advanced28 min studyLesson 639 in Mathematics

Janson inequalities and subgraph counts

Use independent underlying choices and an explicit overlap sum to bound the absence of dependent patterns.

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What you’ll understand

  • Identify containment events in a product probability space.
  • Compute the ordered distinct-pair Delta convention without dropping a factor of two.
  • Apply an absence bound and a lower-tail bound within their stated hypotheses.

Subgraph indicators overlap, so independent-sum concentration is often unavailable. Janson inequalities exploit a more specific structure: each occurrence requires a fixed set of underlying independent choices to be present. They retain the overlap information in a joint-probability sum. The resulting absence and lower-tail estimates can be exponentially stronger than a second-moment estimate, but they are not general inequalities for every family of dependent events.

State the product-space containment setup

Let R be a random subset of a finite ground set, each element included independently with its own fixed probability. For each fixed support S_i, let A_i mean that all elements of S_i belong to R. Count occurrences with X equal to the sum of their indicators and denote its mean by mu. In a graph application the ground elements are possible edges, not vertices, unless the random experiment actually samples vertices. Each support is the edge set of a particular pattern copy.

Join two distinct supports when they intersect and define Delta by summing their joint probabilities over ordered distinct intersecting pairs. Thus a dependent unordered pair contributes twice. The summand is the probability that the whole union of the supports is present, not their covariance and not the product of their marginal probabilities. Diagonal terms are excluded from Delta. Some references use an unordered convention or include diagonal terms, changing formulas; fix the definition before importing a theorem’s constants.

μ=∑iP(Ai),Δ=∑i≠jSi∩Sj≠∅P(Ai∩Aj),P(X=0)≤exp⁡(−μ+Δ/2)\mu=\sum_i\mathbb P(A_i),\qquad \Delta=\sum_{\substack{i\ne j\\S_i\cap S_j\ne\varnothing}}\mathbb P(A_i\cap A_j),\qquad \mathbb P(X=0)\le\exp(-\mu+\Delta/2)
Delta uses ordered distinct pairs in the independent-element containment model; cap the upper bound at one if its exponent is positive.↔ Scroll wide equations horizontally.

Count shared random choices, not visual overlap

Two triangles that share one vertex have disjoint edge supports, so their occurrence indicators are independent in G(n,p). Two triangles sharing an edge have a five-edge union and joint probability p to the fifth power. On four vertices there are four triangles and six unordered pairs of distinct triangles; every pair shares one edge. Consequently Delta has twelve, not six, equal joint-probability terms. This elementary case is an effective way to catch a missing factor of two.

When Delta is much smaller than mu, the absence bound resembles an independent rare-event exponential. When overlap is large, the first bound can become uninformative even though the count is usually positive. One useful lower-tail form incorporates both the diagonal scale mu and the overlap scale Delta in its denominator. Assuming mu>0, it applies to deviations t between zero and mu, including t=mu for absence. If mu=0 then nonnegativity forces X=0 almost surely, which is handled separately without a denominator formula. The denominator is not simply the variance, and no claim is made that Delta equals the entire dependence structure.

P(X≤μ−t)≤exp⁡ ⁣(−t22(μ+Δ)),0≤t≤μ\mathbb P(X\le\mu-t)\le\exp\!\left(-\frac{t^2}{2(\mu+\Delta)}\right),\qquad0\le t\le\mu
This lower-tail version assumes mu>0 and uses the same containment setup and ordered, diagonal-free Delta convention; it is not an upper-tail inequality.↔ Scroll wide equations horizontally.

Use the inequality as a certificate rather than a distribution

The absence upper bound need not be close to the actual absence probability at finite size. In particular, treating overlapping pattern indicators as independent produces a different number, often with no justified inequality direction unless a correlation theorem is also invoked. An exact enumeration for a tiny graph can audit the calculation without being a proof for larger graphs. For asymptotic use, one must count every relevant intersection type and compare its contribution on the same scale as the squared mean or the intended exponential rate.

Containment is also a real restriction. An induced-copy event requires some edges present and other edges absent, so it is not directly a positive containment event on the original edge ground set. Arbitrary color-pattern events or martingale differences likewise do not automatically satisfy this setup. A more general theorem may apply, but its assumptions must be checked separately. Here both examples use only fixed triangles requiring their edges, with no absence constraints, making the product-space interpretation exact.

Absence of triangles on four vertices

Let X count all triangles in G(4,1/2). Compute mu, the ordered Delta, and the first Janson upper bound on P(X=0).

  1. There are four triangle supports, each with probability (1/2)3=1/8(1/2)^3=1/8. Hence μ=1/2\mu=1/2.

  2. Every distinct pair shares an edge, and there are twelve ordered pairs. Their unions have five edges, so Δ=12(1/2)5=3/8\Delta=12(1/2)^5=3/8.

  3. Janson gives P(X=0)≤exp⁡(−1/2+3/16)=e−5/16≈0.731616\mathbb P(X=0)\le\exp(-1/2+3/16)=e^{-5/16}\approx0.731616. Direct enumeration gives 41 triangle-free graphs among 64 equally likely graphs, or 41/64≈0.640625, consistent with the bound.

The result

The exact probability 41/64 lies below the rigorous upper bound e⁻⁵⧸¹⁶. Using an unordered Delta without changing the coefficient would produce the wrong certificate.

Explore the model

Interactive lab

Interactive lab · probability

An independent pattern-count comparison

Pause and Reset, then set p = 0.12 and n = 4. Predict independent zero-count probability 0.88⁴, approximately 0.5997. Reset and set both p = 0.13 and n = 4 to obtain 0.87⁴, approximately 0.5729. These bracket the independent benchmark (7/8)⁴≈0.5862 for four indicators of marginal probability 1/8, which is not selectable exactly on the p grid. The actual overlapping-triangle absence probability is 41/64≈0.6406. The lab shows no graph overlaps and cannot compute Delta.

An independent pattern-count comparisonThese 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 forMatching marginal pattern probabilities does not reproduce the law of an overlapping pattern count.

Check your understanding

Two specified distinct triangles share one edge in G(n,p). What is their total contribution to the ordered Delta used here?

Try it yourself

Count only two specified triangles sharing an edge in G(4,1/2). Find their exact no-occurrence probability and compare it with the first Janson bound.

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

Give me a hint

Use inclusion-exclusion for just two events, then count both orders in Delta.

Compare with the solution

Each event has probability 1/8 and their intersection has probability 1/32. Thus exact absence probability is 1−1/8−1/8+1/32=25/32=0.781251-1/8-1/8+1/32=25/32=0.78125. The mean is 1/4 and ordered Delta is 2/32=1/162/32=1/16. Janson gives exp⁡(−1/4+1/32)=e−7/32≈0.803523\exp(-1/4+1/32)=e^{-7/32}\approx0.803523, which is indeed larger than 25/32. The independent-events guess would instead be (7/8)2=49/64(7/8)^2=49/64, a different value. The discrepancy comes from the shared edge, not from sharing vertex labels alone.

Take it with you

  • Janson’s containment setup requires independent underlying element choices.
  • Ordered, diagonal-free Delta sums joint probabilities of overlapping patterns.
  • Absence and lower-tail bounds must retain their conventions, direction, and event hypotheses.

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