advanced34 min studyLesson 667 in Mathematics

Quadratic variation and martingale brackets

Separate realized squared increments from their predictable conditional means and derive both square-compensation identities.

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What you’ll understand

  • Define the optional square bracket and predictable angle bracket in discrete time.
  • Prove the square-minus-bracket martingale identities with L2 hypotheses.
  • Compute a random realized bracket and compare it with a predictable compensator.

A martingale has zero conditional drift but can still accumulate uncertainty. Squared increments measure that accumulation. In discrete time there are two useful processes: the realized sum of squared increments and the sum of their conditional expectations before each step. They are often denoted by square and angle brackets respectively. Some texts use different terminology, so we define the conventions explicitly. The two processes need not agree path by path, even though they have equal expectations and each compensates the martingale's square.

Realized variation sees the current increment

Assume throughout this lesson a real martingale starting at zero whose value is square integrable at every finite time. Define its square bracket as the sum of the squares of its discrete increments. It is adapted, increasing, and integrable at every finite time. Its time-k increment generally depends on the new observation at k and need not be known at k minus one. The word optional here emphasizes that it uses the realized increment rather than its previous-information average.

Expanding the change in the square of the martingale gives twice the previous value times the new increment, plus the squared increment. Subtracting the realized square-bracket increment removes the second term. The remaining product has conditional mean zero because the earlier value is known and the increment is conditionally centered. Cauchy–Schwarz makes the product integrable. Thus the martingale square minus its realized square bracket is an integrable martingale, although the square bracket itself usually is not predictable.

[M]n=∑k=1n(ΔMk)2,Δ(M2−[M])k=2Mk−1ΔMk,M2−[M] is a martingale[M]_n=\sum_{k=1}^n(\Delta M_k)^2,\qquad \Delta(M^2-[M])_k=2M_{k-1}\Delta M_k,\qquad M^2-[M]\text{ is a martingale}
The bracket starts at zero. This is a discrete-time identity with real L2 martingale increments, not a claim about arbitrary pathwise limits along continuous-time partitions.↔ Scroll wide equations horizontally.

Predictable variation averages before the step

The angle bracket sums the conditional squared increments given the information immediately before each one. Every summand is nonnegative, integrable, and measurable before its step, so the process is predictable and increasing. Conditional centering of the difference between each realized square and its conditional expectation shows that square bracket minus angle bracket is a martingale. Combining this with the first square-compensation identity gives the martingale square minus angle bracket identity.

Equivalently the angle bracket is the predictable increasing component in the Doob decomposition of the squared martingale, normalized to start at zero. Uniqueness of that decomposition makes this compensator unique up to almost-sure equality at all integer times. At each fixed time the expected square equals the expected angle bracket, since the starting value was zero. Equal expectations do not imply equality on individual paths; the difference is itself a centered martingale whose randomness may be substantial.

⟨M⟩n=∑k=1nE[(ΔMk)2∣Fk−1],Mn2−⟨M⟩n is a martingale,EMn2=E⟨M⟩n=E[M]n\langle M\rangle_n=\sum_{k=1}^n\mathbb E[(\Delta M_k)^2\mid\mathcal F_{k-1}],\qquad M_n^2-\langle M\rangle_n\text{ is a martingale},\qquad \mathbb E M_n^2=\mathbb E\langle M\rangle_n=\mathbb E[M]_n
The first equality defines our predictable bracket convention. Its increment is a conditional variance because the martingale difference has conditional mean zero.↔ Scroll wide equations horizontally.

When the two brackets happen to agree

For a fair sign walk, every squared increment is one, so both brackets equal deterministic time. More generally, a predictable amplitude multiplied by a fair sign has squared increment equal to the amplitude squared, already known before the step. The two brackets again agree, but can now both be random because the amplitude depends on the past. These examples should not be generalized to arbitrary increments whose conditional squared size fluctuates after the new observation.

The brackets summarize different aspects of data. The realized bracket can be computed from the observed martingale path; the predictable bracket requires a conditional model. Their difference can diagnose fluctuations around a conditional variance forecast, but agreement of expectations does not validate the forecast model by itself. In later limit theorems the predictable bracket describes the variance accumulated before each increment, while an additional small-jump condition prevents one exceptional increment from determining the entire terminal distribution.

An asymmetric mean-zero increment

Let independent increments take value 33 with probability 1/41/4 and value −1-1 with probability 3/43/4, and let M be their partial sum. Compute both brackets through two steps and inspect the path (3,−1)(3,-1).

  1. The increment mean is 3/4−3/4=03/4-3/4=0 and its second moment is 9/4+3/4=39/4+3/4=3. Thus ⟨M⟩n=3n\langle M\rangle_n=3n and its value at time two is six on every path.

  2. The four paths (−1,−1),(−1,3),(3,−1),(3,3)(-1,-1),(-1,3),(3,-1),(3,3) have probabilities 9,3,3,19,3,3,1 divided by sixteen. Their realized brackets at time two are 2,10,10,182,10,10,18, averaging (18+30+30+18)/16=6(18+30+30+18)/16=6.

  3. On the path (3,−1)(3,-1) the terminal position is two and its square is four. The realized compensated square is 4−10=−64-10=-6, while the predictable compensated square is 4−6=−24-6=-2. Both processes are martingales, but their pathwise values differ.

The result

The predictable bracket is six and the realized bracket can be two, ten, or eighteen. Their equal mean is a compensation identity, not pathwise equality.

Explore the model

Interactive lab

Interactive lab · probability

One binomial law encodes the asymmetric walk

Pause and Reset, set both p=0.25,n=2p=0.25,n=2, and predict count mean 0.50.5 and variance 0.3750.375. The transformed position M2=4K−2M_2=4K-2 then has mean zero and variance six. Reset and set both p=0.25,n=1p=0.25,n=1, predicting transformed variance three. The instrument shows exact counts; compute the realized squared increments and their predictable averages separately, since endpoints alone do not generally determine a bracket. Control changes retain time; Reset clears both parameters and time.

One binomial law encodes the asymmetric walkThese 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 forA terminal second moment checks expected accumulated variation, while distinguishing the two brackets requires their increment definitions.

Check your understanding

What is true about the predictable angle bracket of a real square-integrable martingale starting at zero?

Try it yourself

Let independent fair signs drive M1=ξ1M_1=\xi_1 and ΔM2=H2ξ2\Delta M_2=H_2\xi_2, where H2=1H_2=1 after a positive first sign and H2=2H_2=2 after a negative first sign. Determine both brackets at time two and their expected value.

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 square of a fair sign is one, so the new squared increment is already known after the first sign.

Compare with the solution

Both brackets equal 1+H221+H_2^2, which is two on the positive-first branch and five on the negative-first branch. Their expectation is (2+5)/2=7/2(2+5)/2=7/2, also the terminal second moment by orthogonality. This example has random brackets that nevertheless agree pathwise; the previous asymmetric-increment example shows that such agreement is an extra feature, not the general bracket theorem.

Take it with you

  • The square bracket records realized squared increments.
  • The angle bracket records their previous-information conditional means.
  • Both compensate the martingale square, but only the angle bracket is required to be predictable.

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