Why Volatility Scales the Wiener Process in Geometric Brownian Motion

Summary

The note clarifies the role of volatility in the geometric Brownian motion model, whose price increments contain both a drift term and a Wiener-process term. A standard Wiener process has zero-mean increments with variance proportional to elapsed time. It therefore supplies standardized randomness, rather than an asset-specific volatility level.

Multiplying its increment by sigma scales the random shock so the modeled asset has the desired volatility: over a time interval, the shock’s standard deviation scales with sigma and the square root of elapsed time. The document also sketches a connection to the central limit theorem, though its statement of the normal distribution’s spread is imprecise: it gives sigma times the square root of elapsed time as a standard deviation, not a variance. The core point is that Wiener-process variance is fixed by its definition, while sigma sets the model’s scale.

Key ideas

• A standard Wiener increment has variance equal to the length of its time interval.
• The volatility parameter scales standardized Wiener noise to match the asset’s modeled volatility.
• Over an interval, the shock’s standard deviation scales with volatility and the square root of elapsed time.
• The note’s central limit theorem explanation is suggestive, but its distribution notation confuses standard deviation with variance.

Tags

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


# 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?

## Answer by Matthew Gunn (score 4)

https://quant.stackexchange.com/a/35078

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.

## Answer by David Addison (score 0)

https://quant.stackexchange.com/a/35092

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.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.