Classical risk model ($U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$)

The classical risk model has surplus $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, where $N_t$ is a Poisson process of rate $\lambda$, independent claim sizes are positive, and premiums flow in at constant rate $c$. The relative safety loading is $\rho =c/(\lambda \mathbb{E}{X}_1)-1$. Ruin is the first time surplus becomes negative.

In a classical risk model, $T_j$ is the time of the $j$th claim. The displayed probability restricts ruin to the first $n$ claims and increases to the ultimate ruin probability. Conditioning on the first arrival and severity gives ${\psi }_{n+1}(u)=\mathbb{E}[{\overline{F}}_X(u+c{T}_1)+{\int }_0^{u+c{T}_1}{\psi }_n(u+c{T}_1-x)f_X(x)dx]$, with ${\psi }_0=0$. For exponential distribution claims of mean $\mu$ and relative safety loading $\rho$, put $d=1+\rho$ and $h=d+1$. Then ${\psi }_1(u)=e^{-u/\mu }/h$ and ${\psi }_2(u)=e^{-u/\mu }(h^{-1}+(u/\mu )h^{-2}+d{h}^{-3})$. The recursion follows because a surviving first claim leaves fresh independent future arrivals and claims. The formulas follow by integrating the rate-$\lambda$ first-arrival density; the claim-arrival rate cancels with $c=d\lambda \mu$.

Starting from capital $u$, ruin occurs exactly when the aggregate loss exceeds $u$. Positive relative safety loading and the strong law of large numbers make $S(t)-ct$ tend to minus infinity, so its supremum is finite almost surely. For exponential distribution claims of rate $\alpha$, $L$ has an atom $1-\lambda /(\alpha c)$ at zero, and, conditional on $L>0$, has exponential distribution of rate $\alpha -\lambda /c$. The ultimate ruin probability determines this mixture directly.

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.

A nonnegative lifetime or claim law is new better than used in expectation if its mean residual life at every age with positive survival is at most its original mean. The displayed equivalent inequality says that its equilibrium claim-size distribution is no larger in stochastic order than the original law. For an Erlang distribution of shape two and rate one, the two survival functions are $(1+x/2)e^{-x}$ and $(1+x)e^{-x}$, so the condition holds.

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.

The probability that the insurance surplus in a classical risk model becomes negative at some finite time, starting from capital $u\ge 0$. Ultimate survival has probability $1-\psi (u)$. A finite-horizon ruin probability instead restricts the crossing time to a specified time interval.

For rate-$\alpha$ exponential distribution claims and $c>\lambda /\alpha$, the ruin integro-differential equation reduces to ${\psi }^{\prime \prime }+(\alpha -\lambda /c){\psi }^{\prime }=0$. Differentiate the convolution and use its equation $A^{\prime }=\alpha \psi -\alpha A$ to eliminate it. Positive relative safety loading implies $\psi (u)\to 0$ as $u\to \infty$, and the zero-capital survival probability gives $\psi (0)=\lambda /(\alpha c)$. These two conditions determine the displayed solution. The adjustment coefficient is $\alpha -\lambda /c$.

In the classical risk model, condition on the first arrival of the Poisson process. If available capital just before it is $v$, ruin has probability ${\overline{F}}_X(v)+{\int }_0^v{f}_X(x)\psi (v-x)dx$. Integrate this expression at $v=u+ct$ against $\lambda e^{-\lambda t}$ and change variables from $t$ to $v$. Differentiating the lower limit gives the displayed equation. The claim-density convolution is continuous for bounded $\psi$, so the first-arrival representation also supplies the differentiability required by the equation.

