advanced28 min studyLesson 632 in Mathematics

Second moments and Paley–Zygmund

Control the cost of overlap and turn first moments into quantitative lower bounds.

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What you’ll understand

  • Compute second moments by separating diagonal, independent, and overlapping indicator pairs.
  • Derive the second-moment existence bound using Cauchy–Schwarz.
  • Apply Paley–Zygmund with the correct threshold and moment hypotheses.

A large expected count can hide an almost-always empty random object. Second moments test whether that mean comes from broadly distributed mass or exceptional outcomes. In combinatorics, the calculation often becomes an inventory of overlapping pairs of configurations. The method does not demand independence, but it demands honest bookkeeping: dependencies are paid for through joint probabilities rather than discarded.

Why a first moment can mislead

Consider a variable equal to n2n^2 with probability 1/n1/n and zero otherwise. Its expectation is n, yet its probability of being positive tends to zero. The second moment is n3n^3, so the square of the mean is small relative to the second moment. This ratio detects the rare, excessively large outcomes that invalidate an existence-with-high-probability inference. A useful second-moment argument must therefore compare scales, not merely show that both moments are finite.

For a sum of indicators, expand its square before estimating. The diagonal terms are the indicators themselves, because squaring a zero-one variable changes nothing. Distinct pairs contribute joint occurrence probabilities, with ordered pairs counted twice relative to unordered pairs. If two configurations use disjoint sets of underlying independent random choices, their joint probability factors. Sharing vertices does not always mean sharing random choices: triangles in an independent-edge graph are independent when they share only a vertex.

EX2=∑iEIi+∑i≠jE(IiIj),Var⁡(X)=EX2−(EX)2\mathbb E X^2=\sum_i\mathbb E I_i+\sum_{i\ne j}\mathbb E(I_iI_j),\qquad \operatorname{Var}(X)=\mathbb E X^2-(\mathbb E X)^2
The second sum is over ordered distinct pairs; consistent counting is essential when classifying overlaps.↔ Scroll wide equations horizontally.

Extract mass above zero or above a fraction of the mean

For a nonnegative square-integrable variable with positive mean, write its expectation using only the event where it is positive. Cauchy–Schwarz bounds that expectation by the square root of the second moment times the probability of positivity. Rearranging gives a lower bound, not an upper bound, on existence. Equality of the first two relevant moment scales is favorable. If the variance divided by the squared mean tends to zero, this existence lower bound tends to one.

Paley–Zygmund retains the same idea above a nonzero threshold. For a parameter theta between zero and one, the contribution from values at most theta times the mean is at most that threshold. The remaining contribution must carry at least the fraction one minus theta of the mean, and Cauchy–Schwarz controls it. We use the event strictly greater than the threshold. When the variable is integer valued, translate that event carefully; a threshold equal to one corresponds to at least two, not at least one.

P(X>θEX)≥(1−θ)2(EX)2EX2,0≤θ<1,0<EX2<∞\mathbb P(X>\theta\mathbb E X)\ge(1-\theta)^2\frac{(\mathbb E X)^2}{\mathbb E X^2},\qquad 0\le\theta<1,\quad 0<\mathbb E X^2<\infty
Nonnegativity and a positive mean are assumed; theta zero recovers the second-moment existence bound.↔ Scroll wide equations horizontally.

From pair counts to concentration

An alternative presentation applies Chebyshev to the deviation equal to the mean. Since X=0 is such a deviation, a small relative variance forces positivity with high probability. More generally it forces X divided by its mean to approach one in probability. This is stronger than merely having a nonvanishing chance of finding an object. For a sequence of combinatorial counts, both the diagonal contribution and every class of correlated pairs must be negligible at the squared-mean scale.

The method can fail even when the desired conclusion is true. Counting a very clustered family of witnesses may produce a huge second moment; changing the witness count, conditioning on coarse information, or using a weighted count can improve matters. Failure of a particular moment bound is not a counterexample to existence. In the binomial calculation below, independence makes the variance easy, allowing the direction and thresholds of the inequalities to be inspected without a long overlap classification.

Twenty opportunities and a quantitative lower bound

Let X count successes in twenty independent Bernoulli trials with success probability 0.1. Compute second-moment lower bounds for X>0 and X>1.

  1. The mean is 20(0.1)=220(0.1)=2 and variance is 20(0.1)(0.9)=1.820(0.1)(0.9)=1.8. Thus EX2=1.8+4=5.8\mathbb E X^2=1.8+4=5.8.

  2. With theta zero the lower bound is 4/5.8=20/294/5.8=20/29, approximately 0.689655, for at least one success.

  3. For X>1 choose theta=1/2. Paley–Zygmund gives (1/2)2(4/5.8)=5/29(1/2)^2(4/5.8)=5/29, approximately 0.172414. Since X is integer valued, this bounds the probability of at least two successes.

The result

The guaranteed probabilities are at least 20/29 for a positive count and 5/29 for a count of two or more. Both are lower bounds rather than exact binomial tails.

Explore the model

Interactive lab

Interactive lab · probability

Moments compared with exact binomial bars

Pause and Reset, then set p = 0.10 and n = 20. Predict mean 2 and variance 1.8, and compare the leftmost bar with P(X>0) = 1−0.9²⁰, approximately 0.8784. The second-moment lower bound is only 20/29. Reset and set both p = 0.20 and n = 20 to predict a stronger positivity guarantee. The displayed distribution is exact for independent trials, not for overlapping subgraph indicators.

Moments compared with exact binomial barsThese 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 forMoment bounds can be valid and useful without reproducing the exact distribution.

Check your understanding

For a nonnegative X with mean 3 and second moment 12, what does Paley–Zygmund give at theta=1/3?

Try it yourself

For X binomial with twelve trials and p=1/4, compute the second-moment bound for X>0 and the Paley–Zygmund bound for X>2.

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

Give me a hint

Find variance first. For the second event, choose theta so that theta times the mean equals two.

Compare with the solution

The mean is three, variance is 12(1/4)(3/4)=9/412(1/4)(3/4)=9/4, and second moment is 9+9/4=45/49+9/4=45/4. The positivity lower bound is 9/(45/4)=4/59/(45/4)=4/5. For X>2, theta is 2/3, giving (1/3)2(4/5)=4/45(1/3)^2(4/5)=4/45. The latter event means at least three successes, including X=3; it does not mean at least two. These guarantees do not require computing any binomial coefficient once the moments are known, although the independence of the Bernoulli trials was used to obtain the stated variance.

Take it with you

  • Large means require overlap or variance control before implying typical existence.
  • Distinct ordered indicator pairs and diagonal terms play different roles.
  • Paley–Zygmund bounds mass strictly above a chosen fraction of a positive mean.

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