Introduction

The Brownian Motion, a.k.a. Wiener process, is a foundational block in quantitative finance

Used extensively for

- Pricing of Derivative Products
- Risk Management and Hedging
- Portfolio Construction

Basic Definition

A Wiener process \({W_t}\) in continous time satisfes the following properties for \(t\geq0\).

1. \(W_0 = 0\).
2. If \(t_1<t_2\leq t_3<t_4\), then the increments \(W_{t_2}-W_{t_1}\) and \(W_{t_4} - W_{t_3}\) are independent.
3. For any given \(t_1\) and \(t_2\) with \(t_2 > t_1\), the increment \(W_{t_2}-W_{t_1}\) is normally distributed with mean \(0\) and variance \(t_2-t_1\).

Details are available in the course script Financial Models by Prof. Dr. J.M.Schumacher.

Asmnt Part A: Simulation of Brownian Motion Paths

Plot two sample paths of a discrete-time standard random walk. That is, plot two trajectories of the process defned by the recursion \(X_{k+1} = X_k + Z_k\) where \(X_0 = 0\) and the \(Z_k\)'s are independent standard normal variables. Take 200 steps.

The MATLab function \(cumsum\) helps avoid generating a time loop.

MATLAB Code

N=200; %number of steps
%initial values for the paths

path1(1)=0;
path2(1)=0;

%generating the paths
path1(2:N+1)=cumsum(randn(N,1));
path2(2:N+1)=cumsum(randn(N,1));

%output
figure 
hold on
plot([0:1:200],path1)
plot([0:1:200],path2,'r')
xlabel('Steps')
legend('path1','path2')

Here are two Brownian Motion Paths

Asmnt Part B: Covaraience of overlaping motions

Let \(W_t\) be a Wiener process. Show that

\[Cov(W_{t_1} ;W_{t_2}) = min(t_1; t_2)\]

For every step in your reasoning, indicate which property of the Wiener process you use.

Solution

• Assume that \(t_0 < t_1 < t_2\) where \(t_0=0\).

• Then we can consider that

\(W_{t_1}=W_{t_1}−W_{t_0})\)

\(W_{t_2}=(W_{t_1}-W_{t_0})+(W_{t_2}-W_{t_1})\) where \(W_{t_0}=0\) from property (i) of a Wiener process.

What we did so far is to show that each of the random variables can be written as Brownian Motion increments. Those increments are

• independent due to (ii)
• normally distributed due to (iii),
• with zero mean
• variance that is equal to the time interval between the two random variables.

What we have is that \(W_{t_0}=0\), \(W_{t_1}\sim N(0,t_1)\) and \(W_{t_2}-W_{t_1}\sim N(0,t_2-t_1)\)

The sum of two independent and normally distributed random variables is - normally distributed - with variance equal to the sum of the variances of the initial random variables:

\[W_{t_2}\sim N(0,t_2)\]

Working out the math we have

\[Cov(W_{t_1},W_{t_2})\\ =E(W_{t_1}W_{t_2})-E(W_{t_1})E(W_{t_2})\\ =E(W_{t_1}W_{t_2})=E(W_{t_1}(W_{t_1}+(W_{t_2}-W_{t_1}))\\ =E((W_{t_1})^2)+E((W_{t_1}(W_{t_2}-W_{t_1}))\\ =E((W_{t_1})^2)+E(((W_{t_1}-W_{t_0})(W_{t_2}-W_{t_1}))\\ =E((W_{t_1})^2)=Var(W_{t_1})=t_1\]

Analogously, for \(t_0<t_2<t_1\), then \(Cov(W_{t_1},W_{t_2})=E(W_{t_2}^2)=t_2\).

Conclusion

\[Cov(W_{t_1},W_{t_2})=min(t_1,t_2)\].