 Solution

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 $\gamma >0$.