Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 28 (Actuarial Statistics)

Let $G(z)={\sum }_{n\ge 0}{p}_n{z}^n$ be the probability generating function. For $|z|<1$ the series can be differentiated term by term: $\sum n{p}_n|z{|}^{n-1}$ is finite, even without a finite claim-count expected value. Multiply the Panjer claim-count class recurrence by $n$ and put $m=n-1$. ThenHenceThis argument is valid throughout the open unit disk. Values at boundary points may be obtained by a limit when the required derivatives exist; one need not assume $\mathbb{E}N<\infty$ to establish the identity.

For a Poisson distribution of parameter $\lambda$, its probability generating function is $\exp \{\lambda (z-1)\}$. Applying the law of total expectation to the Poisson mixture givesThusFor $|z|\le 1$ the absolute value of the integrand is bounded by one, so this identity always exists as a Laplace transform of the positive mixing variable. An extension to $z>1$ requires the corresponding moment-generating function to be finite; the conditional computation itself does not guarantee positive exponential moments.

The exponential distribution with rate $\beta >0$ has moment-generating function $\beta /(\beta -t)$ for $t<\beta$. Shifting it by $\alpha$ multiplies the moment-generating function by $e^{\alpha t}$. Therefore the shifted-exponential Poisson mixture has probability generating functionThe probability generating function converges and is analytic for $|z|<\beta +1$, and in particular is finite for real $0\le z<\beta +1$.

Write $p=\beta /(\beta +1)$ and $q=1/(\beta +1)$, so $p+q=1$. The factor $e^{\alpha (z-1)}$ is the probability generating function of $V\sim \mathrm{Poisson}(\alpha )$, while $p/(1-qz)$ is that of a geometric distribution on $\{0,1,\dots \}$, with $\mathbb{P}(W=j)=p{q}^j$. Choose $V$ and $W$ independently. The product rule for the probability generating function of a sum of independent counts then proves $N\stackrel{d}{=}V+W$. The support-zero convention for $W$ is essential here.

The convolution of independent random variables gives the finite sumTaking the logarithmic derivative of the probability generating function givesCompare coefficients of $z^{n-1}$, for $n\ge 2$. The left side is $(\beta +1)n{p}_n-(n-1)p_{n-1}$, and the right side is $\{1+\alpha (\beta +1)\}{p}_{n-1}-\alpha p_{n-2}$. Hence the shifted-exponential Poisson count recursion isThe constant and linear coefficients supply the starting valuesThese two values determine every subsequent probability by the recursion. The negative second term does not mean that the distribution has negative probabilities: the convolution formula exhibits every $p_n$ as a sum of nonnegative quantities.

At $\alpha =0$ the independent Poisson distribution component is identically zero and the count has a geometric distribution. Both the finite sum and the recursion reduce to $p_n=q{p}_{n-1}$. Consequently the Panjer claim-count class parameters areAlthough the shifted model initially has positive $\alpha$, this zero-shift boundary case is well defined.

For an aggregate claims model, let $X$ denote one claim size, with expected value $\mu$ and variance ${\sigma }^2$. The total $S$ is zero when the claim count is zero. Given $N=n$, independence givesThe law of total expectation and law of total variance, together with $\mathbb{E}N=\mathrm{Var}(N)=\lambda$ for a Poisson distribution, implyThe raw second moment appears because the random count itself contributes $\lambda {\mu }^2$ to the variance.

Conditional on $N$, the moment-generating function of the sum is $M_X(t{)}^N$. Averaging with the Poisson distribution therefore gives the compound Poisson distribution transformThis is valid wherever $M_X(t)$ is finite. Finite first and second moments alone do not ensure positive exponential moments; for positive claims the corresponding Laplace transform always exists.

