Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 28 / 4

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.