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For an additive arithmetic function $f$, one standard form of the Turán-Kubilius inequality usesIt states, with an absolute constant,For real $f$, this also bounds the variance under the discrete uniform distribution, because centering at the actual expectation minimizes the mean square.

For the prime omega function, $f(p^k)=1$. The Mertens theorem for reciprocal primes states ${\sum }_{p\le N}1/p=\log \log N+O(1)$. The contribution of higher prime powers is bounded by ${\sum }_{p}1/(p(p-1))\le 1$, as proved in 1(a). It follows that $A_{\omega }(N)=\log \log N+O(1)$ and $B_{\omega }(N{)}^2\ll \log \log N$. Writing $L=\log \log N$, the Turán-Kubilius inequality and Chebyshev inequality therefore yieldThe bounded error in $A_{\omega }-L$ is absorbed by, for example, using the threshold $0.005L$ around $A_{\omega }$ for sufficiently large $N$.

For the multiplication table problem, call a pair $(a,b)\in [N{]}^2$ good when $\omega (a),\omega (b)\ge 0.99L$ and $\omega (\gcd (a,b))\le 0.1L$. The preceding Chebyshev inequality excludes only $O(N^2/L)$ pairs by either of the first two conditions. For the greatest common divisor condition, linearity of expectation under uniform sampling of the two coordinates givesMarkov inequality therefore excludes only another $O(N^2/L)$ pairs.

The prime factors of a product form the union of those of its factors. Consequently, for every good pair,On the other hand, apply the proved concentration estimate at $N^2$. Since $\log \log (N^2)=L+\log 2$, for large $N$ every integer $m\le N^2$ with $\omega (m)\ge 1.88L$ is exceptional there. There are only $O(N^2/L)$ such integers. Products of good pairs lie in this exceptional set; products having only bad representations are at most as numerous as the bad pairs. This proves the multiplication-table upper bound, using distinct prime factors alone:

If $T>n$, either $T_R\le n$, or both stopping times exceed $n$. Therefore the maximal bound for a nonnegative martingale and the Markov inequality give, for $R\ge 1$,The right-hand side is minimized at $R=\sigma \sqrt{n}/\tau$. Whenever the claimed bound is nontrivial, $2\tau /(\sigma \sqrt{n})<1$, this choice has $R>2$ and is allowed under the corrected overshoot hypothesis. Substituting gives the square-root tail bound for martingale absorption:When the displayed upper bound is at least one, the inequality follows from $\mathbb{P}(T>n)\le 1$. This handles every $n$ without using the impossible small-threshold hypothesis.

For every $\theta \ge 0$, the Markov inequality and independence give the Chernoff boundOptimize over nonnegative $\theta$. For $a\ge \mu$, part (a) identifies this supremum with $I(a)$, yielding the finite-$n$ estimateIf $I(a)=\infty$, use parameters with arbitrarily large $\theta a-\psi (\theta )$; the probability must then be zero. Taking the logarithmic upper limit proves the upper bound in the Cramér theorem:For $a<\mu$, the nonnegative-parameter supremum is zero, giving the correct trivial tail-rate upper bound. Exponential Markov bounds and the product moment-generating function supply the entire upper-bound argument.

Let $\mu$ be the given Brownian motion law on $C[0,1]$. Choose $\alpha <\beta <1/2$ and an integer $p$ with $p(1/2-\beta )>1$. For the largest adjacent level-$n$ increment $A_n$, Brownian scaling and the Markov inequality givewhere $c_p=\mathbb{E}|\mathcal{N}(0,1){|}^p<\infty$. These probabilities are summable, so the Borel-Cantelli lemma gives a finite random $K$ such that $A_n\le K{2}^{-n\beta }$ for every $n$, after absorbing the finitely many exceptional levels.

For $s<t$, choose $m$ with $2^{-m}\le t-s<2^{1-m}$. Their left dyadic approximations at level $m$ differ by at most two grid steps; each finer approximation adds at most one adjacent increment. Continuity and dyadic increment chaining therefore giveThus paths are $\beta$-Hölder and hence $\alpha$-Hölder on $[0,1]$. This direct proof supplies the relevant Brownian Hölder regularity, also obtainable from the Kolmogorov continuity theorem.