For the independent portfolios put $\Lambda ={\lambda }_1+{\lambda }_2>0$ and $w_i={\lambda }_i/\Lambda$. Their Poisson distribution counts add to a Poisson distribution with parameter $\Lambda$. By Poisson-multinomial conditioning, conditional on the total count $n$ the first risk count has a binomial distribution with parameters $n,w_1$, and the second is the remaining count. Thus one can generate the same total loss by drawing $n$ independent risk labels with these weights, then drawing each claim from its label's law. The merged severity has mixture distributionIt follows that $T$ has a compound Poisson distribution with count parameter $\Lambda$ and this severity law. This is the fixed-year version of Poisson superposition of insurance portfolios. Its expected value and variance areAlternatively, multiplying the two independent aggregate Laplace transforms yields $\exp \{\Lambda (w_1{M}_{X_1}(t)+w_2{M}_{X_2}(t)-1)\}$ wherever finite, confirming the same compound Poisson distribution.

For the retained compound Poisson aggregate under per-claim reinsurance, replace each claim $X$ by its retained payment $g(X)$. Assume the retention is measurable, with $0\le g(x)\le x$ as usual. The count parameter remains $\Lambda$, and the severity law is the pushforward measure of the mixture severity under $g$. Thus the retained compound Poisson aggregate $T_I$ has a compound Poisson distribution with that transformed severity andIts moment-generating function is $\exp \{{\sum }_i{\lambda }_i(M_{g(X_i)}(t)-1)\}$ on its finite domain. For a general retention $g$, zero retained payments can occur and are allowed as compound-Poisson marks; they may equivalently be removed by Poisson thinning. The two specified contracts retain strictly positive payments for strictly positive claims.

For quota share reinsurance the insurer retains the same fraction of each claim, soThe annual retained loss is consequently $T_I=\alpha T$. Each transformed risk severity has probability density function $f_i(y/\alpha )/\alpha$ for $y>0$. The mixed transformed severity, with the same weights ${\lambda }_i/({\lambda }_1+{\lambda }_2)$, still gives a compound Poisson distribution. Scaling the expected value and variance givesThis agrees with the general retained-claim formulas because $\mathbb{E}g(X_i)=\alpha {\mu }_i$ and $\mathbb{E}[g(X_i{)}^2]={\alpha }^2({\sigma }_i^2+{\mu }_i^2)$.

For excess of loss reinsurance the insurer pays each claim up to its retention level:The cap applies separately to every claim. In particular the retained annual loss is ${\sum }_j\min (X_j,M)$, rather than a single cap on the annual total.

Let $F_i$ be the cumulative distribution function for the claim size on risk $i$, and put ${\overline{F}}_i=1-F_i$. The retained severity on that risk has the original probability density function on $0<y<M$ and an atom of a measure at $M$ of mass ${\overline{F}}_i(M)$. Thus $T_I$ has a compound Poisson distribution with rate ${\lambda }_1+{\lambda }_2$ and the mixture of these capped severity laws. The mixture's mass at $M$ is ${\sum }_i{\lambda }_i{\overline{F}}_i(M)/({\lambda }_1+{\lambda }_2)$.

For the capped claim moments, use the tail integral formula for moments. Since $\mathbb{P}(\min (X_i,M)>x)={\overline{F}}_i(x)$ for $0\le x<M$ and is zero for $x\ge M$,Substitution into the compound Poisson distribution moment formulas givesEquivalently, the integrals are ${\int }_0^{M}x{f}_i(x)dx+M{\overline{F}}_i(M)$ and ${\int }_0^M{x}^2{f}_i(x)dx+M^2{\overline{F}}_i(M)$. The annual variance uses the retained raw second moments; subtracting their squared means would omit the variation in the Poisson distribution count.

