= Solution
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>:
$$
\alpha(s)=\frac{K(s)}s,\qquad K(s)=\log\mathbb E[e^{sX}].
$$
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 $X\geq b$ implies $e^{sX}\geq e^{sb}$:
$$
\mathbb P(X\geq b)\leq e^{-sb}\mathbb E[e^{sX}]
=e^{s(\alpha(s)-b)}.
$$
\b[Optimize over the admissible positive values of $s$] to obtain
$$
\boxed{\mathbb P(X\geq b)\leq
\inf_{s>0:\,K(s)<\infty}\exp\{K(s)-sb\}.}
$$
For independent sources $X_1,\ldots,X_N$, <moment-generating functions> multiply and their <effective bandwidths> add. Thus a total capacity $c$ has the one-time overflow bound $\mathbb P(\sum_iX_i\geq c)\leq\exp\{s(\sum_i\alpha_i(s)-c)\}$. In particular, for <independent and identically distributed random variables>,
$$
\mathbb P\!\left(\frac1N\sum_{i=1}^NX_i\geq c\right)
\leq\exp\{-N\sup_{s>0}[sc-K(s)]\}.
$$
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 $K(0)=0$, $K'(0)=\mathbb E[X]$ and $K''(0)=\operatorname{Var}(X)$. Therefore the \b[small-parameter limit is the mean traffic rate]:
$$
\boxed{\lim_{s\downarrow0}\alpha(s)=\mathbb E[X]},
\qquad \alpha(s)=\mathbb E[X]+\frac{s}{2}\operatorname{Var}(X)+O(s^2).
$$
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 $M=\operatorname{ess\,sup}X$, the <essential supremum>. If $M<\infty$, $X\leq M$ almost surely gives $\alpha(s)\leq M$. For every $a<M$, $p_a=\mathbb P(X\geq a)>0$, and
$$
\alpha(s)\geq a+\frac{\log p_a}{s}.
$$
Taking a lower limit and then $a\uparrow M$ proves the \b[large-parameter limit is peak traffic]:
$$
\boxed{\lim_{s\to\infty}\alpha(s)=\operatorname{ess\,sup}X.}
$$
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.
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