I'm trying to get my head around how a Brownian motion is formed from a simple random walk. I've seen two similar methods used:

enter image description here

enter image description here

Why has one approach used $\frac{1}{\sqrt{k}}$ and the other hasn't? How are they both valid? The second approach suggests $\frac{1}{\sqrt{k}}$ was added so that the resulting Brownian motion followed a normal distribution by the central limit theorem. Is this still the case for the first approach?


2 Answers 2


Not sure about the correctness of the first approach, but second approach uses $1 /\sqrt k$ to scale the variance of the total sum by $k$. So the difference of two processes (say $W_t$ and $W_{t+\Delta t}$) generated by the random walk would have a variation of $\Delta t$, which satisfies one of conditions needed to get a Wiener's process.


There is a very simple elementary derivation of a normal distribution version of an option pricing formula using the concept of the fundamental interaction between buy and sell orders (or up and down operators).

The idea is to establish a relationship between 'steps' and price displacement.

Consider a hypothetical stock chart where a stock has moved some amount $a$ (usually a small percentage) over some time duration $t_1$ (often measured in years). It can be visualized as a triangle, with the vertices being $t_1$ and $p_1$.

$p_1$ is the present price of the stock and the displacement is $a*p_1$

Denote $t_2$ as the time until expiration

Consider a discrete sum of up $u$ and down $d$ stock orders denoted by


Each 'up' and 'down' represents a 'time unit'. Adding the 'ups' and 'downs' gives the sum of units.

The second part of the fundamental interaction is the difference between 'ups' and 'downs':


This means that the difference between 'ups' and 'downs' gives a function in terms of a new displacement, $p_2-p_1$ where $p_2 \ge p_1$.

The pair of linear equations solves for $u$ and $d$:



Consider the proportional relation between two displacements, the base one with $t_1$ and our new one, $p_2-p_1$


Solving for $x$ gives the needed function in terms $p_2-p_1$, which is plugged into $u$:


When $p_2=p_1$, the stock is unchanged, meaning that the number of 'up' units is equal to the 'down' ones.

What we've done is establish a relationship between displacement of price and 'up' and 'down' units.

'Up' and 'down' units, analogous to tossing a coin, also obey a normal distribution:

$\mu_1 +\sigma_1 = \frac{t_2}{2t_1}+\frac{1}{2}\sqrt{\frac{t_2}{t_1}}$

There is also $\mu_2,\sigma_2$ for the price.

$\mu_2 = p_1 e^{r*t} $

(this is because if $p_2=p_1$ the stock is unchanged, hence $\mu_1 = \frac{t_2}{2t_1}$ meaning that the number of 'up' units is the same as 'down', resulting in no displacement.

We have to find $\sigma_2$

Because of the equivalence between units and price displacement, the $\sigma_2$ can be solved by setting $p_2=p_1+\sigma_2$

From the equivalence:

$\mu_1 +\sigma_1=u$

We have:

$\sqrt{\frac{t_2}{t_1}}=\frac{t_1 \sigma_2}{a p_1}$

Rearranging gives the classic result: $\sigma_2=a p_1 \sqrt{t_2}$ setting $t_1=1$ (for a single year and $t_2$ is the fraction of the year)

This helped me understand random walk better


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.