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For independent comparable risks observed over $n\ge 2$ years, estimate the expected process variance $v$ by the average within-risk sample variance. The sample variance $B$ of the risk means estimates $a+v/n$, where $a$ is the variance of hypothetical means. Thus $B-\hat{v}/n$ is unbiased for $a$; truncating at zero produces a practical nonnegative estimate but introduces bias. Use the grand mean for $m$ and plug these estimates into the Bühlmann credibility premium $\hat{Z}{\overline{X}}_s+(1-\hat{Z})\hat{m}$. At least two risks are needed for $B$; a single year cannot estimate within-risk variance this way.

In the Gamma prior for an inverse-gamma scale model, assume $k>2$. The variance of hypothetical means is $a=\alpha /[{\lambda }^2(k-1{)}^2]$, and the expected process variance is $v=\alpha (\alpha +1)/[{\lambda }^2(k-1{)}^2(k-2)]$. The Bühlmann credibility factor is $Z=n/(n+v/a)$, giving the displayed expression. The optimal affine credibility estimate is $(1-Z)\alpha /[\lambda (k-1)]+Z\overline{X}$. Its fixed-weight arithmetic-mean form differs from the reciprocal-based posterior mean.

Given a latent risk parameter $\Theta$, observations are independent and identically distributed random variables with conditional expectation $m(\Theta )$ and conditional variance $v(\Theta )$. The Bühlmann credibility premium estimates $m(\Theta )$ by an affine function of past observations under mean squared error. Its population parameters are $m=\mathbb{E}m(\Theta )$, expected process variance $v=\mathbb{E}v(\Theta )$, and variance of hypothetical means $a=\mathrm{Var}(m(\Theta ))$.

The experience weight is $Z=Wa/(Wa+v)$, where $W$ is total observed exposure, $a$ the variance of hypothetical means, and $v$ the expected process variance. With positive $a$, it is $W/(W+v/a)$. More exposure raises the weight; larger process noise lowers it. If $a=0$, the unknown conditional mean is constant and one can set $Z=0$.

Conditional on a common latent variable $\Theta$, annual averages $X_j$ are independent, have common conditional expectation $\mu (\Theta )$, and have conditional variance ${\sigma }^2(\Theta )/m_j$ for known positive exposures $m_j$. Its structural parameters are $m_0=\mathbb{E}\mu (\Theta )$, expected process variance $v=\mathbb{E}{\sigma }^2(\Theta )$ and variance of hypothetical means $a=\mathrm{Var}(\mu (\Theta ))$. It generalizes the equal-exposure Bühlmann model.

The credibility factor in the Bühlmann model is the weight $Z=na/(na+v)$ placed on the sample mean after $n$ observations. The remaining weight is placed on the population expected value. More experience or larger variance of hypothetical means increases $Z$; larger expected process variance decreases it.

A credibility factor is the weight assigned to a particular risk's observed experience when blending it with a population or manual premium. In the Bühlmann model, the blend is the best affine mean-square predictor, rather than automatically the full posterior mean.

Let $v=\mathbb{E}[{\sigma }^2(\theta )]$, the expected process variance, and assume the displayed second moments are finite. The law of total expectation gives $\mathbb{E}{X}_i=m$. The law of total variance and law of total covariance giveFor the second formula, the conditional covariance is zero by conditional independence, while the covariance of the conditional means is $\mathrm{Var}\mu (\theta )=a$, the variance of hypothetical means. Likewisebecause $\mathbb{E}[X_i-\mu (\theta )\mid \theta ]=0$.

To derive the Bühlmann credibility premium, minimize $\mathbb{E}[(\mu (\theta )-b_0-{\sum }_i{b}_i{X}_i{)}^2]$. Differentiating in $b_0$ sets the predictor's mean to $m$. The remaining normal equations areFor $v>0$, subtracting any two rows forces all $b_i$ to be equal, then gives $b_i=a/(v+na)$. Thus the premium and factor areThe same affine premium predicts $X_{n+1}$: conditional independence makes $\mathbb{E}[(X_{n+1}-\hat{\mu }{)}^2]=v+\mathbb{E}[(\mu (\theta )-\hat{\mu }{)}^2]$, so minimizers agree. If $v=0$, every year's amount equals $\mu (\theta )$ almost surely, and $Z=1$ gives the exact risk mean even though the coefficient vector need not be unique.

For fixed $a>0,n$, $\partial Z/\partial v=-na/(na+v{)}^2<0$. Thus more within-risk variation decreases credibility: individual experience becomes less informative, $Z\to 0$ as $v\to \infty$, and the premium tends to $m$. Increasing the number of observed years has the opposite effect.

For numerical estimation from a balanced panel, assume the $k$ risks are independent draws from the same risk population, with comparable exposures, and $n,k\ge 2$. SetConditionally, each within-risk sample variance has mean ${\sigma }^2({\theta }_s)$; hence $\mathbb{E}\hat{v}=v$. The risk sample means have variance $a+v/n$, so $\mathbb{E}B=a+v/n$. Therefore $\tilde{a}=B-\hat{v}/n$ is an unbiased method of moments estimate of $a$. For a usable nonnegative variance estimate take $\hat{a}=\max (0,\tilde{a})$, noting that this truncation sacrifices exact unbiasedness.

The balanced-panel method-of-moments credibility estimate for risk $s$ is consequentlyIf $\hat{a}=0$ and $\hat{v}>0$, use zero credibility. If both estimates are zero, the within and between sums of squares are zero and all the observed amounts coincide; any blend gives that same premium. With only one year per risk or only one risk, these within/between estimators are unavailable, so external structural estimates or additional assumptions are necessary.

