Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 37 (Actuarial Statistics)

Write $m_j=\mathbb{E}{Y}^j$ and $M=M_Y$. The derivatives of the cumulant-generating function areSince $M(0)=1$ and $M^{(j)}(0)=m_j$, the first three cumulants areThe last equality follows by expanding the cube and taking expectations. We use the paper's unstandardized third central moment throughout. The usual skewness coefficient divides this quantity by $\mathrm{Var}(Y{)}^{3/2}$; its sign is the same whenever the variance is positive.

There is a minor domain qualification: finite ordinary moments do not guarantee a finite moment-generating function for positive arguments. The differentiations are valid in a neighborhood of zero when a positive exponential moment exists. For positive variables with the stated finite moments, all formulas also follow from left derivatives at zero, because $Y^j{e}^{tY}\le Y^j$ for $t\le 0$ and dominated convergence applies. Thus no extra exponential-moment assumption is needed for the moment identities themselves.

For the random sum of independent claims, condition on $N=k$. Independence of the claims and independence from the count giveTaking expectations and then logarithms yields the random-sum transform identity and the cumulant composition for a random sum:Here $G_N$ is the probability generating function. These identities hold wherever the transforms are finite, in particular for $t\le 0$ with positive claims.

For reuse in the three cases, write $m_j=\mathbb{E}{X}_1^j$ and $a_j={\kappa }_{N,j}$. Applying the chain rule three times givesor equivalentlyThis expresses the requested third central moment in raw claim moments, while keeping the contribution of claim-count variability explicit.

For the Poisson distribution, $G_N(z)=e^{\lambda (z-1)}$, so the aggregate has a compound Poisson distribution andThree differentiations immediately giveThe first two cumulants are $\lambda m_1$ and $\lambda m_2$. If $\lambda >0$ and claim sizes are strictly positive, $m_3>0$, so the paper's unstandardized skewness is strictly positive. For the degenerate count $\lambda =0$, the aggregate is zero and the third central moment is zero; strict positivity presumes a positive claim intensity.

Put $q=1-p$. The count has the zero-based geometric distribution, whose probability generating function is $G_N(z)=p/(1-qz)$. Hence the aggregate has a compound geometric distribution, withThis transform requires $q{M}_{X_1}(t)<1$; it holds for every $t\le 0$. To differentiate explicitly, let $g(t)=M_{X_1}(t)$. ThenandAt zero, $g=1$ and $g^{(j)}=m_j$, givingEvery term is positive for $0<p<1$ and strictly positive claim sizes. Therefore the unstandardized skewness is positive, even if the claim-size third central moment itself is negative. The zero-based convention matters: there is no factor $M_{X_1}(t)$ in the numerator of this aggregate transform.

For the binomial distribution, $G_N(z)=(1-p+pz{)}^n$. Thus the compound binomial distribution hasSet $h(t)=1-p+p{M}_{X_1}(t)$. Since $h(0)=1$ and $h^{(j)}(0)=p{m}_j$, differentiating $n\log h$ three times givesThe first two cumulants, for comparison, are $np{m}_1$ and $n(p{m}_2-p^2{m}_1^2)$.

Choose deterministic claim sizes $X_i=d>0$. Then $S=dN$, so $m_j=d^j$ andThis is strictly negative for $p>1/2$ and $n\ge 1$. A concrete negative-skewness example is $n=1$, $p=3/4$, $d=1$, for whichThere is no contradiction with positive claims: bounded Bernoulli-type totals can have more mass near their upper value and a longer left tail.

For excess of loss reinsurance with retention $M$, the direct insurer caps its payment at $M$ and the reinsurer pays the excess. ThusLet $I_i=1_{\{X_i>M\}}$ and $\alpha =\mathbb{P}(X_i>M)=1-F_{X_1}(M)\in (0,1)$. These are independent Bernoulli random variables, independent of $N$. Since $N_R={\sum }_{i=1}^N{I}_i$, its conditional law given $N=n$ is a binomial distribution. Conditioning on $N$ therefore gives the Bernoulli thinning identityHere the probability generating function may be evaluated on $0\le z\le 1$, and the identity extends wherever the transforms converge.

