Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 39 (Actuarial Statistics)

Use the usual aggregate claims model assumption that the claim count $N$ is independent of the sequence of severities. Conditional on $N=n$, the probability generating function of the sum is the product of the $n$ severity generating functions. The empty sum has value zero. Thus the law of total expectation gives the random-sum transform identityfor $|z|<1$, with the usual continuous interpretation at $z=1$.

For a claim count distribution in the Panjer claim-count class, multiply the count recurrence by $n{z}^{n-1}$ and sum. With $m=n-1$,Consequently $(1-az)G_N^{\prime }=(a+b)G_N$, orThe chain rule applied to the composition then gives

For $r\ge 1$, compare the coefficients of $z^{r-1}$. The two products giveMove the $j=0$ term to the left and combine the remaining coefficients, since $a(r-j)+(a+b)j=ar+bj$. The Panjer recursion with zero severities is thereforeFor the usual nondegenerate proper Panjer laws, $a<1$, so $1-a{f}_0>0$. If a degenerate count $N\equiv 0$ is represented by redundant constants making this denominator zero, the aggregate is identically zero and must be handled directly rather than dividing by zero.

Because severities are nonnegative, the aggregate is zero exactly when every included severity is zero. Count-severity independence gives the initializationIn particular $g_0$ generally exceeds $p_0$ when zero severities are possible.

The count-severity independence is necessary and is implicit, rather than explicitly stated, in the printed random-sum setup. For a concrete counterexample without it, take independent Bernoulli-$1/2$ variables $X_j$ and put $N=X_1$. Then $S=X_1$, so $G_S(z)=(1+z)/2$, whereas $G_N(G_X(z))=(3+z)/4$. Independence of the $X_j$ among themselves alone cannot justify the requested formula.

Within the approximation that a factory has either zero or one accident, let $N$ be the accident count. The factories' independence gives $N\sim \mathrm{Binomial}(m,p)$. Given an accident, its cluster of claims has a Poisson distribution of mean $\lambda$, independently across factories. Its probability generating function is $G_X(z)=e^{\lambda (z-1)}$. Thus this is a binomial accident portfolio with Poisson clusters, withA factory can contribute no claims either because it has no accident or because its accident's Poisson cluster is empty. Hence the no-claim probability differs from the no-accident probability $q^m$.

For $0\le n\le m$, the accident probability mass function is $p_n=\left(\genfrac{}{}{0ex}{}{m}{n}\right)p^n{q}^{m-n}$. For $1\le n\le m$,Therefore its Panjer claim-count class parameters areAt $n=m+1$ the multiplier is zero, and afterwards both successive masses are zero, so the recurrence holds for all $n\ge 1$. Finally, $f_0=e^{-\lambda }$ and $G_N(z)=(q+pz{)}^m$ reproduce the same initialization $g_0=G_N(f_0)$.

Let $Y=(\log X-\mu )/\sigma$ and write $\varphi$ for the standard normal density, with $\sigma >0$. The change of variables converts the density integral intoComplete the square using $e^{k\sigma y}\varphi (y)=e^{k^2{\sigma }^2/2}\varphi (y-k\sigma )$. Integration gives the truncated lognormal momentHere $\Phi$ is the standard normal distribution function; take $\log 0=-\infty$ and $\log \infty =\infty$. The identity holds for every real $k$, including the full-interval limits, because the shifted Gaussian integral remains finite.

The limited excess of loss reinsurance payment is $L(X)=\min \{(X-M{)}_{+},A\}$. It is zero below $M$, equals $X-M$ for $M<X<M+A$, and equals $A$ above $M+A$. Since the lognormal distribution has a continuous density,Put $d_j(t)=(\log t-\mu -j{\sigma }^2)/\sigma$, $j=0,1$, and $\overline{\Phi }=1-\Phi$. Applying the moment identity with $k=1$ and $k=0$ gives the conditional lognormal layer recoveryThe denominator is $\mathbb{P}(X>M)$, and payments outside that event are zero. The three numerator terms are respectively the truncated first moment, the attachment subtracted from partial payments, and the capped payments on claims above the layer. This proves the conditional expectation including the limit on each reinsurer payment.

