# Geometric Brownian Motion: Why is the Wiener process multiplied by volatility?

Below is the stochastic differential equation of the Geometric Brownian Motion:

$$dS_t = S_t \mu dt + S_t\sigma dW_t$$

My understanding of the Wiener process is that the volatility component of an asset price is already captured there. Why is $\sigma$ being multiplied with the Wiener process?

If you examine a standard definition of a Wiener process, $W_t - W_0$ follows the normal distribution with mean zero and variance $t$. If you want the variance to be something else, you have to scale it.

• Yes,we want to be able to model stocks with different volatility. And the parameter $\sigma$ allow us to do that. – Alex C Jul 11 '17 at 2:24

To understand why $W_t$ is scaled by standard deviation, I think it helps to reference the central limit theorem.

Assumes $x_1,...,x_n$ are i.i.d. variables with mean, $\mu$, and variance, $\sigma^2$. Then a random sampling $\forall \, x_n \in X$ will converge to the distribution as $n \to \infty$ . In this case, $W_t$ simply represents the process which encompasses all $x$ in a standardized distribution.

Therefore:

$\mu dt + \sigma dW_t \sim \ {\mathcal {N}}\left[\mu\,\Delta t,\sigma \sqrt{\Delta t} \right]$

where $\mathcal {N}$ is cumulative density function.