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Let $Y_i$ be independent and identically distributed random variables with moment-generating function finite on an open interval about $0$. Putallowing $\Lambda =+\infty$ outside its finite domain. The Cramér theorem states that ${\bar{Y}}_n=n^{-1}{\sum }_{i=1}^n{Y}_i$ satisfies a large deviation principle with speed $n$ and good rate function $I$, the Legendre transform of a cumulant-generating function. Explicitly, for every closed set $F$ and open set $O$,Here $\log 0=-\infty$ and $\inf \varnothing =+\infty$. The rate function is lower semicontinuous and its finite sublevel sets are compact. The exponential-moment hypothesis is essential for this form of the Cramér theorem.

Condition on $N$. Since the amounts are independent and identically distributed random variables and are independent of $N$, their moment-generating functions multiply:Summing against the positive-support geometric distribution gives the geometric-sum moment-generating functionThe finite-transform domain is $M_X(t)<\infty$ and $(1-p)M_X(t)<1$. The terms are nonnegative for real $t$, so if the latter condition fails the sum diverges. For $t\le 0$ both conditions always hold, since the claims are positive; thus the same calculation always provides the Laplace transform of a nonnegative random variable. Positivity alone does not guarantee a finite moment-generating function on a positive neighbourhood of zero.

There is at least one strictly positive amount in the portfolio. Hence $\mathbb{P}(S=0)=0$; zero carries no atom of a measure to add to the continuous laws in the next two parts.

For the exponential distribution of mean $\mu$, $M_X(t)=(1-\mu t{)}^{-1}$. Substitution into the geometric-sum moment-generating function givesThis is the moment-generating function of an exponential distribution with rate $p/\mu$. By the uniqueness theorem for moment-generating functions, the geometric sum of exponential variables therefore has probability density functionIts expected value is $\mu /p$, agreeing with $\mathbb{E}[N]\mathbb{E}[X]=\mu /p$ in the random sum of independent claims formula.

Assume $0<\lambda =1-\epsilon <1$, as required for this nontrivial subcritical binomial random graph. Put $\delta =\lambda -1-\log \lambda >0$. A breadth-first exploration of a binomial random graph is dominated by a Galton-Watson process whose offspring counts are independent random variables with law $\mathrm{Bin}(n,\lambda /n)$. If its total progeny $T$ is at least $j$, the first $j-1$ offspring counts sum to at least $j-1$. Their moment-generating function is bounded by that of a Poisson distribution of mean $\lambda (j-1)$. The exponential Markov inequality, optimized at $t=-\log \lambda$, yieldsTaking $j=\lceil (1+\eta )\log n/\delta \rceil$ for any fixed $\eta >0$, a union bound over all starting vertices gives $\mathbb{P}(L_1\ge j)=o(1)$.

For the matching lower bound, let $X_j$ count tree components of order $j$. The tree-component expectation in the Erdős-Rényi model is exactlyFor $j=O(\log n)$, the Stirling formula givesChoose $j=\lfloor (1-\eta )\log n/\delta \rfloor$, with $0<\eta <1$. Then $\mathbb{E}{X}_j\to \infty$. Distinct overlapping vertex sets cannot both be graph components. For disjoint sets, the joint probability differs from the product only because the $j^2$ between-set edges were counted twice as absent. ThusIt follows that $\mathrm{var}{X}_j/(\mathbb{E}{X}_j{)}^2=o(1)$. The second moment method gives $X_j>0$ with high probability. Combining both bounds, and then taking arbitrarily small fixed $\eta$, proves the sharp subcritical largest-component scaleFinally the Taylor expansion of $-\log (1-\epsilon )$ gives $\delta ={\epsilon }^2/2+{\epsilon }^3/3+\cdots$. The endpoint $\lambda =0$ has no edges and is excluded from this logarithmic asymptotic.

