Brownian Motion

Let's model a random motion happening around 0,with equal probability to go left and right with same amount each step.

$\begin{array}{}{X}_i=+\mathrm{\Delta \; <!--\; greek\; capital\; letter\; delta\; -->}x\to \; <!--\; rightwards\; arrow\; -->P\left(X_i\right)=\frac{1}{2}\\ X_i=-\mathrm{\Delta \; <!--\; greek\; capital\; letter\; delta\; -->}{x}_n\to \; <!--\; rightwards\; arrow\; -->P\left(X_i\right)=\frac{1}{2}\\ X=X_1+X_2+\dots \; <!--\; horizontal\; ellipsis\; -->+X_n\end{array}$

$\begin{array}{}⟨\; <!--\; mathematical\; left\; angle\; bracket\; -->X⟩\; <!--\; mathematical\; right\; angle\; bracket\; -->=⟨\; <!--\; mathematical\; left\; angle\; bracket\; -->X_1⟩\; <!--\; mathematical\; right\; angle\; bracket\; -->+\dots \; <!--\; horizontal\; ellipsis\; -->+⟨\; <!--\; mathematical\; left\; angle\; bracket\; -->X_n⟩\; <!--\; mathematical\; right\; angle\; bracket\; -->\\ ⟨\; <!--\; mathematical\; left\; angle\; bracket\; -->X_i⟩\; <!--\; mathematical\; right\; angle\; bracket\; -->=\frac{1}{2}\times \; <!--\; multiplication\; sign\; -->\left(-\mathrm{\Delta \; <!--\; greek\; capital\; letter\; delta\; -->}x\right)+\frac{1}{2}\times \; <!--\; multiplication\; sign\; -->\left(\mathrm{\Delta \; <!--\; greek\; capital\; letter\; delta\; -->}x\right)=0\\ ⟨\; <!--\; mathematical\; left\; angle\; bracket\; -->X⟩\; <!--\; mathematical\; right\; angle\; bracket\; -->=0\end{array}$

So average is zero

On the other hand variance is not zero, since

$\begin{array}{}var\left\{X_i\right\}=⟨\; <!--\; mathematical\; left\; angle\; bracket\; -->X_i^2⟩\; <!--\; mathematical\; right\; angle\; bracket\; -->-⟨\; <!--\; mathematical\; left\; angle\; bracket\; -->X_i{⟩\; <!--\; mathematical\; right\; angle\; bracket\; -->}^2=\left(\frac{1}{2}\right)\times \; <!--\; multiplication\; sign\; -->{\left(\mathrm{\Delta \; <!--\; greek\; capital\; letter\; delta\; -->}x\right)}^2+\frac{1}{2}\times \; <!--\; multiplication\; sign\; -->{\left(-\mathrm{\Delta \; <!--\; greek\; capital\; letter\; delta\; -->}x\right)}^2\\ var\left\{x_1\right\}=var\left\{x_2\right\}=\dots \; <!--\; horizontal\; ellipsis\; -->=var\left\{X_n\right\}={\left(\mathrm{\Delta \; <!--\; greek\; capital\; letter\; delta\; -->}x\right)}^2\end{array}$

Since each step statistically independent The sum of variance is

$⟨\; <!--\; mathematical\; left\; angle\; bracket\; -->X^2⟩\; <!--\; mathematical\; right\; angle\; bracket\; -->=\underset{e^0=1}{\overset{n}{\sum \; <!--\; n-ary\; summation\; -->}}var\left\{X_i\right\}=n\mathrm{\Delta \; <!--\; greek\; capital\; letter\; delta\; -->}{x}^2$

Since total duration of walk is

$\begin{array}{}t=n\mathrm{\Delta \; <!--\; greek\; capital\; letter\; delta\; -->}t\Rightarrow \; <!--\; rightwards\; double\; arrow\; -->n=\frac{t}{\mathrm{\Delta \; <!--\; greek\; capital\; letter\; delta\; -->}t}\\ ⟨\; <!--\; mathematical\; left\; angle\; bracket\; -->X^2⟩\; <!--\; mathematical\; right\; angle\; bracket\; -->=\left(\frac{\mathrm{\Delta \; <!--\; greek\; capital\; letter\; delta\; -->}{x}^2}{\mathrm{\Delta \; <!--\; greek\; capital\; letter\; delta\; -->}t}\right)t\end{array}$

So Brownian motion happens round 0 and with time it diffuses. That is the variance increases with number of steps. The term in front of time is equal to the twice of diffusion coefficient