Actuarial statistics

Actuarial statistics uses probability theory and statistical inference to quantify insurance claims, premiums, and the uncertainty of future losses. Aggregate claims model describes portfolios, reinsurance redistributes losses, and the Bühlmann model combines experience with population information.

A policyholder's future insurance premiums depend on a finite discount level, updated according to whether a claim is submitted in each year. With independent yearly accidents and repair costs, and a claim decision depending only on the current level and cost, the level is a Discrete-time Markov chain. Accident probability and submitted-claim probability differ when the policyholder pays small repair costs without claiming.

In a no claims discount system with discounts $0,\alpha ,\beta$, a reported claim resets next year's discount to zero. With no later accidents, the next two premiums after reporting are $C+(1-\alpha )C$. At the zero level, not reporting instead gives $(1-\alpha )C+(1-\beta )C$; at either other level it gives $2(1-\beta )C$. Their differences give the displayed repair-cost thresholds. A fully reimbursed loss above its threshold should be reported under this undiscounted two-year rule. Current premiums cancel. Different thresholds generally produce state-dependent submitted-claim probabilities, even with a common accident mechanism.

Let a policyholder have one fixed exponential distribution intensity $\Lambda$ of rate $\nu$, with conditionally independent annual Poisson distribution counts. Its one-year marginal count is geometric, but averaging that count does not make the discount process a Markov chain. In the three-level rule, start at zero discount. At time three, the histories $0,\alpha ,\beta ,\beta$ and $0,0,\alpha ,\beta$ both end at the same top level. The first means three no-claim years, giving next no-claim probability $(\nu +3)/(\nu +4)$. The second means a positive count followed by two zero counts, giving $(\nu +2)/(\nu +4)$: divide the integrals of $(1-e^{-\lambda })e^{-(\nu +3)\lambda }$ and $(1-e^{-\lambda })e^{-(\nu +2)\lambda }$. These are distinct, so the Markov property fails when only the current discount is recorded. Conditioning on the intensity, or recording its Bayesian posterior, restores the required predictive information.

Suppose annual claim counts are independent with geometric distribution mass $p{q}^j$, $j\ge 0$. In a no claims discount system with discounts $0,\alpha ,\beta$, a claim returns either of the first two levels to zero; at the top, one claim sends the level to $\alpha$ and two or more send it to zero. The transition matrix is $\left(\begin{array}{ccc}q& p& 0\\ q& 0& p\\ q^2& pq& p\end{array}\right)$. Solving $\pi P=\pi$ gives the displayed stationary distribution. With full premium $c$, the stationary expected value of the premium is $c[1-(\alpha pq+\beta p^2)/(1-p^{2}q)]$. Independent annual sampling is essential; a persistent random intensity does not justify this marginal transition matrix.

For a three-level birth-death chain whose down-or-stay claim probabilities are $p_1,p_2,p_3$, with each in $(0,1)$, detailed balance is ${\pi }_1(1-p_1)={\pi }_2{p}_2$ and ${\pi }_2(1-p_2)={\pi }_3{p}_3$. Normalizing gives the displayed stationary distribution. The finite chain is irreducible and has endpoint holding probabilities, so it is aperiodic and its level proportions converge to this distribution from any initial level.

Let $b$ be the undiscounted annual premium and $d_i$ the discount fraction at level $i$. Compare the next two levels $i_r^0$ after no current claim with $i_r^1$ after a current claim, assuming no later accidents. Their total additional premium is the displayed threshold. A repair cost above it makes a claim advantageous under this two-year rule. For discounts $0,\alpha ,\beta$ and one-level moves with reflecting endpoints, the thresholds at levels one, two, and three are respectively $b\beta$, $b(2\beta -\alpha )$, and $b(\beta -\alpha )$. The premium scale must be specified or fixed as the monetary unit.

For a no claims discount system with discounts $0,\alpha ,\beta$, assume a policyholder starts at $\beta$, the first claim lowers next year's discount to $\alpha$, and a second claim in the same year lowers it to zero. With no further losses and undiscounted future premiums, claiming for the first loss costs $c(\beta -\alpha )$ in extra premium. Once that claim is submitted, the second claim adds $c\alpha$ next year and $c(\beta -\alpha )$ the year after, a total $c\beta$. Thus the second decision uses the incremental threshold, not the cost of both claims relative to no claim. For independent unit-rate exponential distribution losses the probabilities are $e^{-c(\beta -\alpha )}$ and $e^{-c\beta }$, respectively; the latter remains the conditional probability given submission of the first claim.