The full-measure subset $E_{\alpha }=C^{\alpha }[0,1]$ is measurable in $C[0,1]$, because continuity identifies its seminorm with a countable supremum:Its finiteness event is a countable union of countable intersections of coordinate-measurable events. The sigma-algebra specified on $E_{\alpha }$ is precisely the trace of the coordinate sigma-algebra on $C[0,1]$. Define $\nu (A)=\mu (A)$ for trace-measurable $A\subseteq E_{\alpha }$; equivalently restrict the Brownian law to its full-measure subset. Since $\mu (E_{\alpha })=1$, no renormalization is needed, and all finite-dimensional coordinate laws are unchanged. This is the required probability measure:The construction uses the requested coordinate sigma-algebra and makes no assumption about a separable Hölder-norm topology.

Write $J_{n,h}$ for the squared integral norm in question. The scaling property of the Fourier transform gives $\mathcal{F}{K}_h(u)=\mathcal{F}K(hu)$, which vanishes outside $[-1/h,1/h]$. Also $|\mathcal{F}K(v)|\le \Vert K{\Vert }_1$. The condition $\int K=1$ does not imply $\Vert K{\Vert }_1=1$, because $K$ need not be nonnegative.

Part (b), now for observations of $Y$, gives $\mathbb{E}|{\phi }_n^Y(u)-{\phi }^Y(u){|}^2\le 1/n$. Apply the Tonelli theorem to this nonnegative integrand, use the bound on $1/{\phi }_{ϵ}$, and change variables $v=hu$:The characteristic function ${\phi }_{ϵ}$ is continuous and nonzero on the compact interval of integration, so ${\tilde{M}}_h<\infty$ for every $h>0$. The Markov inequality gives, for every $z>0$,Consequently $J_{n,h}^{1/2}=O_{\mathbb{P}}({\tilde{M}}_h/\sqrt{nh})$, including for any deterministic bandwidth sequence $h=h_n>0$. The stochastic order follows directly from this uniform probability bound. Independence of $X$ and $ϵ$ gives ${\phi }^Y={\phi }^X{\phi }_{ϵ}$, explaining why division by ${\phi }_{ϵ}$ is the Fourier deconvolution operation.

Let $m=\mathbb{E}\Vert X_1{\Vert }_2$ and $q=\sqrt{n}K\ge 2$. The Lipschitz continuity of cosine and the Cauchy-Schwarz inequality implyThe centered expectation term contributes $nm$; omitting it would leave the Lipschitz bound incomplete. In particular $S_n$ has continuous sample paths on the compact cube, so its maximum exists and is measurable.

PutChoose a Cartesian finite net $G\subset [-K,K{]}^d$ whose coordinate spacings are at most $\eta$, including the endpoints. Every point of the cube is within Euclidean norm distance $\sqrt{d}\eta$ of $G$, andOn the event $L_n\le n{q}^a$, replacing any $u$ by a nearby grid point changes $S_n$ by at most $\sqrt{dn}$. Since $R>8\sqrt{d}$ and $\log (n{K}^2)\ge \log 4$, this is less than $\frac{R}{4}\sqrt{n\log (n{K}^2)}$. ThereforeThe Markov inequality bounds the first event by $2m{q}^{-a}$. At a fixed $v$, the summands are centered, independent and identically distributed, and have absolute value at most $2$. The supplied Hoeffding inequality, with $M=2$, bounds each grid event byA union bound now givesThe choice of $a$ balances the cost of a large random Lipschitz constant against the covering number of the cube: $d(1+a)-R^2/64=-a$. Consequently $C=2m+2\cdot 4^d$ works, withThis uses only the first moment $m$ and has a constant independent of $n,K,R$. The finite net is deterministic even though the Lipschitz constant is random.

An effective bandwidth measures how much capacity a random traffic source requires when unusually large demands matter. A mean-rate description loses burstiness; a peak-rate description can waste capacity when peaks are rare. The parameter $s>0$ in an effective bandwidth controls the emphasis placed on such bursts through the moment-generating function:Here $X$ is the traffic in a chosen time unit. If $X(t)$ instead measures traffic over length $t$, the corresponding rate is $\alpha (s,t)=\log \mathbb{E}[e^{sX(t)}]/(st)$. With independent stationary increments, this is independent of $t$ whenever the moment-generating function exists. More generally both the time scale and dependence structure matter.

