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The Bennett rate function has Taylor expansion $\varphi (u)=u^2/2-u^3/6+O(u^4)$ near zero. It is the optimized exponent in the Chernoff bound based on the moment-generating function estimate for the Bennett inequality.

Suppose independent centered sub-exponential random variables satisfy $\log \mathbb{E}{e}^{\lambda X_i}\le v_i{\lambda }^2/(2(1-b|\lambda |))$ for $|\lambda |<1/b$, and put $v={\sum }_i{v}_i$. Independence adds the log moment-generating functions. The Chernoff bound with $\lambda =t/(v+bt)$ gives the displayed two-sided tail. If each variable has common scale $K$, one may take $v_i\le C{K}^2$ and $b\le CK$, giving $\mathbb{P}(|n^{-1}{\sum }_i{X}_i|>t)\le 2e^{-cn\min \{t^2/K^2,t/K\}}$. This form of Bernstein's inequality allows unbounded variables with factorial moment control, including the centered square of a sub-Gaussian random variable.

If $\mathbb{E}{e}^{tX}\le e^{{\sigma }^2{t}^2/2}$, then $X^2-\mathbb{E}{X}^2$ is a sub-exponential random variable with scale at most a universal constant times ${\sigma }^2$. The Chernoff bound gives $\mathbb{P}(|X|>x)\le 2e^{-x^2/(2{\sigma }^2)}$; integrating this tail bounds $\mathbb{E}|X{|}^{2q}$ by $2(2{\sigma }^2{)}^{q}q!$. Centering preserves a factorial moment bound, so the Bernstein bound for independent sub-exponential variables apply to independent averages of these squares.

For $V\sim {\chi }_n^2$ and $0<a<1$, apply the Chernoff bound to $e^{-sV}$ using $\mathbb{E}{e}^{-sV}=(1+2s{)}^{-n/2}$. Optimization at $s=(a^{-1}-1)/2$ proves the displayed estimate. It remains useful when the additive lower threshold in the chi-squared concentration inequality is nonpositive. For $0<r\le 1$, choosing $a=e^{-4r}$ gives a tail at most $e^{-n{r}^2}$ because $e^{-4r}-1+4r\ge 2r^2$.

For $V\sim {\chi }_n^2$ and $x>0$, the upper tail is at most $e^{-x}$ above $n+2\sqrt{nx}+2x$, and the lower tail is at most $e^{-x}$ below $n-2\sqrt{nx}$. Indeed $\log \mathbb{E}{e}^{s(V-n)}=-ns-(n/2)\log (1-2s)\le n{s}^2/(1-2s)$ for $0<s<1/2$. The Chernoff bound with $s=\sqrt{x/n}/(1+2\sqrt{x/n})$ proves the upper estimate. For the lower estimate, $\log \mathbb{E}{e}^{-s(V-n)}\le n{s}^2$ for $s>0$, and optimization gives the result. The lower threshold can be negative; the chi-squared Chernoff lower-tail bound provides an always-positive alternative.

For independent identically distributed real variables with everywhere finite cumulant-generating function, let $\mu$ be the mean. If $a\ge \mu$ and $\mathbb{P}(X_1>a)>0$, then $\mathbb{P}(S_n/n\in [a,a+\epsilon )\mid S_n/n\ge a)\to 1$ for each $\epsilon >0$. The Legendre transform of a cumulant-generating function is finite at $a$, and a positive exponential parameter gives strictly greater cost at $a+\epsilon$. A Chernoff bound for the farther tail divided by the Cramér theorem lower estimate at $a$ decays exponentially. The support condition both ensures positive conditioning probabilities and avoids a vacuous conditioning event.

For a nondecreasing process $A$ with stationary increments and independent increments, suppose $\mathbb{E}{e}^{\theta A(t)}=e^{t\Lambda (\theta )}$. If $\Lambda (\theta )\le C\theta$, then $e^{\theta (A(t)-Ct)}$ is a nonnegative supermartingale. Applying the maximal inequality for a nonnegative supermartingale up to time $T$ and then letting $T\to \infty$ givesThe stationary workload of a stable queue fed by $A$ has this supremum distribution, using the past input and stationary increments and independent increments. For independent flows, the constraint is ${\sum }_j{a}_j(\theta )\le C$, where $a_j(\theta )={\Lambda }_j(\theta )/\theta$ is the long-time effective bandwidth. The result bounds a continuous-time supremum directly; a one-time Chernoff bound by itself does not do that.

