1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Hi Karry, welcome to Quant.SE. What have you tried? $\endgroup$ – Bob Jansen Sep 27 '19 at 18:32
4
$\begingroup$

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, $\int_t^T 1 \mathrm{d}W_s = \sum\limits_{i=0}^{n-1} 1 (W_{s_{i+1}}-W_{s_i}) = B_{s_n}-B_{s_0} = W_T-W_t$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.