Brownian Motion

Theory

• Conditional Probability

• Convergence. (Rate \(O_p\))

  • Derive from Chebyshev’s inequality

  • \[ \bar{X_n} - \mu = O(\frac{1}{\sqrt{n}}) \]

  • \[ P(|\bar{X_n} - \mu| > \frac{M}{\sqrt{n}}) <= \frac{\mu^2}{M^2} < \epsilon \]

Wiener Process

\[ P(B(t)<= y | B(0) = x) = \Phi(\frac{y-x}{\sqrt t}) \]

Using Markov property, this equals:

\[ P(B(t)-B(0) <= y-x \]

We have \(P\{B(s+t) \leq\) \(y \mid B(s)=x\}=P\{B(s+t)-B(s) \leq y-x\}=\Phi\left(\frac{y-x}{\sqrt{t}}\right)\) By part (b), \(\operatorname{Cov}\{B(s), B(t)\}=\min (s, t)\). for \(s, t \geq 0\).

Definition

Brownian motion is a stochastic process {\(B(t) ; t>=0\)} with properties:

• Every increment \(B(s+t)-B(s)\) is normal distributed

• For every pair of disjoint time intervals, the increments are independent

In variance Principle

Example 16.2

Evaluate \(E[e^{\lambda B(t)}]\) for an arbitrary constant \(\lambda\) and Brownian motion B(T)

Solution: Use MGF of normal dist N(0,t), sub in the values. \(e^{\lambda^2 t/2}\)

Example 16.3

Find the variance of linear combinations of \(B(t)\), using

\[ \sum_{i=1}^{n} \sum_{j=1}^{n} a_ia_j min{t_i,t_j} \]

Solution: the variance of \(\sum_{i=1}^{n}a_iB(t_i)\) is \(Var(\sum_{i=1}^{n} a_iB_i)\)

\[ =Cov(\sum_{i}^{n} a_iB_i,\sum_{j}^{n} a_jB_j),using ，Cov(B_s,B_t) = min(t_s,t_i) \]

Example 16.5

a. s<t = e^(s-t) s>t = e^(t-s)