# Integrating Brownian Motion [closed]

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.

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

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 $$t=s_0. Then, by definition, $$\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$$.