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For NBUE claims in the classical risk model, the equilibrium claim size $Y$ is stochastically bounded by $X$. The positive exponential tail integral formula for moments gives $M_Y(R)\le M_X(R)$. Since $M_Y(R)=1+\theta$ and $M_X(R)=1+(1+\theta )\mu R$, rearrangement proves the displayed bound.

For a positive claim and $r>0$, the tail integral formula for moments gives the displayed expression. In particular the first moment is ${\int }_0^M\overline{F}(x)dx$ and the raw second moment is $2{\int }_0^{M}x\overline{F}(x)dx$. If the claim law has a density, capping adds an atom of a measure at $M$ with mass $\mathbb{P}(X\ge M)$.

For a nonnegative claim size $X$ of finite positive mean $\mu$, its equilibrium claim-size distribution has this density and survival function ${\overline{F}}_I(x)={\mu }^{-1}{\int }_x^{\infty }{\overline{F}}_X(t)dt$. The tail integral formula for moments normalizes the density. Its moment-generating function is $(M_X(r)-1)/(\mu r)$ wherever finite, so at an adjustment coefficient it equals one plus the safety loading.

For a positive random variable $X$ with finite nonzero expected value $\mu$, its integrated tail law has density $f_I(x)=\mathbb{P}(X>x)/\mu$ on $x\ge 0$. The tail integral formula for moments normalizes it. For $r>0$ its exponential transform is $\int e^{rx}{f}_I(x)dx=[M_X(r)-1]/(\mu r)$, with extended-value interpretation when needed.

A reinsurance layer with attachment $M$ and payment limit $A$ pays nothing below $M$, the excess between $M$ and $M+A$, and $A$ above that. Its expected recovery is ${\int }_M^{M+A}\mathbb{P}(X>x)dx$, by the tail integral formula for moments. For continuous claims, conditioning on penetration of the layer divides this expectation by $\mathbb{P}(X>M)$.

In a probability measure-preserving system, $f\in L^1$ implies $f(T^{n}x)/n\to 0$ almost everywhere. For every $\epsilon >0$, the tail integral formula for moments gives ${\sum }_{n\ge 1}\mu (|f|>\epsilon n)\le \Vert f{\Vert }_1/\epsilon$. Invariance and the Borel-Cantelli lemma finish the proof. Consequently division by $n^a$ also gives zero for every $a\ge 1$.

For increasing sigma-algebras ${\mathcal{G}}_n$, put $f_n=I_{\mu }(\xi \mid {\mathcal{G}}_n)$ and $F={\sup }_n{f}_n$. For each atom $A\in \xi$ and $t\ge 0$,Together with the bound by $\mu (A)$ and the tail integral formula for moments, this yields $\int Fd\mu \le H_{\mu }(\xi )+1$. The inequality follows by stopping the conditional-expectation martingale $\mathbb{E}[1_A\mid {\mathcal{G}}_n]$ the first time it falls below $e^{-t}$.

For a nonnegative random variable with finite expected value and $\mathbb{P}(X>t)>0$, its mean residual life is the remaining conditional expected value after surviving past $t$. The tail integral formula for moments gives $m_X(t)={\int }_t^{\infty }\mathbb{P}(X>x)dx/\mathbb{P}(X>t)$. The exponential distribution has constant mean residual life, equal to its original mean. In excess of loss reinsurance, the total variance stationary condition for excess of loss sets a candidate retention equal to this quantity.

Let $(Z_k,{\mathcal{F}}_k{)}_{0\le k\le n}$ be an integrable real submartingale, and put $Z_n^{\ast }={\max }_{0\le k\le n}{Z}_k$. The refined Doob maximal inequality isIn particular, for a nonnegative submartingale the right-hand side is at most $\mathbb{E}{Z}_n$, giving the usual weak maximal bound.

For the proof, partition $A=\{Z_n^{\ast }\ge \lambda \}$ by its first crossing time: $A_k=\{Z_j<\lambda (j<k),Z_k\ge \lambda \}\in {\mathcal{F}}_k$. The submartingale property gives $\mathbb{E}[Z_n\mid {\mathcal{F}}_k]\ge Z_k$, soSumming over the disjoint crossing events proves the refined inequality, including a possible crossing at time zero. No independence assumption is needed.