The Chernoff bound follows directly from the Markov inequality, since $e^{sX}$ is nonnegative and $X\ge b$ implies $e^{sX}\ge e^{sb}$:Optimize over the admissible positive values of $s$ to obtainFor independent sources $X_1,\dots ,X_N$, moment-generating functions multiply and their effective bandwidths add. Thus a total capacity $c$ has the one-time overflow bound $\mathbb{P}({\sum }_i{X}_i\ge c)\le \exp \{s({\sum }_i{\alpha }_i(s)-c)\}$. In particular, for independent and identically distributed random variables,This is the positive-parameter part of the Legendre transform of a cumulant-generating function. The size of its exponent describes how a service margin reduces the chance of a large aggregate demand. A buffer-overflow event involves traffic over many intervals; a one-time Chernoff bound alone is not automatically a bound on that event. A time-dependent model and, for example, bounds on the union of relevant interval events are then needed.

Assume a finite moment-generating function in a neighborhood of zero. Differentiating the cumulant-generating function gives $K(0)=0$, $K^{\prime }(0)=\mathbb{E}[X]$ and $K^{\prime \prime }(0)=\mathrm{Var}(X)$. Therefore the small-parameter limit is the mean traffic rate:The variance term shows how an effective bandwidth charges extra capacity for burstiness. Also $K$ is a convex function with $K(0)=0$, so $K(s)/s$ is nondecreasing for $s>0$ in its domain.

For the other limit, let $M=\mathrm{esssup}X$, the essential supremum. If $M<\infty$, $X\le M$ almost surely gives $\alpha (s)\le M$. For every $a<M$, $p_a=\mathbb{P}(X\ge a)>0$, andTaking a lower limit and then $a\uparrow M$ proves the large-parameter limit is peak traffic:If $X$ has no finite upper bound and all positive exponential moments exist, the same lower bound for arbitrarily large $a$ makes the limit infinite. If positive exponential moments cease to exist at a finite parameter, the effective bandwidth is infinite beyond that domain, and the infinite limit is understood in this extended sense. For example, a normal distribution with mean $m$ and variance $v$ has $\alpha (s)=m+sv/2$: its unbounded upper tail has no finite peak. Thus effective bandwidth interpolates between mean and peak demand while permitting independent sources to be aggregated by addition.

Put $M_k=S_k-mk$ as above. Centering the linear interpolation also interpolates the centered values: for $u\in [0,1]$,The absolute value of this convex combination is at most the larger endpoint absolute value. For every fixed $A>0$, part (c) therefore givesHence there is uniform convergence on compacts in probability to the deterministic path $h(t)=mt$, by Markov inequality. Equip $C([0,\infty ),\mathbb{R})$ with its standard compact-open topology, metrized byFor each finite number of terms their suprema converge to zero in probability, and the remaining tail is bounded deterministically by ${\sum }_{j>J}{2}^{-j}$. Thus $d(S^{(N)},h)\to 0$ in probability. Convergence in probability to a deterministic point implies weak convergence of probability measures, so the fluid limit isHere ${\delta }_h$ is the Dirac measure concentrated on that continuous path. The topology is locally uniform convergence; no assertion of uniform convergence over the entire unbounded half-line is needed. The interpolation is intended for integers $k\ge 0$, including the initial interval $[0,1]$. If the PDF's ${\mathbb{Z}}^{+}$ is interpreted as strictly positive integers, that initial piece is omitted from the displayed definition and must be supplied by the same formula using $S_0=0$.

For a kernel for density estimation $K$ with integral one and a smoothing bandwidth $h>0$, the univariate kernel density estimator isIts expected value isThis is the convolution identity. It holds wherever the integral is defined, in particular everywhere for the bounded kernel used below. For a nonnegative kernel for density estimation, Tonelli theorem also gives the identity as an extended nonnegative integral. A signed kernel for density estimation instead requires absolute integrability for this interchange.

