The Stratonovich Integral¶

1. Concept Definition¶

The Stratonovich integral is a stochastic integral defined using midpoint sampling:

The circle notation \(\circ\,dB_t\) distinguishes it from the Itô integral, which uses left-endpoint sampling. Both integrals are mathematically valid, but they assign different values to the same formal expression.

The core example that illustrates the difference is

The two integrals differ by \(\frac{1}{2}t\), which is exactly half the quadratic variation of Brownian motion on \([0,t]\). The midpoint rule picks up this additional contribution because it evaluates the integrand symmetrically—partially in the future—rather than strictly in the past.

In Stratonovich form, stochastic differentials obey the classical chain rule:

This makes Stratonovich calculus the natural choice in physics and geometric modeling, where coordinate invariance and classical differential rules are essential.

2. Why Midpoint Sampling Changes the Limit¶

The choice of evaluation point in the Riemann sum determines which stochastic integral we obtain. We illustrate with the integrand \(f(t) = B_t\).

flowchart TD
    A["Partition 0 = t_0 < t_1 < ... < t_n = T"] --> B["Choose evaluation point on each interval"]
    B --> C["Left endpoint t_j"]
    B --> D["Midpoint ½(t_j + t_(j+1))"]

    C --> E["Use B_{t_j}"]
    E --> F["Σ B_{t_j}(B_{t_(j+1)} - B_{t_j})"]
    F --> G["Itô integral ½(B_T² - T)"]

    D --> H["Use ½(B_{t_j} + B_{t_(j+1)})"]
    H --> I["Σ ½(B_{t_j} + B_{t_(j+1)})(B_{t_(j+1)} - B_{t_j})"]
    I --> J["Stratonovich integral ½B_T²"]

    G --> K["Excludes quadratic variation"]
    J --> L["Captures half of quadratic variation"]

Left-endpoint (Itô)¶

Evaluating at the left endpoint \(t_j^* = t_j\):

Since Brownian increments are independent of the past and have mean zero, each term has \(\mathbb{E}[B_{t_j}(B_{t_{j+1}} - B_{t_j})] = 0\). This reflects the martingale property: the integral has zero mean.

Right-endpoint (for contrast only)¶

Evaluating at the right endpoint \(t_j^* = t_{j+1}\):

Writing \(B_{t_{j+1}} = B_{t_j} + (B_{t_{j+1}} - B_{t_j})\) and expanding:

Summing over all intervals: the right-endpoint sum picks up the full quadratic variation \(T\).

Midpoint (Stratonovich)¶

The midpoint rule evaluates at the average of left and right endpoints:

This is the average of the left-endpoint and right-endpoint evaluations, so it picks up exactly half of the quadratic variation. For \(\int_0^T B_t \circ dB_t\), the expected correction is \(T/2\):

This is exactly the result that the classical chain rule applied to \(\frac{d}{dt}\frac{1}{2}B_t^2 = B_t\,\frac{dB_t}{dt}\) would give—if we pretended \(B_t\) were differentiable.

3. The Stratonovich Chain Rule¶

The principal advantage of Stratonovich calculus is that stochastic differentials obey the same formal chain rule as ordinary calculus. For \(f \in C^2\) and an Itô process \(X_t\):

Stratonovich chain rule: \(df(X_t) = f'(X_t) \circ dX_t\)

Itô's formula: \(df(X_t) = f'(X_t)\,dX_t + \frac{1}{2}f''(X_t)\,(dX_t)^2\)

The Stratonovich form has no second-order correction. This is why Stratonovich integrals appear naturally when passing from smooth-noise models to white-noise limits (Wong-Zakai theorem), and in geometric problems where the chain rule must behave classically under coordinate changes.

4. Conversion Formula¶

The two integrals are related by a correction term equal to half the quadratic covariation between the integrand and the Brownian motion.

Suppose \(X_t\) satisfies \(dX_t = b(t,X_t)\,dt + \sigma(t,X_t)\,dW_t\). Then:

where \(\sigma(s,X_s)\) is the diffusion coefficient of \(X_t\). Equivalently, using quadratic covariation notation:

Note: the correction term depends on \(\sigma(s,X_s)\)—how strongly the Brownian noise affects the process \(X_t\). If \(f\) does not depend on \(X_t\) (deterministic integrand), then \(\partial f/\partial x = 0\) and the two integrals coincide.