The expected value premium principle charges the expected value of the covered loss plus a proportional margin. For a nonnegative random variable $L$ with finite expected value, and a fixed dimensionless loading $\theta \ge 0$, it setsHere $\theta =0$ gives the expected claim cost alone; $\theta >0$ gives a positive margin when $\mathbb{E}[L]>0$. The loading is a fraction of the expected value, not of the largest possible claim size. Specify the covered period and use a consistent monetary valuation date. The rule itself does not determine $\theta$. ^[ref].

Consider an explicitly toy one-year policy: it pays £1,000 with probability $p=0.1$, and zero otherwise. Write $L=1000B$, where $B$ has the Bernoulli distribution with parameter $p$. The expected value is $1000(0.1)=\pounds 100$. A loading $\theta =0.2$ adds £20, making the premium £120. Neither the £100 expected value nor the £120 premium is a possible realized claim: the actual payment is £0 or £1,000. The first diagram separates these quantities. All numerical examples and both diagrams here are exact consequences of invented assumptions, not measured insurance data.

Why pay £120 for a loss whose expected value is only £100? Conditional on the insurer meeting its obligation, the policyholder replaces an uncertain £1,000 bill with a known £120 payment. The insurer can combine many such policies, so a few claims are funded by premiums from the whole pool. Winkel develops this motivation and the importance of independence in §15.2 of ^[ref].

To see the pooling mechanism precisely, suppose there are $n$ toy policies with mutually independent Bernoulli random variables $B_1,\dots ,B_n$, each of parameter $0.1$. Every claim size is £1,000, with at most one claim per policy. The aggregate claims model isUsing linearity of expectation and addition of variances for independent random variables, the expected value and standard deviation of this binomial distribution aggregate areThus 100 policies have expected aggregate claims £10,000 and aggregate standard deviation £3,000; 10,000 policies have expected aggregate claims £1,000,000 and aggregate standard deviation £30,000. Total uncertainty grows, but much more slowly than the expected value. Per policy, $S_n/n$ has expected value £100 and standard deviation $\pounds 300/\sqrt{n}$. Equivalently, the aggregate coefficient of variation, its standard deviation divided by its positive expected value, isIt falls from $300\%$ for one policy to $30\%$ for 100 policies and $3\%$ for 10,000. This is the useful beauty of independent pooling: an erratic individual cost becomes a comparatively stable average cost. The law of large numbers describes the corresponding convergence of the average toward its expected value.

At the same fixed loading $\theta =0.2$, total premium is $\Pi (S_n)=120n$, with expected surplus before other outgoings $120n-100n=20n$. The loading margin divided by the aggregate standard deviation is $20n/(300\sqrt{n})=\sqrt{n}/15$. This ratio rises with $n$ under our independence assumption; it is not a specified loss probability. For $n=100$, premiums total £12,000, but 13 claims cost £13,000. The one-year probability that claims exceed collected premiums is exactlyA positive loading therefore leaves about a $19.8\%$ chance of a one-year claims deficit in this toy pool. A claims deficit need not mean insolvency if capital is available. This calculation assumes premiums are collected before claims and ignores expenses and investment income; it is not an ultimate ruin probability.

The expected value premium principle also has a revealing limitation. With a common fixed $\theta$, linearity of expectation gives $\Pi (L_1+L_2)=\Pi (L_1)+\Pi (L_2)$ even without independence. Pooling therefore does not automatically reduce the premium per policy under this rule, and two losses with the same expected value receive the same premium regardless of their variance or large-loss probabilities. Any decision to change the loading after pooling requires information beyond the formula.

Dependence can remove the stability benefit. For arbitrary square-integrable losses, the aggregate variance includes their covariances:In the extreme toy common-shock model, every policy has $L_i=1000B$ for the same Bernoulli random variable $B$. Each policy still has expected value £100, so total premium is still $120n$, but $S_n=1000nB$ has standard deviation $300n$ and coefficient of variation $3$ for every $n$. With probability $0.1$, all policies claim together and losses exceed premiums by $880n$. The second diagram compares this model with independent pooling. Real shared events can invalidate independence; Winkel uses flood claims as an example in §15.2 of Actuarial Science II. The common-shock curve is an illustrative limiting model, not a fit to flood data.