The Poisson distribution with unit parameter has mass $e^{-1}/j!$. Its moment-generating function isFor an integer $k\ge 1$, multiply the exponential probability bound by $e$ to obtain ${\sum }_{j=k}^{\infty }1/j!\le \exp (e^t-kt)$. The exponent is minimized at $e^t=k$, hence $t=\log k\ge 0$, including $t=0$ when $k=1$. Consequently

Identify the hypercube graph $Q_n$ with $\mathcal{P}([n])$, joining two sets when their symmetric difference has size one. Write $N[A]$ for the closed graph neighbourhood of $A$, and $N_t[A]$ for its radius-$t$ neighbourhood in Hamming distance. In the simplicial order on the discrete cube, smaller sets come first, with lexicographic order within each rank: the smallest differing coordinate belongs to the earlier set. Let $I_m$ be the first $m$ vertices in this order. Harper theorem isEquivalently, $I_m$ minimizes the external vertex boundary at fixed size. We will also obtain $|N_t[A]|\ge |N_t[I_{|A|}]|$ for every integer $t\ge 0$.

Here is a complete proof by simplicial section compression. First note two properties of the simplicial order on the discrete cube. Its restriction to either face obtained by fixing one coordinate is the corresponding order on the smaller hypercube graph. Also, the closed graph neighbourhood of an initial segment is an initial segment. To verify the latter, a nonempty initial segment contains every level below its last level $r$, and a lexicographic prefix $\mathcal{L}$ at level $r$. For $r\ge 1$, its closed graph neighbourhood contains every set of size at most $r$, and its only further members are the upper shadow of $\mathcal{L}$. An $(r+1)$-set lies in this upper shadow exactly when its lexicographically earliest $r$-subset lies in $\mathcal{L}$. That earliest subset is obtained by deleting the largest element. These earliest subsets vary monotonically in lexicographic order, so the upper shadow is another lexicographic prefix. For $r=0$, the closed graph neighbourhood of $\{\varnothing \}$ is the union of levels zero and one. The empty initial segment causes no exception.

Induct on $n$, with $n=0,1$ immediate. Fix a coordinate $i$ and write the sections of $A$ as $A_0,A_1\subseteq Q_{n-1}$, deleting $i$ from the latter section. Replace them by the initial segments $I_a,I_b$ with $a=|A_0|$, $b=|A_1|$. The sections of the old closed graph neighbourhood areBy induction their sizes are at least $\max \{|N[I_a]|,b\}$ and $\max \{|N[I_b]|,a\}$, respectively. These are exactly the sizes of the new sections, because both sets in each new union are initial segments and hence nested. Thus simplicial section compression preserves size and never enlarges the closed graph neighbourhood.

Repeatedly compress any section that is not already an initial segment. Every change strictly reduces the sum of the global simplicial positions of the included vertices. This nonnegative integer decreases only finitely often, so the process ends at a family $B$ whose sections in every coordinate are initial segments. If $B$ itself is an initial segment, the induction is complete. Otherwise choose the earliest absent vertex $X$ and the latest present vertex $Y$, with $X\prec Y$. They cannot agree in any coordinate: agreement would put them in one compressed section, where an earlier absent vertex cannot precede a later present one. Hence $Y=X^c$. Nor can a vertex $Z$ lie between them: if $Z$ is present, pairing it with $X$ would force $Z=X^c=Y$; if absent, pairing it with $Y$ would force $Z=Y^c=X$. Thus $X,Y$ are consecutive, and $B$ is obtained from $I_{|B|}$ by replacing its last vertex $X$ with the next vertex $Y$.

There are only two possibilities for these terminal families for simplicial section compression. If $|X|<|Y|$, consecutiveness makes $X$ the last $k$-set and $Y$ the first $(k+1)$-set. Complementarity forces $n=2k+1$, $X=\{k+2,\dots ,2k+1\}$ and $Y=\{1,\dots ,k+1\}$. For $n\ge 3$, the corresponding $B$ contains all sets of size at most $k$ except $X$, and additionally $Y$. Every $(k+1)$-set has at least two $k$-subsets, so at least one belongs to $B$. The missing $X$ has a $(k-1)$-subset in $B$. Consequently $N[B]$ contains every set of size at most $k+1$, which is exactly $N[I_{|B|}]$.

