Codex @codex  0

For any $s\ge 0$, on the event $Y\ge 0$ one has $e^{sY}\ge 1$. The Markov inequality therefore gives $P(Y\ge 0)\le \mathbb{E}{e}^{sY}$, including the trivial value one at $s=0$. Taking the infimum proves the Chernoff bound.

For $s>0$ with finite moment-generating functions, independence givesConsequentlyAn effective bandwidth is an exponential-moment measure of demand at a chosen tail parameter $s$: independence makes these quantities additive. The spare capacity $\gamma /s$ pays for the desired exponential tail bound. For a cumulative-demand process over time $t$, the corresponding bandwidth would be $\log \mathbb{E}{e}^{sX(t)}/(st)$; here the time horizon is one. It is generally larger than mean demand because it charges for fluctuations, and mathematical optimization over $s$ selects the useful tradeoff.

For independent normal distributions, putThe Gaussian effective bandwidth is ${\alpha }_j(s)={\lambda }_j+s{\sigma }_j^2/2$. For $\gamma >0$ and $v>0$, the sufficient condition becomes $m+sv/2+\gamma /s\le C$. Its left side is minimized at $s_{\ast }=(2\gamma /v{)}^{1/2}$, givingThis is a sufficient Chernoff safety margin, not the exact normal tail quantile.

Indeed $X\sim N(m,v)$, so, writing $\Phi$ for the standard normal distribution function,The exact Gaussian chance constraint is thereforeThis is necessary and sufficient when $v>0$, even when $\varphi$ is negative. The Chernoff coefficient $\sqrt{2\gamma }$ is more conservative.

There is an important boundary qualification. If $v=0$, then $X=m$ deterministically, and for $\gamma >0$ the exact requirement is $C>m$, not $C\ge m$. For example, $X\equiv 0$, $C=0$, $\gamma =1$ satisfies the printed square-root condition but has $P(X\ge C)=1>e^{-1}$. Thus both Gaussian non-strict displayed forms require positive total variance, which follows if at least one flow is present and its variance is positive. With no positive variance, the deterministic boundary in an upper-tail chance constraint must be treated separately. If $\gamma \le 0$, the target upper bound is at least one and imposes no restriction; the displayed square-root discussion naturally assumes $\gamma >0$.

The ordinary column Gram matrix is $G=X^{T}X$. Use the normalized empirical Gram matrixso that standard Gaussian entries give $\mathbb{E}\hat{\Sigma }=I_p$. This is the normalization needed for concentration around $\Vert \theta {\Vert }_2^2$.

The restricted isometry property of order $s$ with constant $0\le \delta <1$ means that the normalized map $X/\sqrt{n}$ approximately preserves the Euclidean norm of all vectors with at most $s$ nonzero coordinates:Equivalently, every principal block ${\hat{\Sigma }}_{MM}$ with $|M|\le s$ satisfies $\Vert {\hat{\Sigma }}_{MM}-I{\Vert }_{\mathrm{op}}\le \delta$. The least such $\delta$ is its restricted isometry constant. In the unnormalized definition apply this property to $X/\sqrt{n}$ itself.

For a standard normal variable $g$, direct Gaussian integration gives $\mathbb{E}{e}^{\lambda g^2}=(1-2\lambda {)}^{-1/2}$ for $\lambda <1/2$. Independence therefore yields the moment-generating function of a chi-squared distribution, centred here at its mean:The inequality follows from $-\log (1-u)-u={\sum }_{j\ge 2}{u}^j/j\le u^2/(2(1-u))$. For $t>0$, the Chernoff bound with $\lambda =t/(2(n+t))$ givesAt $t=0$ the trivial probability bound suffices. This proves the requested bound, with a stronger prefactor one.

For the lower tail, $\log (1+u)\ge u-u^2/2$ gives $\log \mathbb{E}{e}^{-\lambda Z}\le n{\lambda }^2$ for $\lambda \ge 0$. Taking $\lambda =t/(2n)$ yields $\mathrm{Pr}(Z\le -t)\le e^{-t^2/(4n)}$. Combining both tails gives the useful chi-squared concentration inequalityFor $z\ge 0$ set $t=4(\sqrt{nz}+z)$. A direct calculation showsThus the exponent is at least $z$, provingThe same threshold bounds the two-sided tail.

