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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]

Variance and covariance obey their centred-product definitions, and the variance of a finite sum is the sum of individual variances plus twice the unordered pairwise covariances (Variance, standard deviation, and covariance on a finite probability space, Variance of a finite sum as the sum of all variances and covariances).

Verification

technique · direct

1.1

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 $E [T] = (\binom{n}{3}) p^{3}$ .

L2L3L4

1.2

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 $p^{5} - p^{6}$ .

L1L2L5algebra

1.3

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.

2.1

Sum the individual variances and twice the unordered covariances to obtain the formula. For $n < 3$ all relevant binomial coefficients vanish; for $p = 0, 1$ the variance is zero.

step 1.1step 1.2step 1.3L5∎

Depends on

Used by

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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