The positive reinsurer marks have the conditional law $W\stackrel{d}{=}{X}_1-M\mid X_1>M$. HenceThe independent-mark representation is justified as follows. Conditional on any fixed count and pattern of exceedance indicators, the positive excesses are independent with this same conditional law; that law depends neither on the number of exceedances nor on their locations. Averaging over the indicator pattern thus leaves the positive mark sequence independent of $N_R$. We may extend it by additional independent copies if needed, giving $S_R={\sum }_{j=1}^{N_R}{W}_j$ with the stated independence.

Now write $q=1-p$. For exponential distribution claim sizes of mean $\mu$,Thus $W$ is exponential with the original mean $\mu$, by memorylessness of the exponential distribution. The original zero-based geometric distribution count has $G_N(z)=p/(1-qz)$. ConsequentlyThis is the geometric count under Bernoulli thinning, withFinally use the Laplace transform of a nonnegative random variable, keeping the mass at zero. For $t\ge 0$,The last expression is the transform of a mixture distribution: mass $p_R$ at zero and, with probability $q_R$, an exponential distribution of rate $p_R/\mu$. Uniqueness of the Laplace transform identifies this zero-based geometric sum of exponential variables:Equivalently, $\mathbb{P}(S_R=0)=p_R$ and its positive density is $q_R(p_R/\mu )e^{-p_{R}s/\mu }$. Therefore the requested exceedance probability isThe prefactor is essential: the aggregate is zero whenever no claim crosses the retention.

In the classical risk model, let $N(t)$ be the Poisson process of claim arrivals, independent of the independent claim sizes. The reserve starts at $u$ and earns premiums continuously at rate $c$, soHere $\psi$ is the ultimate ruin probability and $\varphi =1-\psi$ is the survival probability. The relative safety loading is $\theta =(c-\lambda \mu )/(\lambda \mu )>0$.

To derive the survival integro-differential equation, condition on the first claim time $T$, which has the exponential distribution of rate $\lambda$. Immediately before that claim the available capital is $u+cT$. A claim $x\le u+cT$ leaves survival probability $\varphi (u+cT-x)$, since the future claim process restarts with the same independent law; a larger claim causes ruin immediately. The first-claim decomposition for survival probability is thereforePut $r=\lambda /c$ and $H(v)={\int }_0^v\varphi (v-x)f(x)dx$. Changing variables to $v=u+ct$ yieldsThis representation justifies differentiation. The function $H$ is continuous: extend $\varphi$ by zero to negative arguments, then use continuity of translations of the integrable density $f$ in $L^1$ to control its convolution with the bounded function $\varphi$. Differentiating the lower endpoint of the displayed integral gives ${\varphi }^{\prime }=r\varphi -rH$. Since $c=(1+\theta )\lambda \mu$,This deduction uses the first-arrival law rather than taking the desired differential equation as an assumption.

For $f(x)=4x{e}^{-2x}$, the claim law is an Erlang distribution of shape two and rate two. Its mean is $\mu =4{\int }_0^{\infty }{x}^2{e}^{-2x}dx=1$. With $\theta =2$, we have $r=1/3$. Changing the convolution variable to $t=u-x$ givesHence the first requested specialization isLet $J(u)={\int }_0^u\varphi (t)e^{2t}dt$. Differentiation under the finite integral gives $I^{\prime }=J$ and $J^{\prime }=\varphi e^{2u}$. Differentiating the preceding equation and using $4e^{-2u}I=\varphi -3{\varphi }^{\prime }$ then givesThus the second identity isDifferentiate once more:The preceding second-derivative equation gives $8e^{-2u}J=-6{\varphi }^{\prime \prime }-10{\varphi }^{\prime }+4\varphi$. Substitution cancels the undifferentiated terms and provesThis is the Erlang claim-size differential equation for survival probability with $\beta =2$ and $r=1/3$.

