Itô Isometry¶

1. Concept Definition¶

On a simple adapted process, \(\int_0^T \beta\,dW\) is a finite sum of products \(\beta_{t_k}(W_{t_{k+1}}-W_{t_k})\) -- independent, mean-zero, with variance \(\beta_{t_k}^2(t_{k+1}-t_k)\). Summing the variances gives \(\sum \beta_{t_k}^2(t_{k+1}-t_k)\), which is exactly a Riemann sum for \(\int_0^T \beta^2\,dt\). Passing to the \(L^2\)-limit preserves this identity, producing the Itô isometry: for an adapted, square-integrable process \(\beta(t)\),

In words: the \(L^2\) norm of the stochastic integral equals the \(L^2\) norm of the integrand in the space \(\mathcal{L}^2([0,T])\).

This identity plays the same role for stochastic integrals that Parseval's identity plays for Fourier series: energy is preserved under the transform. Stochastic integration is an isometry between two Hilbert spaces:

Distances in \(\mathcal{L}^2([0,T])\) (measured by \(\mathbb{E}[\int \beta^2\,dt]\)) are preserved as distances in \(L^2(\Omega)\) (measured by \(\mathbb{E}[(\int \beta\,dW)^2]\)). This Hilbert space structure is what makes the \(L^2\)-extension of the integral possible.

2. Explanation¶

Why the identity holds¶

Recall (see § Construction of the Itô Integral): for a simple adapted process \(\beta(t) = \beta_i\) on \([t_i, t_{i+1})\), squaring the stochastic sum \(\sum_i \beta_i \Delta W_i\) produces a diagonal piece and a cross piece. Cross terms vanish in expectation because each future increment \(\Delta W_j\) is independent of \(\mathcal{F}_{t_j}\) with mean zero; diagonal terms reduce to \(\sum_i \mathbb{E}[\beta_i^2]\,(t_{i+1}-t_i)\), which is the Riemann sum for \(\mathbb{E}[\int_0^T \beta^2\,dt]\). The proof for simple processes is given in full there, and the identity extends to all of \(\mathcal{L}^2([0,T])\) by continuity.

The conceptual content: orthogonality¶

The two facts driving the identity are:

• Brownian increments on disjoint intervals are orthogonal in \(L^2(\Omega)\) — independent with mean zero.
• \(\mathbb{E}[(\Delta W_i)^2] = \Delta t_i\), the quadratic-variation fact \([W]_t = t\) (canonical home: § Quadratic Variation).

Orthogonality kills cross terms; quadratic variation converts \((\Delta W)^2\) into \(\Delta t\). The combination is what gives stochastic integration a Hilbert-space (isometry) structure.

Connection to the L² extension¶

The isometry is not just an interesting identity—it is the mechanism of construction. Recall (see § Construction of the Itô Integral): the three-step extension (define on simple processes, prove isometry, complete in \(L^2(\Omega)\)) fails without the isometry because step 2 (Cauchy approximation) requires it. The isometry is the stochastic analogue of Parseval's identity: energy in equals energy out.

3. Diagram¶

flowchart LR
    A["Brownian increments ΔW_i orthogonal in L²(Ω)"] --> B["Multiply by strategy β_i"]
    B --> C["Square the stochastic sum"]
    C --> D["Cross terms vanish (independence + mean zero)"]
    C --> E["Diagonal terms survive: (ΔW_i)² ≈ Δt_i"]
    E --> F["Riemann sum of β² → ∫ β² dt"]
    D --> G["E[(∫ β dW)²] = E[∫ β² dt]"]
    F --> G

4. Examples¶

Example 1: Deterministic integrand β(t) = t¶

Since \(\beta\) is deterministic, the Itô isometry gives:

A deterministic integrand applied to a Gaussian process gives a Gaussian result, so:

Example 2: Random integrand β(t) = t Wₜ¶

The integral is not Gaussian since \(\beta(t) = tW_t\) is a random integrand.