For the centred unit-width uniform kernel, substitute $u=(x-y)/h$:Put $M={\sup }_z|f^{\prime \prime }(z)|$. The Taylor theorem with Lagrange remainder givesThe odd function term integrates to zero, while ${\int }_{-1/2}^{1/2}{u}^{2}du=1/12$. Hence the bias of a kernel density estimator obeysThis is a second-order pointwise bias bound for kernel density estimation. Existence and boundedness of the second derivative suffice for the stated remainder; continuity of that derivative is not an extra assumption.

For concentration, writeThe $B_i$ are independent and identically distributed random variables with a Bernoulli distribution, and ${\hat{f}}_h(x)=(nh{)}^{-1}{\sum }_i{B}_i$. The required concentration bound follows from the Hoeffding lemma. We prove its Bernoulli case directly. Withwe have $\psi (0)={\psi }^{\prime }(0)=0$ and ${\psi }^{\prime \prime }(s)=p_s(1-p_s)\le 1/4$, where $p_s=p{e}^s/(1-p+p{e}^s)$ is the exponentially tilted success probability. Therefore $\psi (s)\le s^2/8$ for every real $s$. If $p=0$ or $p=1$, the centred variable is identically zero and the same bound holds directly.

Independence and the exponential form of the Markov inequality now give, for $s>0$,Choose $s=4ht$. Applying the same argument to the lower tail and adding the probabilities proves the Hoeffding inequality

To obtain the requested variance constant, use the additional bound that a probability is at most one. Set $Z={\hat{f}}_h(x)-\mathbb{E}{\hat{f}}_h(x)$ and $a=2n{h}^2$. The tail integral formula for moments, justified by Tonelli theorem, yieldsThusThe clipping at one is essential for this integration argument to give that constant. It is an instance of a second moment bound from a clipped Gaussian tail. The exact Bernoulli distribution calculation also gives $\mathrm{Var}({\hat{f}}_h(x))=p(1-p)/(n{h}^2)\le 1/(4n{h}^2)$, an even stronger bound.

Let $R$ interchange the two complete graphs, sending $v_i$ to $v_i^{\prime }$ and fixing $w$. This involution preserves the transition matrix of the simple random walk. A mirror coupling under an involution starts $X$ at $v_i$, keeps $Y_t=R(X_t)$ until the hitting time $\tau$ of $w$, and then makes the two walks move identically. Each coordinate has the correct transition probabilities, and they coalesce precisely at $\tau$. The coupling inequality for total variation gives

There is a genuine error in the printed expected hitting time bound, as well as an unspecified initial law. For $n\ge 2$, write $a={\mathbb{E}}_{v_1}\tau$ and $b={\mathbb{E}}_{v_i}\tau$ for $i\ne 1$. From every clique vertex there is a positive chance of reaching $w$ within two steps, uniformly over the finite state space for each fixed $n$, so $\tau$ has finite expected value. First-step analysis, using the degree of a vertex $n$ at $v_1$ and $n-1$ at the other clique vertices, givesSolving these linear equations yieldsIn particular even the attachment vertex exceeds the printed bound by one. The valid uniform replacement is ${\max }_x{\mathbb{E}}_x\tau \le n^2$, the hitting time of the bridge vertex between two cliques formula.

We must also handle arbitrary starting states, including $w$, before applying the mirror argument. Identify related vertices and consider the lumped Markov chain $Z$ on $\{1,\dots ,n,w\}$. Its transition matrix $Q$ has $Q(1,j)=1/n$ for $j\ne 1$, $Q(1,w)=1/n$, $Q(i,j)=1/(n-1)$ for $i\ne 1$ and $j\ne i$, and $Q(w,1)=1$, with all other entries zero. Here the row for $i\ne 1$ only ranges over $j\in \{1,\dots ,n\}$.