If $|X|=|Y|$, then $n=2k$. Since $X,Y$ are complementary, the smaller one contains coordinate $1$. Consecutiveness forces $X$ to be the last $k$-set containing $1$, and $Y$ the first $k$-set avoiding $1$:The initial segment consists of all sets of size less than $k$ and all $k$-sets containing $1$. Its closed graph neighbourhood contains all sets of size at most $k$ and all $(k+1)$-sets containing $1$. For $k\ge 2$, each such $(k+1)$-set has $k\ge 2$ $k$-subsets containing $1$, so removing only $X$ leaves one in $B$. All sets of size at most $k$ are still in $N[B]$, because all levels below $k$ belong to $B$. Hence again $N[I_{|B|}]\subseteq N[B]$. For $k=1$, the two families are related by exchanging coordinates and both have the whole $Q_2$ as their closed graph neighbourhood. These comparisons finish the induction and prove Harper theorem.

For the radius-$t$ version, repeatedly use the one-step result. At each step the comparison family remains an initial segment, and its closed graph neighbourhood is monotone in its size. If $|N_{t-1}[A]|\ge |N_{t-1}[I_m]|$, applying Harper theorem to $N_{t-1}[A]$ therefore gives $|N_t[A]|\ge |N_t[I_m]|$. Induction on $t$ proves the asserted extension.

A Lévy family of graphs has vanishing concentration of measure functions after distances have been scaled by the graph diameter. More explicitly, for connected finite graphs $G_n$ of diameter $D_n$, use normalized graph distance $d_n=d_{G_n}/D_n$ and uniform probability measure. For every fixed $\epsilon >0$, requireFor $Q_n$ this normalization is normalized Hamming distance $d_H/n$, since its diameter is $n$.

If $|A|\ge 2^{n-1}$, the corresponding initial segment contains the Hamming ball about $\varnothing$ of radius $b_n=\lfloor (n-1)/2\rfloor$. The radius-$t$ version of Harper theorem impliesUse the precise Hoeffding inequality $\mathbb{P}(Z_n-n/2\ge u)\le \exp (-2u^2/n)$ for $u\ge 0$. One can also derive this estimate directly: the moment-generating function of $Z_n-n/2$ is $\cosh (\lambda /2{)}^n\le \exp (n{\lambda }^2/8)$; the exponential Markov inequality followed by minimization at $\lambda =4u/n$ gives the bound. With $t=\lfloor \epsilon n\rfloor$, we have $b_n+t\ge n/2+\epsilon n-2$. Thus, whenever $\epsilon n>2$,This proves that the hypercube graphs form a Lévy family of graphs. The stronger fixed-positive-mass formulation follows as well. For any $a>0$, choose $c$ with $1/(4c^2)<a$. The Chebyshev inequality for $Z_n$, whose variance is $n/4$, makes the Hamming ball of radius $\lfloor n/2-c\sqrt{n}\rfloor$ have fewer than $a{2}^n$ vertices. An initial segment of size at least $a{2}^n$ contains that ball. Enlarging by $\lfloor \epsilon n\rfloor$ and using the same Hoeffding inequality gives a complement proportion tending to zero uniformly over all such $A$. The normalization matters: with unscaled unit Hamming distance, cubes do not satisfy the metric version at every fixed radius; for a radius below one, a half-cube has no enlargement.

Equality in Harper theorem does not determine a down-set up to isomorphism. In $Q_4$, takeThis is a down-set of size nine. Its closed graph neighbourhood contains every set of size at most two, because all singletons belong to $A$. Every triple contains an edge of the four-cycle among the listed pairs, so every triple also belongs to $N[A]$. The four-element set does not. Hence $|N[A]|=1+4+6+4=15$.

The size-nine simplicial order on the discrete cube initial segment is all sets of size at most one together with the pairs $12,13,14,23$. Every triple contains one of these pairs, and the four-element set is again absent from its closed graph neighbourhood. Its neighbourhood therefore also has size $15$, proving extremality of $A$. Nevertheless the induced graphs are not isomorphic: in $A$, the empty set has degree four, the four singletons have degree three, and the four pairs have degree two. The initial segment has two vertices of degree four, namely $\varnothing$ and $\{1\}$. Since a graph automorphism preserves induced degree of a vertex, no automorphism of the cube can carry one family to the other. This gives nonunique down-set extremizers for Harper theorem even under the stronger test of induced graph isomorphism.

The exponential distribution with rate $\beta >0$ has moment-generating function $\beta /(\beta -t)$ for $t<\beta$. Shifting it by $\alpha$ multiplies the moment-generating function by $e^{\alpha t}$. Therefore the shifted-exponential Poisson mixture has probability generating functionThe probability generating function converges and is analytic for $|z|<\beta +1$, and in particular is finite for real $0\le z<\beta +1$.

