advanced31 min studyLesson 56 in Mathematics

Itô integrals and nonanticipating accumulation

Construct integration against Brownian motion from left-endpoint decisions and use the isometry to control stochastic accumulation.

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What you’ll understand

  • Form a stochastic integral for an elementary predictable process.
  • Apply the Itô isometry under square-integrability assumptions.
  • Explain why future-dependent and endpoint sampling rules change the problem.

Imagine choosing an exposure before each random disturbance arrives. Multiply each exposure by the subsequent Brownian increment and add. This produces the elementary Itô integral. The timing of the choice is essential: using a future disturbance to select the exposure would create a different construction. Since Brownian paths have unbounded variation, ordinary pathwise integration is not generally available. The Itô isometry replaces that missing route with a mean-square estimate that both defines the integral and explains its uncertainty.

Choose the coefficient before the next increment

For a deterministic partition, an elementary predictable integrand holds a random coefficient Hᵢ constant on each interval (tᵢ,tᵢ₊₁]. The coefficient is measurable using information at tᵢ, before the increment it multiplies is revealed. The integral is the finite sum of those products. Brownian future increments have conditional mean zero, so each product has zero expectation when its integrability is adequate. This temporal ordering is the nonanticipating feature that makes the resulting accumulation a martingale.

A coefficient can still be random and depend strongly on the past. For example, selecting Hᵢ from the current Brownian position is allowed. Selecting it from the sign of the next Brownian increment is not allowed in this elementary Itô construction. Adaptedness alone can require measurability qualifications in full generality; a standard sufficient setting uses predictable or progressively measurable square-integrable processes. The definition is not a blanket rule that every arbitrary time-indexed random coefficient can be integrated against Brownian motion.

Ht=∑iHi1(ti,ti+1](t),∫0THt dBt=∑iHi(Bti+1−Bti)H_t=\sum_iH_i\mathbf1_{(t_i,t_{i+1}]}(t),\qquad\int_0^T H_t\,dB_t=\sum_iH_i(B_{t_{i+1}}-B_{t_i})
Hᵢ is F(tᵢ)-measurable, Brownian increments after tᵢ are independent of F(tᵢ), and the elementary integrand has finite expected integrated square.↔ Scroll wide equations horizontally.

The mean-square identity constructs the limit

When the finite sum is squared, the expected cross terms vanish. For a later increment, condition on its starting information; everything multiplying that increment is already known and the increment has conditional mean zero. A diagonal term contributes the expected squared coefficient times the interval length. Summing yields the Itô isometry. The right side is an ordinary integral of a nonnegative expectation, giving a computable measure of stochastic energy without assigning a derivative to a Brownian sample path.

Approximate a predictable square-integrable H by elementary processes in the norm given by expected integrated square. The isometry guarantees that their stochastic integrals form a Cauchy sequence in L2, so completeness supplies a limit independent of the chosen approximation. This is a definition through controlled approximation, not an instruction to integrate a wildly oscillating drawn curve. Under these assumptions the integral has mean zero, and its second moment equals its variance. Locally square-integrable settings can extend the theory by localization, but need additional care.

E ⁣[(∫0THt dBt)2]=E ⁣[∫0THt2 dt]<∞\mathbb E\!\left[\left(\int_0^T H_t\,dB_t\right)^2\right]=\mathbb E\!\left[\int_0^T H_t^2\,dt\right]<\infty
H is predictable or progressively measurable in the standard square-integrable construction, and B is Brownian motion relative to the filtration. The integral also has expectation zero.↔ Scroll wide equations horizontally.

An endpoint convention leaves a finite correction

For the integrand Hₜ=Bₜ, the left-endpoint sum has a simple identity obtained by expanding each squared increment. The sum of B(tᵢ) times its next increment equals half the terminal squared displacement minus half the sum of increment squares. Quadratic variation makes the latter term approach T, so the integral is one-half of Bₜ squared minus T. Its expectation is zero, as the isometry-based construction requires. Omitting T would produce a nonzero mean and expose the error.

Right-endpoint sums differ from left-endpoint sums by the sum of squared increments, which tends to T rather than zero. A midpoint convention leads to Stratonovich integration, with its own transformation rule and hypotheses. These constructions are related but cannot be interchanged silently. Nor can the formal coefficient B_T be treated as a known constant throughout [0,T] in an Itô integral: before T it contains future information. Different anticipation theories exist, but they are outside the nonanticipating model developed here.

∫0TBt dBt=12(BT2−T),E ⁣[(∫0TBt dBt)2]=∫0Tt dt=T22\int_0^T B_t\,dB_t=\frac12(B_T^2-T),\qquad\mathbb E\!\left[\left(\int_0^T B_t\,dB_t\right)^2\right]=\int_0^T t\,dt=\frac{T^2}{2}
The first equality is the Itô identity obtained from left sums and quadratic variation. E[Bₜ²]=t supplies the isometry calculation in the second equality.↔ Scroll wide equations horizontally.

Two predeclared exposure intervals

Let H=2 on (0,1] and H=−1 on (1,3]. Find I=∫03Ht dBtI=\int_0^3H_t\,dB_t, its mean, and its variance.

  1. Apply the elementary definition: I=2(B1−B0)−(B3−B1)=3B1−B3I=2(B_1-B_0)-(B_3-B_1)=3B_1-B_3, using B₀=0. The coefficients are deterministic and therefore predictable.

  2. The two underlying increments B₁ and B₃−B₁ are independent centered normals with variances 1 and 2. Hence the mean of I is zero and its variance is 2²×1+(−1)²×2=6.

  3. Verify the same answer by the isometry: ∫03Ht2 dt=4×1+1×2=6\int_0^3H_t^2\,dt=4\times1+1\times2=6. Therefore I has law N(0,6). Treating B₁ and B₃ as independent in the equivalent expression 3B₁−B₃ would incorrectly give variance 12 instead of 6.

The result

The integral is 3B₁−B₃ with mean zero and variance 6. Independent increments or the Itô isometry give consistent results.

Explore the model

Interactive lab

Interactive lab · probability

Finite-step variance before the stochastic limit

Keep p at 1/2 and increase n. The exact binomial bars encode the endpoint 2K−n of a symmetric walk, whose variance is n. A deterministic exposure multiplier c would multiply that variance by c²; calculate the change before adjusting a later hypothetical interval. The lab does not sample Brownian increments, accept an integrand, or compute an Itô integral.

Finite-step variance before the stochastic limitThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. 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 forIndependent centered increments add variances with squared coefficients, the finite-step pattern generalized by the Itô isometry.

Check your understanding

Which elementary coefficient choice is valid for the ordinary nonanticipating Itô integral on (tᵢ,tᵢ₊₁]?

Try it yourself

For deterministic Hₜ=t on [0,2], compute the mean and variance of ∫02t dBt\int_0^2t\,dB_t.

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

Give me a hint

The deterministic integrand is predictable. Integrate its square, not its absolute value or its ordinary integral squared.

Compare with the solution

The mean is zero. The Itô isometry gives variance ∫02t2 dt=8/3\int_0^2t^2\,dt=8/3. Because the integrand is deterministic, the integral is Gaussian with law N(0,8/3). This is different from the ordinary integral of t, which is 2, and from multiplying B₂ by a single coefficient chosen after the fact.

Take it with you

  • Itô coefficients use information available before the increment they multiply.
  • The square-integrable integral is defined by an L2 limit controlled by the Itô isometry.
  • Brownian quadratic variation makes endpoint conventions differ by a nonvanishing correction.

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