For $n\ge 3$ and $j\in \{2,\dots ,n\}$, direct two-step calculation givesAll these entries are at least $1/(2n)$. Thus $Q^2(z,\cdot )\ge \alpha \nu (\cdot )$ for every $z$, where $\nu$ is the uniform distribution on a finite set $\{2,\dots ,n\}$ and $\alpha =(n-1)/(2n)\ge 1/3$. This is a Doeblin condition. At each two-step block couple the two quotient endpoints to the same sample from $\nu$ with probability $\alpha$, using the residual probability distributions otherwise. The first successful block has expected value at most $1/\alpha$, so the alignment time $A$ has $\mathbb{E}A\le 6$. Lift each endpoint coupling of probability distributions to the original walks by their conditional two-step path laws; this preserves every marginal path law. The success test uses fresh block randomness and so permits restarting the walks with their original transition matrix at the aligned endpoints.

At alignment the original walks are equal or related. Use identical transitions in the first case and the mirror coupling under an involution in the second. The resulting coalescing coupling has coalescence time $T$ satisfying $\mathbb{E}T\le 6+n^2$ for every pair of initial states. Therefore Markov inequality and the pairwise mixing diameter implyFor the hint's particular starting states, the quotient one-step laws from two different clique indices have a common component of mass at least $1-2/n$: for two ordinary indices it is $(n-2)/(n-1)$; for one attachment index it is $(n-2)/n$. A maximal coupling therefore aligns those indices with probability $1-O(1/n)$, as suggested. The two-step argument above additionally covers $w$.

The nonlazy walk is an aperiodic Markov chain for $n\ge 3$, because the connected graph contains a triangle. For $n=1,2$ it is a periodic Markov chain on a bipartite graph; each bipartition class has stationary mass $1/2$, and the walk stays on a single class at each time. Hence $d(t)\ge 1/2$ and the $1/4$ mixing time is infinite. The asymptotic conclusion is valid for $n\ge 3$.

Put $L={\sup }_x\Vert \nabla f(x){\Vert }_{\infty }$, the uniform bound on the supremum norm of the gradient. The fundamental theorem of calculus on a line segment, together with the inner product inequality $|⟨a,b⟩|\le \Vert a{\Vert }_{\infty }\Vert b{\Vert }_1$, givesThus $f$ is Lipschitz continuous with respect to the L1 norm. The Gaussian random variables comprising $X$ and $Y$ have finite exponential moments of their absolute values. More explicitly, for $c\ge 0$, Hölder's inequality gives $\mathbb{E}{e}^{c\Vert X{\Vert }_1}\le {\prod }_{j=1}^n(\mathbb{E}{e}^{nc|X_j|}{)}^{1/n}<\infty$. Consequently $\mathbb{E}f(X)$ and every exponential expression below are finite, including when the covariance matrix is singular.

Use the convex function $\Psi (v)=e^{\lambda |v|}$. Since $Y$ is an independent copy of $X$, Jensen inequality for the conditional expectation givesFor $0\le \theta \le \pi /2$, define the Gaussian rotation of independent copiesWriting $\Sigma =\mathrm{Cov}(X)$, both rotated covariance matrices are $\Sigma$ and their cross-covariance is zero. The pair has a centered multivariate normal distribution, so its components are independent and $(U_{\theta },V_{\theta })$ has the same probability law as $(X,Y)$. This calculation uses no inverse of $\Sigma$ and therefore also proves the assertion for degenerate multivariate normal distributions.

The path $\theta \mapsto U_{\theta }$ runs from $Y$ to $X$. The chain rule and the fundamental theorem of calculus give the pathwise identityApply Jensen inequality to the uniform probability measure on $[0,\pi /2]$. Then use Tonelli theorem and the Gaussian rotation of independent copies to obtainThis proves the Gaussian rotation interpolation inequality in the required form:The stated almost-sure domination now gives an exponential moment of an absolute standard normal variable bound. If $a=\lambda \pi /2$, completing the square in the standard normal density yieldswhere $\Phi$ is the standard normal distribution function. Hence Markov inequality gives, for every $\lambda >0$,The exponent is minimized at $\lambda =4u/{\pi }^2>0$. Therefore the resulting Gaussian tail bound isNo independence of $Z$ from $X$ or $Y$ is needed: only the given almost-sure domination is used.

