Moment tail bounds and tightness

What you will learn

Derive Markov and Chebyshev bounds through pointwise certificates, account for slack and construct an equality law under a fixed mean.

Prerequisites: Transforming distributions and losing information · Variance and standard deviation

A guarantee from deliberately incomplete information

Suppose a toy settlement loss L is nonnegative, measured in fake chips, and its population mean is exactly 3. We ask about the inclusive event L≥6. The mean alone does not specify the loss distribution. A constant loss of 3 never reaches 6, while a lottery paying loss 6 with probability one half and loss zero otherwise reaches it half the time. Both means are 3. Our task is to identify a statement valid for every nonnegative law with that mean, rather than guess which of these laws generated observations. The given mean is a model assumption, not a sample estimate or a guarantee about future trading losses.

Throughout this lesson all displayed workshop tickets are equally likely, all moments are population moments, and thresholds are strictly positive. We assume no independent repetition, normal distribution or poker strategy. Markov requires a nonnegative variable and finite expectation. Chebyshev requires finite variance but permits signed scores. A probability upper bound is a ceiling, not an exact probability and not a promise that an event never occurs. Additional support information can make a bound stronger; discarding information can make a correct guarantee less useful.

Build a pointwise certificate before averaging

Define I to be one on L≥a and zero otherwise, with a>0. On a tail outcome, L/a≥1=I. Off the tail, I=0 and L/a≥0 because L is nonnegative. Thus I≤L/a on every outcome. Multiplying by outcome probabilities and adding preserves the inequality: P(L≥a)=E[I]≤E[L]/a. Combine this with the elementary probability ceiling one to report min(1,E[L]/a). This proof works for unequal masses as well as equal tickets. The hard step is the pointwise comparison, not dividing two numbers.

The difference L/a−I is a nonnegative slack on each outcome. Its average equals E[L]/a−P(L≥a), the gap in the raw Markov bound. Off-tail positive losses consume mean without contributing to the event. Tail losses strictly above a also consume extra mean without adding more than one to the indicator. Equality therefore requires all positive-mass losses to be zero or exactly a. This explains the two-point extremal construction and diagnoses why many fully specified distributions fall well below the ceiling. If the raw bound exceeds one, the clipped bound has a different gap; do not confuse that clipped gap with average pointwise slack.

Construct a tight law without changing the mean

Begin with eight loss tickets 0,0,0,0,3,3,9,9. Their sum is 24 and their mean is 3. At a=6, two tickets qualify, so the actual tail probability is 2/8=1/4. Markov gives 3/6=1/2. Each loss-3 ticket contributes slack 1/2, and each loss-9 ticket contributes slack 1/2. The four zero tickets contribute none. Average slack is therefore 2/8=1/4, exactly the raw bound minus the actual tail probability. Independently, tail losses contribute 18/8 to the mean and off-tail losses contribute 6/8; their sum is 3.

Move one chip from each loss-9 ticket to each loss-3 ticket, repeating three times. The final tickets are 0,0,0,0,6,6,6,6. Each transfer preserves the sum and nonnegativity, so the Markov ceiling stays one half. Now four tickets qualify, actual probability equals one half, and every slack is zero. This constructs a witness that no smaller universal ceiling can follow from only nonnegativity and mean 3 at threshold 6. It does not claim every distribution with mean 3 has this tail. At threshold strictly greater than 6 this particular witness has probability zero. Inclusive boundary language is essential to the equality claim.

Turn squared deviations into a tail bound

For a signed score X with mean μ and variance v, introduce Z=(X−μ)². This transformed variable is nonnegative and E[Z]=v. For d>0, the events |X−μ|≥d and Z≥d² are identical. Apply the pointwise argument to Z to obtain P(|X−μ|≥d)≤min(1,v/d²). This is Chebyshev’s inequality. The denominator is squared because Z has squared-score units. Dividing variance by d would leave units in a quantity supposed to be a probability. No independence assumption is needed for a single score.

Take scores −4,0,0,0,0,0,0,4 on eight tickets. Their mean is zero; their squared sum is 32, so variance is 4. At d=4 exactly two tickets qualify and the bound is 4/16=1/4, attained. At d=2 the same two tickets qualify, but the bound clips to 1 and is uninformative. Zero variance instead forces X=μ with probability one, giving zero tail probability for every positive d. A two-sided ceiling also bounds either one-sided tail by set inclusion, but halving it requires extra justification such as symmetry; Chebyshev alone does not provide equal tail masses.

Manipulate the certificate and audit its limits

In the workshop select a donor ticket and a distinct receiver to transfer one loss chip. The total is fixed at 24 and no ticket may become negative. Build the equality witness yourself, then change the threshold from 6 to 3 or 9. Classify all eight tickets using the current inclusive cutoff and enter both the exact tail probability and the clipped Markov ceiling. Finally decide whether the raw certificate has zero average slack. This last question distinguishes attaining the pointwise bound from merely matching a ceiling clipped at one. Every edit clears the proposed classification and answers so that old work cannot certify a changed model.

Use the ledger to explain every nonzero slack before retrying. Hint and reset controls remain available, and the current draft lasts for this visit. A failed audit should tell you whether event membership, probability, bound or tightness is wrong. Transfer order can change intermediate tails even though the final mean never changes. For a signed net payoff, first identify a valid nonnegative transformation or a known lower support bound; applying Markov directly can fail because negative outcomes cancel positive contributions in the mean. Neither inequality estimates a real loss law, establishes an investment policy, nor substitutes an estimated moment for a known population moment.

Hands-on: Build a tight tail certificate

Transfer conserved loss chips, classify inclusive tail tickets and audit probability, ceiling and zero slack.

Common mistakes

• Reporting a ceiling as an exact probability.
• Applying Markov directly to a signed variable.
• Forgetting to square the Chebyshev distance.
• Replacing an inclusive cutoff by a strict event in an equality example.
• Claiming clipped equality proves zero raw slack.

Keep these ideas

• Prove the outcome-level inequality before averaging.
• Markov needs nonnegativity and a positive threshold.
• Slack accounts for off-tail loss and tail overshoots.
• Chebyshev applies Markov to squared centered scores.
• Moment bounds express guarantees over laws, not fitted tail forecasts.

Further reading

• MIT 6.1200J Spring 2024, Lecture 24: Large Deviations ↗

  Read PDF pages 2 through 5: Markov proof, nonnegativity, shifted bounds, looseness and Chebyshev via squared deviations. Original conserved-ticket example, slack audit and equality construction; no source exercises, Chernoff section or video review claimed. Checked 2026-10-24.