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Keep $y$ fixed. Integral bounds for the decreasing function $1/t$ give $H_M=\log M+O(1)$, so $H_M-H_{y-1}=\log M+O_y(1)$. ConsequentlyThe posterior mean thus grows like $M/\log M$ instead of approaching a data-determined limit. Even the posterior distribution mass below a fixed endpoint $n$ is $(H_n-H_{y-1})/(H_M-H_{y-1})\to 0$. Hence increasing the supposedly innocuous prior cutoff moves the inference to larger and larger endpoints. In an integral approximation its median is of order $\sqrt{yM}$, also depending strongly on the cutoff. This explains the sensitivity of the entire inference, not just of its posterior mean.

Multiplying the reciprocal improper prior by the likelihood function givesThus this discrete uniform endpoint posterior is proper. Its exact cumulative probability is $S_y^{-1}{\sum }_{k=y}^n{k}^{-2}$. Integral approximation gives $S_y\approx {\int }_y^{\infty }{t}^{-2}dt=1/y$ and ${\sum }_{k=y}^n{k}^{-2}\approx {\int }_y^n{t}^{-2}dt=1/y-1/n$, soThese approximations are useful when $y$ is moderately large and are not exact discrete identities. For example the exact untruncated median at $y=100$ is $199$. Although its quantiles are finite, its posterior mean is infinite, since its expectation numerator is the divergent sum ${\sum }_{N=y}^{\infty }1/N$.

The posterior distribution has a long right tail. Its posterior mean is pulled upward by relatively rare large endpoints and is sensitive to the artificial upper cutoff; its untruncated posterior mean does not exist. The median is a central quantile that stays finite as the cutoff is removed and is much less sensitive to extreme draws. It is also the Bayes estimator under absolute-error loss, whereas a mean corresponds to squared-error loss when the relevant expectation exists.

The median better represents the typical posterior endpoint in this strongly right-skewed distribution. It need not be preferable for every decision problem, since the loss function matters. For comparison, exact summation of the truncated kernel at cutoff $5000$ gives median $195$ and mean about $397.67$; the displayed simulation summaries are Monte Carlo estimates, not exact posterior calculations.

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.

Multiplying the Poisson distribution likelihood by the Jeffreys prior gives the posterior density kernelThus, in shape-rate convention,It is a proper posterior distribution even when $y_i=0$, provided $E_i>0$. The posterior mean and posterior variance are $(y_i+1/2)/E_i$ and $(y_i+1/2)/E_i^2$. This is Poisson-gamma conjugacy with unequal exposures, with the zero-rate prior interpreted as an improper kernel.

The new prior density contributes the factor ${\lambda }_i^{15}{e}^{-16{\lambda }_i}$. Multiplication by the Poisson distribution likelihood givesIts posterior mean displays the resulting shrinkage:The observed risk ratio $y_i/E_i$ is pulled towards one, most strongly for small exposures. The posterior variance is $(y_i+16)/(E_i+16{)}^2$.

For retained Markov chain Monte Carlo draws ${\theta }^{(1)},\dots ,{\theta }^{(S)}$, calculate both the parameter average $\hat{m}=S^{-1}{\sum }_s{\theta }^{(s)}$ and the average deviance ${\hat{D}}_{\mathrm{av}}=S^{-1}{\sum }_{s}D({\theta }^{(s)})$. Two estimates areThe first uses the approximate equality of the posterior mean and maximum-likelihood estimator. The second subtracts the mean $p$ of the chi-squared distribution deviance excess. These are different operations: evaluation after averaging versus averaging after evaluation. Both rely on a well-mixed chain and the regular locally flat-prior approximation. The smallest sampled deviance is an upper bound on the true minimum, but merely taking that minimum can be inefficient in high dimension because few draws approach the mode closely.

The PDF gives an inverse-gamma distribution with shape $k$ and scale $\theta$, and a gamma distribution prior with shape $\alpha$ and rate $\lambda$. Write $\mu (\theta )=\theta /(k-1)$. The supplied conditional moments, together with $\mathbb{E}\theta =\alpha /\lambda$ and $\mathbb{E}{\theta }^2=\alpha (\alpha +1)/{\lambda }^2$, give the three structural quantities of the Bühlmann model:Here $v$ is the variance of hypothetical means and $s^2$ the expected process variance. The law of total covariance and conditional independence of different years implyAll these quantities are finite because $k>2$.