In a neutral Moran model, a population has $N$ labelled individuals; reproduction copies one individual's type to a uniformly chosen other individual, keeping population size fixed. One convenient time convention puts independent rate-$1/2$ Poisson arrows on each ordered pair of distinct individuals. Backwards in time, two sampled ancestral lineages merge whenever one copies the other, at total rate $1$ per unordered pair. In the convention where each individual reproduces at rate $1$, this pair rate is $2/(N-1)$ if the parent cannot replace itself, and time must be rescaled accordingly.

Kingman's coalescent on $n$ labels is the continuous-time Markov chain on partitions of $\{1,\dots ,n\}$, initially all singleton blocks, in which each pair of distinct blocks merges at rate $1$. With $j$ blocks, the waiting time is exponential of rate $\left(\genfrac{}{}{0ex}{}{j}{2}\right)$, and the merging pair is uniform. Kingman's infinite coalescent is the consistent partition process on $\mathbb{N}$ whose restriction to each finite label set has this law; consistency and the extension theorem construct it. These time conventions are part of the definition.

Let $B_t^{(n)}$ be the number of blocks in the first $n$ labels. The expected time to reach at most $m$ blocks isMarkov inequality gives $\mathbb{P}(B_t^{(n)}>m)\le 2/(mt)$. As $n$ increases these counts increase to the total number of infinite-coalescent blocks, so the same bound holds for that number. Letting $m\to \infty$ shows it is finite almost surely for each $t>0$. Apply this simultaneously to positive rational times; monotonicity of the block count then covers every positive real time. Hence Kingman's infinite coalescent comes down from infinity almost surely.

The Cauchy-Schwarz inequality is $|\mathbb{E}(XY){|}^2\le \mathbb{E}(X^2)\mathbb{E}(Y^2)$. Markov inequality says that for $X\ge 0$ and $a>0$, $\mathbb{P}(X\ge a)\le \mathbb{E}X/a$.

Jensen inequality states that for convex $\varphi$,Let $\mu =\mathbb{E}X$. A supporting line at $\mu$ gives $\varphi (x)\ge \varphi (\mu )+c(x-\mu )$. Taking expectations makes the linear term vanish.

Use natural logarithms. A $K_t$ graph minor is represented by $t$ disjoint nonempty branch sets of a graph minor, each inducing a connected graph, with an edge between every pair. The complete graph minor density threshold is defined over nonempty finite graphs byIt uses $e(G)/|G|$, half the average degree of a vertex; the nonempty convention excludes the vacuous density inequality for the empty graph. The PDF has the intended definition with a universal implication over $G$; the TeX transcription corrupts that definition.

Probabilistic lower bound. Put $N=\lfloor t\sqrt{\log t}/4\rfloor$ and take the binomial random graph $G(N,1/2)$. Its edge count has mean $N(N-1)/4$. The Chernoff bound impliesWe show that the probability of a $K_t$ graph minor tends to zero. Ignore connectedness, which only enlarges the set of possible models, and assign each of the $N$ vertices one of $t$ branch labels or an unused label. There are at most $(t+1{)}^N$ assignments.

For any assignment with $t$ nonempty branch sets, at least $t/2$ of them have size at most $2N/t\le \sqrt{\log t}/2$. Between any two such small sets $A,B$, the probability of no edge isThe absence or presence of edges between different pairs of branch sets depends on disjoint sets of random edges, so these events are independent. For this assignment the probability that all small pairs are adjacent is at mostThe union bound over assignments is therefore at mostsince the positive term is $O(t(\log t{)}^{3/2})$ and the negative term has order $t^{7/4}$. There is consequently a graph with no $K_t$ graph minor and with $e(G)/N\ge N/8$. Any constant having the universal forcing property must exceed this ratio. Since $N/8\ge t\sqrt{\log t}/64$ for large $t$,

The auxiliary density lemma contains a typo. Both the original PDF and the TeX print $e(G)\le 11k|G|$ in the hint. This cannot imply its conclusion: for $k\ge 1$, an edgeless graph has only edgeless graph minors, and none satisfies $2\delta (H)\ge |H|+4k-1$. We use the intended dense minor with bounded order and high minimum degree lemma with the hypothesis $e(G)\ge 11k|G|$ for nonempty $G$. The upper-bound argument below depends on that corrected supplied lemma; the literal printed hint is false.

