Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 43 (Actuarial Statistics)

Write $P(z)=\mathbb{E}[z^N]$, $F(z)=\mathbb{E}[z^{X_1}]$ and $G(z)=\mathbb{E}[z^{S_N}]$ for the three probability generating functions. Since both the claim count and every claim size are positive, $P(0)=F(0)=G(0)=0$. Conditioning on the count and using independence gives the random-sum transform identityThe crucial point is that the count recurrence begins at $n=2$. Multiplying it by $n{z}^{n-1}$ and summing, we obtainConsequently $(1-az)P^{\prime }(z)=(a+b)P(z)+p_1$. The chain rule applied to the aggregate probability generating function therefore givesThese identities hold inside the unit disk, or as identities of formal power series. Since $F(0)=0$, every coefficient of the composition depends on only finitely many count probabilities.

Compare the coefficient of $z^{k-1}$. The product $F{G}^{\prime }$ contributes ${\sum }_{j=1}^{k-1}(k-j)f_j{g}_{k-j}$, and $F^{\prime }G$ contributes ${\sum }_{j=1}^{k-1}j{f}_j{g}_{k-j}$. HenceThe required aggregate recursion for zero-truncated Panjer counts is initialized by $g_0=0$ and $g_1=p_1{f}_1$, and for every $k\ge 1$ it isThe empty sum at $k=1$ is zero. Every term on the right is known or has a smaller aggregate index. Omitting $p_1{f}_k$ would incorrectly apply the usual Panjer recursion with a zero initial value, producing zero for every aggregate probability.

For the zero-truncated Poisson distribution, $p_n/p_{n-1}=\lambda /n$ for $n\ge 2$, so $a=0$, $b=\lambda$ and $p_1=\lambda /(e^{\lambda }-1)$, with $\lambda >0$. ThusAs a direct probability generating function check, $P(z)=(e^{\lambda z}-1)/(e^{\lambda }-1)$ and therefore $G(z)=(e^{\lambda F(z)}-1)/(e^{\lambda }-1)$; differentiation reproduces this recursion.

The claim-size exponential distribution has expected value $\mu >0$ and moment-generating functionConditional on $N_i=k$, independence makes the aggregate moment-generating function equal to $M_X(t{)}^k$. Averaging over the Poisson distribution of the count givesPut $\Lambda ={\sum }_{i=1}^n{\lambda }_i$. For fixed intensities, independence between risks givesThis is precisely the transform of a compound Poisson distribution. Indeed the addition of independent Poisson random variables makes the total count Poisson distributed with parameter $\Lambda$, and every claim in the pooled portfolio has the same exponential distribution. Thus the portfolio aggregate is compound Poisson with intensity ${\sum }_i{\lambda }_i$ and exponential claim sizes of mean $\mu$.

Now regard the individual intensities as independent gamma distributed quantities, with shape $m\ge 1$ and rate $\alpha$. Their moment-generating functions are $(\alpha /(\alpha -t){)}^m$ for $t<\alpha$. Multiplying these transforms proves additivity of independent gamma distributions with a common rate:Conditional on the individual intensities, the total count $N$ has Poisson distribution with parameter $\Lambda$. Because this conditional law depends only on their sum, it is also the law of $N$ conditional on $\Lambda$. Integrating its probability mass function against the gamma distribution density yieldsThus the total count has a negative binomial distribution with shape $nm$ and success probability $\alpha /(\alpha +1)$, using the convention that counts failures before the specified number of successes. Its probability generating function is $(\alpha /(\alpha +1-z){)}^{nm}$.

Conditional on $\Lambda$, the claim sizes retain their common exponential distribution and the amount $S$ has a compound Poisson distribution of intensity $\Lambda$. Therefore the unconditional amount has a compound mixed Poisson distribution with gamma mixing intensity of shape $nm$ and rate $\alpha$. To make the mixture explicit, average the conditional aggregate moment-generating function:The displayed domain ensures both the severity transform and the intensity transform are finite. The mixing distribution describes the Poisson intensity, not the claim sizes. In particular, the aggregate has an atom at zero of mass $\mathbb{P}(N=0)=(\alpha /(\alpha +1){)}^{nm}$.

In the classical risk model, the insurer's surplus iswhere $u\ge 0$ is initial capital, $N(t)$ is a Poisson process of rate $\lambda >0$, and the claim sizes are positive independent and identically distributed random variables with expected value $\mu$, independent of the claim-arrival process. Premium flows in at constant rate $c$. A positive relative safety loading $\theta$ meansso premium income exceeds the expected claim cost per unit time. The ruin stopping time is $\tau =\inf \{t\ge 0:U(t)<0\}$, with $\tau =\infty$ if ruin never occurs, and the ultimate ruin probability is $\psi (u)=\mathbb{P}(\tau <\infty )$.