Write $p=\beta /(\beta +1)$ and $q=1/(\beta +1)$, so $p+q=1$. The factor $e^{\alpha (z-1)}$ is the probability generating function of $V\sim \mathrm{Poisson}(\alpha )$, while $p/(1-qz)$ is that of a geometric distribution on $\{0,1,\dots \}$, with $\mathbb{P}(W=j)=p{q}^j$. Choose $V$ and $W$ independently. The product rule for the probability generating function of a sum of independent counts then proves $N\stackrel{d}{=}V+W$. The support-zero convention for $W$ is essential here.

The convolution of independent random variables gives the finite sumTaking the logarithmic derivative of the probability generating function givesCompare coefficients of $z^{n-1}$, for $n\ge 2$. The left side is $(\beta +1)n{p}_n-(n-1)p_{n-1}$, and the right side is $\{1+\alpha (\beta +1)\}{p}_{n-1}-\alpha p_{n-2}$. Hence the shifted-exponential Poisson count recursion isThe constant and linear coefficients supply the starting valuesThese two values determine every subsequent probability by the recursion. The negative second term does not mean that the distribution has negative probabilities: the convolution formula exhibits every $p_n$ as a sum of nonnegative quantities.

At $\alpha =0$ the independent Poisson distribution component is identically zero and the count has a geometric distribution. Both the finite sum and the recursion reduce to $p_n=q{p}_{n-1}$. Consequently the Panjer claim-count class parameters areAlthough the shifted model initially has positive $\alpha$, this zero-shift boundary case is well defined.

For a Poisson distribution of parameter $\lambda$, its probability generating function is $\exp \{\lambda (z-1)\}$. Applying the law of total expectation to the Poisson mixture givesThusFor $|z|\le 1$ the absolute value of the integrand is bounded by one, so this identity always exists as a Laplace transform of the positive mixing variable. An extension to $z>1$ requires the corresponding moment-generating function to be finite; the conditional computation itself does not guarantee positive exponential moments.

For an aggregate claims model, let $X$ denote one claim size, with expected value $\mu$ and variance ${\sigma }^2$. The total $S$ is zero when the claim count is zero. Given $N=n$, independence givesThe law of total expectation and law of total variance, together with $\mathbb{E}N=\mathrm{Var}(N)=\lambda$ for a Poisson distribution, implyThe raw second moment appears because the random count itself contributes $\lambda {\mu }^2$ to the variance.

Conditional on $N$, the moment-generating function of the sum is $M_X(t{)}^N$. Averaging with the Poisson distribution therefore gives the compound Poisson distribution transformThis is valid wherever $M_X(t)$ is finite. Finite first and second moments alone do not ensure positive exponential moments; for positive claims the corresponding Laplace transform always exists.

For the independent portfolios put $\Lambda ={\lambda }_1+{\lambda }_2>0$ and $w_i={\lambda }_i/\Lambda$. Their Poisson distribution counts add to a Poisson distribution with parameter $\Lambda$. By Poisson-multinomial conditioning, conditional on the total count $n$ the first risk count has a binomial distribution with parameters $n,w_1$, and the second is the remaining count. Thus one can generate the same total loss by drawing $n$ independent risk labels with these weights, then drawing each claim from its label's law. The merged severity has mixture distributionIt follows that $T$ has a compound Poisson distribution with count parameter $\Lambda$ and this severity law. This is the fixed-year version of Poisson superposition of insurance portfolios. Its expected value and variance areAlternatively, multiplying the two independent aggregate Laplace transforms yields $\exp \{\Lambda (w_1{M}_{X_1}(t)+w_2{M}_{X_2}(t)-1)\}$ wherever finite, confirming the same compound Poisson distribution.

For the retained compound Poisson aggregate under per-claim reinsurance, replace each claim $X$ by its retained payment $g(X)$. Assume the retention is measurable, with $0\le g(x)\le x$ as usual. The count parameter remains $\Lambda$, and the severity law is the pushforward measure of the mixture severity under $g$. Thus the retained compound Poisson aggregate $T_I$ has a compound Poisson distribution with that transformed severity andIts moment-generating function is $\exp \{{\sum }_i{\lambda }_i(M_{g(X_i)}(t)-1)\}$ on its finite domain. For a general retention $g$, zero retained payments can occur and are allowed as compound-Poisson marks; they may equivalently be removed by Poisson thinning. The two specified contracts retain strictly positive payments for strictly positive claims.

