Brownian Motion

Keith A. Lewis

January 26, 2025

Abstract
Continuous time random walk

Brownian motion is a stochastic process (B_t)_{t\ge0} with increments that are stationary, independent, and normal. It is the continous time limit of a properly scaled random walk.

stationary
t\mapsto B_{t + u} - B_u has the same law as t\mapsto B_t - B_0, u\ge0.
independent
If 0\le t_0 < \cdots < t_n then B_{t_0}, B_{t_1} - B_{t_0}, , B_{t_n} - B_{t_{n-1}} are independent.
normal
The increment B_{t + u} - B_t is normally distributed.

We also require paths are right-continuous almost surely.

Standard Brownian motion has E[B_1] = 0 and \operatorname{Var}(B_1) = 1.

Norbert Wiener (1923) showed if Z_n are independent, standard normal random variables then B_t = Z_0 t + \sqrt{2}\sum_{n=1}^\infty Z_n\frac{\sin\pi n t}{\pi n} satisfies these properties.

Exercise. A stochastic process (X_t) with increments that are stationary, independent, and normal has the form X_t = \mu t + \sigma B_t for some constants \mu and \sigma where (B_t) is standard Brownian motion.

Hint. Define B_t = (X_t - E[X_1])/\operatorname{Var}(X_1). Show (B_t) is standard Brownian motion.

Reflection Principal

Let \overline{B}_t = \max_{s\le t} B_s be the running maximum of Brownian motion and \underline{B}_t = \min_{s\le t} B_s be the running minimum.

Exercise. Show -\underline{B}_t = \overline{-B}_t.

Solution Use -\min\{x\in X\} = \max\{-x\in X\} for X\subset\boldsymbol{R}.

Fix a \ge 0 and define \tau_a = \min \{t\ge 0\mid B_t > a\} to be the first hitting time at level a. Note B_{\tau_a} = a almost surely. Brownian motion reflected at a, B^a_t, is defined as B^a_t = \begin{cases} B_t, &t\le \tau_a \\ a + (a - B_t), &t\ge \tau_a \\ \end{cases} Reflected Brownian motion switches to -B_t once level a is hit. Note a - B_{\tau_a} = 0.

Exercise. Reflected Brownian motion is Brownian motion.

Lemma. (Reflection Principal) For any integrable function f we have E[f(B_t) 1(\overline{B}_t > a)] = E[f(B_t) 1(B_t > a)] + E[f(2a - B_t) 1(B_t > a)].

Proof. Since either B_t > a or B_t \le a E[f(B_t) 1(\overline{B}_t > a)] = E[f(B_t) 1(B_t > a, \overline{B}_t > a)] + E[f(B_t) 1(B_t \le a, \overline{B}_t > a)] Since B_t > a implies \overline{B}_t > a) we have the first term on the right hand side E[f(B_t) 1(B_t > a, \overline{B}_t > a)] = E[f(B_t) 1(B_t > a)]. Since (B^a_t) is Brownian motion E[f(B_t) 1(B_t \le a, \overline{B}_t > a] = E[f(B^a_t) 1(B^a_t \le a, \overline{B^a}_t > a]. If \overline{B^a}_t > a we are on a reflected path so B^a_t = 2a - B_t and \begin{aligned} E[f(B^a_t) 1(B^a_t \le a, \overline{B^a}_t > a] &= E[f(2a - B_t) 1(2a - B_t \le a, \overline{B^a}_t > a)] \\ &= E[f(2a - B_t) 1(B_t \ge a, \overline{B^a}_t > a] \\ \end{aligned} Since B_t \ge a implies \overline{B^a}_t > a we have the second term on the right hand side E[f(2a - B_t) 1(B_t \ge a, \overline{B^a}_t > a] = E[f(2a - B_t) 1(B_t \ge a)].

Exercise. Show P(\overline{B}_t > a) = 2P(B_t > a).

Solution Use f(x) = 1.

Exercise. What are the corresponding formulas for E[f(B_t) 1(\overline{B}_t \le a)], E[f(B_t) 1(\underline{B}_t > a)], and E[f(B_t) 1(\underline{B}_t \le a)]?

Girsanov

If (B_t) is standard Brownian motion with Wiener measure P then B^\alpha_t = B_t - \alpha t, \alpha\in\boldsymbol{R}, is Brownian motion with respect to P^\alpha where E_t[dP^\alpha/dP] = e^{\alpha B_t - \alpha^2t/2}.

Exercise. Show E[e^{\alpha B_t - \alpha^2t/2} g(B_{t_1} - \alpha t_1, \ldots)] = E[g(B_{t_1}, \ldots)] for any integrable function g and t \ge t_1,\ldots.

Hint. B_t, B_{t_1}, \ldots, are jointly normal.

Exercise. If u > t show E_t[E_u[dP^\alpha /dP]] = E_t[dP^\alpha /dP].

Hint. e^{\alpha B_t - \alpha^2t/2} is a martingale.

Let F_t = fe^{\sigma B_t - \sigma^2t/2}. An up-and-in barrier option (Mer75?) with barrier h pays \phi(F_t) 1(\overline{F}_t > h) at expiration for some payoff function \phi. Using Girsanov’s theorem E[\phi(F_t) 1(\overline{F}_t > h)] = E^\alpha [\phi(F^\alpha _t) 1(\overline{F^\alpha }_t > h)] where F^\alpha _t = f e^{\sigma B^\alpha _t - \sigma^2t/2} = f e^{\sigma (B_t - \alpha t) - \sigma^2t/2}. Note \overline{F^\alpha }_t > h if and only if \max_{s\le t} F^\alpha _s > h.

Exercise. Show \overline{F^\alpha }_t > h if and only if \max_{s\le t} B_s - (\alpha + \sigma/2)s > (\log h/f)/\sigma.

Solution Since F^\alpha _t = f e^{\sigma B^\alpha _t - \sigma^2t/2} = f e^{\sigma (B_t - \alpha t) - \sigma^2t/2} we have F^\alpha _t > h if and only if B_t - (\alpha + \sigma/2)t > (\log h/f)/\sigma. Take the max over s\le t.

Exercise. Find a closed form solution for the value of an up-and-in barrier option.

Hint. Use \alpha = -\sigma /2.

Solution From above, \overline{F^\alpha }_t > h if and only if \overline{B_t} > (\log h/f)/\sigma so we can apply the reflection principal.