Constructing a Brownian motion from a Simple Random Walk

Question

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:

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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?

Answer 1

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.

Answer 2

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

$u+d=t_2/t_1$

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':

$u-d=f(p_2-p_1)$

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 $d$:

$u=\frac{1}{2}\left(\frac{t_2}{t_1}+f(p_2-p_1)\right)$

$d=\frac{1}{2}\left(\frac{t_2}{t_1}-f(p_2-p_1)\right)$

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

$\frac{a*p}{t_1}=\frac{p_2-p_1}{x}$

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

$u=\frac{1}{2}\left(\frac{t_2}{t_1}+\frac{t_1(p_2-p_1)}{a*p_1}\right)$

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