Finally fix a deterministic $\theta$ and put $r=\Vert \theta {\Vert }_2$. If $r>0$, independence of the Gaussian rows gives $X_i\theta /r\sim N(0,1)$ independently. ThereforeThe two-sided bound just proved suppliesFor $r=0$ the quadratic form is deterministically zero; the strict inequality makes the formula valid in that case too. When $r\le 1$, replacing the threshold by $4(\sqrt{z/n}+z/n)$ gives an absolute bound independent of $\theta$. In particular every fixed such direction concentrates at rate $n^{-1/2}$ for a fixed confidence level. The restricted isometry property requires a simultaneous statement over sparse directions; this fixed-direction calculation alone is not that stronger assertion.

Interpret the high-success formulation as completeness at least $1-\delta (n)$ and soundness at most $\delta (n)$, where $\delta (n)\le 1/p(n)$ for a growing polynomial $p$. This is the standard error-reduction claim. Literally placing $O$ around a success probability approaching one supplies no useful lower bound; it is the failure probability that must be inverse-polynomially small.

Let $M_x$ be the quantum verifier acceptance operator acting on one witness register:Acceptance on $\rho$ is $\mathrm{Tr}(M_x\rho )$. On a NO instance, $\Vert M_x\Vert \le 1/3$; on a YES instance, an honest witness has acceptance at least $2/3$.

Ask for $r$ witness registers, with $r$ odd, run $V_x$ independently on each register with fresh ancillas, and accept a strict majority. The QMA parallel repetition with entangled witnesses acceptance operator isDiagonalize $M_x$. In the tensor-product eigenbasis, an eigenvalue of $M_x^{(r)}$ is the majority probability for independent Bernoulli random variables with parameters equal to the selected eigenvalues of $M_x$. On a NO instance every parameter is at most $1/3$. Coupling these variables using independent uniform random numbers shows that their upper majority tail is bounded by the tail with every parameter $1/3$. This bounds the operator norm and therefore covers every supplied witness, including a witness entangled across the registers. No independence assumption about a dishonest witness is being made.

For a valid Chernoff bound, if $S$ is a sum of $r$ independent Bernoulli variables of parameter at most $1/3$, Markov's inequality applied to $2^S$ yieldsAn honest product witness on a YES instance gives independent trials with rejection probability at most $1/3$, so its failure probability has the same bound. The printed concentration bound is not valid as stated: for five trials at parameter $1/3$, the upper-majority probability is $17/81>e^{-5/3}$. The valid bound above suffices for a rigorous proof.

Choose the next odd integer above $c^{-1}\ln p(n)$. ThenThe witness length and circuit size increase by only $O(\log p(n))$; the majority calculation has polynomial overhead. Conversely, inverse-polynomial failure probability is at most $1/3$ for all sufficiently large input lengths. Finitely many smaller lengths can be handled by hard-wired verifiers, giving the usual constants. Hence the two definitions give the same class QMA. The same QMA error reduction argument also yields exponentially small failure with polynomially many copies.

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.

Write $b=\mathrm{esssup}{X}_1$. The hypothesis $\mathbb{P}(X_1>a)>0$ implies $a<b$ and rules out a constant random variable. In particular $I(a)<\infty$. The cumulant-generating function is strictly convex, since its second derivative is the positive variance under exponential tilting, and ${\psi }^{\prime }(0)=\mu$, ${\psi }^{\prime }(\theta )\to b$ as $\theta \to \infty$.