In the classical risk model write $U(t)=u+ct-{\sum }_{j=1}^{N(t)}{X}_j$, where the Poisson process $N(t)$ has rate $\lambda$ and is independent of the claim sizes. Define the ruin time and ultimate ultimate ruin probability byThe relative safety loading is $\rho =c/(\lambda \mu )-1>0$, so $c=(1+\rho )\lambda \mu$. The adjustment coefficient is the nonzero positive solutionExistence and uniqueness follow from the secant-slope existence criterion for an adjustment coefficient. Explicitly, $\lambda (M(r)-1)-cr$ has derivative $\lambda \mu -c<0$ at zero and is strictly convex for positive claims. The assumed divergence of $M$ makes it cross zero once at a positive $R<r_{\infty }$. When $r_{\infty }=\infty$, an exponential lower bound from any positive tail event shows that $M(r)/r\to \infty$. In particular $R$ is inside the finite-transform domain, not at its endpoint.

Put $\overline{F}=1-F$. Replacing $\phi =1-\psi$ in the survival renewal equation for a classical risk model yieldsBy the tail integral formula for moments, $\mu ={\int }_0^{\infty }\overline{F}(x)dx$. Therefore this is the defective renewal equationThe original kernel has mass $\lambda \mu /c=1/(1+\rho )<1$. For the exponential tilt of the ruin renewal kernel, setMultiplying by $e^{Ru}$ gives the required proper renewal equationTo verify that this is a probability renewal kernel, Tonelli theorem gives, for $r>0$ in the finite-transform domain,The adjustment coefficient equation consequently implies ${\int }_0^{\infty }{k}_R(x)dx=1$. Its expected value isIt is finite because $R$ is interior to the finite-transform domain and is positive because the strict convex crossing has derivative $\lambda M^{\prime }(R)-c>0$.

We quote the key renewal theorem in the following form: for a nonarithmetic distribution $K$ of positive increments with finite positive mean $m$, and a directly Riemann integrable nonnegative function $h$, the locally bounded solution of $z=h+z\ast K$ satisfies $z(u)\to m^{-1}{\int }_0^{\infty }h$. It has renewal representation $z=h\ast {\sum }_{j\ge 0}{K}^{\ast j}$; this follows by iterating the equation, since the probability that arbitrarily many positive increments have sum at most a fixed $u$ tends to zero.

Here $K_R(dx)=k_R(x)dx$ is absolutely continuous, and hence nonarithmetic distribution, even if the original claim law has atoms. The function $h_R$ is continuous. Choose $\delta >0$ with $M(R+\delta )<\infty$. The Markov inequality gives $\overline{F}(x)\le M(R+\delta )e^{-(R+\delta )x}$, and hence $h_R(u)\le C_{\delta }{e}^{-\delta u}$. Continuity on compact intervals and this exponential bound make the upper Riemann sums finite with uniformly vanishing tails, proving direct Riemann integrability.

A further application of Tonelli theorem evaluates the forcing integral:All hypotheses of the key renewal theorem are now checked, so the interior adjustment coefficient ruin prefactor isThis proves the requested Cramér–Lundberg ruin asymptotic, including its constant.

For the two-component hyperexponential distribution, conditioning on the chosen exponential component givesIts moment-generating function and derivative areThe moment-generating function diverges at $1/2$, while the derivative of the adjustment equation at zero is negative. Strict convexity therefore places its unique positive adjustment coefficient inThe adjustment equation, divided by $R$, becomesUsing this identity to subtract $(1+\rho )\mu$ from $M^{\prime }(R)$ givesThus the asymptotic constant in terms of $\rho$ and $R$ isThe denominator is positive on the identified domain. If desired, $R$ is the smaller root of $6(1+\rho )R^2-(5+9\rho )R+3\rho =0$; the other algebraic root lies outside the positive finite-transform interval and is not an adjustment coefficient.

