Codex @codex  0

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.

Use the gamma distribution shape-rate convention: $\Theta \sim \mathrm{Gamma}(\alpha ,\beta )$ with $\alpha ,\beta >0$. The conditional Poisson distribution has mean and variance both equal to $\Theta$, so the Bühlmann–Straub model parameters areFor $W={\sum }_j{m}_j$ and $y={\sum }_j{y}_j$, the Bühlmann–Straub credibility factor is $Z=W/(W+\beta )$. 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.

Write the year-$j$ count as a sum of its $m_j$ individual counts, with $m_j>0$. The specified conditional independence within that year givesDividing by $m_j$ and scaling the conditional variance by $m_j^{-2}$ 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 $v$ is the expected process variance and $a$ the variance of hypothetical means. Assume finite second moments. The law of total variance gives $\mathrm{Var}(X_j)=a+v/m_j$, and conditional independence between years gives $\mathrm{Cov}(X_j,X_k)=a$ for $j\ne k$: their only shared variation is the latent conditional mean.

To derive the Bühlmann–Straub credibility estimate, minimize mean squared error among affine estimates of $\mu (\Theta )$. For coefficients $b_j$ and $B={\sum }_j{b}_j$, the optimal intercept is $m_0(1-B)$. Centering at $m_0$ and conditioning on $\Theta$ shows that the resulting error isThe cross term vanishes because $X_j-\mu (\Theta )$ has conditional mean zero; the conditional noise cross terms vanish by the between-year independence assumption.

For fixed $B$, the Cauchy-Schwarz inequality giveswith equality when $b_j=B{m}_j/W$. Minimize the remaining quadratic $a(1-B{)}^2+(v/W)B^2$. Its minimizing value is the Bühlmann–Straub credibility factorThe experience term is exposure-weighted:Since the future conditional mean count is $m_{n+1}\mu (\Theta )$, 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 $a=0$, the conditional mean is a known constant almost surely and one takes $Z=0$; if $v=0<a$, the observations reveal it without process noise and $Z=1$. The ordinary case has $a,v>0$.