In the classical risk model with density $3e^{-4x}+e^{-2x}/2$ and relative safety loading $3/5$, the claim mean is $5/16$ and $\lambda /c=2$. The convolution terms $A_a(u)={\int }_0^u\varphi (u-x)e^{-ax}dx$ satisfy $A_a^{\prime }=\varphi -a{A}_a$. Applying $(D+4)(D+2)$ to ${\varphi }^{\prime }=2\varphi -6A_4-A_2$ yields ${\varphi }^{(3)}+4{\varphi }^{\prime \prime }+3{\varphi }^{\prime }=0$. Its solution is fixed by $\varphi (\infty )=1$, $\varphi (0)=3/8$ and ${\varphi }^{\prime }(0)=3/4$, giving the displayed probability. The derivative condition comes from the original integral equation and excludes spurious solutions of the higher-order equation.

For positive claims with finite variance and $c\le \lambda \mu$, the classical risk model has ruin probability one from any finite capital. At claim times, surplus increments are independent copies of $cT-X$ with mean $c/\lambda -\mu$. Negative mean sends their partial sums to minus infinity by the strong law of large numbers. At zero mean, the increments have finite nonzero variance. For every fixed $K$, the central limit theorem gives limiting probability $1/2$ of a partial sum below $-K$. The probability of unboundedness below is therefore at least $1/2$; as a tail event it has probability zero or one by the Kolmogorov zero-one law, and hence one.

Independent positively loaded portfolios with zero initial capitals have separate joint survival probability ${\prod }_i(1-{\lambda }_i{\mu }_i/c_i)$. Pool their claims and premium incomes to obtain zero-capital merged survival $1-({\sum }_i{\lambda }_i{\mu }_i)/({\sum }_i{c}_i)$. This is the premium-weighted average of the individual survival probabilities and is at least their product. The event of aggregate solvency permits transfers of surplus between the original portfolios.

The ultimate survival probability satisfies $\phi (u)=\phi (0)+(\lambda /c){\int }_0^u\phi (u-x)(1-F(x))dx$. Condition on the first arrival of the Poisson process, differentiate the resulting exponentially weighted integral, and integrate the convolution derivative equation from zero. Tonelli theorem converts the claim-density convolution to the tail convolution. The kernel mass is $\lambda \mu /c<1$, so this is a defective renewal equation.

In the classical risk model with premium rate $c>0$ and claim Poisson process rate $\lambda$, the first-claim decomposition for survival probability implies the displayed equation for ultimate survival probability. Change the first-claim integral to an integral over available capital and differentiate its lower limit. The bounded survival probability function convolved with the integrable claim probability density function is continuous, so this also establishes the needed differentiability. In particular ${\phi }^{\prime }(0)=(\lambda /c)\phi (0)$.

For an Erlang distribution claim density $f(x)={\beta }^{2}x{e}^{-\beta x}$ and $r=\lambda /c$, write $H=\varphi \ast f$. The survival integro-differential equation gives ${\varphi }^{\prime }=r(\varphi -H)$. The convolution satisfies $(D+\beta {)}^{2}H={\beta }^2\varphi$, because $H={\beta }^2{e}^{-\beta u}{\int }_0^u(u-t)\varphi (t)e^{\beta t}dt$. Applying $(D+\beta {)}^2$ to the first equation eliminates $H$ and gives the displayed third-order equation. The original integral equation also supplies ${\varphi }^{\prime }(0)=r\varphi (0)$ and ${\varphi }^{\prime \prime }(0)=r^2\varphi (0)$; these conditions must not be lost when solving the higher-order equation.

For a mixture distribution of exponential distributions, introduce one convolution state $A_j(u)={\int }_0^u\phi (z)e^{-{\beta }_j(u-z)}dz$ for each rate ${\beta }_j$. It satisfies $A_j^{\prime }=\phi -{\beta }_j{A}_j$. Together with the survival integro-differential equation these form a constant-coefficient first-order system; applying the differential operators $D+{\beta }_j$ eliminates the convolution states. For equal mixing weights at rates one and one half, with $r=\lambda /c$, the result is ${\phi }^{\prime \prime \prime }+(3/2-r){\phi }^{\prime \prime }+(1/2-3r/4){\phi }^{\prime }=0$. The original integral equation supplies initial conditions lost in elimination.

