With a Poisson-gamma conjugacy with unequal exposures model, and , so the Bühlmann–Straub credibility factor is . The credibility estimate simplifies to , exactly the posterior mean and hence the Bayes estimator under squared error loss. The agreement is exact because that posterior mean is already affine in the exposure-weighted data.
Past exam of the mathematics course of the University of Cambridge 2006 iii Paper 43 4 Solution Created 2026-10-03 Updated 2026-10-14
Let be the total exposure, and define the prior distribution expected value, prior variance and average binomial noise bySince the prior distribution has a probability density function on , and . All relevant moments exist because the parameter and observations are bounded. The binomial distribution givesThe law of total expectation, law of total variance and law of total covariance consequently yieldIn particular, conditional independence of the annual observations does not imply unconditional independence: they share the uncertain parameter.
To derive the optimal credibility estimate among affine functions of the observations, put and write . The conditional errors have zero conditional expectation, conditional variance , and zero pairwise conditional covariances. They are also uncorrelated with every integrable function of . Therefore the joint mean squared error isFor fixed slopes, the first term is uniquely minimized by . For fixed sum , the Cauchy-Schwarz inequality giveswith equality exactly when . Thus the remaining optimization is the strictly convex quadraticIts derivative vanishes at . Consequently the Bühlmann–Straub credibility estimate isThe individual coefficients are and . The strict quadratic minimizations establish global optimality and uniqueness, not just necessary equations. Here is the variance of hypothetical means, is the expected process variance, and is the Bühlmann–Straub credibility factor. More exposure gives more weight to the observed claim fraction. Since , the estimate remains between the observed pooled fraction and the prior expected value. If one separately permits a degenerate prior, leads to the constant estimate , interpreted as .
For the uniform distribution prior on ,With two years, put and . Then andTo compare with the Bayes estimator under squared error loss, use conditional independence to multiply the two binomial likelihood functions. Terms independent of cancel on normalization, and the uniform prior makes the posterior density proportional to . Hence Beta-binomial conjugacy givesFinally, for an arbitrary reported value , conditional quadratic loss decomposes asThe unique minimum is the posterior mean. Therefore the affine credibility estimate and the exact Bayesian estimate coincide in this case. The agreement illustrates exact beta-binomial credibility with unequal exposures; for a general prior distribution, minimizing over affine functions of the observations need not recover the unrestricted posterior mean.
Past exam of the mathematics course of the University of Cambridge 2009 iii Paper 37 4 Solution Created 2026-10-03 Updated 2026-10-14
Let the Bühlmann–Straub model structural parameters beHere is the variance of hypothetical means and is the expected process variance. Assume the structural second moments are finite and the exposures are positive, as required by this least-squares problem. Write . Conditional independence separates the squared prediction error into squared conditional bias and conditional variance. The law of total expectation givesIndeed, conditional on , the prediction has mean and variance ; there are no cross-covariances because the annual observations are conditionally independent.
For fixed slope coefficients the optimal intercept is . For fixed sum , the Cauchy-Schwarz inequality giveswith equality when . Thus the remaining minimization is the scalar quadraticDifferentiation gives . In the nondegenerate case , 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 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 and , the unknown conditional mean is the known constant , so take . If and , each observation already equals almost surely, so gives exact prediction. If both vanish, all observations and the target equal almost surely; use the constant predictor rather than the undefined ratio .
In the approximate normal model, , , , and . Substitution into the derived credibility estimate givesWe now derive, rather than assume, the Bayes estimator under squared error loss. For positive , the likelihood and normal distribution prior give a posterior density proportional toThe coefficient of is , and the coefficient in the linear term is . Completing the square proves the normal-normal conjugacy with unequal exposures formulawithFor any decision , the posterior quadratic loss isIts unique minimizer is the posterior mean . 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.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 31 4 a Solution Created 2026-10-03 Updated 2026-10-06
Use the gamma distribution shape-rate convention: with . The conditional Poisson distribution has mean and variance both equal to , so the Bühlmann–Straub model parameters areFor and , the Bühlmann–Straub credibility factor is . Thusand the required expected-count estimate isThe total exposure determines how much information the observed count carries; the number of years alone is not the appropriate denominator.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 31 4 Solution 2026-10-06
Write the year- count as a sum of its individual counts, with . The specified conditional independence within that year givesDividing by and scaling the conditional variance by therefore givesThis is the Bühlmann–Straub model: larger exposures reduce the process noise in a year's average.
Define the population parameters and total observed exposure byHere is the expected process variance and the variance of hypothetical means. Assume finite second moments. The law of total variance gives , and conditional independence between years gives for : their only shared variation is the latent conditional mean.
To derive the Bühlmann–Straub credibility estimate, minimize mean squared error among affine estimates of . For coefficients and , the optimal intercept is . Centering at and conditioning on shows that the resulting error isThe cross term vanishes because has conditional mean zero; the conditional noise cross terms vanish by the between-year independence assumption.
For fixed , the Cauchy-Schwarz inequality giveswith equality when . Minimize the remaining quadratic . Its minimizing value is the Bühlmann–Straub credibility factorThe experience term is exposure-weighted:Since the future conditional mean count is , multiplying the optimal estimate by the known future exposure givesIt is the best affine linear least-squares projection, not a claim that the exact conditional expectation given all data is always affine. If , the conditional mean is a known constant almost surely and one takes ; if , the observations reveal it without process noise and . The ordinary case has .