advanced28 min studyLesson 633 in Mathematics

Random graph thresholds and components

Separate edge, triangle, giant-component, and connectivity scales in independent-edge graphs.

On this page

What you’ll understand

  • State threshold quantifiers for sequences of edge probabilities.
  • Verify the triangle threshold using first and second moments.
  • Interpret sparse branching approximations without confusing a giant component with connectivity.

A threshold is an asymptotic statement about a sequence of probability spaces. It is not a claim that one finite graph switches behavior at an exact parameter. Different graph properties live on different scales: an edge can appear long before a triangle, a giant component, or connectivity. Careful quantifiers explain which changes are established and which critical regimes need a separate analysis.

Define the model before discussing the scale

In G(n,p), the vertex set is labeled and every possible edge is included independently with probability p, which may depend on n. For an increasing graph property, a threshold function f means probability tends to zero whenever p/f tends to zero, and tends to one whenever p/f tends to infinity. Neither clause specifies the limiting probability at a fixed constant multiple of f. A sharp threshold requires additional relative-window information and should not be inferred merely from the existence of a threshold function.

For containing an edge, the number of edges is binomial with mean approximately n2p/2n^2p/2. If this mean tends to zero, Markov gives absence with high probability. If it tends to infinity, the exact empty-graph probability is at most the exponential of minus that mean, which tends to zero. Thus the edge threshold is of order n−2n^{-2}. This elementary example already demonstrates why the same parameter p must be interpreted together with the number of available opportunities.

P(G(n,p) has no edges)=(1−p)(n2)≤exp⁡ ⁣(−(n2)p)\mathbb P(G(n,p)\text{ has no edges})=(1-p)^{\binom n2}\le\exp\!\left(-\binom n2p\right)
Independence is used in the product formula; a diverging expected edge count then forces at least one edge with high probability.↔ Scroll wide equations horizontally.

Triangles reveal why overlap must be counted

Let X count triangles. Its mean is the number of vertex triples times the probability of their three edges. Below p of order 1/n, the mean tends to zero. Above that scale, a second-moment calculation works: distinct triangles sharing no edge have independent indicators, even if they share a vertex. The correlated pairs share exactly one edge, hence require five edges in their union. There are order n to the fourth such pairs, while the diagonal contributes at most the mean.

Dividing these contributions by the squared mean gives a bound of order the reciprocal of n3p3n^3p^3 plus the reciprocal of n2pn^2p. Both terms vanish whenever np tends to infinity. This establishes the threshold 1/n without claiming the probability is near one at every fixed multiple of 1/n. In the constant-multiple regime the number of triangles can have a nontrivial limiting law. The first and second moments deliberately leave that transition window unresolved.

EX=(n3)p3,Var⁡(X)(EX)2=O ⁣(1n3p3+1n2p)\mathbb E X=\binom n3p^3,\qquad \frac{\operatorname{Var}(X)}{(\mathbb E X)^2}=O\!\left(\frac1{n^3p^3}+\frac1{n^2p}\right)
The asymptotic variance bound comes from diagonal triangles and pairs sharing an edge; it applies as n grows and p>0.↔ Scroll wide equations horizontally.

Components require a different exploration

For p=c/n with fixed c, an early component exploration resembles a branching process with approximately Poisson(c) offspring. This approximation describes a small exploration before collisions become significant; it is not an identity for an entire finite component. The resulting established random-graph theorem says that for fixed c<1 every component has logarithmic size with high probability, whereas fixed c>1 yields a unique component with a positive limiting vertex fraction. The critical case c=1 needs its own scaling analysis.

The giant fraction for c>1 is the positive solution rho of ρ=1−e−cρ\rho=1-e^{-c\rho}; the zero solution must not be mistaken for the supercritical survival probability. Even a giant need not contain every vertex. A particular vertex is isolated with exact probability (1−p)n−1(1-p)^{n-1}, so the expected number of isolates is easy to calculate. Connectivity occurs near the larger scale log(n)/n. An expected isolate count alone does not prove typical disconnection, but it exposes why a mean degree near one is no certificate of connectivity.

Mean degree one does not mean connected

In G(101,0.01), calculate the expected edge count, expected degree of one vertex, and expected number of isolated vertices.

  1. There are (1012)=5050\binom{101}{2}=5050 candidate edges. Linearity gives an expected edge count of 50.5.

  2. A specified vertex has 100 incident independent edge trials, so its expected degree is one. It is isolated exactly when all these trials fail, with probability 0.991000.99^{100}.

  3. Summing the isolate indicators gives 101(0.99)100≈36.9693101(0.99)^{100}\approx36.9693. The isolate events are dependent, but independence between those indicators is unnecessary for this expectation.

The result

Expected edges: 50.5; expected degree: 1; expected isolates: approximately 36.9693. These expectations do not determine a finite-sample connectivity probability.

Explore the model

Interactive lab

Interactive lab · probability

Inspect a vertex degree marginal

Pause and Reset, then set p = 0.10 and n = 10. Interpret the ten trials as possible neighbors of one vertex in G(11,0.1). Predict mean degree one and isolate probability 0.9¹⁰, approximately 0.3487. Reset and set both p = 0.20 and n = 10 to compare the zero-degree bar. This is one degree marginal; playback does not construct a graph, discover components, or establish any asymptotic threshold.

Inspect a vertex degree marginalThese 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 degree distribution is local information; a component or connectivity claim needs additional graph structure.

Check your understanding

Which statement is part of saying that 1/n is a threshold for containing a triangle in G(n,p)?

Try it yourself

For G(51,0.04), compute the expected degree and expected number of isolates. Then decide whether p=n^(-3/2) is below or above the triangle threshold as n grows.

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

Give me a hint

For isolates, keep the probability of all fifty incident edges being absent. For the asymptotic question, compute np.

Compare with the solution

The expected degree is 50(0.04)=250(0.04)=2. The expected isolate count is 51(0.96)50≈6.6241851(0.96)^{50}\approx6.62418, obtained by summing fifty-one dependent isolate indicators. Neither expectation itself is a connectivity probability. For p equal to n−3/2n^{-3/2}, np equals n−1/2n^{-1/2} and tends to zero. The triangle mean is asymptotic to n3p3/6=n−3/2/6n^3p^3/6=n^{-3/2}/6, which tends to zero, so Markov establishes triangle absence with probability tending to one. The graph can nevertheless have many edges, because its expected edge count is of order the square root of n.

Take it with you

  • Threshold statements quantify over sequences below and above a scale.
  • Triangles depend on shared edges, not merely shared vertices.
  • A supercritical giant component and full connectivity occur at different scales.

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