Solution: Quadratic Variation by Simulation

Exercise: Quadratic Variation by Simulation

Part 1 — Single-path quadratic variation

# Python
import numpy as np

rng = np.random.default_rng(42)

def simulate_path(n, T=1.0):
    dt = T / n
    dW = rng.normal(0, np.sqrt(dt), size=n)
    return dW

for n in [10, 100, 1000, 10000]:
    dW = simulate_path(n)
    Q_n = np.sum(dW ** 2)
    print(f"n = {n:5d}   Q_n = {Q_n:.4f}")
# n =    10   Q_n = 0.8761
# n =   100   Q_n = 0.9635
# n =  1000   Q_n = 0.9975
# n = 10000   Q_n = 0.9942

$Q_n$ hovers around $T = 1$ already at $n = 100$ and is within a fraction of a percent at $n = 10{,}000$ . The convergence is clear and fast.

Part 2 — First-order variation

# Python
rng = np.random.default_rng(42)
for n in [10, 100, 1000, 10000]:
    dW = simulate_path(n)
    V_n = np.sum(np.abs(dW))
    print(f"n = {n:5d}   V_n = {V_n:.4f}")
# n =    10   V_n = 2.4067
# n =   100   V_n = 7.9212
# n =  1000   V_n = 25.2168
# n = 10000   V_n = 79.7416

$V_n$ grows roughly as $\sqrt{n}$ : going from $n = 100$ to $n = 10{,}000$ (a factor of 100) increases $V_n$ by a factor of about 10. This is consistent with the theoretical result

\mathbb{E}[V_n] = n \cdot \mathbb{E}[|\Delta W|] = n \cdot \sqrt{\tfrac{2\Delta t}{\pi}} = \sqrt{\tfrac{2n}{\pi}} \cdot \sqrt{T}

(using

\mathbb{E}[|Z|] = \sqrt{2/\pi}

for

Z \sim \mathcal{N}(0, 1)

), which diverges as

n \to \infty

. Total variation is infinite in the refinement limit — this is the rigorous fact behind the claim that

W

is not of bounded variation.

Part 3 — Empirical distribution of $Q_n$

# Python
rng = np.random.default_rng(0)
N = 1000
n = 1000
T = 1.0

Q_samples = np.empty(N)
for k in range(N):
    dW = rng.normal(0, np.sqrt(T / n), size=n)
    Q_samples[k] = np.sum(dW ** 2)

mean_Q = Q_samples.mean()
std_Q = Q_samples.std(ddof=1)
theoretical_std = np.sqrt(2 * T**2 / n)

print(f"Empirical mean(Q_n) = {mean_Q:.4f}   (theoretical 1.0000)")
print(f"Empirical std(Q_n)  = {std_Q:.4f}   (theoretical {theoretical_std:.4f})")
# Empirical mean(Q_n) = 1.0012   (theoretical 1.0000)
# Empirical std(Q_n)  = 0.0446   (theoretical 0.0447)

The mean is indistinguishable from

T = 1

and the standard deviation matches

\sqrt{2T^2/n} = \sqrt{2/1000} \approx 0.0447

to three decimal places. At finite

n

Q_n

is a random quantity — it fluctuates around

T

with spread

\sqrt{2T^2/n}

. The convergence

Q_n \to T

L^2

is exactly the statement that this spread vanishes as

n \to \infty

Part 4 — Why both statements are consistent

The two sums behave differently because of how they weight increments of typical size $\sqrt{\Delta t}$ . Each $|\Delta W_i| \sim \sqrt{\Delta t}$ , so:

First-order variation: $V_n = \sum |\Delta W_i| \sim n \cdot \sqrt{\Delta t} = \sqrt{n}\cdot\sqrt{T} \to \infty$ .
Quadratic variation: $Q_n = \sum (\Delta W_i)^2 \sim n \cdot \Delta t = T$ (finite).

The squaring converts a term of size

\sqrt{\Delta t}

into one of size

\Delta t

, which when summed

n = T/\Delta t

times yields exactly

T

. The absolute value does not, so the sum of absolute increments blows up. This is the defining signature of a rough path: infinite variation at first order, finite variation at order 2.

Takeaways