Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 43 (Actuarial Statistics)

Let $P(z)={\sum }_{n\ge 0}{p}_n{z}^n$, $F(z)={\sum }_{j\ge 1}{f}_j{z}^j$, and $G(z)={\sum }_{k\ge 0}{g}_k{z}^k$ be the count, claim-size and aggregate probability generating functions. Independence in the aggregate claims model givesThe Panjer claim-count class recurrence can be written $n{p}_n=(an+b)p_{n-1}$. Multiply by $z^{n-1}$ and sum to obtainUsing the chain rule in $G=P\circ F$ then yieldsThe coefficient of $z^{k-1}$ on the left is $k{g}_k-a{\sum }_{j=1}^k{f}_j(k-j)g_{k-j}$, where the $j=k$ term is zero. On the right it is $(a+b){\sum }_{j=1}^{k}j{f}_j{g}_{k-j}$. Rearranging gives the Panjer recursionThe initial value is $p_0$ because all claim sizes are positive: zero total loss occurs exactly when $N=0$. Only already computed aggregate probabilities occur on the right. The count normalization determines $p_0$ from $a,b$ as well: with $w_0=1$ and $w_n={\prod }_{j=1}^n(a+b/j)$, we have $p_n=p_0{w}_n$ and $p_0=({\sum }_{n\ge 0}{w}_n{)}^{-1}$. The assumed valid count law makes this sum finite.

Finally $p_0$ must be positive. If it were zero, the recurrence would make every $p_n$ zero, contradicting normalization. At $n=1$ the recurrence therefore givesThis proves the sign constraint without assuming any particular count family. The three parts below also provide explicit initial probabilities for the aggregate recursion.

If $a+b=0$, then $p_1=(a+b)p_0=0$. Applying the count recurrence successively gives $p_n=0$ for every $n\ge 1$. Normalization forcesThus the claim count distribution is a point mass at zero, and the aggregate claims model also has $S=0$ almost surely. In particular the recursion starts at $g_0=1$ and gives no positive-loss probabilities.

With $a=0$ and $b>0$, the recurrence is $p_n=(b/n)p_{n-1}$. Iteration gives $p_n=p_0{b}^n/n!$. Since ${\sum }_{n\ge 0}{b}^n/n!=e^b$, normalization givesHence the claim count distribution is a Poisson distribution of mean $b$, and the Panjer recursion begins at $g_0=e^{-b}$.

Here $a+b/n=a(n+m-1)/n$, so iteration givesThe negative-binomial series ${\sum }_{n\ge 0}\left(\genfrac{}{}{0ex}{}{m+n-1}{n}\right)a^n=(1-a{)}^{-m}$, valid for $0<a<1$, normalizes this toThis is the negative binomial distribution counting failures before the $m$th success, with success probability $1-a$. The convention is specified because a negative-binomial variable is sometimes defined to count trials instead. Its count expected value is $ma/(1-a)$, and the Panjer recursion begins at $g_0=(1-a{)}^m$. For $m=1$ this reduces to the failure-count geometric distribution.

Under quota share reinsurance, the direct insurer retains a fixed fraction $\alpha$ of each claim and the reinsurer pays the other fraction. Thus their annual claim totals are $\alpha S$ and $(1-\alpha )S$. Under excess of loss reinsurance with retention $d$, the direct insurer pays $\min (X_j,d)$ on each individual claim and the reinsurer pays $(X_j-d{)}_{+}$. Under stop loss reinsurance with annual retention $D$, the direct insurer pays $\min (S,D)$ and the reinsurer pays $(S-D{)}_{+}$. The last two contracts differ because one caps each claim and the other caps the aggregate.

Conditioning on the annual Poisson distribution count gives the compound Poisson distribution transform:This equality is understood on the domain where the moment-generating function is finite. By the expected value premium principle, annual premium income is $c=(1+\theta )\lambda \mu$. Without reinsurance, final capital is $W_0=u+c-S$, and therefore

For the quota-share contract, the reinsurer's expected claim cost is $(1-\alpha )\lambda \mu$, so its annual premium isThe direct insurer keeps its original premium income, pays this reinsurance premium and retains only $\alpha S$. ThusandThe reinsurer's loading is applied to the ceded expected loss, not to the direct insurer's original premium.

