Use $F(t)=\mathrm{Pr}(T>t)$, a nonincreasing right-continuous survivor function. The empirical likelihood treats an exact event as a probability mass rather than maximizing an unrestricted probability density function at isolated data points. Under a noninformative observation mechanism, the event-time likelihood contributions areThe second, third and fourth rows account for right censoring, left censoring and interval censoring, respectively. A censoring mechanism with unknown dependence on $T$ cannot simply be dropped; these are the event-time factors under independent censoring or another ignorable mechanism.

There is a useful finite-dimensional reduction before optimization. Let $A_i$ be the set in which observation $i$ tells us its event time lies. Partition the time axis using all observed endpoints, keeping endpoint singletons when required. Within each cell, membership in every $A_i$ is constant. Assign cell masses $w_j\ge 0$, with ${\sum }_j{w}_j=1$, including a tail cell when necessary, and let $A_{ij}$ be the cell-membership indicator. ThenMass locations within a cell are not identifiable and need not be separate parameters. Repeated observation sets can be grouped and their factors raised to their counts. Cells with the same membership column can also be combined. If one column is componentwise dominated by another, transferring its mass to the dominating column weakly increases every factor; hence an unconstrained maximizing distribution can omit the dominated cell. This proves likelihood support reduction under censoring directly. The remaining log-likelihood is a concave function of the probability masses on a simplex, often far smaller than the original unrestricted distribution problem. Added constraints can restrict which transfers remain permissible.

For right censoring alone, take a grid of all distinct observation times $s_j$ and initially allow a discrete hazardLet $d_j$ be the number of events at $s_j$ and $n_j=\#\{i:x_i\ge s_j\}$ the risk set immediately before that time. For tied recorded times, use the convention that events precede censoring, so those censored at $s_j$ remain in this risk set. The mass at $s_j$ is $q_j{\prod }_{k<j}(1-q_k)$, and the probability of surviving past $s_j$ is ${\prod }_{k\le j}(1-q_k)$. Substitution into the exact-event and right-censoring factors yieldsThe exponent $n_j-d_j$ counts all observed individuals whose information requires survival beyond that jump. In particular, censoring observations affect the risk set even though they give no event factor.

Each factor can now be maximized separately. For $0<d_j<n_j$, its derivative is $d_j/q_j-(n_j-d_j)/(1-q_j)$, giving ${\hat{q}}_j=d_j/n_j$; the second derivative is negative. The same result holds at the boundaries: ${\hat{q}}_j=0$ if $d_j=0$, and ${\hat{q}}_j=1$ if $d_j=n_j$. The Kaplan–Meier estimator is thereforeOnly event times contribute nontrivial factors. Residual mass after the last observation may be placed anywhere later without changing the empirical likelihood. A terminal plateau is therefore not evidence of permanent survival; the tail is unobserved.

For a constrained Kaplan–Meier estimator with $F(t)=p$, first split the support cells at $t$. In the probability-mass representation the exact optimization problem isThe last sum excludes a singleton mass at $t$, because $F(t)=\mathrm{Pr}(T>t)$. This is a concave maximization with linear equality constraints, so it directly gives a constrained nonparametric maximum-likelihood estimator. It also handles a time lying between observed events or beyond all follow-up.

An equivalent hazard formulation adds $t$ to the grid, permits zero-event jumps when needed, and imposesFor an interior solution using only positive-event jumps before $t$, add $\lambda \{{\sum }_{s_j\le t}\log (1-q_j)-\log p\}$ to the log-likelihood. Differentiating givesThe multiplier is found by solving the scalar equationwith denominators and fitted hazards in their admissible ranges. On an admissible event-only interior branch, this product is increasing in $\lambda$, making a one-dimensional root search straightforward.

A constraint can require mass at a time with no observed event. The event-only multiplier equation must not be used blindly. To include boundary cases, put $u_j=-\log (1-q_j)\ge 0$. Before $t$, the constraint becomes ${\sum }_j{u}_j=-\log p$, and each contribution isFor $d_j>0$ its multiplier equation again gives $q_j=d_j/(n_j+\lambda )$. For $d_j=0$, its derivative is $-n_j$, so the maximum has $u_j=0$ when $\lambda >-n_j$, and may have $u_j>0$ when $\lambda =-n_j$. Such a jump supplies mass needed by the constraint; an unobserved tail with $n_j=0$ is included by the same rule. For example, if every observation is right-censored after $t$ and there are no events, the ordinary estimate has $F(t)=1$. To impose $p<1$, put mass $1-p$ at $t$ and the remaining $p$ after all censoring times. The maximized likelihood is $p^n$, although there are no event-time factors at all. The mass-constrained optimization above automatically accounts for this possibility and for any remaining tail nonuniqueness.

