{wiki}

The second moment method uses a <random variable>'s <expected value> and <second moment> to establish that a counted or measured quantity is positive with substantial <probability>. It is useful when a positive average alone could be caused by extremely large values on very rare outcomes. Controlling the <variance> rules out too much of that concentration in rare outcomes.

For a nonnegative <random variable> $N$ with $m=\mathbb EN>0$ and finite <second moment>, the <second-moment bound for positivity> gives
$$
\boxed{\mathbb P(N>0)\geq\frac{(\mathbb EN)^2}{\mathbb E N^2}=\frac{m^2}{m^2+\operatorname{Var}(N)}.}
$$
Equivalently, $\mathbb P(N=0)\leq\operatorname{Var}(N)/(m^2+\operatorname{Var}(N))$. The method also includes the simpler <Chebyshev's inequality> bound $\mathbb P(N=0)\leq\operatorname{Var}(N)/m^2$. Thus a <sequence> of such <random variables> with $\operatorname{Var}(N)/(\mathbb EN)^2\to0$ has $N>0$ with <probability> tending to one. These statements require no <independence> assumption. The variance argument and its use for random counts are developed in Zhao's lecture notes. https://ocw.mit.edu/courses/18-226-probabilistic-methods-in-combinatorics-fall-2022/mit18_226_f22_lec06-07.pdf[Yufei Zhao, MIT 18.226, §4.1, corollaries 4.1.7–4.1.8: variance and non-existence]{ref}.

For an original toy count, let $N$ count successes in ten <independent> <Bernoulli trials> with success <probability> $p=1/10$. The <binomial distribution> gives
$$
\mathbb EN=1,\qquad \operatorname{Var}(N)=9/10,\qquad \mathbb E N^2=19/10.
$$
Consequently \b[$\mathbb P(N>0)\geq10/19\approx0.526316$]. The exact <probability> is $1-(9/10)^{10}\approx0.651322$. The <first moment method> and <union bound> only give the upper bound $\mathbb P(N>0)\leq\min\{1,\mathbb EN\}=1$ here. First and second moments therefore answer complementary existence questions.

\Image[/second-moment-count-positivity-and-dependence.png]
{title=The <second moment method> certifies a positive count, while <statistical dependence> can change the conclusion despite an unchanged <mean>}
{description=Left: exact <binomial distribution> probabilities of at least one success and second-moment lower bounds for ten <independent> trials as the success probability varies. Right: the same ten trials with $p=1/10$ either occur independently or share one common <Bernoulli> outcome. Both counts have <mean> one, but the perfectly linked count has much larger <variance> and only $1/10$ probability of being positive. All data are original specified probability models.}

For a <sum> of <indicator random variables>, the <variance of a sum> is
$$
\operatorname{Var}\left(\sum_i I_i\right)=\sum_i\operatorname{Var}(I_i)+2\sum_{i<j}\operatorname{Cov}(I_i,I_j).
$$
<Independence> makes the <covariances> vanish, but the formula itself also covers dependent counts. For the same ten success probabilities, let every trial share a single <Bernoulli random variable> $B$: then $N=10B$, with <mean> one, <variance> nine, and $\mathbb P(N>0)=1/10$. The second-moment lower bound is exactly $1/10$. Ignoring the positive <covariances> would produce the false independent-count conclusion. Zhao's discussion of sums of indicators explains why overlapping counted structures require these pairwise terms. https://ocw.mit.edu/courses/18-226-probabilistic-methods-in-combinatorics-fall-2022/mit18_226_f22_lec06-07.pdf[Zhao, MIT 18.226, §4.1: indicator covariances and Bernoulli counts]{ref}.


 Second moment method (source code)