advanced35 min studyLesson 668 in Mathematics

Discrete stochastic integrals and isometries

Construct predictable martingale transforms, derive their energy identity, and expose the failure of anticipative weights.

On this page

What you’ll understand

  • Construct a discrete integral with predictable coefficients.
  • State an energy condition that makes the integral an L2 martingale.
  • Compute the isometry and transformed brackets on an exact finite tree.

A discrete stochastic integral is a sum of increments weighted by decisions made before those increments are observed. Predictability supplies zero conditional drift, while an explicit energy condition supplies integrability and controls fluctuations. Unlike an ordinary pathwise integral, its most useful continuity estimate is an identity of expected squares. The predictable bracket specifies the relevant weighting of the integrand's squared size. This construction is a rigorous finite-sum precursor to stochastic integration, without importing continuous-time assumptions or limits prematurely.

The integrand is a pre-increment rule

Let M be a real square-integrable discrete martingale, starting at zero, and let H at step k be measurable for the information at k minus one. Define its integral as the cumulative sum of H times the corresponding martingale increments, starting at zero. Each product is measurable at the current time. If H is bounded on every finite horizon, conditional centering and integrability follow immediately. Unbounded coefficients are also allowed, but require a condition that controls the actual weighted increments rather than just the coefficients in isolation.

For a fixed finite horizon, assume the expected sum of squared coefficients times predictable bracket increments is finite. Taking conditional expectations shows this equals the sum of expected squares of the weighted martingale increments. Each product is therefore square integrable and has conditional mean zero. The resulting integral process is an L2 martingale. Merely saying that H is adapted or measurable eventually does not suffice: a coefficient that sees its own upcoming increment can alter its conditional mean.

In=(H⋅M)n=∑k=1nHkΔMk,Hk is Fk−1-measurable,E∑k=1NHk2Δ⟨M⟩k<∞I_n=(H\mathbin{\cdot}M)_n=\sum_{k=1}^nH_k\Delta M_k,\qquad H_k\text{ is }\mathcal F_{k-1}\text{-measurable},\qquad \mathbb E\sum_{k=1}^NH_k^2\Delta\langle M\rangle_k<\infty
The coefficient is measurable before its increment. The displayed finite-horizon energy condition ensures every weighted increment lies in L2.↔ Scroll wide equations horizontally.

Orthogonal increments yield an isometry

Expand the square of the finite integral. Cross terms between different steps have expectation zero because the later weighted increment has zero conditional mean and the earlier one is already measurable. The remaining diagonal terms are expectations of squared weighted increments. Since the coefficient is known before the step, conditioning moves its square outside and replaces the increment square by the predictable bracket increment. This proves the discrete isometry exactly, with no approximation or independence assumption beyond the martingale conditions.

The same calculation at each step identifies the predictable bracket of the integral as the sum of the squared coefficients against the predictable bracket of M. Its realized square bracket instead sums the squared coefficients against the realized square bracket. These are pathwise finite-sum identities under the stated conventions. They show why a coefficient's ordinary unweighted L2 norm is not always the appropriate energy: the driving martingale may have history-dependent conditional variance, so some times or branches carry more variation than others.

EIN2=E∑k=1NHk2Δ⟨M⟩k,⟨I⟩n=∑k=1nHk2Δ⟨M⟩k,[I]n=∑k=1nHk2(ΔMk)2\mathbb E I_N^2=\mathbb E\sum_{k=1}^NH_k^2\Delta\langle M\rangle_k,\qquad \langle I\rangle_n=\sum_{k=1}^nH_k^2\Delta\langle M\rangle_k,\qquad [I]_n=\sum_{k=1}^nH_k^2(\Delta M_k)^2
The expected-square isometry and the two bracket identities have different roles. Predictable measurability is essential for replacing the increment square by its conditional mean.↔ Scroll wide equations horizontally.

Infinite sums need finite total energy

If the expected total energy summed over all steps is finite, the integral partial sums are L2 bounded and hence converge almost surely and in L2. The isometry then identifies the terminal second moment and the mean-square tails. This is a useful completion procedure: approximating predictable integrands in the bracket-weighted energy norm makes their integral terminal values converge in L2. It does not claim convergence from a bound on individual coefficients when infinitely many increments continue to contribute energy.

A stopped martingale is a special integral with the predictable indicator that stopping has not happened before the current step. This connects optional sampling with stochastic integration. Another special case uses the previous martingale value as coefficient and yields half the difference between the current square and realized bracket, by expanding squares. That coefficient is predictable but can be unbounded; applying an L2 isometry to it still requires the corresponding higher-energy integrability. An algebraically valid pathwise sum and an L2 theorem should not be conflated.

A branch-dependent two-step integral

Let M be a fair sign walk. Set H1=1H_1=1 and H2=2H_2=2 after a positive first sign, but H2=−1H_2=-1 after a negative first sign. Compute the integral's terminal second moment in two independent ways.

  1. In path order (+,+),(+,−),(−,+),(−,−)(+,+),(+,-),(-,+),(-,-), the terminal integral ξ1+H2ξ2\xi_1+H_2\xi_2 takes values 3,−1,−2,03,-1,-2,0, all with probability one quarter. Their mean is zero.

  2. Direct squaring gives (9+1+4+0)/4=7/2(9+1+4+0)/4=7/2. The underlying walk has predictable bracket increment one at both steps.

  3. The integrand energy is therefore E(1+H22)=1+(4+1)/2=7/2\mathbb E(1+H_2^2)=1+(4+1)/2=7/2, exactly matching the direct second moment. Its predictable bracket is five after a positive first sign and two after a negative one.

The result

The integral remains a martingale despite its history-dependent coefficient. Both direct enumeration and the bracket-weighted isometry give terminal variance seven halves.

Explore the model

Interactive lab

Interactive lab · probability

A sign square removes sign uncertainty

Pause and Reset, set both p=0.5,n=1p=0.5,n=1, and map the two outcomes to signs 2K−12K-1. Predict mean zero but squared value one on both outcomes. Reset and set both p=0.5,n=2p=0.5,n=2 to compare the terminal sign-sum variance two with two units of increment energy. The lab shows unweighted binomial laws, not the branch-dependent integral above; enumerate that tree on paper. Changing controls retains elapsed time, so Reset restores both controls and clock before each comparison.

A sign square removes sign uncertaintyThese 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 forPredictable multiplication preserves conditional centering, and expected squared increments add to the integral's energy.

Check your understanding

Which hypothesis makes the displayed finite-horizon stochastic-integral isometry valid for an L2 martingale driver?

Try it yourself

Let a fair sign walk have N increments. Compare the predictable constant coefficient Hk=2H_k=2 with the anticipative coefficient H~k=ξk\widetilde H_k=\xi_k. Find each sum's mean and second moment, and say which is covered by the isometry.

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 predictable sum is twice the walk. Each anticipative summand is a sign multiplied by itself.

Compare with the solution

The predictable integral is 2SN2S_N, with mean zero and second moment 4N4N, matching energy ∑k=1N4=4N\sum_{k=1}^N4=4N. The anticipative sum equals N identically, so its mean is N and second moment is N2N^2. It is not covered by the predictable isometry; inserting its coefficients into the energy formula would incorrectly produce N. For N equal to one the second moments happen to match, but the nonzero mean already reveals the failure.

Take it with you

  • A discrete integrand is known before each increment it weights.
  • The energy norm uses the driver's predictable bracket.
  • Finite total expected energy gives an L2 terminal integral; anticipation can invalidate the entire construction.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.