Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 45 (Actuarial Statistics)

Condition on the claim count distribution. If $\mu =\mathbb{E}{X}_1$ and $v=\mathrm{Var}(X_1)$, then independence of the count and the independent and identically distributed random variables givesThe empty sum is zero. The law of total expectation and law of total variance therefore prove the random sum of independent claims formulasFor the variance calculation assume the count and severity have finite second moments; the expectation identity needs the corresponding integrability. Nonnegative claims also permit the expectation identity with extended nonnegative values.

Represent each policy payment as $S_i=B_i{Y}_i$, where $B_i$ is a Bernoulli random variable of parameter $q_i$, independent of its severity. Equivalently, $Y_i$ specifies the payment distribution conditional on a claim. ThenIndependence of the policies makes their variances add. PutThe heterogeneous individual claims model consequently has

For the Poisson approximation, define the severity mixture distributionLet ${\phi }_i(t)=\mathbb{E}[e^{it{Y}_i}]$ be the individual characteristic functions. Conditioning on each Poisson count and then multiplying transforms of the independent policy payments givesThis is Poisson superposition of insurance portfolios. By the uniqueness theorem for characteristic functions,with count and severities independent. Its compound Poisson distribution has distribution functionwhere $F^{\ast k}$ is the convolution distribution of $k$ independent severities and $F^{\ast 0}$ is the distribution concentrated at zero. Characteristic functions avoid assuming that severity moment-generating functions exist. Since $\mathbb{E}Z=M/Q$ and $\mathbb{E}{Z}^2=A/Q$, the random-sum formulas give

Use $T_{\mathrm{B}}$ to distinguish the second, compound-binomial approximation from the Poisson one. It has $K_{\mathrm{B}}\sim \mathrm{Bin}(n,p)$ with $p=Q/n$ and the same severity law $F$. Thus $\mathbb{E}{K}_{\mathrm{B}}=Q$ and $\mathrm{Var}(K_{\mathrm{B}})=Q(1-p)$. Applying the random sum of independent claims identities again givesHence

Both approximations match the exact expected value. The Cauchy-Schwarz inequality gives $M^2\le nC$, proving the variance comparison for compound portfolio approximations:The excess variances are respectively $C-M^2/n$ and $C$. The compound-binomial model is therefore closer in variance; equality with the exact variance occurs when all $q_i{\mu }_i$ are equal. This variance comparison is not a general ordering of approximation errors or tail probabilities.

The Poisson replacement is useful when all claim probabilities are small: it replaces the individual restriction of at most one claim by an unrestricted Poisson count, and the variance error is of second order in those probabilities. The compound-binomial replacement retains a count bounded by $n$, but pools both frequencies and severity laws and can lose policy-specific information. If all $q_i$ and all $F_i$ agree, it is exactly the original portfolio law. In a heterogeneous portfolio, agreement of means, or even variances, does not make either approximate law exact. Neither approximation is guaranteed accurate merely by having many policies.

The claim law is an Erlang distribution of shape two and rate $2/\beta$. Direct integration, using ${\int }_0^{\infty }x{e}^{-ax}dx=a^{-2}$ for $a>0$, gives its moment-generating functionDifferentiating at zero gives $\mathbb{E}X=\beta$. The positive relative safety loading condition for the classical risk model is thereforeThe adjustment coefficient is the positive root in the finite-transform domain ofSet $z=\beta R/2$ and $k=2c/(\lambda \beta )>2$. For $R>0$, division by $z$ reduces the equation toThe two roots are $z=(2k-1\pm \sqrt{1+4k})/(2k)$. The minus root lies strictly between zero and one: positivity follows from $(2k-1{)}^2-(1+4k)=4k(k-2)>0$, and the numerator is smaller than $2k$. The plus root exceeds one and is outside the moment-generating function's domain. Consequently the adjustment coefficient for shape-two Erlang claims isThe zero root of the original equation is not an adjustment coefficient. The increasing secant slope $(M_X(r)-1)/r$ on $(0,2/\beta )$ also establishes uniqueness of the positive root.