Its characteristic polynomial factors as $z(3z+8)(z+1)$. The limit $\varphi (u)\to 1$ therefore makes the general bounded solutionThe supplied zero-capital survival probability is $\varphi (0)=2/3$, so $A+B=-1/3$. We also need the initial derivative carried by the original integral equation: $I(0)=0$ implies ${\varphi }^{\prime }(0)=\varphi (0)/3=2/9$, giving $-A-(8/3)B=2/9$. Solving gives $A=-2/5$ and $B=1/15$. Therefore the survival and ruin probabilities areThe extra initial derivative is essential: $\varphi (0)$ and the limit at infinity alone would leave one free parameter in the third-order equation. The result has $\psi (0)=1/3$ and tends to zero. It is positive and decreasing, since ${\psi }^{\prime }(u)=-(2/5)e^{-u}+(8/45)e^{-8u/3}<0$ for $u\ge 0$. These signs agree with its interpretation as an ultimate ruin probability.

Let the Bühlmann–Straub model structural parameters beHere $a$ is the variance of hypothetical means and $v$ is the expected process variance. Assume the structural second moments are finite and the exposures $m_j$ are positive, as required by this least-squares problem. Write $A={\sum }_j{a}_j$. Conditional independence separates the squared prediction error into squared conditional bias and conditional variance. The law of total expectation givesIndeed, conditional on $\Theta$, the prediction has mean $a_0+A\mu (\Theta )$ and variance ${\sigma }^2(\Theta ){\sum }_j{a}_j^2/m_j$; there are no cross-covariances because the annual observations are conditionally independent.

For fixed slope coefficients the optimal intercept is $a_0=(1-A)m_0$. For fixed sum $A$, the Cauchy-Schwarz inequality giveswith equality when $a_j=A{m}_j/W$. Thus the remaining minimization is the scalar quadraticDifferentiation gives $A=Z=aW/(aW+v)$. In the nondegenerate case $a,v>0$, the quadratic is strictly convex and the equality condition determines the unique slope coefficients. The Bühlmann–Straub credibility factor and premium are thereforeandThis derives the Bühlmann–Straub credibility estimate from its minimization criterion. Increasing exposure increases the weight on experience; increasing process noise decreases it. No exposure for year $n+1$ appears because the estimate concerns the expected claim per car, not the variance of next year's observed average.

The degenerate cases have a direct interpretation. If $a=0$ and $v>0$, the unknown conditional mean is the known constant $m_0$, so take $Z=0$. If $v=0$ and $a>0$, each observation already equals $\mu (\Theta )$ almost surely, so $Z=1$ gives exact prediction. If both vanish, all observations and the target equal $m_0$ almost surely; use the constant predictor rather than the undefined ratio $0/0$.

In the approximate normal model, $\mu (\Theta )=\Theta$, $m_0=\mu$, $\mathrm{Var}(\Theta )=a$, and ${\sigma }^2(\Theta )=v$. Substitution into the derived credibility estimate givesWe now derive, rather than assume, the Bayes estimator under squared error loss. For positive $a,v$, the likelihood and normal distribution prior give a posterior density proportional toThe coefficient of ${\theta }^2$ is $1/a+W/v$, and the coefficient in the linear term is $\mu /a+{\sum }_j{m}_j{X}_j/v$. Completing the square proves the normal-normal conjugacy with unequal exposures formulawithFor any decision $d$, the posterior quadratic loss isIts unique minimizer is the posterior mean $d=m_{\ast }$. Hence the Bayesian and credibility estimates coincide exactly in this normal model:The reason is that the posterior mean is already affine in the observations, so restricting prediction to affine functions loses nothing. In the general Bühlmann–Straub model, the posterior mean need not be affine; the credibility estimate is then only the best affine squared-error predictor. The zero-variance limits of the normal formulas agree with the degenerate cases above.