Effective bandwidth describes the exponential-moment traffic rate at a specified time scale and tail parameter. Independent sources have additive effective bandwidths. The Chernoff bound connects their sum to overflow bounds. For a single time unit, the small-parameter limit is the expected value and the large-parameter limit is the essential supremum, with appropriate moment or extended-value conventions.

For a normal distribution with mean $m$, positive variance $v$, and $\gamma >0$, the target $\mathbb{P}(X\ge C)\le e^{-\gamma }$ is equivalent to the displayed constraint, where $\Phi$ is the standard normal distribution function. This follows from $\mathbb{P}(X\ge C)=1-\Phi ((C-m)/\sqrt{v})$ and strict monotonicity of $\Phi$. Unlike the sufficient Chernoff bound margin $\sqrt{2\gamma v}$, this constraint is necessary as well as sufficient. The exact quantile can be negative when $e^{-\gamma }>1/2$.

Applying Markov's inequality to the positive random variable $e^{tX}$ givesHere $M_X(t)=\mathbb{E}[e^{tX}]$ is the moment-generating function, and the useful bound requires it to be finite at the chosen $t$. Optimizing over $t>0$ gives the Chernoff bound method: an exponential function weights large values more strongly before applying the same average-budget argument.

For example, let $N$ have the Poisson distribution with parameter one. Its expected value and variance are both one, and its moment-generating function is $M_N(t)=\exp (e^t-1)$. At threshold four, choosing $t=\log 4$ givesThis $t$ minimizes the exponent $e^t-1-4t$ because its derivative is $e^t-4$, with positive second derivative. The direct Markov's inequality bound is $1/4$, and the two-sided Chebyshev's inequality bound, applied to a deviation of at least three from the mean one, is $1/9$. The exact Poisson distribution tail probability is $1-e^{-1}(1+1+1/2+1/6)\approx 0.0189882$. These compare three valid bounds obtained from progressively different information; none is an exact tail probability formula for general random variables.

Let a slotted queue have stationary input, constant service $c$, and finite stationary queue workload $Q={\sup }_{n\ge 0}(A_n-cn)$. Write $K_n(\theta )=\log \mathbb{E}{e}^{\theta A_n}$ for its past cumulative arrivals. The union bound and Chernoff bound give the displayed inequality for $B>0$. It requires no temporal independence. If each $K_n(\theta )$ is finite and $K_n(\theta )/n\to \kappa (\theta )<c\theta$, the series is finite: its tail is bounded by a geometric series. Consequently ${\mathrm{limsup}}_{L\to \infty }{L}^{-1}\log \mathbb{P}(Q>Lb)\le -\theta b$ for $b>0$. The condition is that the asymptotic effective bandwidth $\kappa (\theta )/\theta$ be less than service capacity.

In a uniform linearized chord diagram, $\mathrm{Pr}(\sqrt{n}{R}_1\ge x)=(1-x^2/n{)}^n\to e^{-x^2}$. Conditional on the right endpoints, $d_1(n)=2+{\sum }_{i=2}^n{1}_{\{L_i\le R_1\}}$, with independent success probabilities $R_1/R_i$. A Chernoff bound gives $R_i=(1+o(1))\sqrt{i/n}$ uniformly for $i\ge n^{1/10}$. The early indicators are $o(\sqrt{n})$, and the remaining conditional mean is $(2+o(1))n{R}_1$. Their conditional variance is $O_{\mathbb{P}}(\sqrt{n})$. Hence $d_1(n)/\sqrt{n}-2\sqrt{n}{R}_1\to 0$ in probability, proving the displayed tail limit. A self-loop's contribution is included in the constant two.

Let $X_N\in {\mathbb{R}}^d$ and $a_N\to \infty$. Define the scaled cumulant-generating functionSuppose its pointwise limit $\Lambda$ exists, is lower semicontinuous, has $0$ in the interior of its effective domain, and is essentially smooth: it is differentiable on that interior, which is nonempty, and its gradient norm diverges at every finite boundary point approached from the interior. Then $X_N$ satisfies a large deviation principle with good rate function given by the Legendre-Fenchel transformThe Chernoff bound supplies the upper estimate. The lower estimate uses exponential tilting at a parameter whose gradient selects the required mean; essential smoothness supplies the boundary approximation needed for the full lower bound. A finite boundary slope can leave an affine branch of ${\Lambda }^{\ast }$ outside this argument, so a separate lower-bound proof is then necessary.

