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For any $s\ge 0$, $Y\ge 0$ implies $e^{sY}\ge 1$. Hence $1_{\{Y\ge 0\}}\le e^{sY}$, and taking expected values gives the Chernoff boundAt $s=0$ this is the trivial bound one; an infinite exponential moment merely gives a noninformative bound. For positive $s$ at which all required moment-generating functions are finite, define ${\alpha }_j(s)=s^{-1}\log \mathbb{E}[e^{s{X}_{ji}}]$. Independence of the source random variables impliesApplying the Chernoff bound to $Y=X-C$ givesConsequentlyHere $s>0$ and, for a nontrivial target, $\gamma >0$.

The effective bandwidth ${\alpha }_j(s)$ is the capacity charge of one source for this exponential-moment tail criterion. Independent sources contribute additively. It incorporates variability as well as mean demand: the Jensen inequality gives ${\alpha }_j(s)\ge \mathbb{E}{X}_{ji}$, and, under exponential integrability near zero, ${\alpha }_j(s)\to \mathbb{E}{X}_{ji}$ as $s\downarrow 0$. For bounded traffic its large-$s$ limit is the essential peak, as described by the mean and peak limits of effective bandwidth. Thus larger $s$ gives more weight to rare high demands. This is a one-window demand bound; a queue's overflow probability over many windows requires additional temporal modeling, not just this single-window calculation.

For the normal distribution $X_{ji}\sim N({\lambda }_j,{\sigma }_j^2)$, completing the square in its density gives the moment-generating functionWrite $m={\sum }_j{n}_j{\lambda }_j$ and $v={\sum }_j{n}_j{\sigma }_j^2$ for the aggregate mean and variance. The exponential-moment capacity condition becomesFor $v>0$, the last two terms are minimized at $s=\sqrt{2\gamma /v}$, where their sum is $\sqrt{2\gamma v}$. Therefore Gaussian effective bandwidth yields the sufficient conditionEquivalently, optimizing the exponential bound directly when $C>m$ gives $\exp [-(C-m{)}^2/(2v)]$; when $C\le m$ the best nonnegative-$s$ bound is one. The square-root condition is sufficient rather than necessary because an exponential bound does not equal the exact normal tail.

For the exact condition, independence and multiplication of the normal moment-generating functions show that $X\sim N(m,v)$. For $v>0$, let $\Phi$ be the standard normal distribution function. The normal law is continuous, soSince $\Phi$ is continuous and strictly increasing from zero to one, the exact Gaussian overflow constraint isThe constant depends only on the desired tail exponent $\gamma$, not on the source counts. It is positive when $\gamma >\log 2$, zero when $\gamma =\log 2$, and negative when $0<\gamma <\log 2$. For example, at $\gamma =\log 100$ the exact normal quantile is about $2.32635$, whereas the Chernoff bound uses $\sqrt{2\log 100}\approx 3.03485$; the latter reserves more capacity.

Finally, the positive-variance hypothesis must not be hidden. If $v=0$, then $X=m$ almost surely, and the zero-variance boundary of an overflow constraint gives $\mathbb{P}(X\ge C)=1_{\{m\ge C\}}$. For $\gamma >0$, the exact condition is then $m<C$. In particular $m=C$ does not meet the target even though substituting $v=0$ into the printed non-strict square-root inequalities would allow it. The displayed normal-quantile equivalence applies to nondegenerate Gaussian traffic; the deterministic case requires this strict boundary convention.