5. Numerical Illustration¶

The following script simulates one Brownian path and compares the left-point and midpoint approximations to \(\int_0^T B_t\, dB_t\).

```python import matplotlib.pyplot as plt import numpy as np from dataclasses import dataclass from typing import Optional

@dataclass class BrownianMotionResult: time_steps: np.ndarray time_step_size: float brownian_paths: np.ndarray increments: np.ndarray

class BrownianMotion: DEFAULT_STEPS_PER_YEAR = 252

def __init__(self, maturity_time: float = 1.0, seed: Optional[int] = None):
    if maturity_time <= 0:
        raise ValueError("maturity_time must be positive")
    self.maturity_time = maturity_time
    self.rng = np.random.RandomState(seed)

def simulate(self, num_paths: int = 1,
             num_steps: Optional[int] = None) -> BrownianMotionResult:
    if num_steps is None:
        num_steps = int(self.maturity_time * self.DEFAULT_STEPS_PER_YEAR)
    time_steps = np.linspace(0, self.maturity_time, num_steps + 1)
    dt = time_steps[1] - time_steps[0]
    increments = self.rng.standard_normal((num_paths, num_steps)) * np.sqrt(dt)
    brownian_paths = np.concatenate(
        [np.zeros((num_paths, 1)), increments.cumsum(axis=1)], axis=1
    )
    return BrownianMotionResult(time_steps=time_steps, time_step_size=dt,
                                brownian_paths=brownian_paths, increments=increments)

if name == "main": N = 1000 bm = BrownianMotion(maturity_time=1.0, seed=42) result = bm.simulate(num_paths=1, num_steps=N)

W = result.brownian_paths[0]
dW = result.increments[0]
t = result.time_steps

# Itô integral (left-point): Σ W_{t_j} ΔW_j
ito = np.concatenate(([0.0], np.cumsum(W[:-1] * dW)))

# Stratonovich integral (midpoint): Σ ½(W_{t_j} + W_{t_{j+1}}) ΔW_j
strat = np.concatenate(([0.0], np.cumsum(0.5 * (W[:-1] + W[1:]) * dW)))

# Theoretical closed forms
ito_theory = 0.5 * (W**2 - t)
strat_theory = 0.5 * W**2

difference = strat - ito

plt.figure(figsize=(12, 6))
plt.plot(t, ito,        "b-",  label="Itô (left-point approximation)")
plt.plot(t, ito_theory, "b--", label=r"Exact Itô: $\frac{1}{2}(W_t^2 - t)$")
plt.plot(t, strat,        "r-",  label="Stratonovich (midpoint approximation)")
plt.plot(t, strat_theory, "r--", label=r"Exact Stratonovich: $\frac{1}{2}W_t^2$")
plt.plot(t, difference, "k-",  label=r"Stratonovich $-$ Itô")
plt.plot(t, 0.5 * t,    "k--", label=r"Exact difference: $\frac{1}{2}t$")
plt.legend(loc="upper left", fontsize=9)
plt.xlabel("$t$")
plt.title(r"Itô vs. Stratonovich for $\int_0^t B_s\,dB_s$")
plt.grid(True)
plt.tight_layout()
plt.savefig("./image/ito_and_stratonovich_integration.png", dpi=150)
plt.show()

```

The numerical approximations closely track their theoretical values. The black curve shows the difference between the Stratonovich and Itô approximations; it follows the \(t/2\) line (dashed black), confirming that the gap equals exactly half the quadratic variation.

6. Summary¶

The Stratonovich integral and the Itô integral differ by a correction term equal to half the quadratic covariation between the integrand and the driving Brownian motion. The correction is zero when the integrand does not depend on the stochastic process.

Exercises¶

Exercise 1. Compute the Stratonovich integral \(\int_0^t s \circ dB_s\). Does it differ from the Ito integral \(\int_0^t s\, dB_s\)? Explain why or why not, using the conversion formula.

Exercise 2. Using the conversion formula

compute \(\int_0^t B_s^2 \circ dB_s\) and verify that the result is consistent with the classical chain rule applied to \(g(x) = x^3/3\).

Exercise 3. The Stratonovich chain rule gives \(d(\sin B_t) = \cos(B_t) \circ dB_t\). Convert this to Ito form by finding \(d(\sin B_t)\) using Ito's formula. Identify the drift correction term and verify it matches the conversion formula.

Exercise 4. Consider the Stratonovich SDE \(dX_t = \sigma X_t \circ dB_t\) (no drift in Stratonovich form). Convert this to its equivalent Ito SDE. What drift term appears in the Ito form? Solve the resulting Ito SDE.

Exercise 5. Let \(f(x) = e^x\). Using the Stratonovich chain rule, write \(d(e^{B_t})\) in Stratonovich form. Then convert to Ito form and verify that you recover the standard result from Ito's formula.

Exercise 6. Using the coin-flip approximation with \(n = 10\) and the sequence \(H, T, H, H, T, H, T, T, H, H\), compute both the left-endpoint (Ito) and midpoint (Stratonovich) sums for \(\int_0^1 B_s\, dB_s\). Verify that their difference is approximately \(\frac{1}{2} \cdot 1 = 0.5\) times the quadratic variation sum \(\sum (\Delta B_k)^2\).

Exercise 7. Explain why the Stratonovich integral \(\int_0^t B_s \circ dB_s = \frac{1}{2}B_t^2\) is not a martingale, while the Ito integral \(\int_0^t B_s\, dB_s = \frac{1}{2}(B_t^2 - t)\) is a martingale. What property of the midpoint evaluation causes the martingale property to fail?