For claims inflation by $h=1.1$, set $X_h=hX$. Its log-mean is ${\mu }_h=\mu +\log h$ and its log-standard-deviation stays $\sigma$. The attachment and payment limit stay fixed. Therefore the requested new conditional mean isExplicitly, in the preceding fraction replace $d_j(t)$ by $d_j^{(h)}(t)=(\log t-\mu -\log 1.1-j{\sigma }^2)/\sigma$ and replace $e^{\mu +{\sigma }^2/2}$ by $1.1e^{\mu +{\sigma }^2/2}$. Equivalently, claims inflation of a fixed reinsurance layer givesbecause $L_{M,A}(hX)=h{L}_{M/h,A/h}(X)$ and the new conditioning event is $X>M/h$. Merely multiplying the original conditional mean by $1.1$ would omit both of these threshold changes.

In the classical risk model, the surplus with initial capital $u\ge 0$ iswhere $N(t)$ is a Poisson process of rate $\lambda$, and the nonnegative independent claim sizes, of mean $\mu >0$, are independent of that process. Positive relative safety loading $\rho$ makes expected premium income exceed expected claim outflow. With ruin time $T=\inf \{t\ge 0:U(t)<0\}$, the ultimate ruin probability isThe adjustment coefficient is the positive root $R$ in the finite domain of the claim moment-generating function satisfyingThe zero root is excluded by definition.

Under the stated transform-divergence assumption this positive root exists and is unique. Indeed $F(r)=M_X(r)-1-(1+\rho )\mu r$ has $F(0)=0$, derivative $F^{\prime }(0)=-\rho \mu <0$, and strictly positive second derivative on its positive domain. At a finite endpoint the diverging transform makes $F$ positive eventually. At an infinite endpoint, choose $a>0$ with $\mathbb{P}(X\ge a)>0$; the bound $M_X(r)\ge \mathbb{P}(X\ge a)e^{ar}$ makes it dominate the linear term. Strict convexity then gives exactly one positive crossing. This is the secant-slope existence criterion for an adjustment coefficient.

The Lundberg inequality isFor a short justification, the compound-Poisson transform and the defining equation make $e^{-RU(t)}$ a nonnegative martingale starting at $e^{-Ru}$. Stop it at $T\wedge t$, a bounded stopping time. On $\{T\le t\}$ its value is greater than one. The optional stopping theorem therefore gives $\mathbb{P}(T\le t)\le e^{-Ru}$; let $t$ increase to infinity. No expectation identity at an unbounded ruin time is required.

For an exponential distribution with mean $\mu$, the claim moment-generating function is $M_I(r)=(1-\mu r{)}^{-1}$ for $r<1/\mu$. The calculated adjustment coefficient must solveSubtract one, divide by the nonzero factor $\mu R_I$, and obtain $(1-\mu R_I{)}^{-1}=1+\rho$. ThusIt belongs to $(0,1/\mu )$ for every $\rho >0$, as required by the transform domain. The Poisson intensity cancels because the premium rate is specified using the same relative loading and mean claim cost.

Normalizing the true density givesa gamma distribution of shape $1/2$ and rate $\beta$. Its mean is $1/(2\beta )=\mu$, and the Gamma integral givesPut $x=R/\beta$. Since $\mu \beta =1/2$, the positive adjustment coefficient satisfiesBoth sides are positive in this domain. Squaring, multiplying by $1-x$, and dividing out the zero root $x=0$ yieldsThe discriminant is $(1+\rho {)}^2[(\rho -3{)}^2+16\rho ]=(1+\rho {)}^2(\rho +9)(\rho +1)$. The negative constant term makes one root positive and the other negative. The polynomial is negative at zero and has value four at one, so the positive root lies in $(0,1)$ and is not an extraneous solution introduced by squaring. Therefore the adjustment coefficient for shape-one-half gamma claims is

