Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 31 (Actuarial Statistics)

Condition on the count and use the independence of random variables of the claim sizes. For $n\ge 1$ their joint exponential transform factors, while for $n=0$ the empty sum contributes $1$. Consequently the law of total expectation givesThis random sum of independent claims transform uses both independence assumptions: the count must be independent of the entire claim-size sequence, and the sizes must be mutually independent with the same probability distribution. The probability generating function is interpreted through its defining nonnegative series. The identity holds as a finite moment-generating function wherever that series is finite; outside that domain the expectation and series can agree at $+\infty$. In particular a positive argument may take $M_X(r)$ beyond $1$, so finiteness does not follow merely from the usual unit-disk domain of a probability generating function.

The moment-generating function of a mixture distribution is the mixture of its component transforms, soFor the aggregate, positivity of every claim implies $S=0$ exactly when $N=0$. ChooseIf $q>0$, take $\tilde{N}$ to have the zero-truncated claim-count distribution, namely the conditional law of $N$ given $N>0$. ThenChoose this count independently of a fresh independent claim-size sequence and put $Z={\sum }_{i=1}^{\tilde{N}}{X}_i$. Its probability distribution is that of $S$ conditional on being positive. Thus the hurdle decomposition of a positive random sum givesand an independent Bernoulli random variable $B$ of success probability $q$ realizes $S\stackrel{d}{=}BZ$. This establishes the distributional representation, including that $Z$ is itself a positive random sum of independent claims. If $p=1$, the aggregate is identically zero; set $q=0$ and choose any positive $Z$, for example one claim. Conditioning the count on positivity is then unnecessary and would be undefined.

For the specified geometric distribution on the nonnegative integers,An exponential distribution of expected value $\mu$ has transform $M_X(r)=(1-\mu r{)}^{-1}$. Substitution yieldsHence $Z$ is exponential with rate $1/(2\mu )$ and expected value $2\mu$. This is the geometric sum of exponential variables with a rescaling of the claim mean. Identification can also use the uniqueness theorem for Laplace transforms of nonnegative random variables by taking $r\le 0$.

The resulting distribution function isIts jump of size $1/2$ at zero is important: the aggregate law is not a purely continuous exponential distribution.

Consider the secant slopeThe moment-generating function is continuous on the interior of its finite domain and has right derivative $M^{\prime }(0)=\mu$. Thus $H(r)\to \mu$ as $r\downarrow 0$. For every fixed positive claim amount $x$, the function $(e^{rx}-1)/r$ is strictly increasing in $r>0$: its derivative has numerator $e^{rx}(rx-1)+1>0$, since this numerator starts at zero and has derivative $rx{e}^{rx}$ as a function of $rx$. Taking expected values preserves the strict inequality. Therefore $H$ is continuous and strictly increasing. Equivalently, the strictly convex transform $M$ has strictly increasing secant slopes from the origin.

If $r_{\infty }<\infty$, the assumed blow-up of $M$ gives $H(r)\to \infty$. If $r_{\infty }=\infty$, choose $a>0$ with $d=\mathbb{P}(X_1\ge a)>0$. Such an $a$ exists because the claims are positive. Then $M(r)\ge d{e}^{ar}$, so again $H(r)\to \infty$. This exponential lower bound is needed at an infinite endpoint: mere divergence of $M$ would not, by itself, establish divergence of $M(r)/r$.

The target $(1+\theta )\mu$ strictly exceeds the limiting slope $\mu$. The intermediate value theorem and strict monotonicity therefore giveMultiplying by $R$ gives the defining adjustment coefficient equation. The zero root of the undivided equation is excluded. This is the secant-slope existence criterion for an adjustment coefficient.

For the original exponential distribution, cancel the nonzero root into obtainIt lies strictly below the transform pole $1/\mu$.