The earlier $n^{-1/2}$ relative-variability calculation needs finite variance and the stated model of independent and identically distributed random variables. A finite expected value alone suffices to define the premium but does not imply finite variance. The rule makes no promise of solvency: losses, available capital and the timing of payments must be modelled separately. In the classical risk model, claim arrival rate $\lambda$ and mean claim size $\mu$ give premium rate $c=(1+\theta )\lambda \mu$; relative safety loading is then $\theta$. That model tracks capital through time and asks for ultimate ruin probability, a different question from calculating an expected margin. ^[ref]. Under quota share reinsurance, the reinsurer can apply the expected value premium principle to its ceded fraction $(1-\alpha )S$ using its own loading, which need not equal the direct insurer's loading.

A credibility estimate combines a risk's individual experience with population information. In the basic equal-exposure form, the credibility factor $Z$ weights the observed sample mean, while $1-Z$ weights a prior or collective expected value. The Bühlmann credibility premium is the best affine estimate under mean squared error; an exact Bayesian posterior mean can also have this form, as in the natural conjugate credibility identity.

Bayesian credibility estimates a risk's conditional mean by its Bayesian posterior expected value under squared-error loss. When that posterior mean is an affine combination of exposure-weighted individual experience and a prior mean, the coefficient of the experience is its credibility factor. Exact Bayesian credibility is distinct from merely selecting the best affine estimator when the true posterior mean is nonlinear.

Conditionally independent normal distribution observations of mean $\theta$ and known variance ${\sigma }_1^2$, with normal prior mean $\mu$ and variance ${\sigma }_2^2$, give normal-normal conjugacy with known observation variance. Their posterior mean has the displayed exact credibility estimate form. It minimizes posterior squared-error loss and predicts the conditional mean of the next observation.

For conditionally independent annual averages $X_j\mid \theta \sim N(\theta ,v/m_j)$ and prior $\theta \sim N(\mu ,{\sigma }^2)$, the posterior mean is $Z{\overline{X}}_w+(1-Z)\mu$, with ${\overline{X}}_w={\sum }_j{m}_j{X}_j/W$. Completing the quadratic likelihood gives posterior precision ${\sigma }^{-2}+W/v$. This is an exact Bayesian credibility version of the Bühlmann–Straub credibility estimate. Increasing the between-risk prior variance increases the experience weight; increasing process variance decreases it. The next total pure premium is $m_{n+1}$ times the estimated per-life mean.

For a common gamma distribution (shape $\alpha$, rate $\beta$) frequency prior distribution and conditionally independent Poisson distribution yearly counts of means $w_j\Theta$, observed average amounts $y_j$ determine counts $k_j=w_j{y}_j/c_j$. The posterior distribution shape is $\alpha +{\sum }_j{k}_j$ and rate is $\beta +{\sum }_j{w}_j$. The future per-policy mean is the display, equal to credibility factor $W/(W+\beta )$ times ${\sum }_j(w_j/W)(c_{n+1}/c_j)y_j$ plus the complementary weight times $c_{n+1}\alpha /\beta$. The data must lie on their count lattice; impossible noninteger recovered counts have zero likelihood.

// Destination: probability-theory.bigb

For a common success chance with beta prior and conditionally independent binomial observations of sizes $m_i$, put $M={\sum }_i{m}_i$ and $s={\sum }_i{x}_i$. Beta-binomial conjugacy gives posterior mean $(\alpha +s)/(\alpha +\beta +M)$, equal to $Z(s/M)+(1-Z)\alpha /(\alpha +\beta )$. Multiplying by the next exposure predicts its expected count. The factor tends to one as exposure increases and to zero as prior concentration increases at fixed prior mean.

Given a latent risk parameter $\Theta$, observations are independent and identically distributed random variables with conditional expectation $m(\Theta )$ and conditional variance $v(\Theta )$. The Bühlmann credibility premium estimates $m(\Theta )$ by an affine function of past observations under mean squared error. Its population parameters are $m=\mathbb{E}m(\Theta )$, expected process variance $v=\mathbb{E}v(\Theta )$, and variance of hypothetical means $a=\mathrm{Var}(m(\Theta ))$.

Conditional on a common latent variable $\Theta$, annual averages $X_j$ are independent, have common conditional expectation $\mu (\Theta )$, and have conditional variance ${\sigma }^2(\Theta )/m_j$ for known positive exposures $m_j$. Its structural parameters are $m_0=\mathbb{E}\mu (\Theta )$, expected process variance $v=\mathbb{E}{\sigma }^2(\Theta )$ and variance of hypothetical means $a=\mathrm{Var}(\mu (\Theta ))$. It generalizes the equal-exposure Bühlmann model.