In the Bühlmann model, one policy has a random risk parameter $\Theta$. Conditional on $\Theta$, its annual observations $X_i$ are independent and identically distributed random variables, with mean $m(\Theta )$ and variance $v(\Theta )$. Put $m_0=\mathbb{E}[m(\Theta )]$, let $v=\mathbb{E}[v(\Theta )]$ be the expected process variance, and let $a=\mathrm{Var}(m(\Theta ))$ be the variance of hypothetical means. Assume the required second moments are finite.

The Bühlmann credibility premium is the best affine estimate of $m(\Theta )$ under mean squared error. After choosing the intercept to make the estimate unbiased, write it as $m_0+{\sum }_{i=1}^n{b}_i(X_i-m_0)$. If $B={\sum }_i{b}_i$, the law of total variance and conditional independence give its errorFor fixed $B$, the Cauchy-Schwarz inequality minimizes the last sum at $b_i=B/n$. Differentiating $a(1-B{)}^2+v{B}^2/n$ gives $B=na/(na+v)$. Thus the credibility factor and credibility premium areThe ratio form assumes $a>0$; when $a=0$ and $v>0$, the conditional mean is constant and $Z=0$. Predicting $X_{n+1}$ rather than its conditional mean adds the constant $v$ to the error and leaves the optimal affine estimate unchanged.

Here the latent intensity has a Pareto distribution with shape three and lower bound one. For $r<3$, its moments are $\mathbb{E}[{\Theta }^r]=3{\int }_1^{\infty }{\theta }^{r-4}d\theta =3/(3-r)$. ConsequentlyThe equality for $v$ uses the Poisson distribution conditional variance $v(\Theta )=\Theta$. There are two years of experience, so $Z=1/2$ and the observed sample mean is $k/2$. The Poisson credibility with a shape-three Pareto intensity is thereforeIn particular it equals $2$ when $k=5$.

For the exact Bayesian credibility estimate, the total count conditional on $\Theta =\theta$ has Poisson distribution of mean $2\theta$. Multiplication of its likelihood by the prior distribution density gives a Bayesian posterior density proportional to ${\theta }^{k-4}{e}^{-2\theta }$ on $\theta >1$. At $k=5$, its normalizing integral and first-moment integral areHence the next-year predictive expected value isThe Bühlmann credibility premium optimizes over affine estimates, whereas the Bayesian posterior mean optimizes over all square-integrable estimates. This explains why the two credibility estimates need not agree for this nonconjugate prior.

Let $M={\sum }_{i=1}^n{m}_i$ 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 $(0,1)$, $v>0$ and $w>0$. 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 $A={\sum }_i{a}_i$ and write $X_i=\theta +{\epsilon }_i$. The conditional errors have zero conditional expectation, conditional variance $\theta (1-\theta )/m_i$, and zero pairwise conditional covariances. They are also uncorrelated with every integrable function of $\theta$. Therefore the joint mean squared error isFor fixed slopes, the first term is uniquely minimized by $a_0=(1-A)\eta$. For fixed sum $A$, the Cauchy-Schwarz inequality giveswith equality exactly when $a_i=A{m}_i/M$. Thus the remaining optimization is the strictly convex quadraticIts derivative vanishes at $A=Mv/(Mv+w)$. Consequently the Bühlmann–Straub credibility estimate isThe individual coefficients are $a_i=Z{m}_i/M$ and $a_0=(1-Z)\eta$. The strict quadratic minimizations establish global optimality and uniqueness, not just necessary equations. Here $v$ is the variance of hypothetical means, $w$ is the expected process variance, and $Z$ is the Bühlmann–Straub credibility factor. More exposure gives more weight to the observed claim fraction. Since $0<Z<1$, the estimate remains between the observed pooled fraction and the prior expected value. If one separately permits a degenerate prior, $v=0$ leads to the constant estimate $\eta$, interpreted as $Z=0$.

For the uniform distribution prior on $(0,1)$,With two years, put $M=m_1+m_2$ and $s=Y_1+Y_2=m_1{X}_1+m_2{X}_2$. Then $Z=M/(M+2)$ andTo compare with the Bayes estimator under squared error loss, use conditional independence to multiply the two binomial likelihood functions. Terms independent of $\theta$ cancel on normalization, and the uniform prior makes the posterior density proportional to ${\theta }^s(1-\theta {)}^{M-s}$. Hence Beta-binomial conjugacy givesFinally, for an arbitrary reported value $d$, 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.

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.

The ratio of expected process variance to variance of hypothetical means is $K=s^2/v=(\alpha +1)/(k-2)$, and the credibility factor is $Z=n/(n+K)$. Holding $\alpha ,\lambda$ fixed, an increase in $k$ decreases $K$. Hence the credibility factor increases, approaching one as $k$ tends to infinity. Extending the formula to real $k>2$ makes the sign explicit:For the integer values in the model this establishes the same strict monotonicity. The collective claim expected value $m=\alpha /[\lambda (k-1)]$ also changes, but the conclusion concerns the relative weight of experience: conditional noise declines faster than the variation of the risk means.

Set $h=\mathbb{E}\theta >0$ and $w=\mathrm{Var}(\theta )>0$. Keeping $h$ fixed in the gamma distribution family gives $\alpha =h^2/w$ and $\lambda =h/w$. The ratio of expected process variance to variance of hypothetical means becomesAs the prior variance $w$ decreases, $K$ increases, so the credibility factor decreases. In the limit $w\downarrow 0$, $Z\downarrow 0$: the risk parameter is already essentially known from the prior, leaving little need to use the noisy sample mean. The population claim expected value $h/(k-1)$ remains fixed throughout this comparison.

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.

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.

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 $a,v>0$.