For any coefficients $c_i$, minimizing the mean squared error over the intercept first gives $c_0=m(1-{\sum }_i{c}_i)$. Let $Y=\mu (\theta )-m$ and $W_i=X_i-m$. Expanding the remaining mean squared error givesIts normal equations are $s^2{c}_i+v{\sum }_j{c}_j=v$ for every $i$. Since $s^2>0$, subtracting any two equations shows that all $c_i$ agree. The quadratic part is strictly positive for every nonzero coefficient change, so the solution is the unique global minimizer. Solving givesA credibility estimate combines individual experience with the collective expected value, using an experience weight $Z\in [0,1]$. In this equal-exposure setting, the Bühlmann credibility premium isThus the computed coefficients give exactly a credibility estimate, with a fixed credibility factor determined by known parameters and the number of observed years. This is Bühlmann credibility for inverse-gamma observations.

For the final Bayesian calculation, the likelihood function in $\theta$ is proportional to ${\theta }^{nk}\exp (-\theta {\sum }_i{x}_i^{-1})$. Multiplication by the gamma distribution prior gives the Gamma prior for an inverse-gamma scale updateUnder quadratic loss, the conditional risk of an estimate $d$ is $\mathrm{Var}(\mu (\theta )\mid x)+(d-\mathbb{E}[\mu (\theta )\mid x]{)}^2$. Its minimizer is therefore the posterior mean:This is not a fixed-weight arithmetic-mean credibility estimate. For $n=2$, the observations $(1,3)$ and $(2,2)$ have the same sample mean, but reciprocal sums $4/3$ and $1$, hence different posterior means. For any $n>2$, append the same positive observations to these two samples. For $n=1$, the estimate is $Cx/(1+\lambda x)$, with $C=(\alpha +k)/(k-1)$; its second derivative is $-2C\lambda /(1+\lambda x{)}^3\ne 0$, so it is not affine either. Symmetry in the observations would force equal slopes in any affine representation, so these examples also exclude a general fixed-coefficient linear representation.

One can express the posterior mean as a broader data-dependent blend. With $A={\sum }_i{x}_i^{-1}$ and harmonic mean $H=n/A$, it isThis weights a harmonic-mean-based experience estimate, and the weight itself depends on the observations. It is not the credibility estimate based on $\overline{X}$ and the fixed Bühlmann credibility factor derived above.

A credibility estimate blends the risk's own sample mean with a population or prior expected value. In the basic equal-exposure form it is $Z\overline{X}+(1-Z)\mu$, where the credibility factor $Z\in [0,1]$ is the weight attached to the individual experience. More data ordinarily increase that weight; greater within-risk variation relative to between-risk variation reduces it.

Write $m(\theta )=\mu (\theta )$ for the conditional claim expected value, keeping $\mu$ for the prior-level parameter. Normalization of the probability density function givesAt interior points of its finite domain, differentiation under the integral sign yieldsThis is an instance of the exponential-family derivative identities for natural parameter $-\theta$. The minus sign is present in the PDF and is lost in the converted TeX. The conditional variance is likewise $(\log q{)}^{\prime \prime }(\theta )$.

Differentiate the log of the natural conjugate prior:Integrate on a compact subinterval and then let its endpoints approach those of the parameter interval. The prescribed vanishing of the prior density eliminates the boundary term. Since $m(\theta )\ge 0$, monotone convergence theorem justifies the limit of its expectation, and the identity givesIn particular the assumed proper prior and boundary conditions themselves establish this first-moment integrability.

For $T={\sum }_{i=1}^n{x}_i$ and $\overline{x}=T/n$, conditional independence makes the likelihood function proportional to $q(\theta {)}^{-n}{e}^{-\theta T}$. Multiplication by the prior distribution gives the Bayesian posteriorThus this is a conjugate prior family, with updated parameters