Example 3: Numerical verification¶

The following script verifies the Itô isometry by comparing both sides across three integrands.

```python import numpy as np

T = 3.0 n_per_year = 252 N = int(T * n_per_year) dt = T / N sqrt_dt = np.sqrt(dt)

n_paths = 20000 seed = 42 rng = np.random.default_rng(seed)

t = np.linspace(0.0, T, N + 1)

Brownian motion from fair coins (Donsker approximation)¶

coins = rng.choice([-1, 1], size=(n_paths, N)) dB = coins * sqrt_dt

B = np.zeros((n_paths, N + 1)) B[:, 1:] = np.cumsum(dB, axis=1)

def ito_integral(integrand_values, dB): """Cumulative left-endpoint Itô sum.""" dI = integrand_values * dB I = np.zeros((dI.shape[0], dI.shape[1] + 1)) I[:, 1:] = np.cumsum(dI, axis=1) return I

Three integrands¶

H1 = t[:-1] * B[:, :-1] # H_s = s B_s H2 = t[:-1] * (B[:, :-1] ** 2) # H_s = s B_s^2 H3 = -np.cos(B[:, :-1]) # H_s = -cos(B_s)

I1_T = ito_integral(H1, dB)[:, -1] I2_T = ito_integral(H2, dB)[:, -1] I3_T = ito_integral(H3, dB)[:, -1]

def check_ito_isometry(name, H, I_T, dt): lhs = np.mean(I_T ** 2) rhs = np.mean(np.sum(H ** 2 * dt, axis=1)) rel_err = abs(lhs - rhs) / rhs if rhs != 0 else float("nan") print(f"{name}") print(f" LHS E[(∫ H dB)²] = {lhs:.6f}") print(f" RHS E[∫ H² dt] = {rhs:.6f}") print(f" Relative error = {rel_err:.4%}")

check_ito_isometry("H_s = s B_s (theoretical RHS = T⁴/4 = 20.25)", H1, I1_T, dt) check_ito_isometry("H_s = s B_s²", H2, I2_T, dt) check_ito_isometry("H_s = -cos(B_s)", H3, I3_T, dt) ```

Typical output:

text H_s = s B_s (theoretical RHS = T⁴/4 = 20.25) LHS E[(∫ H dB)²] = 20.738498 RHS E[∫ H² dt] = 20.231128 Relative error = 2.51% H_s = s B_s² LHS E[(∫ H dB)²] = 151.042892 RHS E[∫ H² dt] = 145.266226 Relative error = 3.98% H_s = -cos(B_s) LHS E[(∫ H dB)²] = 1.749029 RHS E[∫ H² dt] = 1.744969 Relative error = 0.23%

Both sides agree closely in all three cases. The residual errors (~2–4% for the first two cases, <1% for the third) arise from two sources: discretization bias from the coin-flip approximation to Brownian motion, and Monte Carlo sampling error from using a finite number of paths. Both errors decrease as \(N\) and n_paths increase.

For Case 1, the theoretical value is \(\mathbb{E}[\int_0^T t^2 B_t^2\, dt] = \int_0^T t^3\, dt = T^4/4 = 20.25\), which the Monte Carlo estimates bracket from above and below, confirming the isometry.

5. Summary¶

• The Itô isometry converts the variance of a stochastic integral into an ordinary (Lebesgue) integral of the squared integrand.
• Cross terms vanish because Brownian increments on disjoint intervals are orthogonal random variables in \(L^2(\Omega)\).
• The identity gives stochastic integration a Hilbert space structure: it is an isometry \(\mathcal{L}^2([0,T]) \to L^2(\Omega)\).
• This structure enables the \(L^2\)-extension of the integral from simple processes to all adapted square-integrable processes.
• Nearly every structural result about Itô integrals—existence, uniqueness, continuity, the martingale property—traces back to this theorem.

Exercises¶

Exercise 1. Compute the variance of the Ito integral \(\int_0^2 (3t + 1)\, dW_t\) using the Ito isometry.

Exercise 2. Let \(\beta(t) = e^{-\alpha t}\) for \(\alpha > 0\). Compute

and find the limit as \(T \to \infty\).

Exercise 3. In the proof of the Ito isometry for simple processes, the cross terms \(\mathbb{E}[\beta_i \beta_j \Delta W_i \Delta W_j] = 0\) for \(i < j\). Explain why this argument relies on the left-endpoint evaluation. What would happen if we used right-endpoint evaluation instead?

Exercise 4. Let \(\beta(t) = W_t^2\). Use the Ito isometry to compute

Hint: You will need \(\mathbb{E}[W_t^4] = 3t^2\).

Exercise 5. The Ito isometry extends to the polarized form: \(\mathbb{E}[\int_0^T H\, dW \cdot \int_0^T K\, dW] = \mathbb{E}[\int_0^T HK\, dt]\). Use this to compute the covariance \(\operatorname{Cov}(\int_0^1 t\, dW_t,\; \int_0^1 t^2\, dW_t)\).

Exercise 6. Consider two simple processes \(H^{(n)}\) and \(H^{(m)}\) that approximate the same integrand \(H \in \mathcal{L}^2([0,T])\). Using the Ito isometry, show that

as \(n, m \to \infty\). Explain why this makes the sequence of integrals a Cauchy sequence in \(L^2(\Omega)\).

Exercise 7. The analogy between the Ito isometry and Parseval's identity states that stochastic integration preserves \(L^2\) norms. In Fourier analysis, Parseval's identity says \(\sum_n |c_n|^2 = \frac{1}{2\pi}\int |f|^2\, dx\). Write a brief comparison: what plays the role of the Fourier coefficients \(c_n\) in the stochastic setting, and what plays the role of the \(L^2\) norm of \(f\)?