Quadratic Variation¶

1. Concept Definition¶

Quadratic variation measures the cumulative size of squared increments of a stochastic process. For a partition \(\Pi = \{0 = t_0 < t_1 < \cdots < t_n = T\}\) with mesh \(\|\Pi\| = \max_k(t_k - t_{k-1})\), the quadratic variation sum of a process \(X\) is

\[ Q(X, \Pi) := \sum_{k=1}^n (X_{t_k} - X_{t_{k-1}})^2 \]

If this sum converges as \(\|\Pi\| \to 0\), the limit is the quadratic variation \([X]_T\). More precisely, the quadratic variation process \([X]_t\) is defined on \([0,t]\) for each \(t \ge 0\).

Two facts are foundational:

\[ \boxed{[W]_t = t} \qquad\text{for standard Brownian motion } W_t \]

\[ \boxed{[A]_T = 0} \qquad\text{for any process of finite (total) variation} \]

These two facts have a single consequence: Brownian motion is categorically different from any smooth or piecewise smooth path. A differentiable function has zero quadratic variation; Brownian motion has quadratic variation equal to \(t\). This distinction is the mechanism behind every correction term in stochastic calculus.

2. Explanation¶

Why quadratic variation matters¶

In ordinary calculus, when expanding \(f(x + h) - f(x)\) by Taylor's theorem, the second-order term \(f''(x)h^2/2\) is negligible because \(h^2 \ll h\) as \(h \to 0\). In stochastic calculus, the second-order term survives because Brownian increments satisfy \((\Delta W)^2 \approx \Delta t\) rather than \((\Delta W)^2 \ll \Delta t\).

This is why Itô's formula has an extra term compared to the ordinary chain rule:

\[ df(W_t) = f'(W_t)\,dW_t + \frac{1}{2}f''(W_t)\,dt \]

The \(\frac{1}{2}f''(W_t)\,dt\) term comes from summing \(f''(W_{t_k})(\Delta W_k)^2 \approx f''(W_{t_k})\Delta t_k \to \int f''(W_t)\,dt\). This is exactly the quadratic variation.

Proof that \([W]_t = t\)¶

We show \(Q(W, \Pi) \to t\) in \(L^2\) (and hence in probability) as \(\|\Pi\| \to 0\).

Write \(\Delta W_k = W_{t_k} - W_{t_{k-1}}\) and \(\Delta t_k = t_k - t_{k-1}\). Since \(\Delta W_k \sim \mathcal{N}(0, \Delta t_k)\) and the increments are independent:

\[ \mathbb{E}[Q(W, \Pi)] = \sum_k \mathbb{E}[(\Delta W_k)^2] = \sum_k \Delta t_k = t \]

For the variance, use \(\operatorname{Var}((\Delta W_k)^2) = 2(\Delta t_k)^2\) (fourth moment of a centred normal):

\[ \operatorname{Var}(Q(W, \Pi)) = \sum_k \operatorname{Var}((\Delta W_k)^2) = 2\sum_k (\Delta t_k)^2 \le 2\|\Pi\| \sum_k \Delta t_k = 2\|\Pi\|\, t \;\to\; 0 \]

Since \(\mathbb{E}[Q(W,\Pi)] = t\) and \(\operatorname{Var}(Q(W,\Pi)) \to 0\), we have \(Q(W,\Pi) \to t\) in \(L^2\). \(\square\)

Why smooth paths have zero quadratic variation¶

For a function \(f: [0,T] \to \mathbb{R}\) with \(|f'(t)| \le M\) for all \(t\), each increment satisfies \(|f(t_k) - f(t_{k-1})| \le M \Delta t_k\), so

\[ Q(f, \Pi) = \sum_k (f(t_k) - f(t_{k-1}))^2 \le M \sum_k |\Delta f_k| \cdot \Delta t_k \le M^2 \sum_k (\Delta t_k)^2 \le M^2 \|\Pi\| T \to 0 \]

More generally, any process of bounded variation has zero quadratic variation.

