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The expected number of triangles in $G (n, p)$ is $(\binom{n}{3}) p^{3}$

Example

Let $T$ be the number of triangles in $G (n, p)$ . Then $E [T] = (\binom{n}{3}) p^{3},$ and $Var (T) = (\binom{n}{3}) (p^{3} - p^{6}) + 2 (\binom{n}{2}) (\binom{n - 2}{2}) (p^{5} - p^{6}) .$

Facts & Assumptions

Given: The random graph $G (n, p)$ and its triangle count $T$ .

[L1]

$G (n, p)$ has independent Bernoulli edge coordinates (The Erdős-Rényi finite random graph $G (n, p)$ ).

[L2]

A prescribed set of $r$ present edges has probability $p^{r}$ (A prescribed set of present and absent edges in $G (n, p)$ has product probability).

[L3]

Indicators count occurrences and expectation is linear (Indicators turn event probabilities, intersections, and finite counts into expectations and products, Expectation is linear for every finite family of random variables, without any independence hypothesis).

[L4]

There are $(\binom{n}{3})$ vertex triples and $(\binom{n}{2})$ vertex pairs (The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $(\binom{n}{k}) : = ∣ [n]^{k} ∣$ ).

[L5]

On the finite random-graph probability space, variance and covariance are the expectations of the centered square and centered product. (Variance, standard deviation, and covariance on a finite probability space)

Verification

technique · direct

1.1L2L3L4

Write $T = \sum_{Q} I_{Q}$ , with $Q$ ranging over three-element vertex sets. Each $I_{Q}$ requires three present edges and has expectation $p^{3}$ , so $E [T] = (\binom{n}{3}) p^{3}$ .

1.2L1L2L5algebra

Each indicator has variance $p^{3} - p^{6}$ . Two distinct triangles are independent unless they share an edge; if they share an edge, their joint occurrence requires five edges and their covariance is $p^{5} - p^{6}$ .

1.3L4

An unordered pair of triangles sharing an edge is determined by the common edge and the two distinct extra vertices, so there are $(\binom{n}{2}) (\binom{n - 2}{2})$ such pairs.

1.4step 1.1step 1.2step 1.3L3L5algebra∎

Put $Y_{Q} = I_{Q} - E I_{Q}$ . By finite linearity [L3], $T - E T = \sum_{Q} Y_{Q}$ . Expanding this finite square and applying [L3] gives $Var (T) = E {(\sum_{Q} Y_{Q})}^{2} = \sum_{Q} E Y_{Q}^{2} + 2 \sum_{{Q, R} : Q \neq R} E [Y_{Q} Y_{R}],$ where the second sum is over unordered pairs of distinct triangles. By [L5] these terms are respectively the individual variances and pairwise covariances. Substituting their values from step 1.2 and their counts from steps 1.1 and 1.3 proves the displayed variance formula. For $n < 3$ the triangle family and the relevant pair counts are empty, so the expectation and variance are zero; for $p = 0, 1$ each variance contribution is zero.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Matousek and J. Vondrak, The Probabilistic Method, Chapters 1 and 3 (standard reference, not scraped)

The expected number of triangles in G(n,p)G(n,p) is (n3)p3(3n​)p3