To compare it with the exponential calculation, write $R_I=2\beta \rho /(1+\rho )$. Their difference issince $9(1+\rho {)}^2-(\rho +9)(\rho +1)=8\rho (1+\rho )>0$. Thus the two proposed Lundberg inequality bounds are ordered asOnly ${\psi }_{\mathrm{true}}(u)\le e^{-Ru}$ is justified for the true claim law. The exponential calculation gives ${\psi }_{\mathrm{exponential}}(u)\le e^{-R_{I}u}$ for a different model, not a valid guaranteed bound on the true ruin probability.

This exponential misspecification of shape-one-half gamma claims is optimistic: the true variance is $1/(2{\beta }^2)=2{\mu }^2$, whereas the assumed exponential variance is ${\mu }^2$. Its tail also decays at rate $\beta$, more slowly than the assumed rate $1/\mu =2\beta$. Equivalently, for $0<r<\beta$, the true transform $(1-2\mu r{)}^{-1/2}$ is larger than $(1-\mu r{)}^{-1}$, since $(1-\mu r{)}^2>1-2\mu r$. This stronger transform growth makes the true adjustment root smaller. Comparing upper bounds alone does not by itself prove an ordering of the two actual ruin probabilities. At $u=0$ both displayed bounds equal one.

A credibility premium estimates a risk's expected future claims by combining its own observed experience with a collective or prior mean. In its basic form $Z\overline{X}+(1-Z)\mu$, the credibility factor $Z\in [0,1]$ is the weight assigned to individual experience. Larger credibility places less weight on pooling with the population. The amounts below are pure expected-claim premiums; no additional expense or profit loading is specified.

Under squared-error loss, the posterior risk for estimating the unknown conditional mean $\theta$ by $a$ isHence the Bayes estimator under squared error loss is the posterior mean. In the equal-year model, conditional independence makes the likelihood proportional to $\exp [-{\sum }_{j=1}^n(X_j-\theta {)}^2/(2{\sigma }_1^2)]$. Combining it with the normal prior and completing the square givesBecause $\mathbb{E}[X_{n+1}\mid \theta ]=\theta$, the required estimator is $m_n$; it also equals the predictive mean $\mathbb{E}[X_{n+1}\mid \mathbf{X}]$. The normal-normal credibility expression isThis is exact Bayesian credibility, not merely the best linear approximation to a possibly nonlinear posterior mean.

For unequal life counts use the usual exposure model: individual-life losses are conditionally independent with variance $v$, and the annual totals are conditionally independent given the same $\theta$. ThenThe dot denotes the stipulated normal approximation. The positive exposures $m_j$ are known. Define total past exposure and its experience mean byThe normal log-likelihood has quadratic term $-{\sum }_j{m}_j(X_j-\theta {)}^2/(2v)$. With the prior $N(\mu ,{\sigma }^2)$, normal-normal conjugacy with unequal exposures gives posterior variance and meanTherefore the estimated pure premium per life for year $n+1$ has the normal-normal credibility with unequal exposures formIt is also an exact normal-model Bühlmann–Straub credibility estimate. The individual experience is exposure-weighted: the unweighted mean of the yearly per-life observations would generally discard the different precisions.

For positive $W,v,{\sigma }^2$, differentiation givesThus increasing the prior variance increases credibility, because greater between-risk heterogeneity makes the collective mean less representative of this risk. Increasing the individual process variance decreases credibility, because a fixed amount of observed experience becomes noisier. The future exposure supplies no new observation, so it does not enter $Z_W$. It only scales the total premium:The independent-life variance scaling is part of the usual exposure interpretation. Without it, a known variance $v$ for one life does not determine the variance of a total: a common pairwise conditional covariance $c$ would add $m_j(m_j-1)c$ to $m_{j}v$. Similarly, cross-year conditional dependence would change the joint likelihood. Those assumptions are implicit in applying the standard credibility model to the final paragraph.