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

• if you have independent increments, identically distributed (when over the same time period), and if the distributions have their first few moments finite, you expect long term behavior to be approximately normal by the central limit theorem. If you can take arbitrarily small increments, then all times can be viewed as long term on some scale. – Aaron Jan 29 '15 at 8:35

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 $t\in[0,1]$.
• Many thanks. I'm reviewing Shreve (Continuous-Time Models p86) whilst reading your answer and I can now see that you've started essentially with a scaled symmetric random walk. Good, makes sense. Two questions if I may. 1) would you have a reference which explains how $(\sqrt{nt})^{-1} \sum_{k=1}^{nt} X_k\rightarrow N(0,1)$? 2) I should know this but why does $\sqrt{t} N(0,1) = N(0,t)$ ? – John Smith Jan 29 '15 at 8:52
• @JohnSmith You're welcome! 1) By the CLT, we have that $N^{-1/2}\sum_{k=1}^NX_k\to N(0,1)$ as $N\to\infty$. If we express $N$ as some integer-valued function, for example, $N(n)=\lfloor nt\rfloor$, the CLT still holds as long as $N\to\infty$. Hence, $(\lfloor nt\rfloor)^{-1/2}\sum_{k=1}^{\lfloor nt\rfloor}X_k\to N(0,1)$ since $\lfloor nt\rfloor\to\infty$ as $n\to\infty$. 2) Suppose that $\xi\sim N(0,1)$, then $\operatorname E[\sqrt t\xi]=0$, $\operatorname{Var}[\sqrt t\xi]=(\sqrt t)^2\operatorname{Var}[\xi]=t$ and $\sqrt t\xi\sim N(0,t)$. – Cm7F7Bb Jan 29 '15 at 9:08