With the extra expenses, the insurer's payment per claim is $Y=X+A$, where $A$ has exponential distribution of expected value $\mu /2$ and is independent of $X$. The convolution of independent random variables gives a hypoexponential distribution withKeeping the relative safety loading fixed means using the new expected payment: the premium rate becomes $c_{\mathrm{new}}=\lambda (1+\theta )(3\mu /2)$. It does not mean keeping the old premium rate fixed. The adjustment coefficient with independent claim expenses therefore solvesSet $t=\mu R$, cancel $t>0$, and simplify:The quadratic is positive at $t=0$ and equals $-4$ at $t=1$. Its leading coefficient is positive, so the smaller root is in $(0,1)$ and the larger root exceeds $1$. Only the smaller root lies in the finite moment-generating function domain. HenceFor $\theta =1$,Thus the new adjustment coefficient is about $9.72\%$ smaller, even though the premium rate has been increased to retain the same relative safety loading. The Lundberg inequality consequently has a slower exponential decay rate as a function of capital.

First derive the survival renewal equation for a classical risk model. Before the first claim, capital rises deterministically. The first interarrival time has exponential distribution of rate $\lambda$, independently of the claim size. Conditioning on its time and size, and using the Markov property after that claim, givesA claim larger than the capital available then causes immediate ruin and contributes zero. There is almost surely a first claim because $\lambda >0$.

Put $g(v)={\int }_0^v\phi (v-x)f(x)dx$. Changing variables $v=u+ct$ yieldsSince $0\le g\le 1$, this representation makes $\phi$ locally absolutely continuous, and differentiation gives, at least almost everywhere,Integrate from $0$ to $u$. The integrands are nonnegative, so Tonelli theorem permits interchange in the double integral. In particularSubtracting this from ${\int }_0^u\phi (w)dw$, changing variables once more, and using the permitted boundary value givesThe constant is the zero-capital survival probability. Positive relative safety loading means $c>\lambda \mu$, so it is positive. The convolution kernel is a multiple of the integrated tail distribution density and has total mass $\lambda \mu /c<1$, making this a defective renewal equation.

For the individual portfolios define ${\rho }_i={\lambda }_i{\mu }_i/c_i$. Under the positive-loading convention of the question, $0<{\rho }_i<1$. At zero capital, portfolio $i$ survives with probability $1-{\rho }_i$. The independence of random variables of the entire portfolio processes makes their ultimate survival events independent. ConsequentlyThe exponential form is not needed for this zero-capital probability under positive loading; only the claim mean enters. For completeness, the company description does not separately repeat positive loading for every portfolio. If nonpositive loadings are allowed, certain ruin with nonpositive loading and finite claim variance instead givesHere the positive part sets a nonpositive factor to zero. To justify the additional case, observe capital at claim times: its independent increments are $c_{i}T-X$, with mean $c_i/{\lambda }_i-{\mu }_i$. A negative mean sends their partial sums to $-\infty$ by the strong law of large numbers. At zero mean, these increments have finite nonzero variance. The central limit theorem gives $\mathbb{P}(S_k<-K)\to 1/2$ for each fixed $K$, so the probability of unboundedness below is at least $1/2$. That event is unchanged by altering finitely many increments and hence is a tail event; the Kolmogorov zero-one law makes its probability one. Ruin therefore occurs almost surely also at zero loading.

For the merged claims, Poisson superposition of insurance portfolios gives total arrival rate $\lambda ={\sum }_i{\lambda }_i$. Each arrival is from portfolio $i$ with probability ${\lambda }_i/\lambda$, independently of other arrival labels. Thus its claim-size mixture distribution has densityThe weights sum to one, so this density integrates to one. An independent transform verification uses the aggregate for one accounting period. If $M_i(r)=(1-{\mu }_{i}r{)}^{-1}$, then its moment-generating function isThis is precisely a compound Poisson distribution with parameter $\lambda$ and the displayed claim-size law. The transform identity can safely be read at $r\le 0$; positive arguments must be below the relevant poles.

