= Solution
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 $X\geq0$ 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 $e^{sX}$:
$$
\mathbb P(X\geq b)\leq e^{-sb}\mathbb E[e^{sX}]=e^{-s[b-\alpha(s)]},\qquad \alpha(s)=\frac{\Lambda(s)}s,\quad s>0.
$$
Optimizing gives \b[$\boxed{\mathbb P(X\geq b)\leq\exp[-\sup_{s>0}\{sb-\Lambda(s)\}]}$], 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 is
$$
\alpha(s)=\mathbb E X+\frac{s}{2}\operatorname{Var}(X)+\frac{s^2}{6}\kappa_3(X)+O(s^3).
$$
\b[For 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 $\alpha(s)\geq\mathbb E X$.
If $X$ is bounded with <essential supremum> $M$, then $\alpha(s)\leq M$. For any $a<M$, the probability $p_a=\mathbb P(X>a)$ is positive and
$$
\alpha(s)\geq a+\frac{\log p_a}{s}.
$$
Let $s\to\infty$ and then $a\uparrow M$. \b[For large $s$, $\boxed{\alpha(s)\to\operatorname{ess\,sup}X}$], 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 $p$ gives
$$
\alpha(s)=\frac1s\log(1-p+pe^{sb}),
$$
which 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 $\alpha_r(s,t)=\log\mathbb E[e^{sA_r(t)}]/(st)$. A resource with rate capacity $C$ then satisfies
$$
\mathbb P\left(\sum_r A_r(t)\geq Ct\right)\leq\exp\left[-st\left(C-\sum_r\alpha_r(s,t)\right)\right].
$$
Thus a sufficient admission condition for a target probability $\varepsilon$ is $\sum_r\alpha_r(s,t)\leq C-\log(1/\varepsilon)/(st)$ 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 $C>\mathbb EX$, and stationary workload
$$
W=\sup_{n\geq0}\left(\sum_{i=1}^nX_i-nC\right),
$$
the <workload Chernoff bound for independent increments> follows from a <union bound>:, whenever $a(s)=e^{\Lambda(s)-sC}<1$,
$$
\mathbb P(W>B)\leq e^{-sB}\sum_{n\geq1}a(s)^n=e^{-sB}\frac{a(s)}{1-a(s)}.
$$
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.
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