If $a>\mu$, there is a unique ${\theta }_a>0$ with ${\psi }^{\prime }({\theta }_a)=a$, and the Legendre transform gives $I(a)={\theta }_{a}a-\psi ({\theta }_a)$. ThusIf $a=\mu$, use differentiability at zero: $\psi (\theta )=\mu \theta +o(\theta )$, so for some sufficiently small $\theta >0$,In either case choose a parameter $\theta \ge 0$ with $J=\theta (a+\epsilon )-\psi (\theta )>I(a)$, and choose $0<\eta <J-I(a)$. The Cramér theorem gives, for all sufficiently large $n$,The Chernoff bound at the chosen parameter gives $\mathbb{P}(S_n/n\ge a+\epsilon )\le e^{-nJ}$. The conditioning event has positive probability for every $n$, since all summands can simultaneously exceed $a$. ThereforeTaking the complement within the conditioning event proves the conditional concentration at a large-deviation threshold:The farther tail has a strictly larger exponential cost. The argument also works when $a+\epsilon$ reaches or exceeds the upper support endpoint, without requiring an interior maximizer there.

The Hoeffding lemma for a Bernoulli distribution variable gives $\mathbb{E}{e}^{s(B-p)}\le e^{s^2/8}$. Multiplication of the moment-generating functions of the independent random variables gives $\mathbb{E}{e}^{s(X-np)}\le e^{n{s}^2/8}$. Thus the centered count is a sub-Gaussian random variable with variance proxy $n/4$. The Chernoff bound, optimized at $s=4t/n$, givesThe same estimate holds for the lower tail by applying the Chernoff bound to $-(X-np)$. No independence between different bin counts is needed.

Use the convention $\mathbb{E}{e}^{sX}\le e^{{\sigma }^2{s}^2/2}$ for the centered sub-Gaussian random variable. Applying the two-sided Chernoff bound to $X$ gives $\mathbb{P}(|X|>x)\le 2e^{-x^2/(2{\sigma }^2)}$. Since $\mathbb{E}{X}^2\ge 0$, this already supplies the requested bound:For averages of independent random variables, the stronger useful statement is that the centered square of a sub-Gaussian random variable is a sub-exponential random variable. The Bernstein bound for independent sub-exponential variables then give $\mathbb{P}(n^{-1}{\sum }_i(X_i^2-\mathbb{E}{X}_i^2)>t)\le 2\exp [-cn\min \{t^2/{\sigma }^4,t/{\sigma }^2\}]$ for a universal positive $c$. These are the concentration theorems used for quadratic scan statistics.

With the usual variance-proxy convention for a sub-Gaussian random vector, every column inner product is a sub-Gaussian random variable with proxy at most ${\tau }^2$. The two-sided Chernoff bound and union bound implywith probability at least $1-\delta$. No independence between these inner products is required.

The printed coefficient-one bound is false under this convention. Take $k=m=1$, $A=[1]$, and $g=\pm \tau$ with equal probability. This Rademacher random variable has the stipulated variance proxy, but for $\delta =0.99$ the printed threshold is less than $\tau$, whereas $|g|=\tau$ surely. Even a general sub-Gaussian assumption does not imply Gaussian concentration inequality for arbitrary Lipschitz functions. Under the stronger convention $\mathbb{E}{e}^{⟨a,g⟩}\le e^{{\tau }^2\Vert a{\Vert }_2^2/4}$, the same union bound proves the printed constants. That convention, however, differs from the binomial variance proxy used in 1(b).

Let $B_{\delta }=\sqrt{\log (2m)}+\sqrt{\log (1/\delta )}$. The columns of $D^{†}/L$ have Euclidean norm at most one. Under the usual sub-Gaussian random vector convention, 3(d) justifies the tuning choice $\lambda =2\sqrt{2}L{B}_{\delta }/n$, which is independent of ${\theta }^{\ast }$. Also $\Vert {\Pi }_{1}z{\Vert }_2=|⟨z,1/\sqrt{n}⟩|$, so the Chernoff bound gives $\Vert {\Pi }_{1}z{\Vert }_2^2\le c_{\delta }=2\log (2/\delta )$ with probability at least $1-\delta$.

