mean I can say any brownian motion process X<sub>t</sub> with parameters is µ,σ is also normally distributed N(µt,σ<sup>2</sup> s)?

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