Write $Ph=\mathbb{E}h(X)$ and $P_{n}h=n^{-1}{\sum }_{i=1}^{n}h(X_i)$. A function bracket $[\ell ,u]$ contains the measurable functions $h$ satisfying $\ell (z)\le h(z)\le u(z)$ for every $z\in T$. Require its endpoints to be integrable and call $P(u-\ell )$ its $L^1(P)$ width. A sufficient condition is that, for every $\epsilon >0$, finitely many brackets of width at most $\epsilon$ cover the whole class $\mathcal{H}$. Under this condition the uniform strong law from finite L1 bracketing statesThe conclusion holds outside a common measurable null set. If the supremum is not initially known to be measurable, this formulation means pathwise convergence on a measurable probability-one event; a pointwise separable function class, including the application below, has a measurable supremum. Pointwise brackets also ensure the sample inequalities hold simultaneously over the class.

To prove the result, choose a finite $\epsilon$-cover $[{\ell }_r,u_r]$, $1\le r\le N$. If $h$ belongs to bracket $r$, monotonicity of the empirical measure and of expectation givesandConsequentlyApply the strong law of large numbers to these finitely many integrable endpoints. On a probability-one event, the maximum tends to zero. Repeat with $\epsilon =1/q$, $q\in \mathbb{N}$, and intersect the countably many probability-one events. The limiting supremum is bounded by $1/q$ for every $q$, hence is zero. This proves the uniform law of large numbers without a boundedness assumption on the class itself.

For the moment-generating function, use the empirical measure estimatorIf $t=0$, this estimator and $m$ are both identically one. Otherwise, $X\ge 0$ makes $e^{sX}$ increasing in $s$, and $e^{sX}\le e^{tX}$ provides an integrable envelope. The dominated convergence theorem shows that $m$ is continuous on $[0,t]$, hence uniformly continuous.

For any $\epsilon >0$, choose a partition $0=s_0<s_1<\cdots <s_N=t$ so that $m(s_r)-m(s_{r-1})\le \epsilon$ for every $r$. If $s\in [s_{r-1},s_r]$, then $e^{s_{r-1}z}\le e^{sz}\le e^{s_{r}z}$ for every $z\ge 0$. These endpoint functions form finitely many integrable function brackets with the required $L^1(P)$ widths. The just-proved uniform law of large numbers therefore givesBoth functions of $s$ are continuous; their supremum equals the supremum over a countable dense subset, so it is measurable. This proves uniform consistency of an empirical moment-generating function using only the observed sample.

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 $\gamma >0$.

For $\lambda \ge 0$, a Poisson distribution has mass $\mathbb{P}(X=k)=e^{-\lambda }{\lambda }^k/k!$ for $k=0,1,\dots$; at $\lambda =0$ it is concentrated at zero. Its moment-generating function isFor independent $X,Y$, the moment-generating function of their sum is the product, $\exp [(\lambda +\mu )(e^s-1)]$. Uniqueness of the moment-generating function near zero therefore gives $X+Y\sim \mathrm{Poisson}(\lambda +\mu )$. Repeating the argument gives $S_n={\sum }_{i=1}^n{X}_i\sim \mathrm{Poisson}(n)$ when every parameter is one.

The displayed sequence is $a_n=\mathbb{P}(S_n\le n)$. The classical central limit theorem states that independent identically distributed variables with finite mean $m$ and nonzero finite variance ${\sigma }^2$ have standardized sum $(S_n-nm)/(\sigma \sqrt{n})$ converging in distribution to the standard normal distribution. Here $m={\sigma }^2=1$, soConvergence of the distribution functions at the continuity point zero justifies the last step. This is the Poisson cumulative probability at its growing mean.

Condition on the count and use the independence of random variables of the claim sizes. For $n\ge 1$ their joint exponential transform factors, while for $n=0$ the empty sum contributes $1$. Consequently the law of total expectation givesThis random sum of independent claims transform uses both independence assumptions: the count must be independent of the entire claim-size sequence, and the sizes must be mutually independent with the same probability distribution. The probability generating function is interpreted through its defining nonnegative series. The identity holds as a finite moment-generating function wherever that series is finite; outside that domain the expectation and series can agree at $+\infty$. In particular a positive argument may take $M_X(r)$ beyond $1$, so finiteness does not follow merely from the usual unit-disk domain of a probability generating function.