Now set $\beta =1$. Under quota share reinsurance, the retained payment is $Y_{\alpha }=\alpha X$. For $0<\alpha \le 1$, the change of variables $x=y/\alpha$ yieldsHere the second gamma parameter is the rate. In particular, $\mathbb{E}{Y}_{\alpha }=\alpha$ and $M_{Y_{\alpha }}(r)=(1-\alpha r/2{)}^{-2}$. At full cession $\alpha =0$ the retained payment instead has a point mass at zero.

The reinsurer's expected annual payments are $\lambda (1-\alpha )$, so its loaded premium is $(1+{\theta }_R)\lambda (1-\alpha )$. Subtract it from the original premium income to obtain the direct insurer's net premium rateIts expected retained claim outflow is $\lambda \alpha$. Thusby the stipulated lower bound on $\alpha$. In particular $c_{\alpha }>0$, so the retained portfolio satisfies the net profitability condition. Its relative safety loading is ${\theta }_R-({\theta }_R-\theta )/\alpha$.

Apply the same adjustment coefficient for shape-two Erlang claims formula with scale $\alpha$ and net premium $c_{\alpha }$. WritinggivesEquivalently it is the positive solution, with $R_{\alpha }<2/\alpha$, of

For fixed positive initial capital, a larger adjustment coefficient reduces the exponential upper bound ${\psi }_{\alpha }(u)\le e^{-R_{\alpha }u}$ from the Lundberg inequality. It also makes the large-capital decay exponent in the Cramér–Lundberg ruin asymptotic larger. These are the reasons for maximizing $R_{\alpha }$ as a ruin-risk criterion; the coefficient alone need not order exact finite-capital ruin probabilities for arbitrary different models.

If ${\theta }_R=\theta$, then $c_{\alpha }=(1+\theta )\lambda \alpha$ and $k_{\alpha }=2(1+\theta )$ is constant. The equal-loading quota share and vanishing retention relation becomesThe numerator is positive for $\theta >0$, so $R_{\alpha }$ decreases strictly with $\alpha$ and tends to infinity as $\alpha \downarrow 0$. There is no maximizing retention in the stipulated interval $0<\alpha \le 1$; the best limit is full cession. If the endpoint $\alpha =0$ is admitted as an idealized contract, all claims are transferred and the net premium income is zero, leaving the direct insurer's capital constant and its ultimate ruin probability zero. At that endpoint the coefficient equation is identically zero, so it no longer defines a unique finite adjustment coefficient. Calling $\alpha =0$ optimal requires this explicit extension of the original admissible set.

The surplus in the classical risk model is $U(t)=u+ct-S(t)$. With ruin defined as a strictly negative surplus, its ultimate ruin probability isLet ${\overline{F}}_X(v)={\int }_v^{\infty }{f}_X(x)dx$. The first arrival time has exponential distribution of rate $\lambda$, independent of the first claim. Before that claim no ruin occurs, because the premium rate is positive. Given its arrival at $t$ and size $x$, available capital just before it is $u+ct$. If $x>u+ct$ ruin is immediate; otherwise the independent increments and fresh claim sizes make subsequent ruin probability $\psi (u+ct-x)$. Thus first-claim conditioning givesThis is the ruin counterpart of the first-claim decomposition for survival probability. DefineThe convolution term is continuous: extend the bounded function $\psi$ by zero to negative arguments and use continuity of translations in $L^1$ for the integrable density $f_X$. The tail function is also continuous. Changing variables to $v=u+ct$ yieldsDifferentiating this exponentially weighted integral establishes differentiability rather than presuming it. The fundamental theorem of calculus givesReversing the integration variable in the convolution therefore proves the ruin integro-differential equation:At zero the derivative is the right derivative; the same representation supplies it.

For the exponential claim law, introduceDifferentiating the explicit convolution gives $A^{\prime }=\alpha \psi -\alpha A$. Henceusing the proved integral equation. Differentiating that equation once more givesso the required ordinary differential equation isPositive relative safety loading means $c>\lambda /\alpha$, so $\kappa =\alpha -\lambda /c>0$. The general solution is $a+b{e}^{-\kappa u}$.