The Itô multiplication table¶

Quadratic variation is the rigorous content behind the heuristic rules used in stochastic calculus. For a multidimensional Brownian motion \(W_t = (W_t^1, \ldots, W_t^m)\):

\[ \boxed{ (dt)^2 = 0, \qquad dt\, dW_t^\alpha = 0, \qquad dW_t^\alpha\, dW_t^\beta = \delta^{\alpha\beta}\, dt } \]

These rules arise from the covariation formula \([W^\alpha, W^\beta]_t = \delta^{\alpha\beta} t\).

Derivation of \(dW^\alpha\, dW^\beta = \delta^{\alpha\beta}\,dt\). For \(\alpha = \beta\): we have just shown \([W^\alpha]_t = t\), so \((dW^\alpha)^2 \to dt\). For \(\alpha \ne \beta\): \(W^\alpha\) and \(W^\beta\) are independent Brownian motions with independent increments \(\Delta W_k^\alpha\) and \(\Delta W_k^\beta\), so

\[ \mathbb{E}\!\left[\sum_k \Delta W_k^\alpha \Delta W_k^\beta\right] = \sum_k \mathbb{E}[\Delta W_k^\alpha]\mathbb{E}[\Delta W_k^\beta] = 0 \]

and the variance of the sum also tends to zero, giving \([W^\alpha, W^\beta]_t = 0\).

Quadratic variation of Itô integrals¶

Let \(M_t = \int_0^t H_s\, dW_s\) with \(H\) square-integrable and adapted. Then \(M\) is a continuous martingale and

\[ \boxed{[M]_t = \int_0^t H_s^2\, ds} \]

More generally, for \(M_t = \int_0^t H_s\, dW_s\) and \(N_t = \int_0^t K_s\, dW_s\):

\[ \boxed{[M, N]_t = \int_0^t H_s K_s\, ds} \]

This generalizes both formulas: setting \(H_s = K_s = 1\) gives \([W,W]_t = t\).

3. Diagram¶

flowchart TD
    A["Partition Π of [0,T]"]
    A --> B["Quadratic variation sum: Q(X, Π) = Σ(ΔX_k)²"]

    B --> C{"Process type?"}

    C --> D["Smooth / finite variation — e.g. f(t) = t, sin(t)"]
    C --> E["Brownian motion W_t"]
    C --> F["Itô integral M_t = ∫H_s dW_s"]

    D --> G["(ΔX)² ~ (Δt)² — sum → 0 — [A]_T = 0"]
    E --> H["(ΔW)² ~ Δt — E[Q] = t, Var[Q] → 0 — [W]_T = T"]
    F --> I["(ΔM)² ~ H²Δt — [M]_T = ∫H_s² ds"]

    H --> J["Itô formula correction: ½f''(W_t)dt"]
    I --> J
    G --> K["Ordinary chain rule — no correction needed"]

