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Why Wiener Process Increments Scale with the Square Root of Time

Article Quant Q&A · Author: jessica
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Summary

The document explains why a Brownian motion increment over an interval of length Δt has a normal distribution with mean zero and variance Δt. It distinguishes an increment over a unit interval, which is standard normal in distribution, from an increment over a general interval, whose standard deviation is the square root of the interval length. This gives the stochastic differential notation dW = Z√Δt, where Z is standard normal, as a distributional representation for a finite step.

The replies point to normally distributed independent increments as a defining property of Brownian motion; one response also mentions the central limit theorem as intuition for a random walk approximation. The discussion is brief and does not provide a formal proof or derive the geometric Brownian motion stock-price equation. It also cautions implicitly that equality in distribution is not literal equality between a realized increment and a newly chosen normal variable, and that infinitesimal SDE notation should not be confused with an ordinary finite difference.

Key ideas

  • A Brownian motion increment from s to t is normally distributed with mean zero and variance t−s.
  • For a unit interval, the increment has the standard normal distribution.
  • An increment over a finite interval Δt can be represented in distribution as a standard normal variable times √Δt.
  • The scaling follows from the defining increment distribution of Brownian motion, while the central limit theorem offers intuition from random walks.
  • The representation is equality in distribution rather than literal equality to an independent variable.

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# Wiener process proof


# Wiener process proof












Can someone prove to me how $dW_t=W_t-W_s$, where $t=s+1$, the difference of the Wiener process eventually equates to $dW_t=z*(dt)^{(1/2)}$ where $z$ is standard normal, $N(0,1)$ in the following SDE: $dS_t=S_t μ dt+S_t σdW_t$?

## Answer by Phil H (score 2)

https://quant.stackexchange.com/a/10499

I think this question might be asking for the central limit theorem.

If we consider a process W which varies as a series of independent random steps, then the Central Limit Theorem tells us that after many steps, the value of W will be normally distributed.

## Answer by Richi Wa (score 1)

https://quant.stackexchange.com/a/10498

The question is not 100% clear.

If you set $X = W_t-W_s$ where $t-s = 1$ then this is equal in distribution to $W_1-W_0$ and the defining property of Brownian motion is that increments are normally distributed.

In the general case $W_t-W_s$ is $N(0,t-s)$, where the second parameter is variance.

If you set $dW_t = W_{t+dt}-W_t = Z \sqrt{dt}$, where $dt>0$ is not infinitesimal but just some positive real number and $Z$ is standard normal (mean $0$ and variance $1$), then the first "$=$" is a definiton and the second "$=$" is "equal in distribution".

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.