Why is Brownian Motion B_t distributed as N(0,t)? Almost all textbooks define a Brownian Motion ($B_t$)using three / four points: $B_0 = 0$; it has stationary independent increments; for every $t>0$, $B_t$ has a normal $N(0,t)$ distribution; it has continuous sample paths.
Could someone please explain,
1 - why is a Brownian Motion $\sim$ Gaussian(0,t)?
I ask these questions because much of the time I find myself stating $\mathbb{E}(B_t) = 0$ or $\mathbb{E}(B_t^2) = t$ without really understanding why. 
I can compute the moments of the distribution to understand why $\mathbb{E}(B_t)=0$,   $\mathbb{E}(B_t^2) = t$ and hence why $\text{Var($B_t$)} = \mathbb{E}(B_t^2) - \mathbb{E}(B_t)^2 = t$, but it all stems from the definition that $B_t \sim N(0,t)$ - and I do not really understand why this is.
Appreciate any clarity.
Many thanks,
John
 A: I will try to give an intuitive explanation. Suppose that $X_1,X_2,...$ are independent and identically distributed random variables with zero means and unit variances. Define
$$
\zeta_n(t)=\frac1{\sqrt n}\sum_{k=1}^{\lfloor nt\rfloor}X_k
$$
for $t\in[0,1]$ and $n\ge1$, where $\lfloor\cdot\rfloor$ is the floor function. It is known that $\zeta_n$ converges in distribution to the Brownian motion as $n\to\infty$ in the space $D[0,1]$ (see, for example, the famous textbook by Billingsley). Let us fix some $t\in[0,1]$ and let us investigate the partial sums. We have that
$$
\frac1{\sqrt n}\sum_{k=1}^{\lfloor nt\rfloor}X_k
=\frac{\sqrt{\lfloor nt\rfloor}}{\sqrt n}\cdot\frac1{\sqrt{\lfloor nt\rfloor}}\sum_{k=1}^{\lfloor nt\rfloor}X_k.
$$
Also, $\sqrt{\lfloor nt\rfloor}/\sqrt n\to\sqrt t$ as $n\to\infty$ and $(\sqrt{\lfloor nt\rfloor})^{-1}\sum_{k=1}^{\lfloor nt\rfloor}X_k\to N(0,1)$ in distribution as $n\to\infty$ by the central limit theorem. Hence,
$$
\frac1{\sqrt n}\sum_{k=1}^{\lfloor nt\rfloor}X_k\to N(0,t)
$$
in distribution as $n\to\infty$ for each $t\in[0,1]$. So it makes sense that the distribution of $B_t$ is $N(0,t)$ for $t\in[0,1]$.
