Using the Itô Isometry to Find a Stochastic Integral’s Variance

For a predictable integrand $H$ with $\mathbb E\int_0^T H_t^2\,dt<\infty$ , the integral $I_T=\int_0^T H_t\,dW_t$ has mean zero. The Itô isometry gives

\operatorname{Var}(I_T)=\mathbb E[I_T^2] =\mathbb E\int_0^T H_t^2\,dt =\int_0^T \mathbb E[H_t^2]\,dt.

Square the integrand before taking the expectation. A random integrand can produce a non-Gaussian integral even though the integrator is Brownian motion.

Why the off-diagonal terms have zero expectation

For a step process with values $H_i$ measurable at the start $t_i$ , write the integral as $\sum_i H_i\Delta W_i$ . If $i<j$ , the product $H_i\Delta W_iH_j$ is measurable at $t_j$ , and $\mathbb E[\Delta W_j\mid\mathcal F_{t_j}]=0$ . Thus the mixed terms have zero expectation. For the diagonal terms,

\mathbb E[(H_i\Delta W_i)^2] =\mathbb E\!\left[H_i^2\mathbb E[(\Delta W_i)^2\mid\mathcal F_{t_i}]\right] =(t_{i+1}-t_i)\mathbb E[H_i^2].

Summing gives the isometry for step processes. Approximating a square-integrable predictable $H$ in mean square extends the identity to its stochastic integral. Tonelli's theorem permits exchanging expectation and integration in the nonnegative squared integrand.

The factor

H_j

can depend on earlier increments, so it generally cannot be separated from the earlier summand by independence. Instead condition on

\mathcal F_{t_j}

. Everything up to the start of interval

j

, including

H_j

, is then known, while the new Brownian increment has conditional mean zero. On the diagonal, the new increment has conditional second moment

t_{j+1}-t_j

. These two observations produce the sum of expected squared coefficients times interval lengths. Square-integrable approximation extends that identity to the integrands used here.

Integrate the expected squared integrand

For an original exercise, find the variance of

\int_0^2(W_t+t)\,dW_t

. Expand

\mathbb E[(W_t+t)^2]=t+2t\mathbb E[W_t]+t^2=t+t^2

. Then

\operatorname{Var}(I_2)=\int_0^2(t+t^2)\,dt =2+\frac83=\frac{14}{3}.

The

W_t

part contributes

2

and the

t

part contributes

8/3

. Their covariance is zero by the bilinear form of the isometry, since

\int_0^2 \mathbb E[tW_t]\,dt=0

. That orthogonality does not imply independence.

For the Brownian part alone, Itô's lemma gives

\int_0^2 W_t\,dW_t=(W_2^2-2)/2

. Since

W_2^2/2

has a chi-squared distribution with one degree of freedom, the variance of this part is

2

, matching the isometry. This supplies a direct example of a Brownian stochastic integral with a non-Gaussian distribution.

Check the variance over simulated paths

import numpy as np

rng = np.random.default_rng(42)
paths, steps, horizon = 20_000, 256, 2.0
dt = horizon / steps
d_w = rng.normal(0.0, np.sqrt(dt), (paths, steps))
w_left = np.cumsum(d_w, axis=1) - d_w
t_left = np.arange(steps) * dt

integrals = np.sum((w_left + t_left) * d_w, axis=1)
print(integrals.mean(), integrals.var(ddof=1), 14 / 3)

The finite grid approximates the integral with left endpoints. Increase the number of paths to reduce sampling noise and refine the grid to improve the time approximation.

Why the off-diagonal terms have zero expectation

Integrate the expected squared integrand

Check the variance over simulated paths

References and next steps