For the Poisson process of claims and relative safety loading $\rho$, the premium rate in the classical risk model is $c=(1+\rho )\lambda \mu$. Dividing the adjustment coefficient equation $\lambda (M(r)-1)=cr$ by $\lambda$ explains why the arrival rate cancels. Interpret $r_{\infty }$ as the upper endpoint of the finite moment-generating function domain, so $M(r)<\infty$ for $0\le r<r_{\infty }$ and its stated divergence occurs at that endpoint.

Set $F(r)=M(r)-1-(1+\rho )\mu r$. ThenDifferentiation is valid inside the finite-transform domain, and its positive neighbourhood gives finite moments of every order. Thus $F$ is strictly convex and negative just to the right of zero. If $r_{\infty }<\infty$, divergence of $M(r)$ makes $F(r)\to \infty$ at the endpoint. If $r_{\infty }=\infty$, choose $a>0$ with $\mathbb{P}(X_1\ge a)>0$; then $M(r)\ge \mathbb{P}(X_1\ge a)e^{ar}$, which eventually exceeds every linear function. Again $F(r)\to \infty$. The intermediate value theorem gives a positive zero.

There is at most one positive zero: strict convexity and $F(0)=0$ make the secant slope $F(r)/r$ strictly increasing for $r>0$. It starts at $-\rho \mu$ and crosses zero exactly once. HenceThis proves the secant-slope existence criterion for an adjustment coefficient in the present setting.

Put $m_j=\mathbb{E}{X}_1^j$. For $u>0$, the positive remainder of the exponential series gives $e^u>1+u+u^2/2$. Apply it to $u=R{X}_1$ and take expectations:Dividing by $R>0$ yieldsKeeping the cubic term similarly givesThe polynomial $g(r)=m_3{r}^2+3m_{2}r-6\rho \mu$ is strictly increasing for $r\ge 0$, is negative at zero and tends to infinity. Let $r_2$ be its unique positive zero. The last inequality says $g(R)<0$, soFinally, $3m_2{r}_1=6\rho \mu$, whence $g(r_1)=m_3{r}_1^2>0$. Monotonicity gives $r_2<r_1$. Thus the requested upper bounds can in fact be sharpened to the strict ordering $0<R<r_2<r_1$. The bounds themselves need not lie in the finite-transform domain; the argument only evaluates the actual transform at $R$ and polynomial moments elsewhere. These are polynomial moment bounds for the adjustment coefficient.

For rate-$\alpha$ exponential distribution claims,The positive root satisfies $1/(\alpha -R)=(1+\rho )/\alpha$ after dividing out $R$, soThe cubic-bound equation reduces to $r_2^2+\alpha r_2-\rho {\alpha }^2=0$, selecting the displayed positive root. In particular $R<\alpha$, as required by the transform domain, for every $\rho >0$.

Write $M(t)=\mathbb{E}{e}^{tX}$, with $M(0)=1$, and differentiate $\kappa (t)=\log M(t)$:Using $M^{(j)}(0)=\mathbb{E}{X}^j$ and $\mu =\mathbb{E}X$ givesThus the first three cumulants are the mean, variance and third central moment. A finite moment-generating function near zero justifies these differentiations. For positive claims with only three finite ordinary moments, the same calculations are valid as left derivatives at zero of the logarithm of the Laplace transform of a nonnegative random variable; positive exponential moments are not needed for the resulting moment identities.

For the independent Poisson distribution count, the random-sum transform identity givesConsequently the compound Poisson distribution hasThese are raw severity moments, not severity cumulants.

The printed gamma density uses rate $\nu$, not scale. Direct integration against $e^{ty}$ givesTherefore the required matching equations for the shifted gamma distribution areFor $\lambda >0$ and nonzero positive claims with finite $m_3$, these equations give positive $\alpha ,\nu$ and the unique solutionThis is the three-cumulant shifted gamma approximation. In particular, its skewness $2/\sqrt{\alpha }$ equals $m_3/(\sqrt{\lambda }{m}_2^{3/2})$, the skewness of $S$.