Premium incomes and initial capitals add. Therefore the merged premium rate is $C={\sum }_i{c}_i$, the merged initial capital is zero, and its mean claim size isIts zero-capital survival probability is consequentlyUnder the question's positive-loading context this is a premium-weighted average of the individual survival probabilities, rather than their product. If arbitrary individual loadings are admitted, recompute the loading of the merged portfolio; its finite-variance mixture law gives the complete formulaAn individually nonpositive loading does not force merged ruin when aggregate premiums still exceed aggregate expected claims. The survival probability under risk pooling is at least their product because a product of numbers in $[0,1]$ is at most each factor. Pooling allows one portfolio's surplus to cover another's deficit, so merged survival does not require every original portfolio to remain solvent separately.

Write the year-$j$ count as a sum of its $m_j$ individual counts, with $m_j>0$. The specified conditional independence within that year givesDividing by $m_j$ and scaling the conditional variance by $m_j^{-2}$ therefore givesThis is the Bühlmann–Straub model: larger exposures reduce the process noise in a year's average.

Define the population parameters and total observed exposure byHere $v$ is the expected process variance and $a$ the variance of hypothetical means. Assume finite second moments. The law of total variance gives $\mathrm{Var}(X_j)=a+v/m_j$, and conditional independence between years gives $\mathrm{Cov}(X_j,X_k)=a$ for $j\ne k$: their only shared variation is the latent conditional mean.

To derive the Bühlmann–Straub credibility estimate, minimize mean squared error among affine estimates of $\mu (\Theta )$. For coefficients $b_j$ and $B={\sum }_j{b}_j$, the optimal intercept is $m_0(1-B)$. Centering at $m_0$ and conditioning on $\Theta$ shows that the resulting error isThe cross term vanishes because $X_j-\mu (\Theta )$ has conditional mean zero; the conditional noise cross terms vanish by the between-year independence assumption.

For fixed $B$, the Cauchy-Schwarz inequality giveswith equality when $b_j=B{m}_j/W$. Minimize the remaining quadratic $a(1-B{)}^2+(v/W)B^2$. Its minimizing value is the Bühlmann–Straub credibility factorThe experience term is exposure-weighted:Since the future conditional mean count is $m_{n+1}\mu (\Theta )$, multiplying the optimal estimate by the known future exposure givesIt is the best affine linear least-squares projection, not a claim that the exact conditional expectation given all data is always affine. If $a=0$, the conditional mean is a known constant almost surely and one takes $Z=0$; if $v=0<a$, the observations reveal it without process noise and $Z=1$. The ordinary case has $a,v>0$.

Use the gamma distribution shape-rate convention: $\Theta \sim \mathrm{Gamma}(\alpha ,\beta )$ with $\alpha ,\beta >0$. The conditional Poisson distribution has mean and variance both equal to $\Theta$, so the Bühlmann–Straub model parameters areFor $W={\sum }_j{m}_j$ and $y={\sum }_j{y}_j$, the Bühlmann–Straub credibility factor is $Z=W/(W+\beta )$. Thusand the required expected-count estimate isThe total exposure determines how much information the observed count carries; the number of years alone is not the appropriate denominator.

Conditional on $\Theta =\theta$, summing the individual independent Poisson counts gives $Y_j\sim \mathrm{Poisson}(m_j\theta )$. The between-year conditional independence makes the likelihood functionMultiplying by the gamma distribution prior density, proportional to ${\theta }^{\alpha -1}{e}^{-\beta \theta }$, gives the Poisson-gamma conjugacy with unequal exposures:For an action $t$, the posterior squared-error loss decomposes asHence the Bayes estimator under squared error loss is the posterior mean, givingFuture counts are independent of the observed years conditional on $\Theta$, so the law of total expectation then gives the posterior predictive expected countThe Bayesian and credibility estimates coincide exactly. This exact Bühlmann–Straub credibility for Poisson-gamma counts occurs because the posterior mean is already affine in the exposure-weighted experience, and therefore belongs to the class over which the credibility estimate minimizes mean squared error.