Maximizing expected exponential utility $\mathbb{E}[-e^{-\beta W}]$ is equivalent to minimizing the exponent $H(\alpha )=\log \mathbb{E}[e^{-\beta W}]$. For exponential distribution claim sizes of mean $\mu$,so on the finite-utility domain $\beta \mu \alpha <1$,For a nontrivial positive arrival rate, differentiation givesThe unique unconstrained stationary retention is thereforeIt is positive and lies below $1/(\beta \mu )$. If ${\alpha }_{\ast }<1$, it is the unique optimal retention in the printed range $0<\alpha <1$. If ${\alpha }_{\ast }\ge 1$, expected utility increases throughout that open interval and its supremum is approached as $\alpha \uparrow 1$; there is no attained maximum in the strictly open range. Allowing the usual no-reinsurance endpoint gives the compact formulaRetentions with $\beta \mu \alpha \ge 1$ have infinite exponential loss moment and expected utility $-\infty$, so they cannot displace the finite stationary solution. The stationary point and full-retention endpoint just described stay within the finite-transform domain whenever they are selected.

For the requested comparisons,Higher reinsurer loading increases optimal retention, up to full retention: reinsurance is more expensive. The direct insurer's loading $\theta$ does not affect the optimizer. It adds only the constant $-\beta \lambda \mu \theta$ to $H$, reflecting the translation property of constant absolute risk aversion. These observations give the quota-share retention under exponential utility rule; $u$ and the positive arrival rate also drop out of the stationary equation. In the degenerate case $\lambda =0$, no claims or premiums occur, $W=u$, and every retained fraction has the same expected utility.

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.

A credibility estimate combines a risk's own observed experience with a collective or prior mean, in the form $Z{\hat{\mu }}_{\mathrm{experience}}+(1-Z){\mu }_{\mathrm{collective}}$. The credibility factor $Z\in [0,1]$ is the weight given to that risk's experience. Bayesian credibility starts with a prior distribution for a latent risk parameter, updates it using observations, and estimates the conditional risk mean by its Bayesian posterior expected value under squared-error loss. In conjugate models this posterior mean can have exactly the displayed credibility form.

For the unequal policy exposures, defineConditional independence of the binomial distribution observations gives the likelihood, up to a factor not depending on $\theta$,Multiplication by the beta distribution prior therefore gives Beta-binomial conjugacy:Both posterior parameters are positive, since $0\le s\le M$ and the prior parameters are positive.

The conditional target is $\mu (\theta )=m_{k+1}\theta$. To verify the quadratic-loss rule directly, for any proposed estimate $d$ and observed data $\mathcal{D}$,The variance term does not depend on $d$, so the unique Bayes estimator under squared error loss is the posterior mean. Applying the beta mean formula givesIt is also the posterior expected future claim count, by the tower property of conditional expectation.

Let $p_0=\alpha /(\alpha +\beta )$ be the collective prior claim probability and $\hat{p}=s/M$ the observed claims per policy. ThenTherefore the exact beta-binomial credibility with unequal exposures isThe individual experience must be weighted by exposure: $s/M$ is the policy-weighted frequency, not the unweighted average of $x_i/m_i$ when the $m_i$ differ. The factor multiplies next-year expected counts after scaling both component probabilities by $m_{k+1}$.

With fixed positive prior parameters and $m_i\ge 1$, total exposure satisfies $M={\sum }_{i=1}^k{m}_i\ge k$. Hence $M\to \infty$ as the number of observed years grows, andThus the credibility estimate places asymptotically all its weight on the observed policy-weighted claim frequency. The assertion concerns the credibility weight; the scale of the predicted number of claims still includes the known next-year exposure $m_{k+1}$.

Write $A=\alpha +\beta$ and keep $p_0=\alpha /A$ fixed. Then $\alpha =p_{0}A$, $\beta =(1-p_0)A$, while the fixed observed years and exposures leave $M$ unchanged. The credibility factor isAs prior concentration $A$ increases without bound,The beta prior variance is $p_0(1-p_0)/(A+1)$, which tends to zero while its mean stays fixed. Increasing $A$ therefore makes the prior increasingly concentrated and reduces the influence of the same finite data. This explains the limiting Bayesian credibility estimate, rather than merely treating $\alpha ,\beta$ as two larger arbitrary parameters.