advanced33 min studyLesson 696 in Mathematics

Martingale Bernstein and Freedman's inequality

Replace independent-sum variance by predictable conditional variance and control crossings with a deterministic variance budget.

On this page

What you’ll understand

  • Define predictable quadratic variation for a centered martingale difference sequence.
  • State a one-sided time-uniform Freedman bound with explicit constants.
  • Distinguish deterministic variance budgets from realized optional squared increments.

Adaptive decisions destroy ordinary independence, but they need not destroy exponential concentration. A martingale difference has conditional mean zero given the information available before it occurs. Its conditional variance measures the uncertainty remaining at that time. Freedman's inequality combines these predictable variances with a one-sided increment bound, extending Bernstein's logic to a process that can change its behaviour in response to the past.

Variance known just before an increment

Let M_k start at zero and be a square-integrable real martingale adapted to a filtration F_k. Write X_k=M_k−M_(k−1), so E[X_k|F_(k−1)]=0. Its predictable quadratic variation V_k is the sum of E[X_j²|F_(j−1)]. Each contribution is known before the corresponding increment, though the accumulated V_k can itself be random. It is generally different from the optional sum of the realized X_j², which is observable only after those increments occur.

Assume X_k≤b almost surely for a deterministic b>0 at every step. This is an upper bound for an upper-tail result; controlling a negative crossing instead requires a bound on −X_k. Conditional Bennett-type estimates produce an exponential supermartingale with a variance compensator. Iterated conditioning replaces the independent product used earlier. The method does not require the increments to be independent, but conditional centering and the almost-sure bound must hold for the actual filtration and adaptive rule, not just for their marginal distributions.

Mk=∑j=1kXj,E[Xj∣Fj−1]=0,Vk=∑j=1kE[Xj2∣Fj−1]M_k=\sum_{j=1}^kX_j,\qquad\mathbb E[X_j\mid\mathcal F_{j-1}]=0,\qquad V_k=\sum_{j=1}^k\mathbb E[X_j^2\mid\mathcal F_{j-1}]
V is predictable quadratic variation; ΣXⱼ² is a different, optional quantity and cannot be substituted without another theorem.↔ Scroll wide equations horizontally.

A crossing event with a fixed variance budget

For deterministic t≥0 and v>0, Freedman bounds the probability that there exists any time k with M_k≥t and V_k≤v by exp(−t²/(2(v+bt/3))). The time-uniform event permits a stopping-time interpretation: a large positive fluctuation occurs before too much predictable variance has accumulated. The same expression also bounds the corresponding joint event at any fixed time. If V_n≤v almost surely for a chosen horizon, it yields an ordinary upper-tail bound there.

One cannot replace v by the observed random value V_n inside the right-hand side and call the result an unconditional probability bound. The theorem compares one event with one deterministic budget. Uniform bounds over many variance levels can be obtained through additional arguments such as peeling, but their cost must be included. Likewise a small realized sum of squared increments does not imply a small predictable variance on that sample path. A process can repeatedly realize modest values while retaining a substantial conditional chance of a larger next increment.

P{∃k≥0:Mk≥t and Vk≤v}≤exp⁡ ⁣[−t22(v+bt/3)]\mathbb P\{\exists k\ge0:M_k\ge t\ \text{and}\ V_k\le v\}\le\exp\!\left[-\frac{t^2}{2(v+bt/3)}\right]
M₀=0, the increments are conditionally centered and satisfy Xₖ≤b almost surely, and t,v are deterministic with t≥0 and v>0.↔ Scroll wide equations horizontally.

Predictable activation preserves centering

Suppose fresh Bernoulli observations B_k of fixed probability p are independent of the past. Before seeing B_k, select an activation A_(k−1) between zero and one using only past information, then take X_k=A_(k−1)(B_k−p). Conditional centering survives because the activation is predictable. Its conditional variance is A_(k−1)²p(1−p), and the upper increment bound is 1−p. Selecting activation after seeing the current B_k would generally break the centering argument.

For activations restricted to zero or one, the predictable variance counts active trials times p(1−p). It does not require all active times to be fixed in advance. Independent-sum Bernstein is recovered when all activations equal one and the horizon is deterministic. This comparison clarifies what adaptive concentration adds: it allows the sampling rule to respond to observed history while retaining a quantified variance budget, rather than pretending the selected data remain an ordinary independent sample.

Eight active trials of an adaptive process

Let fresh Bₖ be Bernoulli(1/4), and use predictable activations Aₖ₋₁ in {0,1}; for example activate the next trial when the preceding outcome was zero. Bound the chance that Mₖ=ΣAⱼ₋₁(Bⱼ−1/4) ever reaches three while at most eight trials have been active.

  1. Conditional on the past, Xₖ has mean zero and variance (3/16)Aₖ₋₁. Its largest possible value is 3/4. Thus use b=3/4 and predictable variance Vₖ=(3/16) times the active-trial count.

  2. At most eight active trials means Vₖ≤8·3/16=3/2. The event in the prompt is therefore exactly a positive crossing with deterministic budget v=3/2, independent of how long inactive periods last.

  3. Freedman's exponent at t=3 is −9/[2(3/2+(3/4)·3/3)]=−2. The crossing probability is at most exp(−2)≈0.135335. The activation must be chosen before the current Bernoulli observation; choosing only its successes would invalidate this calculation.

The result

The time-uniform crossing bound is exp(−2), using predictable variance rather than a realized sum of squares.

Explore the model

Interactive lab

Interactive lab · probability

The fixed-activation special case

Pause and Reset, then set p = 0.25 and n = 8. With every trial active, M=K−2, V=1.5, and b=0.75. Predict that M≥3 means K≥5; the exact upper-tail bars total 0.027298, below the worked Freedman bound exp(−2). Reset and set both p = 0.5 and n = 8: now V=2, b=0.5, and M≥3 means K≥7, whose exact tail is 9/256. This static binomial instrument does not display adaptive paths or verify a time-uniform crossing event.

The fixed-activation special caseThese 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 bounded sums are a special case; adaptive results require the separate predictable-variance argument.

Check your understanding

Which variance quantity appears in the displayed Freedman inequality?

Try it yourself

For one fresh Bernoulli(1/4) variable, let X=B−1/4. Compute its conditional second moment before observing B and its realized square for each outcome. Explain why these are not interchangeable in Freedman.

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 two increment values are −1/4 and 3/4. Average their squares using probabilities 3/4 and 1/4 before comparing with the realized outcomes.

Compare with the solution

The predictable second moment is (3/4)(1/16)+(1/4)(9/16)=3/16(3/4)(1/16)+(1/4)(9/16)=3/16. If B=0, the realized square is 1/16; if B=1, it is 9/16. Thus the predictable value equals neither realized value in general. After several trials their sums differ pathwise, even though expectations can agree under the martingale assumptions. A sequence of zeros would produce small optional squares but would not reduce the fresh observation's conditional variance from 3/16. Freedman's displayed theorem uses the predictable sum and does not authorize replacing it by the smaller realized sum on such a path.

Take it with you

  • Conditional centering and a predictable variance process replace independent-sum factorization.
  • Freedman controls crossings before a deterministic predictable-variance budget is exceeded.
  • Optional squared increments and random plug-in variance levels require separate theorems.

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