4. Examples¶

Example 1: Quadratic variation of \(W_t\) — numerical verification¶

The following script simulates many Brownian paths and plots the quadratic variation sum against the theoretical value \(t\).

```python import numpy as np import matplotlib.pyplot as plt

T = 1.0 N = 500 dt = T / N n_paths = 2000 seed = 42 rng = np.random.default_rng(seed)

t = np.linspace(0, T, N + 1)

dW = rng.standard_normal((n_paths, N)) * np.sqrt(dt) W = np.concatenate([np.zeros((n_paths, 1)), np.cumsum(dW, axis=1)], axis=1)

Cumulative quadratic variation: [W]t = Σ (ΔW_k)²¶

qv = np.cumsum(dW ** 2, axis=1) qv = np.concatenate([np.zeros((n_paths, 1)), qv], axis=1)

fig, axes = plt.subplots(1, 2, figsize=(13, 5))

Left: several sample paths of cumulative QV versus t¶

ax = axes[0] for i in range(10): ax.plot(t, qv[i], "b", alpha=0.4, linewidth=0.8) ax.plot(t, t, "r--", linewidth=2, label=r"\([W]_t = t\) (theoretical)") ax.plot(t, qv.mean(axis=0), "k", linewidth=2, label="Monte Carlo mean") ax.set_title("Cumulative quadratic variation of \(W_t\)") ax.set_xlabel("\(t\)") ax.set_ylabel("\([W]_t\)") ax.legend() ax.grid(True)

Right: QV at T=1 — distribution across paths¶

ax = axes[1] ax.hist(qv[:, -1], bins=60, density=True, alpha=0.7, edgecolor="black") ax.axvline(1.0, color="r", linestyle="--", linewidth=2, label=r"\([W]_T = T = 1\)") ax.axvline(qv[:, -1].mean(), color="k", linestyle="-", linewidth=2, label=f"Monte Carlo mean = {qv[:, -1].mean():.4f}") ax.set_title(r"Distribution of \([W]_1\) across paths") ax.set_xlabel(r"\([W]_1\)") ax.set_ylabel("Density") ax.legend() ax.grid(True)

plt.tight_layout() plt.savefig("./image/quadratic_variation_figure.png", dpi=150) plt.show()

print(f"Monte Carlo mean of [W]_1 : {qv[:, -1].mean():.6f}") print(f"Monte Carlo std of [W]_1 : {qv[:, -1].std():.6f}") print(f"Theoretical value : {T:.6f}") print(f"Theoretical std (N={N}) : {np.sqrt(2 * dt * T):.6f}") ```

Quadratic variation figure

Each sample path of \([W]_t\) (blue) fluctuates around the deterministic line \(t\) (red). The Monte Carlo mean (black) tracks \(t\) closely. The right panel shows that \([W]_1\) concentrates tightly around \(1\)—consistent with \(\operatorname{Var}([W]_T) = 2T \cdot \Delta t \to 0\) as \(\Delta t \to 0\).

Example 2: Quadratic variation of the Ornstein-Uhlenbeck process¶

The OU process \(dX_t = -\theta X_t\,dt + \sigma\,dB_t\) has \(\sigma_t = \sigma\) (constant). Therefore:

\[ [X]_t = \int_0^t \sigma^2\, ds = \sigma^2 t \]

The drift \(-\theta X_t\,dt\) contributes nothing to the quadratic variation. This confirms that quadratic variation detects only the diffusion coefficient, regardless of the drift.

Example 3: Itô's formula from quadratic variation¶

Apply Itô's formula to \(f(x) = x^2\) with \(X_t = W_t\):

\[ W_t^2 = W_0^2 + \int_0^t 2W_s\,dW_s + \int_0^t 1\, d[W]_s = 0 + 2\int_0^t W_s\,dW_s + t \]

where \(d[W]_s = ds\) since \([W]_s = s\).

Rearranging:

\[ \int_0^t W_s\, dW_s = \frac{W_t^2 - t}{2} \]

The \(-t/2\) correction is exactly the quadratic variation of \(W\) on \([0,t]\). In ordinary calculus, \(\int_0^t x\, dx = x^2/2\); in stochastic calculus, the answer is \(W_t^2/2 - t/2\) because the quadratic variation term \(\frac{1}{2} \cdot 2 \cdot [W]_t = t\) must be subtracted.

5. Summary¶

The quadratic variation \([X]_t\) is the \(L^2\) (and probability) limit of the sum of squared increments \(\sum_k (\Delta X_k)^2\) as the partition mesh tends to zero.
