advanced29 min studyLesson 672 in Mathematics

Itô integral construction and isometry

Complete predictable step integrands in mean square and retain the information restriction.

On this page

What you’ll understand

  • Construct the Brownian integral from predictable step processes.
  • Use Itô isometry to compute second moments.
  • Distinguish global square integrability from a localized integral.

A Brownian path is too rough for an ordinary finite-variation integral, but its increments are orthogonal when weighted by information from their past. The Itô integral exploits that orthogonality. We retain the usual complete, right-continuous filtration and require Brownian future increments to be independent of its past. Begin on a finite horizon with predictable integrands square integrable in the joint time-probability measure; the resulting integral is an adapted continuous square-integrable martingale.

Start with information available on time

A simple predictable integrand is constant on each interval (ti,ti+1](t_i,t_{i+1}], with its coefficient measurable at time tit_i. Its integral is the coefficient-weighted sum of the corresponding Brownian increments. Endpoints are chosen this way because the weight cannot inspect the increment it multiplies. Independence and conditional mean zero make cross terms disappear when the square of the sum is averaged. This is a probabilistic construction, not a pathwise prescription that works for arbitrary anticipative weights.

Predictability is measurability with respect to the sigma algebra generated by adapted left-continuous processes. Continuous adapted processes are predictable. A general adapted process need not automatically meet every measurability requirement for integration, so adaptedness should not replace predictability without qualification. Predictable step processes are dense in the predictable square-integrable space. Approximating there gives a controlled norm in which different step approximations converge to the same stochastic integral.

Hs=∑iHi1(ti,ti+1](s),∫0THs dWs=∑iHi(Wti+1−Wti),Hi∈L2(Fti)H_s=\sum_i H_i\mathbf1_{(t_i,t_{i+1}]}(s),\qquad \int_0^T H_s\,dW_s=\sum_iH_i(W_{t_{i+1}}-W_{t_i}),\quad H_i\in L^2(\mathcal F_{t_i})
Each random weight is known before its Brownian increment occurs.↔ Scroll wide equations horizontally.

The isometry completes the definition

Itô isometry equates the integral's second moment with the expected time integral of the squared integrand. It first follows for step processes from the vanished cross terms. If integrands approach one another in that joint square norm, their integrals approach in mean square as well. Completeness therefore defines the integral for the limiting predictable integrand. The identity survives the limit, providing both uniqueness of the construction and a practical way to calculate variances without finding a full distribution.

The mean is zero, and applying the construction up to each intermediate time gives a martingale. A maximal inequality strengthens convergence to convergence in expected squared supremum for the resulting martingales, allowing a continuous version to be selected. This is why the terminal-value completion is consistent with a continuous process, rather than merely an unrelated random variable at each time. The integrand is identified only up to time-probability null sets, and such changes do not alter the integral.

E ⁣[(∫0THs dWs)2]=E∫0THs2 ds,E∫0THs dWs=0\mathbb E\!\left[\left(\int_0^T H_s\,dW_s\right)^2\right]=\mathbb E\int_0^T H_s^2\,ds,\qquad \mathbb E\int_0^T H_s\,dW_s=0
These statements require predictable H with finite expected squared integral on the stated horizon.↔ Scroll wide equations horizontally.

Localization extends the domain cautiously

If the squared time integral is finite almost surely but its expectation may be infinite, stop when that accumulated integral reaches a large level. The stopped integrands satisfy the square-integrable construction, and their integrals agree before the stopping times. Patching produces a continuous local martingale. Without further integrability it need not be a true martingale, so the assertion of zero expectation cannot simply be carried over. Local existence of an integral and global mean conservation are different conclusions.

A deterministic integrand has an additional Gaussian property: its integral is centered Gaussian with variance its squared time integral. A random predictable integrand generally does not produce a Gaussian random variable merely because isometry supplies its second moment. For example, integrating Brownian motion against itself will give a centered quadratic function of the terminal Brownian value. The integral remains a square-integrable martingale at each finite horizon, while its terminal distribution is visibly not a normal distribution.

A genuinely random step integrand

For Brownian motion on the interval from zero to two, let the integrand equal two on (0,1](0,1] and equal W1W_1 on (1,2](1,2]. Compute the stochastic integral, its mean, and its variance.

  1. Both coefficients are measurable at their left endpoints; the second coefficient uses the Brownian value at time one, not its future increment. The integral is 2W1+W1(W2−W1)2W_1+W_1(W_2-W_1).

  2. The future increment is independent of W1W_1 and centered. Consequently the product term has mean zero, and so does the first term. The complete integral therefore has mean zero.

  3. Itô isometry gives variance ∫014 ds+∫12EW12 ds=4+1=5\int_0^1 4\,ds+\int_1^2\mathbb E W_1^2\,ds=4+1=5. Expanding the square gives the same result: the cross term has zero expectation by conditioning at time one.

The result

The terminal integral is centered with variance five. The random coefficient does not prevent the construction, but it prevents an automatic claim of a Gaussian terminal distribution.

Explore the model

Interactive lab

Interactive lab · function

An isometry variance as an ordinary area

Pause/Reset, then set both coefficient to 1 and position to 1. Predict displayed area one third, equal to the variance of the stochastic integral of the deterministic integrand s from zero to one. Reset and set both coefficient to 1 and position to 2; predict area eight thirds. The plotted curve is the squared integrand, not a Brownian path.

An isometry variance as an ordinary areaThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forFor deterministic H(s)=s, Itô isometry reduces variance to the ordinary area under s squared. The lab computes that area exactly and does not sample stochastic integrals.

Check your understanding

Which condition supports the stated square-integrable Itô construction?

Try it yourself

Compute the mean, variance, and distribution of ∫032s dWs\int_0^3 2s\,dW_s. Then explain which distributional conclusion would fail to follow if the integrand were replaced by a random predictable process having the same expected squared time integral.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Integrate 4s24s^2 over the given horizon, and use the special Gaussian property of deterministic integrands.

Compare with the solution

The mean is zero and the variance is 4∫03s2 ds=364\int_0^3s^2\,ds=36. Since the integrand is deterministic, the integral has distribution N(0,36)N(0,36). A random predictable square-integrable integrand with the same expected squared integral would still yield mean zero and variance thirty-six, but those two moments do not imply a Gaussian law. The missing hypothesis is deterministic, rather than merely predictable, dependence of the integrand.

Take it with you

  • Predictable weights cannot inspect the increments they multiply.
  • Itô isometry defines the completion and calculates second moments.
  • Localization yields local martingales without automatic global mean conservation.

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