For a normal distribution $N(\lambda ,{\sigma }^2)$, the logarithm of its moment-generating function is $s\lambda +s^2{\sigma }^2/2$. Dividing by $s>0$ gives its effective bandwidth. Independent demands add these bandwidths. For total mean $m$ and positive total variance $v$, optimizing a Chernoff bound with target $e^{-\gamma }$, $\gamma >0$, yields the sufficient margin $C\ge m+\sqrt{2\gamma v}$.

A short concentration inequality argument explains the bound. For a fixed nonzero vector $u$, the normalized row products $⟨G_a,u⟩/\Vert u{\Vert }_2$ are independent random variables with the standard normal distribution, soUse the moment-generating function of a chi-squared distribution and the Chernoff bound to obtainFor $0<\epsilon <1$, elementary logarithm inequalities bound their sum by $2e^{-kc(\epsilon )}$. Apply the union bound to the $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ nonzero pair differences to get failure probability at most $n(n-1)e^{-kc(\epsilon )}\le \delta$. Independence of different pairs is unnecessary; they use the same map. This combines fixed-vector concentration with a finite-set union bound. ^[ref]

The Herbst argument integrates a logarithmic Sobolev bound on $\lambda H^{\prime }(\lambda )-H(\lambda )$ to control the centered cumulant-generating function $H$, then applies a Chernoff bound to obtain concentration.

The Hoeffding inequality turns finite interval bounds and independence into exponentially small tail probabilities for a sum. It requires neither a normal distribution nor identical probability distributions. Let $X_1,\dots ,X_n$ be independent essentially bounded real random variables, with deterministic finite bounds $a_i\le X_i\le b_i$ almost surely. Define the centered sum and its squared-width budget byFor $V>0$ and $t>0$,The union bound therefore gives $\mathbb{P}(|S_n|\ge t)\le \min \{1,2e^{-2t^2/V}\}$. If $V=0$, every $X_i$ is constant almost surely, so $S_n=0$ and both positive-threshold tail probabilities vanish. The equal-width version is stated in the authored Stanford notes.^[ref]

The quantity $V$ is a sum of squared interval widths, not the variance of $S_n$. Centering changes the endpoints to $a_i-\mathbb{E}{X}_i$ and $b_i-\mathbb{E}{X}_i$, leaving each width unchanged. For example, independent random variables in $[-2,1]$ and $[4,5]$ have $V=3^2+1^2=10$, whatever their expected values or probability distributions. A deviation at least three above their total expected value has probability at most $e^{-18/10}\approx 0.165299$. If all widths equal $b-a>0$, the sample mean ${\overline{X}}_n=n^{-1}{\sum }_i{X}_i$ satisfies

Independence is essential to this general guarantee. Take one Bernoulli random variable $B$ with success probability $1/2$ and set every $X_i=B$. Each lies in $[0,1]$, but for every $n$ the sample mean is either zero or one. Thus $\mathbb{P}(|{\overline{X}}_n-1/2|\ge 1/4)=1$, whereas incorrectly inserting these dependent observations into the formula would give $2e^{-n/8}$, less than one for $n\ge 6$. Some special dependent models have their own valid concentration inequalities; arbitrary statistical dependence cannot be ignored.

For the short proof, apply Markov's inequality to $e^{\lambda S_n}$ with $\lambda >0$, factor the expectation using independence, and apply the Hoeffding lemma to each factor:Minimizing this quadratic polynomial at $\lambda =4t/V$ gives the upper tail probability; applying it to $-S_n$ gives the lower one. This is the Chernoff bound method. The original 1963 article is Probability Inequalities for Sums of Bounded Random Variables by Wassily Hoeffding; its publisher record identifies the article, but its full text was unavailable during this source check. The argument above is a direct derivation rather than a claim to have inspected that article's proof.

The entropy method givesfor $\lambda \ge 0$. Consequently the Chernoff bound yields $\mathbb{P}(Z-\mathbb{E}Z\le -t)\le \exp [-t^2/(2\mathbb{E}Z)]$.