For Brownian motion, \([W]_t = t\), proved by computing that \(\mathbb{E}[Q] = t\) and \(\operatorname{Var}(Q) \to 0\).
For smooth or finite-variation processes, \([A]_t = 0\).
For an Itô integral \(M_t = \int_0^t H_s\,dW_s\), the quadratic variation is \([M]_t = \int_0^t H_s^2\,ds\).
The Itô multiplication table \(dW^\alpha\,dW^\beta = \delta^{\alpha\beta}\,dt\) is the symbolic expression of these quadratic variation identities.
Quadratic variation is the reason Itô's formula differs from the ordinary chain rule: second-order Taylor terms survive in the stochastic setting because \((\Delta W)^2 \approx \Delta t \not\to 0\).

Exercises¶

Exercise 1. Let \(f(t) = t^2\) on \([0,1]\). For a uniform partition with \(n\) subintervals, compute the quadratic variation sum \(Q(f, \Pi) = \sum_{k=1}^n (f(t_k) - f(t_{k-1}))^2\) explicitly. Show that \(Q(f, \Pi) \to 0\) as \(n \to \infty\), confirming that smooth functions have zero quadratic variation.

Solution to Exercise 1

On a uniform partition \(t_k = k/n\), \(\Delta t = 1/n\), and \(f(t_k) = (k/n)^2\). The increments are:

\[ f(t_k) - f(t_{k-1}) = \frac{k^2}{n^2} - \frac{(k-1)^2}{n^2} = \frac{2k - 1}{n^2} \]

The quadratic variation sum is:

\[ Q(f, \Pi) = \sum_{k=1}^n \left(\frac{2k-1}{n^2}\right)^2 = \frac{1}{n^4}\sum_{k=1}^n (2k-1)^2 \]

Using \(\sum_{k=1}^n (2k-1)^2 = \frac{n(2n-1)(2n+1)}{3}\):

\[ Q(f, \Pi) = \frac{(2n-1)(2n+1)}{3n^3} = \frac{4n^2 - 1}{3n^3} \]

As \(n \to \infty\):

\[ Q(f, \Pi) = \frac{4n^2 - 1}{3n^3} \sim \frac{4}{3n} \to 0 \]

This confirms that the smooth function \(f(t) = t^2\) has zero quadratic variation.

Exercise 2. Using the identity \(\operatorname{Var}((\Delta W_k)^2) = 2(\Delta t_k)^2\) for a Gaussian increment, show that

\[ \operatorname{Var}(Q(W, \Pi)) = 2 \sum_{k=1}^n (\Delta t_k)^2 \]

For a uniform partition with \(n\) subintervals on \([0,T]\), evaluate this expression and verify it tends to zero as \(n \to \infty\).

Solution to Exercise 2

Since \(\Delta W_k \sim \mathcal{N}(0, \Delta t_k)\) and increments are independent:

\[ Q(W, \Pi) = \sum_{k=1}^n (\Delta W_k)^2 \]

Each \((\Delta W_k)^2\) is an independent random variable. For a centered Gaussian \(Z \sim \mathcal{N}(0, \sigma^2)\), \(\operatorname{Var}(Z^2) = \mathbb{E}[Z^4] - (\mathbb{E}[Z^2])^2 = 3\sigma^4 - \sigma^4 = 2\sigma^4\).

So \(\operatorname{Var}((\Delta W_k)^2) = 2(\Delta t_k)^2\), and by independence:

\[ \operatorname{Var}(Q(W, \Pi)) = \sum_{k=1}^n \operatorname{Var}((\Delta W_k)^2) = 2\sum_{k=1}^n (\Delta t_k)^2 \]

For a uniform partition on \([0, T]\): \(\Delta t_k = T/n\) for all \(k\):

\[ \operatorname{Var}(Q(W, \Pi)) = 2\sum_{k=1}^n \frac{T^2}{n^2} = 2n \cdot \frac{T^2}{n^2} = \frac{2T^2}{n} \to 0 \quad \text{as } n \to \infty \]

Exercise 3. Let \(X_t = \mu t + \sigma W_t\) be Brownian motion with drift. Compute the quadratic variation \([X]_t\). Does the drift \(\mu t\) contribute to the quadratic variation? Justify your answer.

Solution to Exercise 3

Write \(X_t = \mu t + \sigma W_t\). The increment is \(\Delta X_k = \mu \Delta t_k + \sigma \Delta W_k\). The quadratic variation sum is:

\[ \sum_k (\Delta X_k)^2 = \sum_k (\mu \Delta t_k + \sigma \Delta W_k)^2 \]

\[ = \mu^2 \sum_k (\Delta t_k)^2 + 2\mu\sigma \sum_k \Delta t_k \Delta W_k + \sigma^2 \sum_k (\Delta W_k)^2 \]

