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In the Erdős-Rényi model $G(n,p)$, the vertex set is $[n]$ and every one of the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ possible edges is present independently with probability $p$.

Let $X$ count isolated vertices. If $p\ge (1+\epsilon )\log n/n$, the union bound givesThus $\mathbb{P}(E_n)\to 1$.

Chebyshev inequality states thatNow suppose $p\le (1-\epsilon )\log n/n$. Monotonicity lets us use the largest allowed $p$. Withthe supplied lower exponential estimate showsFor distinct vertices $i,j$,HenceandChebyshev with $t=\mathbb{E}X$ gives $\mathbb{P}(X=0)\to 0$, so $\mathbb{P}(E_n)\to 0$.

For connectivity, the lower-threshold result is immediate because a connected graph has no isolated vertex:For the upper threshold, if a graph is disconnected it has a component with vertex set $S$ of some size $1\le k\le n/2$. In particular, every one of the $k(n-k)$ edges from $S$ to its complement is absent. ThereforeChooseFor $1\le k\le \eta n$, the standard bound $\left(\genfrac{}{}{0ex}{}{n}{k}\right)\le (en/k{)}^k$ givesThe sum of these terms tends to zero. For $\eta n\le k\le n/2$, use $\left(\genfrac{}{}{0ex}{}{n}{k}\right)\le 2^n$ and $k(n-k)\ge \eta n^2/2$:Thus $\mathbb{P}(F_n)\to 1$ above the threshold. Together,This is the connectivity threshold in the Erdős-Rényi model.

Markov inequality says that a nonnegative random variable $Y$ satisfiesChebyshev inequality says that a random variable with finite variance satisfies

Let $X$ count triangles in $G(n,p)$. The triangle count in a binomial random graph calculation givesSince $p\gg n^{-1}$,For sufficiently large $n$, $\mu >200$, and Chebyshev inequality gives

The probability generating function of each $X_i$ is $\exp (z-1)$. Independence makes the generating function of $S_n$ equal to $\exp (n(z-1))$, so the addition of independent Poisson random variables gives $S_n\sim \mathrm{Pois}(n)$.

The Poisson distribution has mean and variance $n$, hence $\mathbb{E}{Y}_n=0$ and $\mathbb{E}[Y_n^2]=1$. Since $Y_n^{-}\le |Y_n|$, Chebyshev inequality gives

Part (iii) and the given harmonic-sum asymptotic show thatFor any fixed $\epsilon >0$, this deterministic ratio lies within $\epsilon /2$ of $1$ for all sufficiently large $n$. The Chebyshev inequality and part (iv) then giveThus

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.

Chebyshev inequality gives $\mathbb{P}(T=0)\le \mathbb{P}(|T-\mathbb{E}T|\ge \mathbb{E}T)\le \mathrm{var}(T)/(\mathbb{E}T{)}^2\to 0$.

Assume without loss of generality that $\mu (f)-\nu (f)=r\sigma$ and let $m=(\mu (f)+\nu (f))/2$. By Chebyshev inequality,Using the event $\{f\ge m\}$ in the variational definition of total variation distance gives

Start the lazy hypercube walk at $0^n$ and writeThis is an eigenfunction with eigenvalue $1-1/n$, soThe stated variance estimates allow the preceding lemma with $\sigma =\sqrt{n}$ andFor $t=\frac{1}{2}n\log n-Cn$, $r^2$ is bounded below by a constant multiple of $e^{2C}$, uniformly for all sufficiently large $n$. Choosing $C=C(\epsilon )$ so that $1-8/r^2>\epsilon$, and absorbing finitely many small $n$ into the constant, proves

Let $N_{\delta }$ count sample points in the intersection of $[0,1{]}^d$ with the ${\ell }^{\infty }$ ball of radius $\delta$ around $x$. That intersection has volume at least ${\delta }^d$, so $N_{\delta }\sim \mathrm{Bin}(n,q)$ with $q\ge M{\delta }^d$. The event $\Vert X_{(L)}-x{\Vert }_{\infty }>\delta$ implies $N_{\delta }<L$. Since $\mathrm{Var}(N_{\delta })\le nq$ and $nq\ge nM{\delta }^d\ge L$, Chebyshev inequality givesTaking the minimum with the trivial bound one proves the result.