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Deriving the Brownian Motion Integral Over a Time Interval

Article Quant Q&A · Author: Karry
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Summary

The document explains how to evaluate the integral of the constant function with respect to standard Brownian motion over an interval. Approximating the Ito integral with a simple process turns it into a sum of Brownian increments; the increments telescope, leaving the change in Brownian motion between the interval’s endpoints.

That difference is normally distributed with mean zero and variance equal to the interval length, so it can also be represented as the square root of that length times a standard normal random variable. The explanation is limited to an integrand equal to one. It does not derive integrals with more general, time-varying or random integrands, which require additional stochastic calculus.

Key ideas

  • The Ito integral of one over an interval equals the Brownian motion increment across that interval.
  • A simple-process approximation expresses the integral as a sum of increments.
  • The sum telescopes to the difference between the endpoint values.
  • The increment has the distribution of the square root of the interval length multiplied by a standard normal variable.

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# Integrating Brownian Motion


# Integrating Brownian Motion












I just wonder how to integrate standard Brownian motion on time interval $(t, T)$.

Let $Z$ be a standard Brownian motion with mean $0$ and standard deviation $1$, with $dZ^2 = dt$. How to derive the expression of the following integral?

$$\int_{t}^{T}dZ$$

Personally, I infer that the above integral equals to $\sqrt{T-t} \, Z$ from other equation, but I don't know how to derive it.

## Answer by Kevin (score 4)

https://quant.stackexchange.com/a/48953

An integral with respect to a stochastic process is the theme of stochastic calculus for which you ought to get an introductory textbook as it is the key to financial models.

A Brownian motion $(W_t)$ is the easiest integrand and typically the first example one encounters. Then, $\int_t^T 1\mathrm{d}W_s=W_T-W_t=W_{T-t}=\sqrt{T-t} Z$ where $Z\sim N(0,1)$.

In your case, the function $f(x)=1$ is a simple function. The Ito integral of simple function with respect to Brownian motion is simply a finite sum which in your case collapses to a telescoping sum. So, the construction of the Ito integral gives the above result directly.

Let $(X_s)$ be a simply process, i.e. $X_s=\sum\limits_{i=0}^{n-1} C_{i}\mathbb{1}_{(s_{i},s_{i+1}]}(s)$ for $\mathcal{F}_{s_i}$-measurable random variables $C_i$ and a partition $t=s_0<s_1<...<s_n=T$. Then, by definition, $\int_t^T X_s \mathrm{d}W_s = \sum\limits_{i=0}^{n-1} C_i (W_{s_{i+1}}-W_{s_i})$. In your case, $C_i=1$ for all $i$. Thus, $\int_t^T 1 \mathrm{d}W_s = \sum\limits_{i=0}^{n-1} 1 (W_{s_{i+1}}-W_{s_i}) = B_{s_n}-B_{s_0} = W_T-W_t$.

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