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For a nonnegative integer-valued random variable $N$, Markov inequality givesConsequently, $\mathbb{E}N<1$ proves that some outcome has $N=0$, while $\mathbb{E}N\to 0$ shows that $N=0$ with probability tending to one.

For the number $X$ of $k$-cliques in a binomial random graph $G(n,p)$, with $0<p<1$ and $\mu =\mathbb{E}X$,Two $k$-sets overlapping in $j$ vertices share $\left(\genfrac{}{}{0ex}{}{j}{2}\right)$ edges. Overlaps of zero or one vertex give independent indicators; $j=k$ includes the diagonal terms. The identity is useful for the second moment method.

Assume $0<\lambda =1-\epsilon <1$, as required for this nontrivial subcritical binomial random graph. Put $\delta =\lambda -1-\log \lambda >0$. A breadth-first exploration of a binomial random graph is dominated by a Galton-Watson process whose offspring counts are independent random variables with law $\mathrm{Bin}(n,\lambda /n)$. If its total progeny $T$ is at least $j$, the first $j-1$ offspring counts sum to at least $j-1$. Their moment-generating function is bounded by that of a Poisson distribution of mean $\lambda (j-1)$. The exponential Markov inequality, optimized at $t=-\log \lambda$, yieldsTaking $j=\lceil (1+\eta )\log n/\delta \rceil$ for any fixed $\eta >0$, a union bound over all starting vertices gives $\mathbb{P}(L_1\ge j)=o(1)$.

For the matching lower bound, let $X_j$ count tree components of order $j$. The tree-component expectation in the Erdős-Rényi model is exactlyFor $j=O(\log n)$, the Stirling formula givesChoose $j=\lfloor (1-\eta )\log n/\delta \rfloor$, with $0<\eta <1$. Then $\mathbb{E}{X}_j\to \infty$. Distinct overlapping vertex sets cannot both be graph components. For disjoint sets, the joint probability differs from the product only because the $j^2$ between-set edges were counted twice as absent. ThusIt follows that $\mathrm{var}{X}_j/(\mathbb{E}{X}_j{)}^2=o(1)$. The second moment method gives $X_j>0$ with high probability. Combining both bounds, and then taking arbitrarily small fixed $\eta$, proves the sharp subcritical largest-component scaleFinally the Taylor expansion of $-\log (1-\epsilon )$ gives $\delta ={\epsilon }^2/2+{\epsilon }^3/3+\cdots$. The endpoint $\lambda =0$ has no edges and is excluded from this logarithmic asymptotic.

The Chebyshev inequality states that for a random variable $Z$ with finite variance and $a>0$,Write ${\mu }_n=\mathbb{E}[T]$. The ratio in the hypothesis presupposes ${\mu }_n>0$ for all sufficiently large $n$. On the event $T=0$, the deviation from the mean equals ${\mu }_n$. Thus the Chebyshev inequality with $a={\mu }_n$ givesThis is the second moment method. It uses only the stipulated relative variance bound; no assumption that triangle indicators are independent is needed.

Let $X_s$ count the copies of the complete graph $K_s$ in the binomial random graph $G_{n,p}$. By linearity of the expected value and independence of distinct edges,Throughout, $n$ is sufficiently large that $k=\lfloor \log \log n\rfloor \ge 2$. We have $k\to \infty$, $k=O(\log \log n)$ and $k\log k=o(\log n)$.

First, since $p=n^{-2/k}$,The first moment method proves $\mathbb{P}(X_{k+1}>0)=o(1)$.

For the existence of a $k$-clique, put $X=X_k$ and $\mu =\mathbb{E}X$. Since $p^{\left(\genfrac{}{}{0ex}{}{k}{2}\right)}=n^{-(k-1)}$,This alone is insufficient: we must control the variance via the overlap formula for the variance of a clique count. Let $I_S$ indicate that a given $k$-set $S$ spans a clique. If $|S\cap T|=j$, the two copies share $\left(\genfrac{}{}{0ex}{}{j}{2}\right)$ edges, soFor $j=0,1$ their edge sets are disjoint and the covariance vanishes, even when they share one vertex. Counting ordered pairs of $k$-sets gives the exact formulaThis includes the diagonal case $j=k$.