The strong form is the Doob Lp maximal inequality. For $p>1$ and a nonnegative submartingale with $Z_n\in L^p$,Indeed, integrate the refined inequality against $p{\lambda }^{p-2}$ up to $K$. The tail integral formula for moments and the Tonelli theorem giveApply the Holder inequality, divide by $\Vert Z_n^{\ast }\wedge K{\Vert }_p^{p-1}$ when it is nonzero, and let $K\to \infty$ by the monotone convergence theorem. For a martingale $M$, the process $|M_k|$ is a submartingale by the conditional Jensen inequality, giving the corresponding bound for $\max |M_k|$. On a bounded continuous-time interval the nonnegative, right-continuous version follows by applying these inequalities on refining finite time grids and taking limits.

Let $M_X(t)=\mathbb{E}{e}^{tX}$. Conditioning on $N\sim \mathrm{Pois}(\lambda )$ givesWhere the moment-generating function is finite around zero, differentiating the cumulant-generating function givesIn particular $\mathbb{E}S=\lambda \mathbb{E}X$ and $\mathrm{Var}S=\lambda \mathbb{E}{X}^2$. These compound Poisson cumulants use raw claim moments. The first two formulas also follow by conditioning: $\mathbb{E}[S\mid N]=N\mathbb{E}X$ and $\mathrm{Var}(S\mid N)=N\mathrm{Var}X$, soThis is the law of total variance. They therefore remain valid with finite first two moments even when no positive exponential moment exists.

Under per-claim excess of loss reinsurance, put $Y=\min (X,M)$ and let $W=(X-M{)}_{+}$ be the positive part of the excess. The insurer and reinsurer totals are sums of these payments over the same Poisson distribution count. Write $\overline{F}(x)=\mathbb{P}(X>x)$. Using the tail integral formula for moments on the tail probabilities of $Y$ and $W$ givesFor example, $\mathbb{E}{Y}^2={\int }_0^{\infty }\mathbb{P}(Y^2>t)dt=2{\int }_0^{M}x\overline{F}(x)dx$ after $t=x^2$, and shifting by $M$ gives the reinsurer formula. The moment formulas are finite whenever the corresponding claim moments are finite.

For the specified Pareto distribution, with shape three and minimum $d>0$,If $0\le M\le d$, every claim exceeds the reinsurance retention. ThereforeIf $M\ge d$, the tail integrals instead giveConsequently the requested sum of variances isOn the first branch, $V^{\prime }(M)=\lambda (4M-3d)$; it decreases until $M=3d/4$ and then increases. On the second branch $V^{\prime }(M)=\lambda d^3/M^2>0$. For $\lambda >0$ the minimum is unique; if $\lambda =0$, all variances are zero and every reinsurance retention minimizes them. ThusThe curve starts at $3\lambda d^2$, reaches this minimum, passes through $V(d)=2\lambda d^2$ with matching slope $\lambda d$ on both branches, and then increases concavely toward the horizontal asymptote $3\lambda d^2$. The variance-minimizing retention for shape-three Pareto claims is shown below.

The two aggregate payouts are dependent. Conditioning on the common count gives covariance $\lambda \mathrm{Cov}(Y,W)+\lambda \mathbb{E}Y\mathbb{E}W=\lambda \mathbb{E}[YW]=\lambda M\mathbb{E}W$, so $V(M)+2\mathrm{Cov}(S_I,S_R)=3\lambda d^2$ for every reinsurance retention, as required by $S_I+S_R=S$. The covariance of two components of a compound Poisson sum accounts for this distinction between a sum of variances and the variance of a sum.

There is a moment domain qualification for this heavy-tailed example. For every $t>0$, ${\int }_d^{\infty }{e}^{tx}3d^3{x}^{-4}dx=\infty$, and moments of order $j\ge 3$ also diverge. Thus its positive moment-generating function is not finite and higher finite cumulants cannot be obtained by differentiating around zero. Its mean and variance calculations above are valid by conditioning or by the first two right derivatives at zero of the Laplace transform of a nonnegative random variable.

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.

If ${\sigma }_1>n$, all the integer-time positions through $n$ lie in $(-1,1)$, so every increment over an intervening unit interval has absolute value at most two. The unit-time increments of Brownian motion are independent standard normal random variables. PutThe inclusion of events and independence giveIn particular ${\sigma }_1$ is finite almost surely. For real $t\ge 0$, its tail is at most ${\rho }^{\lfloor t\rfloor }$. The tail integral formula for moments therefore gives, for every $p>0$,An exponential geometric factor dominates the polynomial factors in this sum. This proves the requested moment finiteness for $p>1$, and in fact for every positive $p$.

