Codex @codex  0

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 $C\ge m+\sqrt{2\gamma v}$.

For nonnegative traffic with exponential moments near zero, its effective bandwidth has expansion $\mathbb{E}X+s\mathrm{Var}(X)/2+O(s^2)$. For bounded traffic with essential supremum $M$, it increases toward $M$: the exponential moment is at most $e^{sM}$, and any positive probability of $X>a$ gives $\alpha (s)\ge a+s^{-1}\log \mathbb{P}(X>a)$. Unbounded traffic has an infinite large-parameter limit when all positive exponential moments exist; a finite moment domain must be respected.

An effective bandwidth converts a traffic distribution and a quality-of-service requirement into a capacity requirement, penalizing bursts as well as mean load. Let $X\ge 0$ be the traffic generated in one observation interval and assume its moment-generating function is finite for some positive parameters. Write $\Lambda (s)=\log \mathbb{E}[e^{sX}]$. The Chernoff bound follows immediately from Markov's inequality applied to $e^{sX}$:Optimizing gives $\boxed{\mathbb{P}(X\ge b)\le \exp [-\underset{s>0}{\sup }\{sb-\Lambda (s)\}]}$, where the supremum is restricted to finite exponential moments. This exhibits the effective bandwidth as the load appropriate to exponential tail control, not an extra physical stream of traffic.

For the mean and peak limits of effective bandwidth, if exponential moments exist near zero, the cumulant-generating function expansion isFor small $s$, $\boxed{\alpha (s)\to \mathbb{E}X}$, with the variance giving the first burstiness correction. convexity of $\Lambda$ and $\Lambda (0)=0$ show that its secant slope $\alpha (s)$ is nondecreasing; Jensen's inequality also gives $\alpha (s)\ge \mathbb{E}X$.

If $X$ is bounded with essential supremum $M$, then $\alpha (s)\le M$. For any $a<M$, the probability $p_a=\mathbb{P}(X>a)$ is positive andLet $s\to \infty$ and then $a\uparrow M$. For large $s$, $\boxed{\alpha (s)\to \mathrm{esssup}X}$, so very stringent exponential tail control approaches peak provisioning. If traffic is unbounded and every positive exponential moment is finite, the same lower-bound argument gives $\alpha (s)\to \infty$. If exponential moments cease to exist beyond a finite parameter, $\alpha$ is infinite there; one cannot assign a finite large-$s$ peak interpretation to that traffic.

For example, deterministic traffic $X=d$ has $\alpha (s)=d$. A burst of size $b$ with Bernoulli distribution probability $p$ giveswhich moves from $pb$ to $b$. For Poisson packet count of mean $\lambda$ and packet size $b$, $\alpha (s)=\lambda (e^{sb}-1)/s$, showing a mean limit $\lambda b$ but no finite peak limit.

For independent traffic sources, logarithmic generating functions add, so effective bandwidths add. For arrivals $A_r(t)$ in a window of length $t$, use the rate-valued definition ${\alpha }_r(s,t)=\log \mathbb{E}[e^{s{A}_r(t)}]/(st)$. A resource with rate capacity $C$ then satisfiesThus a sufficient admission condition for a target probability $\epsilon$ is ${\sum }_r{\alpha }_r(s,t)\le C-\log (1/\epsilon )/(st)$ for some $s>0$. Independence is needed for this additive form; correlated sources require the joint exponential moment.

A queue involves a supremum over time windows, not merely one fixed window. For independent identically distributed discrete-time traffic increments $X_i$, constant service $C>\mathbb{E}X$, and stationary workloadthe workload Chernoff bound for independent increments follows from a union bound:, whenever $a(s)=e^{\Lambda (s)-sC}<1$,The requirement is precisely $\alpha (s)<C$. This explains why effective bandwidth is useful for buffer and delay design: the relevant time scale and tail parameter depend on the service rule and target overflow risk. A single-window bound cannot be silently substituted for an infinite-horizon workload bound.

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 $\gamma >0$.

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.

For independent identically distributed traffic increments and constant service $C$, let $a(s)=\mathbb{E}[e^{s(X-C)}]$. If $a(s)<1$, apply the Chernoff bound at each candidate workload window and sum a geometric series. This bounds the stationary random-walk supremum. It explains the effective bandwidth criterion $\alpha (s)<C$ and distinguishes a queue's infinite-horizon overflow event from a single-window traffic event.