To bound the overlap factor, regard it as the probability that a uniformly chosen $k$-set $T$ intersects a fixed $S$ in exactly $j$ vertices. A union bound over the possible specified $j$-subsets of $S$ yieldsHere $(a{)}_j=a(a-1)\cdots (a-j+1)$ is a falling factorial, and the last bound holds for $n\ge 2k$. ThereforeFor $2\le j\le k$, the concave quadratic $j(k-j+1)$ is minimized at an endpoint. Its endpoint values are $2(k-1)$ and $k$, both at least $k$. Thus every power of $n$ in the last sum is at most $n^{-1}$, andbecause the logarithm of its numerator is $O(k\log k)=o(\log n)$. The second moment method, or Chebyshev's inequality directly, gives $\mathbb{P}(X=0)=o(1)$. Finally, every larger clique contains a $(k+1)$-clique, so the absence of $K_{k+1}$ excludes all larger sizes. The clique number is exactly $k$ with high probability:

Because $X$ is a nonnegative integer, its indicator function satisfies $1_{\{X>0\}}\le X$. Taking expected values proves the upper bound. The finite second moment also gives a finite expected value, by the Cauchy-Schwarz inequality. Applying that inequality to $X$ and $1_{\{X>0\}}$ givesThe assumed positive second moment permits division, yieldingThe lower bound is the basic second moment method; the upper bound also follows from Markov's inequality at threshold one. Integer-valuedness is essential for that upper bound, but not for the lower bound. For example, a constant $X=1/2$ would violate the upper bound.

For events $A_1,\dots ,A_n$, the probabilistic inclusion-exclusion principle isTo prove it, fix an outcome lying in exactly $r$ events. Its total coefficient on the right isby the binomial theorem; outcomes in no event contribute zero. Taking expectations of this pointwise indicator identity proves the formula. Truncating after the pair terms gives the Bonferroni inequalities

For the final bound, let $X={\sum }_i{1}_{A_i}$ and $S=\mathbb{E}X={\sum }_i\mathbb{P}(A_i)$. The intersection assumption givesThe Cauchy-Schwarz inequality applied to $X{1}_{X>0}$ yields $S^2\le \mathbb{E}[X^2]\mathbb{P}(X>0)\le S+n-1$. Thereforeso the required universal constant may be taken as $\boxed{c=2}$. This is a second moment method estimate.

Let $X$ count copies of the complete graph $K_4$ in $G$. Its expected value is

Write $X$ as a sum of indicator random variable. Indicators for two distinct copies are independent when the copies share at most one vertex, since their edge sets are then disjoint. Pairs sharing two vertices have eleven edges in their union, while pairs sharing three have nine. It follows thatThus $\mathrm{var}(X)/{\mu }^2\to 0$. By the second moment method, $X/\mu \to 1$ in probability, so $\mathbb{P}(X\ge 100)\to 1$. This is the sparse clique-count concentration estimate.

If $N=0$, then $|N-\mathbb{E}N|=\mathbb{E}N$. Hence the Chebyshev inequality givesThe assumed limit therefore provesThis is the second moment method in its simplest form.

For fixed $0<\lambda <1$ in $G(n,\lambda /n)$, the largest component of a graph has the displayed order with high probability. A dominating Galton-Watson process gives the tail bound $\mathbb{P}(T\ge j)\le e^{-(\lambda -1-\log \lambda )(j-1)}$ by the exponential Markov inequality. For the lower bound, the tree-component expectation in the Erdős-Rényi model diverges at $j=\lfloor (1-\eta )\log n/(\lambda -1-\log \lambda )\rfloor$. The disjoint-set covariance factor and the exclusion of overlaps give relative variance tending to zero, so the second moment method supplies a tree component of that order.

If $X$ counts triangles in $G(n,p)$, thenandThus $np\to \infty$ implies $\mathrm{var}X/(\mathbb{E}X{)}^2\to 0$, and the second moment method gives $\mathbb{P}(X>0)\to 1$.