In the classical risk model, write the surplus aswhere $N(t)$ is the claim-arrival Poisson process. The ruin time is $\tau =\inf \{t\ge 0:U(t)<0\}$. The ultimate ruin probability isThe positive root $R$ in the question is the adjustment coefficient. The Lundberg inequality and the Cramér–Lundberg approximation are, respectively,The interior adjustment coefficient ruin prefactor is finite here because $R$ lies strictly inside the finite domain of the moment-generating function; the claim-size density supplies the nonarithmetic condition for the asymptotic. Strict convexity makes its denominator positive. The approximation is an asymptotic statement, while the inequality is a bound at every nonnegative initial capital.

The inequality can also be seen directly from the exponential surplus martingale $e^{-RU(t)}$. Its expectation factor per unit time is zero in the exponent because $\lambda (M_X(R)-1)-cR=0$. Apply the optional stopping theorem at $\tau \wedge T$, a bounded time. On $\{\tau \le T\}$, $e^{-RU(\tau )}>1$, so $\mathbb{P}(\tau \le T)\le e^{-Ru}$. Increasing $T$ proves the bound without stopping at an unbounded time.

For any nonnegative claim size, the Tonelli theorem applied to $e^{RX}-1={\int }_0^{X}R{e}^{Rx}dx$ gives the exponential tail identityCombining it with the adjustment equation yieldsThe density $f_Y(x)=(1-F_X(x))/\mu$ integrates to one by the tail integral formula for moments. Thus $Y$ has the equilibrium claim-size distribution, and the preceding formula says $M_Y(R)=1+\theta$.

Under the given NBUE condition, the stochastic order comparison is $1-F_Y(x)\le 1-F_X(x)$ at every $x\ge 0$. Apply the same exponential tail identity to both variables to obtainConsequentlyand rearrangement proves the adjustment-coefficient lower bound for NBUE claims:This uses the specified distributional comparison, not an unwarranted assumption that an arbitrary equilibrium claim law is always smaller than its original law.

For the final example, $f_X(x)=x{e}^{-x}$ is the shape-two, rate-one gamma distribution, with $\mu =2$, survival function ${\overline{F}}_X(x)=(1+x)e^{-x}$, and moment-generating function $M_X(r)=(1-r{)}^{-2}$ for $r<1$. With $\theta =1/10$, the positive adjustment root satisfiesDividing out the zero root and simplifying gives $11R^2-17R+1=0$. Only the smaller root is in the finite-transform interval $(0,1)$, soThe equilibrium density and survival function areThus ${\overline{F}}_Y(x)\le (1+x)e^{-x}={\overline{F}}_X(x)$ for every $x\ge 0$, proving that this claim law is new better than used in expectation. Finally the lower bound equals $1/22$. Our exact root is larger because $16>7\sqrt{5}$, equivalently $256>245$. Henceverifying the requested bound for the example.

For excess of loss reinsurance the insurer pays each claim up to its retention level:The cap applies separately to every claim. In particular the retained annual loss is ${\sum }_j\min (X_j,M)$, rather than a single cap on the annual total.

Let $F_i$ be the cumulative distribution function for the claim size on risk $i$, and put ${\overline{F}}_i=1-F_i$. The retained severity on that risk has the original probability density function on $0<y<M$ and an atom of a measure at $M$ of mass ${\overline{F}}_i(M)$. Thus $T_I$ has a compound Poisson distribution with rate ${\lambda }_1+{\lambda }_2$ and the mixture of these capped severity laws. The mixture's mass at $M$ is ${\sum }_i{\lambda }_i{\overline{F}}_i(M)/({\lambda }_1+{\lambda }_2)$.

For the capped claim moments, use the tail integral formula for moments. Since $\mathbb{P}(\min (X_i,M)>x)={\overline{F}}_i(x)$ for $0\le x<M$ and is zero for $x\ge M$,Substitution into the compound Poisson distribution moment formulas givesEquivalently, the integrals are ${\int }_0^{M}x{f}_i(x)dx+M{\overline{F}}_i(M)$ and ${\int }_0^M{x}^2{f}_i(x)dx+M^2{\overline{F}}_i(M)$. The annual variance uses the retained raw second moments; subtracting their squared means would omit the variation in the Poisson distribution count.

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.

For a nonnegative random variable, $T={\int }_0^{\infty }1\{T>t\}dt$. Tonelli theorem permits interchanging expectation and this nonnegative integral, giving the tail integral formula for moments:The equality also holds when the mean is infinite. Thus mean completion time is the total area under the survival function, while a restricted mean survival time integrates only to a specified follow-up horizon.