The condition at large capital must also be used. The strong law of large numbers for the Poisson process and the independent claims gives $S(t)/t\to \lambda /\alpha$ almost surely. Thus $S(t)-ct\to -\infty$. Its path has only finitely many jumps on each bounded time interval, so its supremum over all time is finite almost surely. Therefore $\psi (u)\to 0$ as $u\to \infty$, forcing $a=0$. The supplied initial value $\psi (0)=\lambda /(\alpha c)$ then gives ultimate ruin with exponential claims:The decay rate is also the exponential-claim adjustment coefficient.

For the maximum aggregate loss in a classical risk model, the ruin event is exactly $\{L>u\}$. Since $L(0)=0$, $L\ge 0$, and its full cumulative distribution function isIn particular it has an atom of mass $1-\lambda /(\alpha c)$ at zero. Conditional on $L>0$, its distribution is exponential with rate $\kappa$; omitting that atom would give an incorrectly normalized law.

Let $Z_m$ be the discount level during policy year $m$, with $Z_0=1$. Under the usual assumption of independent successive annual accidents and repair costs with the same laws each year, the submitted-claim decision depends only on $Z_m$ and that year's cost. Consequently the conditional distribution of $Z_{m+1}$ depends on the past only through $Z_m$: this is the Markov property. The no claims discount system is therefore a time-homogeneous Discrete-time Markov chain. Stationarity of accident probabilities alone, without this independence assumption, would not by itself establish the Markov property.

At level one a claim keeps the policyholder at one, while no submitted claim moves them to two. At level two, a submitted claim moves them to one and no claim to three. At level three, a claim moves them to two and no claim keeps them at three. Thus $p_i$ in the specified transition matrix is the annual probability of a submitted claim at current level $i$, not merely of an accident. A small accident paid privately counts as no claim for these transitions.

Let $b>0$ denote the undiscounted annual premium. It is not specified numerically in the source; costs can instead be measured in units of that premium, giving $b=1$. The actual premium payments at the three levels are $b$, $b(1-\alpha )$, and $b(1-\beta )$. Compare the two future paths, assuming no further accidents, to obtain the two-year premium-loss claim threshold. From level one, a current claim leads to future levels $(1,2)$, whereas no claim leads to $(2,3)$, giving extra premiumFrom level two the two paths are $(1,2)$ and $(3,3)$, and from level three they are $(2,3)$ and $(3,3)$. HenceThese are all strictly positive. The PDF's discount fractions are $\alpha$ and $\beta$; the converted TeX has lost these factors in the percentages.

For a repair cost $X$ with the specified lognormal distribution, $\log X\sim N(\mu ,{\sigma }^2)$. Putwhere $\Phi$ is the standard normal distribution function. Conditional on an accident, the submitted-claim probability is $\mathbb{P}(X>h_i)=\overline{\Phi }(z_i)$. Multiplying by the accident probability givesSet $b=1$ for the premium-unit convention. Without a monetary premium scale, discount percentages alone cannot determine a numerical threshold for a monetary repair cost.

This nearest-neighbour Markov chain is a finite birth-death chain. Its stationary distribution $\pi$ satisfies the endpoint balance equationsThese detailed balance equations also imply the middle-state balance equation. The three-level no claims discount equilibrium is thereforewhereSince each $p_i$ lies strictly between zero and one, this is an irreducible Markov chain. Positive holding probabilities at the endpoints also make it an aperiodic Markov chain. Its unique stationary distribution is the limiting distribution even from the initial level one, and gives the steady-state proportions for a homogeneous population of such policyholders.

A submitted claim at level one conditions the repair cost on $X>h_1$. The accident probability cancels in this conditional expectation. For $z_1=(\log h_1-\mu )/\sigma$, the truncated lognormal moment follows by completing the square:Dividing by $\mathbb{P}(X>h_1)=\overline{\Phi }(z_1)$ yields the expected size of an actual submitted claim:The unconditional expected amount paid per year from level one would instead multiply its tail first moment by $p$; it is not the conditional claim size requested here.