Variation & Martingales

Theorem. For partitions $0 = t_0 < t_1 < \dots < t_n = t$ whose largest gap tends to 0, $\sum_i (W_{t_{i+1}} - W_{t_i})^2 \to t$. The convergence holds in mean square, and with probability 1 if the gaps shrink fast enough, as with the halving partitions used here. We write $[W]_t = t$.

Proof sketch. The sum has mean $\sum_i (t_{i+1} - t_i) = t$. Because the pieces are independent, its variance is $\sum_i 2(t_{i+1} - t_i)^2 \le 2t \cdot \max_i (t_{i+1} - t_i) \to 0$. A quantity whose variance tends to 0 converges to its mean.

By contrast, the total variation $\sum_i \lvert W_{t_{i+1}} - W_{t_i} \rvert$ tends to infinity with probability 1. That is why the Riemann–Stieltjes integral of ordinary calculus can't be used with $dW$.

Reference: Shreve, Stochastic Calculus for Finance II, §3.4.

import numpy as np
rng = np.random.default_rng(4)
N = 2 ** 16
W = np.concatenate([[0], np.cumsum(rng.normal(0, np.sqrt(1 / N), N))])   # one path on [0, 1]
for n in (4, 256, 65_536):
    dW = np.diff(W[:: N // n])                        # rises over n equal pieces
    print(f"n = {n:>6}: sum |dW| = {np.abs(dW).sum():8.2f}   sum dW^2 = {(dW ** 2).sum():.4f}")

Output when we ran it:

n =      4: sum |dW| =     1.90   sum dW^2 = 1.0518
n =    256: sum |dW| =    12.22   sum dW^2 = 0.9188
n =  65536: sum |dW| =   204.54   sum dW^2 = 0.9988

Definition. A process $M_t$ is a martingale with respect to the history (filtration) $\mathcal F_t$ if it depends only on the history up to time $t$, has $\mathbb E\lvert M_t \rvert < \infty$, and satisfies $\mathbb E[M_t \mid \mathcal F_s] = M_s$ for all $s \le t$.

Examples from Brownian motion: $W_t$, $W_t^2 - t$, and $\exp(\lambda W_t - \tfrac12 \lambda^2 t)$ for any number $\lambda$. The last one reappears as the Girsanov weight in section 6.2.

Lévy's characterisation. A continuous martingale $M$ with $M_0 = 0$ for which $M_t^2 - t$ is also a martingale must be a Brownian motion. The two martingales in this section are therefore enough to pin Brownian motion down completely.

Reference: Shreve, Stochastic Calculus for Finance II, §2.3 and §3.3; Lévy's characterisation is §4.6.3.

import numpy as np
rng = np.random.default_rng(5)
s, W_s = 0.4, 0.3                                    # freeze time and the value there
W_1 = W_s + rng.normal(0, np.sqrt(1 - s), 100_000)   # possible values at t = 1
print(f"E[W_1]       = {W_1.mean():.3f}  (W_s = {W_s})")
print(f"E[W_1^2]     = {(W_1 ** 2).mean():.3f}  (W_s^2 + 1 - s = {W_s ** 2 + 1 - s:.3f})")
print(f"E[W_1^2 - 1] = {(W_1 ** 2 - 1).mean():.3f}  (W_s^2 - s = {W_s ** 2 - s:.3f})")

Output when we ran it:

E[W_1]       = 0.301  (W_s = 0.3)
E[W_1^2]     = 0.689  (W_s^2 + 1 - s = 0.690)
E[W_1^2 - 1] = -0.311  (W_s^2 - s = -0.310)

Precise statement. Let $M_t = \max_{s \le t} W_s$. For $a > 0$ and $b \le a$: $P(M_t \ge a,\ W_t \le b) = P(W_t \ge 2a - b)$. Setting $b = a$ and adding $P(W_t \ge a)$ gives the formula in this section.

Why it works: the strong Markov property. At the first hitting time $\tau_a = \min\{t : W_t = a\}$, which is a random time, the future $W_{\tau_a + s} - a$ is a fresh Brownian motion, independent of everything before $\tau_a$. Its mirror image is therefore just as likely.

A surprising consequence. $\tau_a$ has density $\dfrac{a}{\sqrt{2\pi t^3}}\, e^{-a^2/2t}$. Every level is reached eventually, with probability 1, yet the average waiting time is infinite.

References: Shreve, Stochastic Calculus for Finance II, §3.7. Mörters & Peres, Brownian Motion, §2.2.

import numpy as np
from math import erf, sqrt
rng = np.random.default_rng(6)
paths, steps, a = 10_000, 1_000, 1.0
W = np.cumsum(rng.normal(0, sqrt(1 / steps), (paths, steps)), axis=1)
reached = (W.max(axis=1) >= a).mean()                 # ever touched level a
ended_above = (W[:, -1] >= a).mean()
theory = 2 * (1 - 0.5 * (1 + erf(a / sqrt(2))))
print(f"reached a: {reached:.3f}   2 x ended above: {2 * ended_above:.3f}   theory: {theory:.3f}")

Output when we ran it:

reached a: 0.302   2 x ended above: 0.314   theory: 0.317