In a classical risk model, before the first claim at time $t$, available capital is $u+ct$. A claim of size $x\le u+ct$ leaves future survival probability $\phi (u+ct-x)$ by the Markov property. Integrating over the independent first-arrival exponential distribution and claim density gives $\phi (u)={\int }_0^{\infty }\lambda e^{-\lambda t}{\int }_0^{u+ct}\phi (u+ct-x)f(x)dxdt$.

In a classical risk model with positive relative safety loading, zero initial capital has ultimate survival probability $1-\lambda \mu /c$, where $\lambda$ is the claim-arrival rate, $\mu$ the claim expected value and $c$ the premium rate. It depends on the claim law through its mean alone. It is not the event of never receiving a claim: premium accumulates between arrivals.

The adjustment coefficient is a positive root of $\lambda (M_X(r)-1)=cr$ in the classical risk model. When the moment-generating function is finite at $R$, the process $\exp (R[{\sum }_{j=1}^{N_t}{X}_j-ct])$ is a continuous-time martingale. It yields the Lundberg inequality and, under the relevant tilted integrability, the Cramér–Lundberg ruin asymptotic.

The gamma distribution of shape one half and rate $\beta$ has moment-generating function $(1-r/\beta {)}^{-1/2}$. With mean $\mu =1/(2\beta )$ and positive relative safety loading $\rho$, write $x=R/\beta$. Dividing out the zero root after squaring the adjustment equation gives $(1+\rho {)}^2{x}^2-(1+\rho )(\rho -3)x-4\rho =0$. Its unique positive root lies between zero and one, so it belongs to the transform domain and gives the display.

Replacing a shape-one-half gamma claim law by an exponential distribution with the same mean overestimates the adjustment coefficient. Indeed $R_I-R=\beta [3(1+\rho )-\sqrt{(\rho +9)(\rho +1)}]/[2(1+\rho )]>0$ because the difference of the squared terms is $8\rho (1+\rho )$. The resulting putative Lundberg inequality bound $e^{-R_{I}u}$ is smaller than the valid true-model bound $e^{-Ru}$ for $u>0$; the former is not justified for the true claim law. The true variance is twice the assumed variance, and its exponential tail decays more slowly.

For positive claims with a finite positive moment-generating function neighbourhood, use $e^u>1+u+u^2/2$ and $e^u>1+u+u^2/2+u^3/6$ for $u>0$ in the defining adjustment coefficient equation. They imply the displayed strict upper bounds, where $r_2$ is the positive quadratic root. Its defining polynomial is negative at zero and positive at $r_1$, proving $r_2<r_1$. Higher exponential-series truncations similarly yield upper bounds from more raw moments.

Fix positive claim expected value $\mu$ and relative safety loading $\theta >0$ in a classical risk model. Assume the random claim law has an adjustment coefficient $R_X>0$. By Jensen inequality, $M_X(r)\ge e^{\mu r}$. Thus $e^{\mu R_X}-1-(1+\theta )\mu R_X\le 0$. The expression on the left is strictly convex, has negative derivative at zero and tends to infinity, so its unique positive zero $R_{\mu }$ satisfies $R_X\le R_{\mu }$. If the claim size is not almost surely constant, strict Jensen inequality gives $R_X<R_{\mu }$. Consequently the Lundberg inequality upper bound $e^{-R_{\mu }u}$ for deterministic claims is smaller for $u>0$. Comparing these upper bounds alone does not establish an ordering of actual ultimate ruin probabilities.

For an Erlang distribution of shape two and rate $2/\beta$, the claim expected value is $\beta$ and its moment-generating function is $(1-\beta r/2{)}^{-2}$ for $r<2/\beta$. The adjustment coefficient equation is $\lambda [(1-\beta R/2{)}^{-2}-1]=cR$. With $z=\beta R/2$, divide out the zero root to get $k(1-z{)}^2=2-z$. The smaller quadratic root lies in $(0,1)$ when $c>\lambda \beta$; the larger root lies beyond the finite-transform domain and must be rejected. This proves the displayed coefficient.