In the classical risk model write $U(t)=u+ct-{\sum }_{j=1}^{N(t)}{X}_j$, where the Poisson process $N(t)$ has rate $\lambda$ and is independent of the claim sizes. Define the ruin time and ultimate ultimate ruin probability byThe relative safety loading is $\rho =c/(\lambda \mu )-1>0$, so $c=(1+\rho )\lambda \mu$. The adjustment coefficient is the nonzero positive solutionExistence and uniqueness follow from the secant-slope existence criterion for an adjustment coefficient. Explicitly, $\lambda (M(r)-1)-cr$ has derivative $\lambda \mu -c<0$ at zero and is strictly convex for positive claims. The assumed divergence of $M$ makes it cross zero once at a positive $R<r_{\infty }$. When $r_{\infty }=\infty$, an exponential lower bound from any positive tail event shows that $M(r)/r\to \infty$. In particular $R$ is inside the finite-transform domain, not at its endpoint.

Put $\overline{F}=1-F$. Replacing $\phi =1-\psi$ in the survival renewal equation for a classical risk model yieldsBy the tail integral formula for moments, $\mu ={\int }_0^{\infty }\overline{F}(x)dx$. Therefore this is the defective renewal equationThe original kernel has mass $\lambda \mu /c=1/(1+\rho )<1$. For the exponential tilt of the ruin renewal kernel, setMultiplying by $e^{Ru}$ gives the required proper renewal equationTo verify that this is a probability renewal kernel, Tonelli theorem gives, for $r>0$ in the finite-transform domain,The adjustment coefficient equation consequently implies ${\int }_0^{\infty }{k}_R(x)dx=1$. Its expected value isIt is finite because $R$ is interior to the finite-transform domain and is positive because the strict convex crossing has derivative $\lambda M^{\prime }(R)-c>0$.

We quote the key renewal theorem in the following form: for a nonarithmetic distribution $K$ of positive increments with finite positive mean $m$, and a directly Riemann integrable nonnegative function $h$, the locally bounded solution of $z=h+z\ast K$ satisfies $z(u)\to m^{-1}{\int }_0^{\infty }h$. It has renewal representation $z=h\ast {\sum }_{j\ge 0}{K}^{\ast j}$; this follows by iterating the equation, since the probability that arbitrarily many positive increments have sum at most a fixed $u$ tends to zero.

Here $K_R(dx)=k_R(x)dx$ is absolutely continuous, and hence nonarithmetic distribution, even if the original claim law has atoms. The function $h_R$ is continuous. Choose $\delta >0$ with $M(R+\delta )<\infty$. The Markov inequality gives $\overline{F}(x)\le M(R+\delta )e^{-(R+\delta )x}$, and hence $h_R(u)\le C_{\delta }{e}^{-\delta u}$. Continuity on compact intervals and this exponential bound make the upper Riemann sums finite with uniformly vanishing tails, proving direct Riemann integrability.

A further application of Tonelli theorem evaluates the forcing integral:All hypotheses of the key renewal theorem are now checked, so the interior adjustment coefficient ruin prefactor isThis proves the requested Cramér–Lundberg ruin asymptotic, including its constant.

For the two-component hyperexponential distribution, conditioning on the chosen exponential component givesIts moment-generating function and derivative areThe moment-generating function diverges at $1/2$, while the derivative of the adjustment equation at zero is negative. Strict convexity therefore places its unique positive adjustment coefficient inThe adjustment equation, divided by $R$, becomesUsing this identity to subtract $(1+\rho )\mu$ from $M^{\prime }(R)$ givesThus the asymptotic constant in terms of $\rho$ and $R$ isThe denominator is positive on the identified domain. If desired, $R$ is the smaller root of $6(1+\rho )R^2-(5+9\rho )R+3\rho =0$; the other algebraic root lies outside the positive finite-transform interval and is not an adjustment coefficient.

Write $m(\lambda )=\mathbb{E}(X_i\mid \lambda )$ and $v(\lambda )=\mathrm{Var}(X_i\mid \lambda )$. The Bühlmann model uses the finite structural parametersHere $v$ is the expected process variance and $a$ is the variance of hypothetical means. The target is the latent conditional mean $m(\lambda )$, rather than the realized next count. The Bühlmann credibility estimate is the best affine predictor of that target under mean squared error.