The competing normal approximation to a compound Poisson aggregate isTo justify its large-count limit, let ${\phi }_X$ be the claim characteristic function. For fixed severity law with $m_2<\infty$, its expansion is ${\phi }_X(u)=1+i{m}_{1}u-m_2{u}^2/2+o(u^2)$. The characteristic function of $(S-\lambda m_1)/\sqrt{\lambda m_2}$ isThus it converges in distribution to a standard normal distribution as $\lambda \to \infty$.

The normal approximation is simple, uses only $m_1,m_2$, and works well in the centre when the count is large and standardized skewness is small. It imposes symmetry, whereas positive compound-Poisson claims have positive third cumulant. The shifted-gamma approximation reproduces that third cumulant and can better describe moderate right skewness. It requires a finite and reliably known third moment, and matching three cumulants alone does not guarantee accurate extreme tails.

Neither continuous approximation reproduces the genuine atom $\mathbb{P}(S=0)=e^{-\lambda }$, which matters at small $\lambda$. The normal distribution also assigns probability $\Phi (-\sqrt{\lambda }{m}_1/\sqrt{m_2})$ to negative losses. The gamma approximation is bounded below by $k$, but its fitted shift need not be nonnegative: exponential claims of mean $\mu$ give $k=-\lambda \mu /3$, so it too can assign mass to negative losses. If $k>0$, it instead excludes some genuinely possible small losses. These support and atom discrepancies prevent either approximation from being a universal replacement for the exact aggregate law.

Condition on the claim count, defining the empty sum to be zero. Independence gives $\mathbb{E}[S\mid N=k]=k\mathbb{E}{X}_1$. By the law of total expectation,For nonnegative claims this also follows from Tonelli theorem; for signed claims finite $\mathbb{E}N$ and $\mathbb{E}|X_1|$ ensure integrability. Conditional independence likewise givesThis random-sum transform identity holds wherever the composed series is finite, or as an extended nonnegative expectation. It is not permission to use an algebraic continuation outside the convergence domain.

For the zero-based geometric distribution, summing the geometric series givesThe exponential distribution of mean $\mu$ has $M_{X_1}(t)=(1-\mu t{)}^{-1}$ for $t<1/\mu$. ThereforeThe stricter bound $t<p/\mu$ is necessary: the geometric series requires $q{M}_{X_1}(t)<1$. At and above $p/\mu$ that series diverges; at and above $1/\mu$ even one claim has an infinite transform. Thus the displayed rational expression represents the moment-generating function only on its stated domain.

Since all claims are strictly positive, $S=0$ exactly when $N=0$, so $\mathbb{P}(S=0)=p$. Splitting the transform givesUniqueness of the moment-generating function identifies a mixture distribution with a zero atom of weight $p$ and an exponential positive component of weight $q$ and rate $p/\mu$. In particular,The density integrates to $q$, not one. One can also verify it directly: conditional on $N=k\ge 1$, the sum has Erlang distribution density $x^{k-1}{e}^{-x/\mu }/[{\mu }^k(k-1)!]$. Multiplying by $p{q}^k$ and summing over $k$ gives $(pq/\mu )e^{-x/\mu }{e}^{qx/\mu }$, the same density. This is the zero-based geometric sum of exponential variables.

Under excess of loss reinsurance, the ceded amount for each claim is $(X_j-M{)}_{+}$, and the monthly reduction is their random sum. The tail integral formula for moments givesConditioning on $N$ again yieldsThe retained payment is ${\sum }_{j=1}^N\min (X_j,M)$; it is generally not $\min (S,M)$.

Under aggregate stop loss reinsurance, the reduction is $(S-\tilde{M}{)}_{+}$. The aggregate tail is $\mathbb{P}(S>x)=q{e}^{-px/\mu }$ for $x\ge 0$, soThe prefactor is positive. Equality with the per-claim recovery is equivalent to $p\tilde{M}=M$, giving the unique reinsurance retentionThis is the relation for equal expected recoveries from excess of loss and stop loss; since $p<1$, the aggregate reinsurance retention exceeds the individual reinsurance retention. These are reductions in claim payouts, before any reinsurance premium is deducted.