Chapter 6
Brownian Motion
Definition 6.0.1. A process \(\{X(t) : t\geq 0\}\) with state space \(\mathbb {R}\) is a Brownian motion with variance parameter \(\sigma ^2\) if
- (i)
- \(X(0) = 0\);
- (ii)
- it has stationary and independent increments: for \(0<t_1<\cdots <t_n\), the increments \(X(t_1) - X(0), X(t_2) - X(t_1), \dots , X(t_n) - X(t_{n-1})\) are independent, and the distribution of \(X(s + t) - X(s)\) does not depend on \(s\);
- (iii)
- for every \(t>0\), \(X(t)\) is normal with mean \(0\) and variance \(\sigma ^2t\).
When \(\sigma ^2 = 1\) it is a standard Brownian motion.
Brownian motion is the limit of a symmetric random walk with small steps taken frequently: if the walk moves \(\pm \Delta x\) every \(\Delta t\) with \(\frac {(\Delta x)^2}{\Delta t} = \sigma ^2\), the central limit theorem makes the position at time \(t\) approximately normal with variance \(\sigma ^2t\). Its paths are continuous but nowhere differentiable.
For a standard Brownian motion, \(X(t)\) has density \(f_t(x) = \frac {1}{\sqrt {2\pi t}}e^{-x^2/(2t)}\), and \(X(t_i) - X(t_{i-1})\) is normal with mean \(0\) and variance \(t_i - t_{i-1}\). The values \(X(t_1) = x_1, \dots , X(t_n) = x_n\) are equivalent to the increments taking the values \(x_1, x_2 - x_1, \dots , x_n - x_{n-1}\), and these are independent, so the joint density is \[f(x_1, \dots , x_n) = f_{t_1}(x_1)f_{t_2 - t_1}(x_2 - x_1)\cdots f_{t_n - t_{n-1}}(x_n - x_{n-1}) .\]
Theorem 6.0.2 (Brownian bridge). For a standard Brownian motion and \(0<s<t\), given \(X(t) = B\), \(X(s)\) is normal with \[E\left [X(s)\mid X(t) = B\right ] = \frac stB,\qquad \var \left [X(s)\mid X(t) = B\right ] = \frac {s(t - s)}{t} .\]
Proof. The conditional density is \[f_{s\mid t}(x\mid B) = \frac {f_s(x)f_{t-s}(B - x)}{f_t(B)} = K\exp \left \{-\frac {x^2}{2s} - \frac {(B - x)^2}{2(t - s)}\right \},\] where \(K\) does not depend on \(x\). Completing the square in \(x\), \[\frac {x^2}{s} + \frac {(B - x)^2}{t - s} = \frac {t}{s(t - s)}\left (x - \frac {Bs}{t}\right )^2 + \text {terms free of } x,\] so \(f_{s\mid t}(x\mid B) = K_1\exp \left \{-\frac {(x - Bs/t)^2}{2s(t - s)/t}\right \}\), a normal density with the stated mean and variance. □
The conditional mean interpolates linearly between \(X(0) = 0\) and \(X(t) = B\), and the conditional variance is largest half-way and zero at both ends.
Example 6.0.3. In a bicycle race between two riders, let \(Y(t)\) be the number of seconds by which the rider who started on the inside is ahead when a fraction \(t\) of the race has been completed, and suppose that \(\{Y(t) : 0\leq t\leq 1\}\) is a Brownian motion with variance parameter \(\sigma ^2\).
- (a)
- If the inside rider leads by \(\sigma \) seconds at the half-way point, what is the probability that she wins?
- (b)
- If the inside rider wins by \(\sigma \) seconds, what is the probability that she was ahead at the half-way point?
Solution.
- (a)
- By independent increments, \(Y(1) - Y(\frac 12)\) is independent of \(Y(\frac 12)\), and by stationarity it is normal with mean \(0\) and variance \(\frac {\sigma ^2}{2}\). So \begin {align*} P\left (Y(1)>0\mid Y(\tfrac 12) = \sigma \right ) &= P\left (Y(1) - Y(\tfrac 12)>-\sigma \right )\\ &= P\left (Z>\frac {-\sigma }{\sigma /\sqrt 2}\right ) = P(Z>-\sqrt 2) = 0.921 . \end {align*}
- (b)
- Standardise: \(X(t) = \frac {Y(t)}{\sigma }\) is a standard Brownian motion, and the event is \(X(\frac 12)>0\) given \(X(1) = 1\). By theorem 6.0.2, \(X(\frac 12)\mid X(1) = 1\) is normal with mean \(\frac 12\) and variance \(\frac {\frac 12\cdot \frac 12}{1} = \frac 14\), so \[P\left (X(\tfrac 12)>0\mid X(1) = 1\right ) = P\left (Z>\frac {0 - \frac 12}{\frac 12}\right ) = P(Z>-1) = 0.841 .\]