As the mesh \(\|\Pi\| \to 0\):

\(\sum_k (\Delta t_k)^2 \le \|\Pi\| \cdot T \to 0\)
\(\sum_k \Delta t_k \Delta W_k \to 0\) in probability (by a similar argument, the variance is bounded by \(\sum_k (\Delta t_k)^2 \cdot \Delta t_k \to 0\))
\(\sum_k (\Delta W_k)^2 \to t\) in \(L^2\)

Therefore \([X]_t = \sigma^2 t\). The drift \(\mu t\) does not contribute to the quadratic variation, because the drift increments \(\mu \Delta t_k\) are of order \(\Delta t_k\), so their squares \(\mu^2 (\Delta t_k)^2\) are of order \((\Delta t_k)^2\) and vanish when summed.

Exercise 4. Let \(M_t = \int_0^t s\, dW_s\). Compute the quadratic variation \([M]_t\) and use it to verify that \(M_t^2 - [M]_t\) is a martingale by computing \(\mathbb{E}[M_t^2 - [M]_t]\).

Solution to Exercise 4

Since \(M_t = \int_0^t s\, dW_s\), the quadratic variation is:

\[ [M]_t = \int_0^t s^2\, ds = \frac{t^3}{3} \]

To verify that \(M_t^2 - [M]_t\) is a martingale, compute its expectation. By the Ito isometry:

\[ \mathbb{E}[M_t^2] = \mathbb{E}\!\left[\int_0^t s^2\, ds\right] = \frac{t^3}{3} \]

Therefore:

\[ \mathbb{E}[M_t^2 - [M]_t] = \frac{t^3}{3} - \frac{t^3}{3} = 0 \]

for all \(t\), which is consistent with \(M_t^2 - [M]_t\) being a martingale (a martingale starting at zero has constant expectation equal to zero).

Exercise 5. Using the Ito multiplication table, compute \((dX_t)^2\) for the Ornstein-Uhlenbeck process \(dX_t = -\theta X_t\, dt + \sigma\, dW_t\). What is \([X]_t\)?

Solution to Exercise 5

The OU process has \(dX_t = -\theta X_t\, dt + \sigma\, dW_t\). Using the multiplication rules:

\[ (dX_t)^2 = (-\theta X_t\, dt + \sigma\, dW_t)^2 \]

Expanding:

\[ = \theta^2 X_t^2 (dt)^2 - 2\theta X_t \sigma\, dt\, dW_t + \sigma^2 (dW_t)^2 \]

Applying \((dt)^2 = 0\), \(dt\, dW_t = 0\), \((dW_t)^2 = dt\):

\[ (dX_t)^2 = \sigma^2\, dt \]

Therefore \(d[X]_t = \sigma^2\, dt\), giving \([X]_t = \sigma^2 t\).

Solution to Exercise 6

Let \(V = \sup_\Pi \sum_k |A_{t_k} - A_{t_{k-1}}| < \infty\) be the total variation of \(A\) on \([0,T]\). For any partition \(\Pi\):

Since \(A_t\) is continuous (and hence uniformly continuous on \([0,T]\)), as the mesh \(\|\Pi\| \to 0\), \(\max_k |\Delta A_k| \to 0\). Since \(V < \infty\) is a fixed finite constant:

\[ \sum_k (\Delta A_k)^2 \le \left(\max_k |\Delta A_k|\right) \cdot V \to 0 \cdot V = 0 \]

Therefore \([A]_T = 0\).

Exercise 7. Apply Ito's formula to \(f(x) = x^4\) with \(X_t = W_t\) to derive

\[ W_t^4 = 4\int_0^t W_s^3\, dW_s + 6\int_0^t W_s^2\, ds \]

Use this identity and the Ito isometry to compute \(\mathbb{E}\!\left[\left(\int_0^t W_s^3\, dW_s\right)^2\right]\). Hint: You will need \(\mathbb{E}[W_s^6] = 15s^3\).

Solution to Exercise 7

Apply Ito's formula to \(f(x) = x^4\) with \(X_t = W_t\). We have \(f'(x) = 4x^3\) and \(f''(x) = 12x^2\):

\[ W_t^4 = 0 + \int_0^t 4W_s^3\, dW_s + \frac{1}{2}\int_0^t 12W_s^2\, ds = 4\int_0^t W_s^3\, dW_s + 6\int_0^t W_s^2\, ds \]

Now compute \(\mathbb{E}[(\int_0^t W_s^3\, dW_s)^2]\) using the Ito isometry:

\[ \mathbb{E}\!\left[\left(\int_0^t W_s^3\, dW_s\right)^2\right] = \mathbb{E}\!\left[\int_0^t W_s^6\, ds\right] = \int_0^t \mathbb{E}[W_s^6]\, ds \]

Using \(\mathbb{E}[W_s^6] = 15s^3\) (the sixth moment of \(\mathcal{N}(0, s)\) is \(15s^3\)):

\[ = \int_0^t 15s^3\, ds = 15 \cdot \frac{t^4}{4} = \frac{15t^4}{4} \]