Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 27 (Actuarial Statistics)

Where the moment-generating function $M_Y(t)=\mathbb{E}{e}^{tY}$ is finite in a neighborhood of zero, define the cumulant-generating function ${\kappa }_Y(t)=\log M_Y(t)$ and the $r$th cumulant by ${\kappa }_r(Y)={\kappa }_Y^{(r)}(0)$. Differentiation givessince $M_Y(0)=1$. Thus the first two cumulants are the mean and variance.

Conditional on $N=n$, independence gives $\mathbb{E}(e^{tT}\mid N=n)=M_V(t{)}^n$. Summing over the Poisson distribution yieldsHence the compound Poisson cumulants giveFor the moment statements, finite second moments suffice even when a moment-generating function does not exist near zero: conditioning gives $\mathrm{Var}T=\mathbb{E}N\mathrm{Var}V+\mathrm{Var}N(\mathbb{E}V{)}^2=\nu \mathbb{E}{V}^2$. This distinguishes the raw second moment from the variance of an individual claim.

For a single flood, the law of total expectation and law of total variance similarly giveCondition on the number of floods and apply the same laws once more. All floods have these moments and are independent of $M$, so the annual total satisfiesThe additional $\nu {\lambda }^2{\mu }^2$ term is the variation in the number of entire flood clusters; it would be lost by treating individual claims as an ordinary Poisson stream.

For the annual count $N={\sum }_{i=1}^M{N}_i$, its probability generating function isThis is the nested Poisson flood count. Differentiating, or conditioning on $M$, gives $\mathbb{E}N=\nu \lambda$ and $\mathrm{Var}N=\nu \lambda (1+\lambda )$. Thus $N$ is not Poisson when $\nu ,\lambda >0$, since its variance exceeds its mean. If either parameter is zero it is identically zero, a degenerate Poisson distribution.

Let $K_i$ be the number of valid claims in flood $i$. Independent validity marking gives $K_i\mid N_i\sim \mathrm{Binomial}(N_i,1-p)$. Its generating function isso Poisson thinning gives $\boxed{K_i\sim \mathrm{Poisson}(\lambda (1-p))}$. With validity independent also of claim amount, each claim contributes expected paid amount $(1-p)\mu$. Therefore the annual expected payment is $\boxed{\nu \lambda (1-p)\mu }$. The independent-mark assumption is needed for this multiplication; a size-dependent invalidity rule would require the joint expected value of amount and validity instead.

Write the surplus process in the classical risk model as $U_t=u+ct-{\sum }_{j=1}^{N_t}{X}_j$, with positive independent claims. Condition on the first claim time, which has exponential distribution of rate $\lambda$. If that claim occurs at $t$, the available capital is $u+ct$ and survival requires its size not to exceed that capital. The independent future process then givesPut $A(v)={\int }_0^v\varphi (v-x)f(x)dx$ and change variables $v=u+ct$. We obtainThe convolution of bounded $\varphi$ with the integrable density is continuous; differentiation of this expression is therefore justified. It proves the survival version of the ruin integro-differential equation:

The density is a hyperexponential distribution with mixture weights $3/4,1/4$ and exponential rates $4,2$. Its total integral is one and its mean isThe relative safety loading means $c=(1+\rho )\lambda {\mu }_X$. Substitution of $\rho =3/5$ gives $c=\lambda /2$ and $\lambda /c=2$.

Set $A_a(u)={\int }_0^u\varphi (u-x)e^{-ax}dx$. Differentiating in its equivalent form ${\int }_0^u\varphi (v)e^{-a(u-v)}dv$ gives $A_a^{\prime }=\varphi -a{A}_a$. The survival equation becomesApply the constant-coefficient operator $(D+4)(D+2)$, where $D=d/du$. Since $(D+4)A_4=\varphi$ and $(D+2)A_2=\varphi$,Expanding and cancelling yields $\boxed{{\varphi }^{\prime \prime \prime }+4{\varphi }^{\prime \prime }+3{\varphi }^{\prime }=0}$. Thus $\varphi (u)=A+B{e}^{-u}+C{e}^{-3u}$.

Positive loading also fixes the limiting boundary value $\varphi (u)\to 1$. Indeed the aggregate-claims process minus premium income tends to minus infinity almost surely by the strong law of large numbers. Its supremum over all time is consequently finite almost surely, since paths have finitely many jumps on bounded intervals. The probability that this supremum exceeds $u$ tends to zero. Hence $A=1$. At zero, the given value is $\varphi (0)=3/8$, and the integral equation supplies the extra condition ${\varphi }^{\prime }(0)=2\varphi (0)=3/4$. Thereforeso $B=-9/16$ and $C=-1/16$. The nonruin probability isThis is the two-exponential-mixture survival probability. It increases from $3/8$ to $1$ and satisfies the original integral equation, not just the differentiated equation; the zero-capital derivative condition prevents introducing extraneous solutions.

A credibility factor is the weight assigned to a particular risk's observed experience when blending it with a population or manual premium. In the Bühlmann model, the blend is the best affine mean-square predictor, rather than automatically the full posterior mean.