If each claim includes an independent expense $A$, replace its payment law by the convolution of independent random variables $X+A$. Its moment-generating function is $M_X(r)M_A(r)$ and its expected value is $\mathbb{E}X+\mathbb{E}A$. Keeping the relative safety loading $\theta$ fixed therefore changes the premium rate as well. The new coefficient solves $M_X(R)M_A(R)-1=(1+\theta )(\mathbb{E}X+\mathbb{E}A)R$, within the common finite-transform domain.

For positive claims with finite nonzero expected value $\mu$, the function $(M_X(r)-1)/r$ is continuous and strictly increasing on the positive finite-transform domain, starting at $\mu$. If $M_X$ diverges at a finite upper endpoint, or is finite for all positive arguments, the secant slope tends to infinity. In the latter case use $M_X(r)\ge \mathbb{P}(X\ge a)e^{ar}$ for some $a>0$ of positive tail probability. Every target $(1+\theta )\mu$ with positive relative safety loading therefore has exactly one positive root.

For a classical risk model with adjustment coefficient $R>0$, the ultimate ruin probability from capital $u\ge 0$ satisfies $\psi (u)\le e^{-Ru}$. Stop the exponential continuous-time martingale at ruin or a finite horizon, bound its value on the ruin event, and then increase the horizon.

In a classical risk model, an adjustment coefficient $R$ makes $\lambda (M_X(R)-1)-cR=0$. The random-sum transform identity and independent increments show that $\mathbb{E}[Z_t\mid {\mathcal{F}}_s]=Z_s$. This nonnegative martingale starts at one. At the ruin stopping time $\tau$, $Z_{\tau }>e^{Ru}$. Applying the optional stopping theorem only to $\tau \wedge t$ yields $\mathbb{P}(\tau \le t)\le e^{-Ru}$; passage to increasing horizons proves the Lundberg inequality without an unjustified expectation identity at an unbounded stopping time.

In the classical risk model with positive relative safety loading $\rho$ and adjustment coefficient $R$, tilting the ruin defective renewal equation gives a proper renewal equation. The key renewal theorem yields $e^{Ru}\psi (u)\to \rho /[R{\int }_0^{\infty }x{e}^{Rx}{f}_I(x)dx]$. The constant is positive if the denominator is finite and zero if it is infinite; the claim-size density provides the nonarithmetic hypothesis.

If the Cramér–Lundberg ruin asymptotic has a finite positive constant, compare it with the leading nonzero term of a finite exponential representation. An adjustment coefficient smaller than that term's decay rate would give limit zero; a larger coefficient would give infinity. Thus the two rates agree and the prefactor is the leading coefficient. Terms with zero coefficient are absent and do not determine a decay rate.

When the adjustment coefficient lies inside the finite domain of the claim moment-generating function, the tilted kernel has finite mean $(\lambda M^{\prime }(R)-c)/(cR)$. The forcing integral is $(c-\lambda \mu )/(cR)$. Their ratio is the constant $C$ in $\psi (u)\sim C{e}^{-Ru}$, by the key renewal theorem. An extra exponential moment controls the forcing tails and proves direct Riemann integrability.

For unit-rate Erlang distribution of shape two claims put $q=\lambda /c<1/2$. The adjustment coefficient satisfies $q(2-R)=(1-R{)}^2$, so $1-R=(q+\sqrt{q^2+4q})/2$. In the interior-adjustment prefactor $(1-2q)/(2q(1-R{)}^{-3}-1)$, substituting $q=(1-R{)}^2/(2-R)$ gives the displayed expression. It is positive for $0<R<1$ and tends to one as the loading tends to zero.

If the expected claim outflow per unit time is $\lambda \mu$, the relative safety loading is $\rho =c/(\lambda \mu )-1$. Positive loading means expected premium income exceeds expected claim outflow. This is the net profit condition in the classical risk model.