= Solution
For every $s\ge0$, the pointwise inequality $\mathbf1_{\{Y\ge0\}}\le e^{sY}$ holds: on the event the exponential is at least one, and off it the indicator is zero. Taking <expected values> and then the infimum proves the <Chernoff bound>
$$
\boxed{\mathbb P(Y\ge0)\le\inf_{s\ge0}\mathbb E[e^{sY}].}
$$
An infinite exponential moment merely gives a vacuous bound at that $s$; no existence assumption on moments is needed for this extended-value inequality.
For a positive $s$ at which the relevant <moment-generating functions> are finite, apply the <Chernoff bound> to $Y=Z-C$. Independence of the sources gives
$$
\mathbb P(Z\ge C)\le e^{-sC}\prod_{j=1}^J\prod_{i=1}^{n_j}\mathbb E[e^{sX_{ji}}]=\exp\left\{s\left[\sum_{j=1}^Jn_j\alpha_j(s)-C\right]\right\}.
$$
Thus
$$
\boxed{\sum_{j=1}^Jn_j\alpha_j(s)\le C-\gamma/s\implies\mathbb P(Z\ge C)\le e^{-\gamma}.}
$$
Classes with $n_j=0$ make no contribution, so no moment assumption on those inactive classes is necessary. The <effective bandwidth> $\alpha_j(s)$ is the exponential-moment capacity demand of one class-$j$ source over this observation interval. Independent sources have additive effective bandwidths; reserving the margin $\gamma/s$ below capacity certifies the desired overflow bound. If $X_{ji}$ describes traffic during one unit of time, this is a bandwidth in traffic per unit time; for a window of length $t$, divide $\log\mathbb E e^{sX(t)}$ by $st$ instead.
To establish <monotonicity and bounds of effective bandwidth>, write $K(s)=\log\mathbb E e^{sX}$ for a representative source. The <Holder inequality> gives, for $0<t<s$ in its moment domain,
$$
\mathbb E e^{tX}=\mathbb E[(e^{sX})^{t/s}]\le(\mathbb E e^{sX})^{t/s},\qquad \frac{K(t)}t\le\frac{K(s)}s.
$$
Equivalently, the <cumulant-generating function> is convex with $K(0)=0$, so its secant slope from zero is nondecreasing. Hence $\alpha(s)$ is nondecreasing, rather than necessarily strictly increasing: if $X$ is deterministic its effective bandwidth is constant.
Assuming the mean is defined and finite, the <Jensen inequality> gives $\mathbb E e^{sX}\ge e^{s\mathbb EX}$, hence $\alpha(s)\ge\mathbb EX$. Let
$$
M=\sup\{x:\mathbb P(X>x)>0\}=\operatorname{ess\,sup}X.
$$
For finite $M$, $X\le M$ almost surely, so $\mathbb E e^{sX}\le e^{sM}$ and $\alpha(s)\le M$. Combining these proves
$$
\boxed{\mathbb EX\le\alpha(s)\le\operatorname{ess\,sup}X,\qquad\alpha(t)\le\alpha(s)\quad(0<t<s).}
$$
The upper bound is vacuous if the <essential supremum> is infinite. For nonnegative traffic these comparisons also make sense with extended values; for an arbitrary real-valued source with undefined mean, the printed mean comparison itself requires an integrability qualification. The positive parameter domain is essential: $\alpha(0)$ is defined only by a limit when the appropriate moment assumptions hold, and the stated monotonicity comparisons concern $s>0$.
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