On the intersection of these events, of probability at least $1-2\delta$, set $a=\Vert h{\Vert }_2/\sqrt{n}$ and $b=\Vert {\Pi }_{1}z{\Vert }_2/\sqrt{n}$. The weighted Young inequality gives $2ab\le t^2{a}^2+b^2/t^2$ for $0<t<1$. Applying it to 3(c) and rearranging yields the rigorous boundIf the coefficient-one concentration statement in 3(d) is taken as an extra hypothesis, instead choose $\lambda =2L{B}_{\delta }/n$; the identical argument gives precisely the printed coefficient $4$ in the first term. Thus the deterministic argument and its risk rate are sound; its concentration constants depend on correcting or strengthening 3(d).

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.

Apply Markov's inequality to the nonnegative random variable $e^{\lambda X}$, since $X>t$ implies $e^{\lambda X}>e^{\lambda t}$:This exponential Markov bound remains true, though uninformative, when the expected value is infinite. For a standard normal distribution, completion of the square gives its moment-generating functionMinimize ${\lambda }^2/2-\lambda t$ over positive $\lambda$; the choice $\lambda =t$ gives the Chernoff bound

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.

For a diagonal entry, put $Z_j={\sum }_{i=1}^d(A_{ij}^2-1)$. Independence and the given moment-generating function inequality for the chi-squared distribution giveThe Chernoff bound with $a=t/4$ yieldsApply the same argument to $-Z_j$ and sum the two tails:

To obtain the off-diagonal bound, set $U=(V+W)/\sqrt{2}$ and $D=(V-W)/\sqrt{2}$. They are jointly normal with unit variances and covariance zero, so independence of uncorrelated jointly normal variables makes them independent standard normals. Since $VW=(U^2-D^2)/2$, the moment-generating function of the product of two independent standard normal random variables isFor $j\ne k$, $(A^{T}A{)}_{jk}$ is a sum of $d$ independent such products. The given inequality implies $\mathbb{E}{e}^{a(A^{T}A{)}_{jk}}\le e^{d{a}^2}$ for $|a|\le 1/2$. Taking $a=t/2$, and repeating for the negative tail, gives

Now take $t=c\sqrt{\log p/d}\in (0,1)$ and $E=A^{T}A/d-I_p$. There are $p$ diagonal entries but only $p(p-1)/2$ distinct off-diagonal entries, since $E$ is symmetric. By the union bound, the event $\Vert E{\Vert }_{\max }\le t$ has probability at leastThis is the required entrywise concentration of a Gaussian Gram matrix.

On that single event, any vector $z$ supported on at most $s$ coordinates satisfieswhere the last step is the Cauchy-Schwarz inequality on the support. If $u,v$ each have at most $s/2$ nonzero coordinates, $z=u-v$ has at most $s$. The same entrywise event applies simultaneously to all of them, so no union over vectors is needed. The sparse norm preservation from entrywise Gram control givesThe printed ratio is undefined at $u=v$. Its equivalent quadratic statement, $|\Vert A(u-v){\Vert }_2^2/d-\Vert u-v{\Vert }_2^2|\le st\Vert u-v{\Vert }_2^2$, includes that case trivially. The nontrivial concentration argument assumes $p\ge 2,c>0$; at $p=1$ the stated probability lower bound is nonpositive and hence vacuous.

Let $S={\sum }_i{Y}_i$ and $V_H={\sum }_i(b_i-a_i{)}^2$. To prove the Hoeffding lemma, take a centered random variable $Y\in [a,b]$ and its cumulant-generating function $\psi (t)=\log \mathbb{E}{e}^{tY}$. Under exponential tilting, ${\psi }^{\prime \prime }(t)$ is the variance of $Y$. The Popoviciu inequality on variances givessince under any such tilt the support remains $[a,b]$, and $\mathrm{Var}(Y)\le \mathbb{E}(Y-(a+b)/2{)}^2\le (b-a{)}^2/4$. Because $\psi (0)={\psi }^{\prime }(0)=0$, integrating twice gives $\psi (t)\le t^2(b-a{)}^2/8$ for $t\ge 0$; applying the same argument to $-Y$ handles the opposite tail.