Put $q=1-p=1/b$ and $L={\log }_{b}n$. For $s=\lceil 2L\rceil$, the expected number of independent $s$-sets isTaking logarithms gives $\log \mathbb{E}{X}_s\le -c_p\log n\log \log n$ for all sufficiently large $n$: the two quadratic-order terms cancel to within $O_p(\log n)$, leaving the negative term $-s\log s$. By the first moment method, with probability tending to one $\alpha (G)\le s-1<2L$. Hence

For the upper bound, an explicit greedy exposure rule suffices; the sharper factor two is not needed here. SetWhile at least $m_0$ vertices remain, construct a colour class by choosing a candidate, retaining its non-neighbours as candidates, and repeating for $r$ choices. Expose only edges incident to chosen vertices. They will all be removed with this colour class. By deferred edge exposure in greedy independent-set colouring, the remaining internal edges are still independent Bernoulli edges conditional on the history. This assertion depends on this particular rule; it is not true for an arbitrary adaptively selected remainder.

If $M_j$ candidates remain after $j$ choices, then conditionallyThe Chernoff bound gives, for $M_j\ge 1/\eta$,On the event that these lower bounds hold, a class started from $m\ge m_0$ hasIndeed $q^r\ge n^{-1}(\log n{)}^8$, while $(1-2/\log n{)}^r$ is bounded below by a positive constant depending on $p$. Thus each conditional failure probability is at most $\exp [-c_p(\log n{)}^4]$. A union bound over at most $nr$ candidate steps shows that all these classes can be formed with probability tending to one. Colour the final fewer than $m_0$ vertices individually. The number of colours is at mostTogether this provesThe failure estimates are summable in $n$. Thus they give both the usual asymptotically-almost-sure assertion and, on any common probability space carrying the sequence, eventual almost-sure validity by the first Borel-Cantelli lemma.

Replace $\delta$ by $\min (\delta ,1)$ if necessary; this only weakens the hypothesis. We prove the result for finite graphs using a random proper partial graph coloring, then a deterministic completion. All constants below may depend on this fixed positive $\delta$.

First record the extension of a partial colouring by neighbour-colour savings. If a partial colouring uses colours from a palette of size $K$, let $R_v$ be the number of coloured neighbours of $v$, $D_v$ their number of distinct colours, and $S_v=R_v-D_v$ the neighbour-colour saving. There are $K-D_v$ available colours and $d(v)-R_v$ uncoloured neighbours. Thusat every vertex suffices for greedy completion: each subsequently coloured neighbour removes at most one available colour while reducing the uncoloured-neighbour count by one.

Call $v$ high-degree if $d(v)\ge (1-\delta /8)\Delta$. Its graph neighbourhood has at leastnonadjacent pairs for all sufficiently large $\Delta$. Assign independently to every vertex one of $C=\lceil \Delta /2\rceil$ tentative colours. Erase the colour of every vertex having a neighbour with the same tentative colour. The remaining colouring is proper; erasure is based on the original assignments and is not an iterative cascade.

For each $v$, define $A_v$ as the number of tentative colours assigned to at least one nonadjacent pair in its graph neighbourhood. Let $B_v$ count those same colours for which at least one neighbour of $v$ loses its colour. Then $X_v=A_v-B_v\ge 0$ counts repeated colours with no erased occurrence among the neighbours, so $S_v\ge X_v$.

For a particular nonadjacent pair $u,w\in \Gamma (v)$, consider the event that both receive one colour and no vertex of $(\Gamma (v)\cup \Gamma (u)\cup \Gamma (w))\setminus \{u,w\}$ receives that colour. This guarantees that its exactly two occurrences in $\Gamma (v)$ both survive. Its probability is at leastHere $C\ge \Delta /2$, $C\le \Delta$, and $\log (1-1/C)\ge -2/C$ for $C\ge 2$. Different pairs cannot be counted by the same colour in this event, since each event specifies exactly two occurrences in the graph neighbourhood. Thereforefor every high-degree vertex. This is the expected colour savings from sparse neighbourhoods.

We need simultaneous concentration, not just a positive expectation. A single tentative-colour change can affect $A_v$ or $B_v$ only in the old and new colour categories, so each function changes by at most two. A witness for $A_v\ge s$ fixes two vertices for each of $s$ colours. A witness for $B_v\ge s$ fixes such a nonadjacent pair and an erased neighbour together with a same-coloured adjacent vertex, at most four vertices per colour. Thus both are certifiable functions with Lipschitz constant $2$ and certificate parameter at most $4$. Both are bounded by $C\le \Delta$.