Probabilistic upper bound and the constant seven. Suppose $G$ is nonempty and $e(G)\ge 7t\sqrt{\log t}|G|$, and setThe corrected lemma gives a graph minor $H$ of order $N\le 11k+2$ with $2\delta (H)\ge N+4k-1$. Since $\delta (H)\le N-1$, we also have $N\ge 4k+1$. For large $k$, every vertex has at most $N/3$ nonneighbours, counting itself, becausethe last inequality uses $N\le 11k+2\le 12k-3$. Moreover any two distinct vertices have at leastcommon neighbours.

Choose $2t$ disjoint random $\ell$-sets $A_1,\dots ,A_{2t}$ from $V(H)$, uniformly, which is possible since $2t\ell <4k-1\le N$ for sufficiently large $t$. For a set $A$, let $U(A)$ be the set of vertices having no neighbour in $A$. For each vertex, the probability that all members of a uniformly sampled $\ell$-set are its nonneighbours is at most $3^{-\ell }$, even when sampling without replacement. By linearity of expected value,Call $A$ good if $|U(A)|\le N{3}^{-19\ell /20}$. The Markov inequality shows that a random set is bad with probability at most $3^{-\ell /20}=o(1)$. Thus the expected number of bad sets among our $2t$ samples is $o(t)$.

Conditional on a fixed good $A_i$, the marginal distribution of $A_j$ is uniform among $\ell$-sets outside $A_i$. For there to be no edge between them, all its members must lie in $U(A_i)$. HenceHere $(N/(N-\ell ){)}^{\ell }=\exp (O({\ell }^2/N))=1+o(1)$. Consequently the expected number of nonadjacent pairs of good sets is at mostsince $(19/20)\log 3>1$. Applying the Markov inequality to both counts shows that there exists a choice with fewer than $t$ bad sets and at most $t$ nonadjacent pairs of good sets. Select $t$ of the good sets; they still have at most $t$ missing adjacencies.

Make each selected set connected by fixing one of its vertices as a root and, for each of its other members, adding one unused common neighbour of that member and the root. This uses at most $t(\ell -1)$ additional vertices. Then for each pair with no original edge, choose one vertex in each set and add an unused common neighbour to one of the sets. The added vertex is joined to that set and to the other set, so it preserves connectedness and repairs their adjacency. At most $t$ such repairs are required.

Throughout this process the total number of occupied vertices is at mostSince every pair has at least $4k-1$ common neighbours, an unused choice always exists. The final sets are disjoint connected branch sets of a graph minor with every pair adjacent. They represent $K_t$ in $H$, and hence in $G$, by transitivity of graph minors. This random branch-set construction of a complete graph minor proves

The sharp linear threshold for a four-vertex complete minor. We prove by induction that a graph on $n\ge 4$ vertices with at least $2n-2$ edges has a $K_4$ graph minor. The case $n=4$ is $K_4$ itself. If some edge $uv$ has at most one common neighbour, an edge contraction reduces the order by one and removes exactly $1+|N(u)\cap N(v)|\le 2$ edges. The contracted graph has at least $2(n-1)-2$ edges, so induction applies.

Otherwise every edge is in at least two triangles in a graph. Choose a vertex $v$ incident with an edge. Each vertex $u\in N(v)$ has at least two neighbours of a vertex in $N(v)$, since these are the common neighbours of $u,v$. Thus $G[N(v)]$ has minimum degree of a graph at least two. A longest path in this finite graph has an endpoint adjacent to an earlier nonconsecutive path vertex, and hence contains a cycle in a graph. Together with $v$, this cycle in a graph gives a wheel graph as a subgraph. Divide its rim into three consecutive nonempty connected arcs; these arcs and $\{v\}$ form four pairwise adjacent branch sets of a graph minor. Therefore

With $2n-3$ edges the answer is no. Take the join of graphs of $K_2$ and an independent set of $n-2$ vertices. It has $1+2(n-2)=2n-3$ edges. Among four disjoint putative branch sets of a graph minor, at most two can contain the two vertices of $K_2$. Any connected branch set avoiding them is a singleton in the independent set. At least two such singleton sets would therefore have no edge between them. This excludes a $K_4$ graph minor for every $n\ge 4$ and proves sharpness.