The law of total expectation and law of total variance give $\mathbb{E}{X}_i=m_0$ and $\mathrm{Var}(X_i)=v+a$. Conditional independence gives $\mathrm{Cov}(X_i,X_j\mid \lambda )=0$ for $i\ne j$, and the law of total covariance therefore yieldsTo derive the optimal predictor, consider $d=A+{\sum }_{i=1}^n{b}_i{X}_i$. Minimizing with respect to the intercept gives $A=m_0(1-{\sum }_i{b}_i)$. Thus $d=m_0+{\sum }_i{b}_i(X_i-m_0)$. The linear least-squares projection normal equations areFor $v>0$, subtraction of any two equations forces all $b_i$ equal. Substituting a common coefficient then gives $b_i=a/(v+na)$. Equivalently, the mean squared error of the centered predictor is $a-2a{\sum }_i{b}_i+v{\sum }_i{b}_i^2+a({\sum }_i{b}_i{)}^2$, a convex quadratic with precisely these normal equations. HenceThe ratio notation assumes $a>0$; the credibility factor formula $na/(na+v)$ also handles $a=0,v>0$. If $a=0$, the target is the constant $m_0$ almost surely. If $v=0,a>0$, one observation already equals $m(\lambda )$ almost surely, and the average gives it exactly. If both vanish, the target and observations are constant.

The result optimizes over affine functions of the observations. It need not equal the unrestricted posterior mean; exact Bayesian inference generally depends on the whole prior and likelihood, whereas the Bühlmann credibility estimate uses these second-moment structural parameters.

Conditional on $\lambda$, a Poisson distribution has both conditional expectation and conditional variance equal to $\lambda$. For the uniform distribution on $(1,3)$,Thus $K=v/a=6$ and the one-year credibility factor is $Z_1=1/7$. With the observed count equal to one, the Bühlmann credibility estimate of the next year's conditional claim expected value isThe estimate gives relatively little weight to one year because the expected process variance is six times the variance of hypothetical means.

The structural parameters remain $m_0=2$, $v=2$ and $a=1/3$, so the credibility factor after $n$ years is $Z_n=n/(n+6)$. Writing $C_n={\sum }_{i=1}^n{x}_i$, the Bühlmann credibility estimate isThe target is $\mathbb{E}(X_{n+1}\mid \lambda )=\lambda$. It is also the best affine predictor of the next year's count, because its extra conditional noise has zero covariance with past observations.

With a new count $x_{n+1}$, the updated Bühlmann credibility estimate is $(C_n+x_{n+1}+12)/(n+7)$. Subtracting the previous estimate gives the sequential Bühlmann credibility updateTherefore the new estimate is strictly smaller exactly whenBecause a claim count is a nonnegative integer, the equivalent set is $x_{n+1}\in \{0,1,\dots ,\lceil {\hat{m}}_n\rceil -1\}$. A count equal to the previous estimate leaves it unchanged; a larger count increases it. The comparison is with the previous credibility estimate, which already combines the prior population mean and the observed mean.

Under quadratic loss, the Bayes estimator under squared error loss of the latent mean $\lambda$ is its posterior mean. For the Poisson-uniform posterior mean, the prior distribution has density $1/2$ on $(1,3)$ and the one-count Poisson distribution likelihood is $\lambda e^{-\lambda }$. Consequently the Bayesian posterior density is proportional to $\lambda e^{-\lambda }$ on that interval, with normalizing integral ${\int }_1^3\lambda e^{-\lambda }d\lambda$.

By conditional independence, $\mathbb{E}(X_2\mid \lambda ,X_1)=\lambda$, so the law of total expectation makes the posterior predictive expected value equal to this same posterior mean. It isThe required antiderivatives are $-(\lambda +1)e^{-\lambda }$ and $-({\lambda }^2+2\lambda +2)e^{-\lambda }$. Evaluating at the two endpoints givesThis estimate differs from $13/7$: the Bühlmann credibility estimate is an optimal affine rule, while the Bayes estimator under squared error loss optimizes over all rules and uses the truncated uniform prior through its exact Bayesian posterior.