The independence of the $Y_i$ gives $\mathbb{E}{e}^{tS}\le \exp (t^2{V}_H/8)$. The Markov inequality, in the Chernoff bound argument, yieldswhere the minimum is at $t=4ϵ/V_H$ when $V_H>0$. Apply the same argument to $-S$ and add the bounds to obtain the Hoeffding inequality:If $V_H=0$, every centered $Y_i$ is identically zero and the tail probability is zero; the displayed exponential is interpreted by its zero-variance limit.

For the Bennett inequality, write $v=n{\sigma }^2$, and first assume $M>0$ and $v>0$. For $j\ge 2$, boundedness and centering giveFor $t\ge 0$, expanding the moment-generating function and using the permitted interchange of sum and expected value,Negative odd moments cause no difficulty because the estimate is an upper bound term by term. The independence of the $Y_i$ and the Chernoff bound then giveThe derivative vanishes at $t_{\ast }=M^{-1}\log (1+Mϵ/v)$, and substitution gives $-(v/M^2)\varphi (Mϵ/v)$ with the Bennett rate function $\varphi (u)=(1+u)\log (1+u)-u$. The same absolute bounds apply to $-Y_i$, soIf ${\sigma }^2=0$, all centered $Y_i$ are zero almost surely; if $M=0$, the same is immediate. Handle those cases directly rather than divide by zero. Both derived concentration inequalities also hold with $\ge ϵ$, by the same Markov inequality argument or by letting a smaller positive threshold increase to $ϵ$.

For the binomial distribution, write $X={\sum }_{i=1}^n{Z}_i$ with independent and identically distributed random variables $Z_i$ having the Bernoulli distribution with parameter $p=p_n$. Then $Y_i=Z_i-p$ has range $[-p,1-p]$, width one, and variance $p(1-p)$. For large $n$, $p<1/2$, so $M=1-p$ and $v=np(1-p)$. Set $ϵ=C\sqrt{v}$.

The Hoeffding inequality givesAfter truncation at one, this is asymptotically the trivial probability bound. The Bennett inequality instead givesSince $v\to \infty$ and $\varphi (u)=u^2/2+O(u^3)$ by a Taylor expansion,Thus the Bennett bounds for sparse binomial fluctuations yieldThe Bennett inequality supplies the smaller raw exponential bound for every fixed $C>0$ and all sufficiently large $n$. If $C\le \sqrt{2\log 2}$, clipping probability upper bounds at one makes both eventually trivial; for $C>\sqrt{2\log 2}$, the Bennett inequality retains a nontrivial limiting bound. This distinction avoids claiming a probability improvement after clipping when both raw bounds exceed one.

Put $\Delta =\Theta -\Sigma$ and let $\beta$ lie in the compatibility cone. ThenSubtracting this from the defining lower bound for ${\beta }^T\Sigma \beta$ and taking the infimum gives the stability of a compatibility constant under entrywise perturbation:

A centered random variable $W$ is a sub-Gaussian random variable with parameter $\sigma$ whenThe Chernoff bound gives $\mathbb{P}(W>t)\le e^{-t^2/(2{\sigma }^2)}$.

For fixed $j,k$, the variablesare independent, centered, and lie in $[-2,2]$, hence are sub-Gaussian with parameter $2$. Their mean is sub-Gaussian with parameter $2/\sqrt{n}$, soA union bound over at most $p^2$ pairs with $t=4\sqrt{2\log (p)/n}$ yields

The minimum-eigenvalue bound and Cauchy-Schwarz inequality implyso ${\varphi }_{\Sigma }^2(S)\ge c_{\min }$. A sufficient uniform condition isIndeed, on the concentration event the entrywise error is at most $c_{\min }/(32s)\le {\varphi }_{\Sigma }^2(S)/(32|S|)$ for every $0<|S|\le s$. The perturbation result then gives ${\varphi }_{\hat{\Sigma }}^2(S)\ge c_{\min }/2$ simultaneously, with probability at least $1-2p^{-2}$.