Assume the stated positive adjustment coefficient $R$ exists and that the claim moment-generating function is finite at $R$. The Lundberg inequality isHere is a proof, including the stopping argument. By conditioning on the Poisson distribution of $N(t)$, the random-sum transform identity givesThe adjustment coefficient equation is $\lambda (M(R)-1)=cR$. Consequently the exponential surplus martingalehas expected value one at every finite time. More strongly, the claim process has independent increments, so for $s\le t$,Thus $Z$ is a nonnegative continuous-time martingale for the filtration generated by arrivals and claim sizes. For a fixed finite horizon $T$, the optional stopping theorem at the bounded stopping time $\tau \wedge T$ gives $\mathbb{E}{Z}_{\tau \wedge T}=1$. This bounded use is legitimate: the stopped values belong to the closed martingale on $[0,T]$ given by conditional expectations of the integrable variable $Z_T$.

On $\{\tau \le T\}$, the surplus is negative, and therefore $C(\tau )-c\tau =u-U(\tau )>u$. By nonnegativity on the complementary event,The ruin events increase to $\{\tau <\infty \}$ as $T$ increases, so continuity of probability gives the claimed Lundberg inequality. No expectation identity at the unbounded stopping time $\tau$ is needed. Positive loading by itself does not guarantee a positive adjustment coefficient for every claim law; the existence assumption is used here.

For the exponential distribution of claim sizes, $M(r)=1/(1-\mu r)$ for $r<1/\mu$. Substitution in the adjustment coefficient equation givesDiscard the zero root and divide by $\mu r$. The unique positive solution iswhich lies strictly inside the finite-transform domain.

For deterministic claim sizes, the claim moment-generating function is $e^{\mu r}$. Define $h(r)=e^{\mu r}-1-(1+\theta )\mu r$. It satisfies $h(0)=0$, $h^{\prime }(0)=-\theta \mu <0$, $h^{\prime \prime }(r)={\mu }^2{e}^{\mu r}>0$, and $h(r)\to \infty$ as $r\to \infty$. Thus its strict convexity gives exactly one positive zero $R_{\mu }$, with $h(r)<0$ for $0<r<R_{\mu }$.

Put $x=\mu R=\theta /(1+\theta )\in (0,1)$. Since $-\log (1-x)>x$, we have $e^x<1/(1-x)$. HenceIt follows thatThus the deterministic-claim Lundberg inequality provides a smaller upper bound at positive capital; at zero capital both bounds equal one. This is a special case of deterministic claims maximize the adjustment coefficient at fixed mean and loading: fixing the mean and loading, variability lowers the exponential decay coefficient relative to deterministic claims. The ordering of these upper bounds alone is not a proof that the actual ultimate ruin probabilities are ordered.

Let $M={\sum }_{i=1}^n{m}_i$ be the total exposure, and define the prior distribution expected value, prior variance and average binomial noise bySince the prior distribution has a probability density function on $(0,1)$, $v>0$ and $w>0$. All relevant moments exist because the parameter and observations are bounded. The binomial distribution givesThe law of total expectation, law of total variance and law of total covariance consequently yieldIn particular, conditional independence of the annual observations does not imply unconditional independence: they share the uncertain parameter.

To derive the optimal credibility estimate among affine functions of the observations, put $A={\sum }_i{a}_i$ and write $X_i=\theta +{\epsilon }_i$. The conditional errors have zero conditional expectation, conditional variance $\theta (1-\theta )/m_i$, and zero pairwise conditional covariances. They are also uncorrelated with every integrable function of $\theta$. Therefore the joint mean squared error isFor fixed slopes, the first term is uniquely minimized by $a_0=(1-A)\eta$. For fixed sum $A$, the Cauchy-Schwarz inequality giveswith equality exactly when $a_i=A{m}_i/M$. Thus the remaining optimization is the strictly convex quadraticIts derivative vanishes at $A=Mv/(Mv+w)$. Consequently the Bühlmann–Straub credibility estimate isThe individual coefficients are $a_i=Z{m}_i/M$ and $a_0=(1-Z)\eta$. The strict quadratic minimizations establish global optimality and uniqueness, not just necessary equations. Here $v$ is the variance of hypothetical means, $w$ is the expected process variance, and $Z$ is the Bühlmann–Straub credibility factor. More exposure gives more weight to the observed claim fraction. Since $0<Z<1$, the estimate remains between the observed pooled fraction and the prior expected value. If one separately permits a degenerate prior, $v=0$ leads to the constant estimate $\eta$, interpreted as $Z=0$.

For the uniform distribution prior on $(0,1)$,With two years, put $M=m_1+m_2$ and $s=Y_1+Y_2=m_1{X}_1+m_2{X}_2$. Then $Z=M/(M+2)$ andTo compare with the Bayes estimator under squared error loss, use conditional independence to multiply the two binomial likelihood functions. Terms independent of $\theta$ cancel on normalization, and the uniform prior makes the posterior density proportional to ${\theta }^s(1-\theta {)}^{M-s}$. Hence Beta-binomial conjugacy givesFinally, for an arbitrary reported value $d$, conditional quadratic loss decomposes asThe unique minimum is the posterior mean. Therefore the affine credibility estimate and the exact Bayesian estimate coincide in this case. The agreement illustrates exact beta-binomial credibility with unequal exposures; for a general prior distribution, minimizing over affine functions of the observations need not recover the unrestricted posterior mean.