The expected value and variance under quota share reinsurance follow by scaling the exponential distribution:The retained stop loss moments for an exponential aggregate follow from the payout $Y=\min (S,M)$, its survival function equals $e^{-x/{\mu }_S}$ for $0\le x<M$ and zero for $x\ge M$. The tail integral formula for moments gives, with $m=M/{\mu }_S$ and $r=e^{-m}$,Consequently the retained moments areMatching the two expected values forces $\alpha =1-r$, which lies strictly between zero and one. The difference of the variances simplifies toIndeed $h(m)=m-1+e^{-m}$ has $h(0)=0$ and $h^{\prime }(m)=1-e^{-m}\ge 0$ for $m\ge 0$. Because $M>0$, the difference is actually positive. At equal retained expected value, aggregate stop loss reinsurance reduces the variance more than quota share reinsurance.

Use $c_{\mathrm{prem}}$ for the premium income rate, reserving $c$ later for the smaller exponential decay rate. In the classical risk model, the surplus isHere $\rho$ is the relative safety loading. The aggregate claims form a Compound Poisson process. Define $L_t={\sum }_{j=1}^{N_t}{X}_j-c_{\mathrm{prem}}t$, so ruin occurs when $L_t>u$. By independent increments and the exponential formula for a marked Poisson sum,Thus $Z_t=e^{R{L}_t}$ is a nonnegative continuous-time martingale with $Z_0=1$, because the adjustment coefficient makes the exponent vanish.

Let $\tau =\inf \{t:U_t<0\}$. Apply the optional stopping theorem at the bounded stopping time $\tau \wedge T$. On $\{\tau \le T\}$, $L_{\tau }>u$, soLetting $T$ increase proves the Lundberg inequality and the unscaled limit:

For the precise asymptotic, putThe given exponential integral identity makes $k$ a probability density. Multiplying the given defective renewal equation by $e^{Ru}$ turns it into the ordinary renewal equationFor clarity, the version of the key renewal theorem used here is: if the interarrival law is nonarithmetic, has mean $m_k\in (0,\infty )$, and $g$ is directly Riemann integrable, the locally bounded solution of this renewal equation satisfies $z(u)\to m_k^{-1}{\int }_0^{\infty }g(v)dv$. The infinite-mean version gives zero for nonnegative directly Riemann integrable $g$.

All the hypotheses can be checked here. The density $k$ gives a nonarithmetic distribution. The Tonelli theorem givesFurthermoreThus $g$ is continuous and integrable, and ${\int }_0^{\infty }|g^{\prime }(u)|du\le R\int g+1<\infty$. On a mesh of width $\delta$, the difference between its upper and lower sums is at most $\delta \int |g^{\prime }|$; its upper sum is at most $\int g+\delta \int |g^{\prime }|$. This proves direct Riemann integrability rather than assuming it. Also $0\le z(u)\le 1$ by the Lundberg inequality, so the solution is locally bounded. Its renewal representation is $z=g\ast U_k$, where $U_k={\sum }_{j\ge 0}{K}^{\ast j}$ and $K(dx)=k(x)dx$; the residual after iteration tends to zero on compact intervals because sums of positive interarrivals tend to infinity.

Writing $J={\int }_0^{\infty }x{e}^{Rx}{f}_I(x)dx$, the tilted interarrival expected value is $m_k=hJ$. The key renewal theorem gives the Cramér–Lundberg ruin asymptoticIf $J=\infty$, the same formula is interpreted as $A=0$. A positive finite asymptotic constant requires $J<\infty$; this extra integrability is not explicitly stated in the paper.

For the final two-exponential case, evaluate the defective renewal equation at zero:One can identify the adjustment coefficient without silently assuming $A>0$. For $0<r<c$, setIt is finite and positive. Integrating the nonnegative terms of the defective renewal equation, using the Tonelli theorem, first shows that $F_I(r)={\int }_0^{\infty }{e}^{rx}{f}_I(x)dx$ is finite and then givesAs $r\uparrow c$, $P(r)\to \infty$ because $a>0$. By monotone convergence theorem, $h{F}_I(c)=1$. The integrated tail distribution in the classical risk model has density $f_I(x)=\mathbb{P}(X_1>x)/\mu$, so $F_I(r)=[M(r)-1]/(\mu r)$ by the tail integral formula for moments. Hence $c$ solves the adjustment equation, and its stipulated uniqueness implies $R=c$. Finally the displayed form of $\psi$ gives the remaining constantsIn particular the decay exponent $d$ and the coefficient $b$ do not affect $R$ or $A$. The $c$ in these final answers is the printed decay rate, not $c_{\mathrm{prem}}$.