The integration-by-parts calculation for the posterior mean also needs vanishing posterior boundary values. Here that property can be checked, rather than assumed. Let $a,b$ be the essential lower and upper endpoints of the claim support. Every nondegenerate tilted claim law has $a<m(\theta )<b$, so the established prior mean satisfies $a<\mu <b$. Supported observations have $\overline{x}\in [a,b]$; hence $a<{\mu }_n<b$. At a finite parameter endpoint the prior's vanishing forces $q(\theta )\to \infty$, because its exponential factor has a finite positive limit; the increased power $q^{-k_n}$ makes the posterior vanish there too. At $\theta \to +\infty$, choose $a<d<{\mu }_n$ with positive base-measure mass below $d$. Then $q(\theta )\ge C{e}^{-d\theta }$ and the unnormalized posterior is at most $C^{-k_n}{e}^{-k_n({\mu }_n-d)\theta }\to 0$. At $\theta \to -\infty$, choose ${\mu }_n<d<b$ with positive mass above $d$ to obtain the corresponding bound $C^{-k_n}{e}^{k_n(d-{\mu }_n)\theta }\to 0$. These bounds also ensure posterior propriety at infinite endpoints. This is the endpoint control for a Laplace-family conjugate posterior.

Apply the previous score integration to the Bayesian posterior. The natural conjugate credibility identity isThus $Z=n/(n+k)$ is the credibility factor. This is an exact posterior mean, rather than an affine approximation to one.

For the specific shape-two gamma distribution, takeThen $m(\theta )=2/\theta$ and $\mathrm{Var}(X\mid \theta )=2/{\theta }^2$. The gamma rate gamma conjugacy prior isthat is, $\Theta \sim \mathrm{Gamma}(2k+1,\text{rate }k\mu )$. Its normalizing constant in the notation of the question is $c(\mu ,k)=\Gamma (2k+1)/(k\mu {)}^{2k+1}$. The gamma distribution density vanishes at both zero and infinity for $k>0$. The posterior has shape $2k+2n+1$ and rate $k\mu +T$, so its conditional-mean posterior mean is again $(k\mu +T)/(k+n)$.

For the last calculation, gamma distribution integration givesThe second identity requires the printed $k>1/2$. Therefore the expected process variance and variance of hypothetical means, denoted by $a,b$ in this question, areConsequently $a/b=k$, and the Bühlmann credibility factor agrees with the exact Bayesian weight:This is the exact Bühlmann credibility for gamma claims, expressed using the reciprocal rate as the random scale. The symbols $a,b$ here are the process and between-risk variances, respectively; their roles should not be interchanged when comparing with other notation for the Bühlmann model.

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.

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.

Fit two hierarchical Bayesian models to the same observations, one with the bounded uniform distribution sampling density and the other with an exponential distribution density. Specify rate versus mean in the exponential model and assign appropriate, scientifically comparable proper prior distributions; that parameter is no longer a literal maximum size. Obtain posterior distributions, for example by Markov chain Monte Carlo, and check convergence.

At the same observational level in both models use the Bayesian deviance $D(\varphi )=-2\log p(\mathbf{y}\mid \varphi )$, excluding prior densities. Retain the same likelihood constants and consistently either condition on breed effects or integrate them out. Estimate $\overline{D}=\mathbb{E}[D(\varphi )\mid \mathbf{y}]$, evaluate $D(\overline{\varphi })$ at the posterior mean, and compute the effective parameter count in DIC:Smaller DIC favours the fit–complexity tradeoff. Supplement it with posterior predictive checks; the uniform model's parameter-dependent support and the hierarchical structure make the criterion a diagnostic rather than an automatic definitive decision.

The first coefficient is the posterior mean log odds of choosing invertebrates rather than fish for smaller alligators at Lake Hancock. It corresponds tofor those invertebrate-to-fish odds. This is not the absolute probability of choosing the second category, because the other categories also enter the normalization.

The size coefficient adds $-1.524$ to the invertebrate-to-fish log odds on changing to the larger class, holding lake fixed. Its common-across-lakes odds ratio isabout a 78 percent reduction in the relative odds. Exponentiating a posterior mean log odds is a geometric summary, not the arithmetic posterior mean of the odds.

There are $(K-1)p=4\cdot 5=20$ free category coefficients plus $I=8$ group intercepts, giving 28 free parameters in the fitted Poisson model. The effective parameter count in DIC, $26.8$, is close to 28 and slightly smaller, consistent with some regularization or incomplete information. Comparing it with 20 would omit the nuisance intercepts. A direct conditional multinomial fit has 20 coefficients but a different observational likelihood, since it conditions on the totals.