For total exposure $W={\sum }_j{m}_j$ and weighted average ${\overline{X}}_w={\sum }_j{m}_j{X}_j/W$, the best affine mean squared error estimate of the conditional mean is $Z{\overline{X}}_w+(1-Z)m_0$, where $Z=Wa/(Wa+v)$. If the coefficients sum to $B$, their error is $a(1-B{)}^2+v{\sum }_j{b}_j^2/m_j$. Cauchy-Schwarz inequality minimizes the latter term at $b_j=B{m}_j/W$, and a one-dimensional quadratic minimization then gives $B=Z$.

With a Poisson-gamma conjugacy with unequal exposures model, $m_0=v=\alpha /\beta$ and $a=\alpha /{\beta }^2$, so the Bühlmann–Straub credibility factor is $W/(W+\beta )$. The credibility estimate simplifies to $(\alpha +{\sum }_j{Y}_j)/(\beta +W)$, exactly the posterior mean and hence the Bayes estimator under squared error loss. The agreement is exact because that posterior mean is already affine in the exposure-weighted data.

The experience weight is $Z=Wa/(Wa+v)$, where $W$ is total observed exposure, $a$ the variance of hypothetical means, and $v$ the expected process variance. With positive $a$, it is $W/(W+v/a)$. More exposure raises the weight; larger process noise lowers it. If $a=0$, the unknown conditional mean is constant and one can set $Z=0$.

The Bühlmann credibility premium is the best affine mean squared error estimate of a risk’s conditional claim expected value. In the Bühlmann model it is $Z\overline{X}+(1-Z)m$, where $Z=na/(na+v)$. The linear least-squares projection equations use $\mathrm{Var}(X_j)=a+v$ and $\mathrm{Cov}(X_i,X_j)=a$ for distinct years.

For independent comparable risks observed over $n\ge 2$ years, estimate the expected process variance $v$ by the average within-risk sample variance. The sample variance $B$ of the risk means estimates $a+v/n$, where $a$ is the variance of hypothetical means. Thus $B-\hat{v}/n$ is unbiased for $a$; truncating at zero produces a practical nonnegative estimate but introduces bias. Use the grand mean for $m$ and plug these estimates into the Bühlmann credibility premium $\hat{Z}{\overline{X}}_s+(1-\hat{Z})\hat{m}$. At least two risks are needed for $B$; a single year cannot estimate within-risk variance this way.

Let a fixed policy intensity have prior density $3{\theta }^{-4}$ on $\theta >1$, and let $n$ annual claim counts be conditionally independent with Poisson distribution mean $\theta$. The expected process variance is $3/2$ and the variance of hypothetical means is $3/4$, so the Bühlmann credibility factor is $n/(n+2)$. A total count $k$ gives the displayed credibility premium. By contrast, the Bayesian posterior density is proportional to ${\theta }^{k-4}{e}^{-n\theta }$ on $\theta >1$. Its mean is the ratio ${\int }_1^{\infty }{\theta }^{k-3}{e}^{-n\theta }d\theta /{\int }_1^{\infty }{\theta }^{k-4}{e}^{-n\theta }d\theta$. For $n=2,k=5$, that Bayesian credibility estimate is $5/3$, whereas the affine estimate is $2$.

In the Gamma prior for an inverse-gamma scale model, assume $k>2$. The variance of hypothetical means is $a=\alpha /[{\lambda }^2(k-1{)}^2]$, and the expected process variance is $v=\alpha (\alpha +1)/[{\lambda }^2(k-1{)}^2(k-2)]$. The Bühlmann credibility factor is $Z=n/(n+v/a)$, giving the displayed expression. The optimal affine credibility estimate is $(1-Z)\alpha /[\lambda (k-1)]+Z\overline{X}$. Its fixed-weight arithmetic-mean form differs from the reciprocal-based posterior mean.

With fixed structural parameters and $K=v/a$, the Bühlmann credibility estimate has form $({\sum }_{i=1}^n{x}_i+K{m}_0)/(n+K)$. The next observation updates it by the displayed rule. Consequently the estimate decreases exactly when the new observation is less than the previous credibility estimate, with strict equality handled separately.

The credibility factor in the Bühlmann model is the weight $Z=na/(na+v)$ placed on the sample mean after $n$ observations. The remaining weight is placed on the population expected value. More experience or larger variance of hypothetical means increases $Z$; larger expected process variance decreases it.

The variance of hypothetical means in the Bühlmann model is the between-risk variance of the conditional claim expected value. It creates the shared covariance $a$ between different years of the same risk and determines how much individual experience should influence the premium.

The expected process variance in the Bühlmann model averages within-risk conditional variance over the population of risks. It measures variation that repeated observations can average out and contributes $v/n$ to the variance of the sample mean.

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.