Let $v=\mathbb{E}[{\sigma }^2(\theta )]$, the expected process variance, and assume the displayed second moments are finite. The law of total expectation gives $\mathbb{E}{X}_i=m$. The law of total variance and law of total covariance giveFor the second formula, the conditional covariance is zero by conditional independence, while the covariance of the conditional means is $\mathrm{Var}\mu (\theta )=a$, the variance of hypothetical means. Likewisebecause $\mathbb{E}[X_i-\mu (\theta )\mid \theta ]=0$.

To derive the Bühlmann credibility premium, minimize $\mathbb{E}[(\mu (\theta )-b_0-{\sum }_i{b}_i{X}_i{)}^2]$. Differentiating in $b_0$ sets the predictor's mean to $m$. The remaining normal equations areFor $v>0$, subtracting any two rows forces all $b_i$ to be equal, then gives $b_i=a/(v+na)$. Thus the premium and factor areThe same affine premium predicts $X_{n+1}$: conditional independence makes $\mathbb{E}[(X_{n+1}-\hat{\mu }{)}^2]=v+\mathbb{E}[(\mu (\theta )-\hat{\mu }{)}^2]$, so minimizers agree. If $v=0$, every year's amount equals $\mu (\theta )$ almost surely, and $Z=1$ gives the exact risk mean even though the coefficient vector need not be unique.

For fixed $a>0,n$, $\partial Z/\partial v=-na/(na+v{)}^2<0$. Thus more within-risk variation decreases credibility: individual experience becomes less informative, $Z\to 0$ as $v\to \infty$, and the premium tends to $m$. Increasing the number of observed years has the opposite effect.

For numerical estimation from a balanced panel, assume the $k$ risks are independent draws from the same risk population, with comparable exposures, and $n,k\ge 2$. SetConditionally, each within-risk sample variance has mean ${\sigma }^2({\theta }_s)$; hence $\mathbb{E}\hat{v}=v$. The risk sample means have variance $a+v/n$, so $\mathbb{E}B=a+v/n$. Therefore $\tilde{a}=B-\hat{v}/n$ is an unbiased method of moments estimate of $a$. For a usable nonnegative variance estimate take $\hat{a}=\max (0,\tilde{a})$, noting that this truncation sacrifices exact unbiasedness.

The balanced-panel method-of-moments credibility estimate for risk $s$ is consequentlyIf $\hat{a}=0$ and $\hat{v}>0$, use zero credibility. If both estimates are zero, the within and between sums of squares are zero and all the observed amounts coincide; any blend gives that same premium. With only one year per risk or only one risk, these within/between estimators are unavailable, so external structural estimates or additional assumptions are necessary.

Record the discount level at the start of each policy year as the state. A no claims discount system is a Discrete-time Markov chain when, conditional on this state, the next year's claim behavior and update rule do not depend on earlier history. Independent annual accident and loss mechanisms with a fixed state-based reporting rule give this property; one should not equate the probability of an accident with the probability of a submitted claim.

The density identifies $Z=(\log L-\mu )/\sigma$ as a standard normal random variable: substitute $x=e^{\mu +\sigma z}$ and $dx=\sigma xdz$. Therefore, for $a>0$,Here $L$ is the numerical repair cost. The character preceding $L$ in the PDF is a currency symbol, not a separate multiplier.

Order states as $0\%,20\%,30\%$. With the stipulated common annual probability $p$ of no submitted claims, the transition matrix isA claim resets any state to zero, while a no-claim year raises the discount one level or keeps the top level fixed. If claim probabilities depend on the state, the corresponding row instead uses $p_i=\mathbb{P}(\text{no submitted claim}\mid \text{state }i)$; the same transition mechanism applies. The supplied single $p$ is a simplifying model assumption.

For a policyholder currently at zero discount, the current premium $C$ has the same value under either decision and cancels in the comparison. If the repair is self-funded and no later accidents occur, the next two premiums are $0.8C$ and $0.7C$, giving a total cost $L+2.5C$ including the current premium. If a claim is made, the next two premiums are $C$ and $0.8C$, giving total cost $2.8C$, with the repair paid by the insurer. Thus the two-year no-claims threshold with reset to zero isAt equality the holder is indifferent; it has probability zero under the continuous density. This assumes full reimbursement and no excess or deductible, as in the stated model.

Put $a=0.3C$. The conditional expected claim is $\mathbb{E}[L\mid L>a]$. Completing the square in the normal integral gives the truncated lognormal momentDividing by the reporting probability gives the expected size of a submitted claim for the full-premium holder:All monetary amounts must use the same unit despite the PDF's mixed currency notation. The other two current levels both have next-two-year no-claim premiums $0.7C+0.7C$, versus $C+0.8C$ after a claim, so their thresholds are $0.4C$. Consequently incorporating this behavioral decision into claim probabilities generally produces state-specific $p_i$, rather than deriving a common $p$ from the severity density alone; no accident-frequency law is provided.