Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 37 / 4

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.