To see exactly what concentration follows, let an integer-valued $Z$ change by at most $L$ per coordinate and have certificates of size at most $rb$. For an input with $Z\ge b$, fix such a certificate $J$. If $y$ has $Z(y)\le a<b$, form a hybrid agreeing with that input on $J$ and with $y$ outside $J$. The hybrid has value at least $b$, so at least $(b-a)/L$ certificate positions differ from $y$. Equal weights on $J$ show that its Talagrand convex distance from $\{Z\le a\}$ is at least $(b-a)/(L\sqrt{rb})$. Talagrand's convex distance inequality therefore gives the two-threshold concentration for certifiable functions,At an integer median $m$, each median-side probability is at least $1/2$. Taking thresholds $m-t,m$ or $m,m+t$ and using $0\le Z\le \Delta$ givesfor $L=2,r=4$. Summing these tails gives $|\mathbb{E}Z-m|=O(\sqrt{\Delta })$. For sufficiently large $\Delta$, depending on $\beta$, it follows thatApply this to $A_v$ and $B_v$. Except with probability at most $8e^{-{\beta }^2\Delta /4096}$, $X_v\ge \mathbb{E}{X}_v-\beta \Delta /2\ge \beta \Delta /2$.

A bad event at $v$ depends only on tentative colours at distance at most two. It is independent of the entire collection of bad events at distance greater than four. Its dependency graph of events consequently has maximum degree at most $2{\Delta }^4$ for large $\Delta$. The Lovász local lemma now makes all high-degree vertices good simultaneously: exponential decay beats this polynomial dependency bound.

The needed local-lemma step can be checked directly. If every bad-event probability is at most $x(1-x{)}^D$, induction on conditioning sets gives $\mathrm{Pr}(E\mid \text{avoiding specified other events})\le x$: split those events into non-neighbours of $E$, independent of it, and at most $D$ neighbours, whose avoidance probability is at least $(1-x{)}^D$ by the induction. Multiplying the resulting conditional avoidance probabilities makes total avoidance positive. Taking $x=1/(D+1)$ works whenever $ep(D+1)\le 1$, since $(1-1/(D+1){)}^D\ge e^{-1}$. Our bound satisfies this criterion once $\Delta \ge {\Delta }_0(\delta )$.

Choose a successful partial colouring and set $K=\lceil (1-\beta /4)\Delta \rceil$. This contains the original palette. High-degree vertices satisfyLow-degree vertices have $K-d(v)\ge (\delta /8-\beta /4)\Delta \ge 1$ after increasing ${\Delta }_0$ if necessary. Greedy completion therefore gives $\chi (H)\le K$. Finally, for $\Delta \ge 8/\beta$,The constants are deliberately conservative; existence of a fixed positive improvement is the point. The same proof uses an external maximum degree bound, so if infinite locally finite graphs are allowed, every finite subgraph has this uniform colouring bound and compactness extends the colouring to the whole graph.

For the stronger triangle-free result, the appropriate brief route is semi-random palette refinement for triangle-free colouring. Start with $K=C_0\Delta /\log \Delta$ colours and repeatedly make small tentative random assignments, retain conflict-free choices, and update the uncoloured vertices' available lists. Triangle-freeness makes each graph neighbourhood independent. Track list size and average residual neighbour congestion per colour, prune unusually congested colours, and use concentration inequalities and the Lovász local lemma to preserve these invariants through the rounds. The ratio of residual congestion to list size decreases until a final completion using the Lovász local lemma or greedy coloring succeeds. Average control and pruning are important: four-cycles are allowed, so assuming tight uniform concentration for every individual colour load would be unjustified. With a sufficiently large absolute $C_0$, this programme gives $\chi (H)=O(\Delta /\log \Delta )$. The one-round fixed-fraction saving proved above alone does not establish that logarithmic improvement.

To show the order is best possible, let $d\to \infty$ through integers and sample $G(n,d/n)$ with $n=d^7$. A Chernoff bound and union bound give maximum degree at most $2d$ with probability tending to one. The expected triangle count is at most $d^3/6=o(n)$, so fewer than $n/2$ vertices need be deleted to remove all triangles with high probability. For $s=\lceil 4n\log d/d\rceil$, the expected number of independent $s$-sets is at mostits logarithm is at most $s[\log (ed/(4\log d))-2\log d+o(1)]$, which tends to minus infinity. The altered triangle-free graph thus has at least $n/2$ vertices and independence number below $s$, giving $\chi \ge (c+o(1))d/\log d$ for an absolute positive $c$. Append a disjoint star graph whose centre has vertex degree $2d$ if needed to make its maximum degree exactly $\Delta =2d$. This proves the triangle-free chromatic lower bound at bounded maximum degree, $\chi =\Omega (\Delta /\log \Delta )$, so the preceding order cannot be improved in general.