advanced34 min studyLesson 665 in Mathematics

Doob maximal and Lp inequalities

Control the largest observed value using terminal moments, with a careful distinction between weak L1 tails and strong Lp estimates.

On this page

What you’ll understand

  • Derive a maximal tail bound by stopping at the first crossing.
  • Apply the strong Doob inequality only for p greater than one.
  • Compute an exact finite-walk maximum and compare it with moment bounds.

A terminal distribution does not directly reveal how large a process became earlier. A martingale can make an excursion and return near zero before the final observation. Doob's inequalities nevertheless control the largest value using moments at the terminal time. The basic nonnegative-submartingale estimate is a tail inequality of weak L1 type. Integrating a sharper version and applying Hölder produces a strong Lp norm bound, but only when the exponent exceeds one. These statements must not be collapsed into a nonexistent strong L1 estimate.

Stop at the first crossing

Let X be a nonnegative integrable submartingale over a finite horizon and let its maximum include every time from zero through the horizon. Stop at the first value at least a positive threshold, using the terminal horizon if no crossing occurs and keeping track of the crossing event separately. On that event the stopped value is at least the threshold. The conditional submartingale sampling inequality says the conditional expected terminal value is at least that stopped value. Integrating over the crossing event gives a refined tail estimate.

Dropping the event indicator from the nonnegative terminal value yields the simpler bound by its full expectation divided by the threshold. For a martingale, the absolute value is a nonnegative submartingale by conditional Jensen; a power of that absolute value with exponent at least one is also a submartingale when integrable. Thus terminal moments bound the probability of any earlier large absolute excursion. The nonnegativity assumption matters in the elementary tail bound because it permits discarding the contribution from the complement of the crossing event.

XN∗=max⁡0≤k≤NXk,λP(XN∗≥λ)≤E[XN1{XN∗≥λ}]≤EXN(λ>0)X_N^*=\max_{0\le k\le N}X_k,\quad \lambda\mathbb P(X_N^*\ge\lambda)\le\mathbb E[X_N\mathbf1_{\{X_N^*\ge\lambda\}}]\le\mathbb E X_N\qquad(\lambda>0)
X is a nonnegative integrable submartingale and the time horizon N is deterministic and finite. A martingale can enter through X = |M| or a suitable convex power.↔ Scroll wide equations horizontally.

Integrating tails needs the exponent gap

For exponent p greater than one, represent the pth moment of the maximum by integrating its tail against p times the threshold to the power p minus one. Insert the refined bound containing the terminal value times the crossing indicator. Tonelli exchanges the nonnegative integral and expectation, and integration up to the realized maximum produces the factor p divided by p minus one. Hölder then separates the terminal value from the maximum. A truncation argument handles a possibly unbounded maximum without assuming its pth moment in advance.

The resulting norm bound compares the maximum to the terminal value with constant p divided by p minus one. Raising both sides to the pth power raises the constant as well. Confusing these two forms is a frequent arithmetic error, especially for the second-moment bound, where the squared constant is four. At p equal to one, the integral used in the proof has a logarithmic divergence and no finite universal constant of this strong form exists. The weak tail inequality remains available.

∥max⁡k≤N∣Mk∣∥p≤pp−1∥MN∥p(p>1),Emax⁡k≤N∣Mk∣2≤4E∣MN∣2\left\|\max_{k\le N}|M_k|\right\|_p\le\frac p{p-1}\|M_N\|_p\quad(p>1),\qquad \mathbb E\max_{k\le N}|M_k|^2\le4\mathbb E|M_N|^2
M is an Lp martingale on the finite horizon. The same strong estimate holds for a nonnegative Lp submartingale. The constant in the moment inequality is the pth power of the norm constant.↔ Scroll wide equations horizontally.

A bound controls excursions without enumerating them

For a centered fair walk of N steps, terminal second moment is N, so the probability of any absolute excursion at least a positive level is at most N divided by that level squared. One can apply the weak inequality directly to the squared walk, obtaining a better tail constant than combining Markov's inequality with the strong second-moment maximum bound. Bounds from different routes need not have identical constants; the available moment and the desired event determine which route is useful.

These estimates are robust to dependence compatible with the martingale property. They are typically not exact and can exceed one, in which case the trivial probability bound one should also be used. Exact enumeration on a short tree clarifies both the information captured and the slack in the inequality. In longer problems, the value is that a single terminal moment controls a whole collection of observation times, avoiding a union bound whose factor would grow with the number of times.

Enumerate the three-step absolute maximum

For a fair sign walk over three steps, compute P(max⁡k≤3∣Sk∣≥2)P(\max_{k\le3}|S_k|\ge2) and the exact second moment of that maximum. Compare with Doob bounds.

  1. Among eight equally likely paths, the all-positive and all-negative paths have absolute maximum three. The two paths whose first two signs agree but whose third reverses have maximum two. The other four paths have maximum one.

  2. Thus the crossing probability is 4/8=1/24/8=1/2 and the maximum squared mean is (2⋅9+2⋅4+4⋅1)/8=15/4(2\cdot9+2\cdot4+4\cdot1)/8=15/4. The terminal second moment is three.

  3. The weak inequality for Sk2S_k^2 gives crossing probability at most 3/43/4. The strong L2 inequality gives maximum squared mean at most 4⋅3=124\cdot3=12, which indeed exceeds 15/415/4.

The result

Both estimates hold with visible slack. The weak squared-process route gives a direct probability bound, while the strong inequality controls the whole maximum's second moment.

Explore the model

Interactive lab

Interactive lab · probability

Terminal moments feed a maximal estimate

Pause and Reset; set both p=0.5,n=3p=0.5,n=3. Predict count variance 3/43/4, so the transformed walk S3=2K−3S_3=2K-3 has variance three. Reset and set both p=0.5,n=2p=0.5,n=2, predicting transformed variance two. Use these endpoint moments in the maximal inequalities. The binomial instrument does not reveal the largest position along a path; the worked enumeration supplies that separate quantity. Controls retain elapsed time until Reset clears both parameters and time.

Terminal moments feed a maximal estimateThese 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 maximal theorem turns terminal moments into pathwise-excursion bounds, but an endpoint histogram alone is not a histogram of maxima.

Check your understanding

For an L2 martingale, which is the strong Doob second-moment bound?

Try it yourself

For a two-step fair walk, compute the expected absolute maximum and compare it with the expected absolute terminal value. What does this say about replacing the p>1 theorem with a p=1 inequality of constant one?

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

Give me a hint

Two paths have matching signs and two have opposite signs; their maxima and endpoints differ.

Compare with the solution

Matching signs give maximum two and terminal absolute value two. Opposite signs give maximum one and terminal absolute value zero. Hence the expected absolute maximum is (2+2+1+1)/4=3/2(2+2+1+1)/4=3/2, while the expected absolute terminal value is one. A constant-one strong L1 assertion already fails on this finite tree. The example alone does not disprove every finite universal constant; that stronger failure is the general endpoint theorem stated above.

Take it with you

  • First-crossing sampling gives weak L1 tail control.
  • Strong Doob norm estimates require p greater than one.
  • Moment constants must be raised to the same power as the norm inequality.

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.