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The Kalman prediction step advances the previous posterior mean and posterior covariance matrix through the state transition:The minus superscript marks information before observing $y_t$. The transpose and matrix multiplication propagate the old uncertainty; $Q_t$ adds the uncertainty of the new process noise. A known control input changes the mean but contributes no additional variance. If the control input itself is uncertain, that uncertainty needs its own model.

For independent normal observations conditional on $\Theta$, with mean $\Theta$ and variances $v/m_j$, and normal prior mean $\mu$ and variance $a$, write $W={\sum }_j{m}_j$. Multiplying likelihood and prior adds precisions $W/v+1/a$. Completing the square gives the displayed posterior distribution. Its posterior mean is $Z{\overline{X}}_w+(1-Z)\mu$, where $Z=aW/(v+aW)$ and ${\overline{X}}_w={\sum }_j{m}_j{X}_j/W$. Thus the Bayes estimator under squared error loss is exactly the Bühlmann–Straub credibility estimate, not merely an approximation to it. Its posterior variance is $av/(v+aW)$.

Conditionally independent normal distribution observations of mean $\theta$ and known variance ${\sigma }_1^2$, with normal prior mean $\mu$ and variance ${\sigma }_2^2$, give normal-normal conjugacy with known observation variance. Their posterior mean has the displayed exact credibility estimate form. It minimizes posterior squared-error loss and predicts the conditional mean of the next observation.

For conditionally independent annual averages $X_j\mid \theta \sim N(\theta ,v/m_j)$ and prior $\theta \sim N(\mu ,{\sigma }^2)$, the posterior mean is $Z{\overline{X}}_w+(1-Z)\mu$, with ${\overline{X}}_w={\sum }_j{m}_j{X}_j/W$. Completing the quadratic likelihood gives posterior precision ${\sigma }^{-2}+W/v$. This is an exact Bayesian credibility version of the Bühlmann–Straub credibility estimate. Increasing the between-risk prior variance increases the experience weight; increasing process variance decreases it. The next total pure premium is $m_{n+1}$ times the estimated per-life mean.

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.

For any prior density $\pi (p)$, the Bayes estimate under squared-error loss is the posterior mean:For the matched beta distribution, the integrals are beta integrals and their ratio isThe posterior mean minimizes posterior expected squared error because $\mathbb{E}[(p-a{)}^2\mid D]=\mathrm{Var}(p\mid D)+(a-\mathbb{E}[p\mid D]{)}^2$. The phrase Bayes point estimate depends on the loss: under absolute-error loss it would be a posterior median, while the posterior mode is a different summary and here equals $(14.8875-1)/(207.75-2)\approx 0.06750$.

The posterior mean is a weighted average of the sample proportion and prior mean:Thus it is pulled from 10% toward the prior's 5%, giving about 7.17%. This is an affine shrinkage estimator for a binomial proportion. The prior carries substantial concentration relative to the 90 observations, so the posterior is also more precise under this model: its standard deviation is about $0.01785$, compared with the binomial plug-in standard error $\sqrt{0.1(0.9)/90}=0.03162$. The posterior 95% credible interval $[0.04076,0.11035]$ is narrower and shifted down compared with the simple Wald interval $[0.03802,0.16198]$ based on $9/90$ alone.

The point estimate shrinks toward the historical mean, and the interval is narrower under the informative prior. A credible interval and a frequentist confidence interval have different probability interpretations; the comparison does not make the prior's relevance automatic.

A beta distribution is a convenient prior for a probability $p\in (0,1)$ and is conjugate to the binomial distribution of the death count. Match its mean $m=0.05$ and variance $v=0.02^2=0.0004$. For $p\sim \mathrm{Beta}(\alpha ,\beta )$,The moment matching for a beta prior formula givesConditional on $p$, use $D\sim \mathrm{Binomial}(90,p)$ and the observed count $D=9$. Multiplying its likelihood $p^9(1-p{)}^{81}$ by the prior density gives the Bayesian posterior through Beta-binomial conjugacy:Under squared-error loss, take its posterior mean as the point estimate. A 95% equal-tail credible interval is given by its 0.025 and 0.975 quantiles. Numerically,Matching two moments does not uniquely determine a prior distribution; the beta family is a justified convenient choice, not a conclusion forced by those moments. Its transfer to this hospital presumes that the historical between-hospital distribution is relevant to the present risk and case mix.

Assume the known variances are positive and write $\bar{x}=n^{-1}{\sum }_i{x}_i$. The likelihood times the normal prior distribution is proportional toCollecting the quadratic and linear terms in $\theta$ completes the square with precision $v^{-1}=n/{\sigma }^2+1/{\tau }^2$ and mean $m=v(n\bar{x}/{\sigma }^2+\mu /{\tau }^2)$. Thus normal-normal conjugacy givesFor quadratic loss, the posterior risk of reporting $a$ is $\mathbb{E}[(a-\theta {)}^2\mid x]=v+(a-m{)}^2$, uniquely minimized at $a=m$. For absolute-error loss, differentiating the posterior risk gives $2F_{\theta \mid x}(a)-1$, so its minimizer is a posterior median. The normal posterior is symmetric about $m$ and has a strictly increasing distribution function, hence that median is uniquely $m$. Therefore both optimal point estimates equal $m$, the posterior mean.

A credibility premium estimates a risk's expected future claims by combining its own observed experience with a collective or prior mean. In its basic form $Z\overline{X}+(1-Z)\mu$, the credibility factor $Z\in [0,1]$ is the weight assigned to individual experience. Larger credibility places less weight on pooling with the population. The amounts below are pure expected-claim premiums; no additional expense or profit loading is specified.

Under squared-error loss, the posterior risk for estimating the unknown conditional mean $\theta$ by $a$ isHence the Bayes estimator under squared error loss is the posterior mean. In the equal-year model, conditional independence makes the likelihood proportional to $\exp [-{\sum }_{j=1}^n(X_j-\theta {)}^2/(2{\sigma }_1^2)]$. Combining it with the normal prior and completing the square givesBecause $\mathbb{E}[X_{n+1}\mid \theta ]=\theta$, the required estimator is $m_n$; it also equals the predictive mean $\mathbb{E}[X_{n+1}\mid \mathbf{X}]$. The normal-normal credibility expression isThis is exact Bayesian credibility, not merely the best linear approximation to a possibly nonlinear posterior mean.

For unequal life counts use the usual exposure model: individual-life losses are conditionally independent with variance $v$, and the annual totals are conditionally independent given the same $\theta$. ThenThe dot denotes the stipulated normal approximation. The positive exposures $m_j$ are known. Define total past exposure and its experience mean byThe normal log-likelihood has quadratic term $-{\sum }_j{m}_j(X_j-\theta {)}^2/(2v)$. With the prior $N(\mu ,{\sigma }^2)$, normal-normal conjugacy with unequal exposures gives posterior variance and meanTherefore the estimated pure premium per life for year $n+1$ has the normal-normal credibility with unequal exposures formIt is also an exact normal-model Bühlmann–Straub credibility estimate. The individual experience is exposure-weighted: the unweighted mean of the yearly per-life observations would generally discard the different precisions.

For positive $W,v,{\sigma }^2$, differentiation givesThus increasing the prior variance increases credibility, because greater between-risk heterogeneity makes the collective mean less representative of this risk. Increasing the individual process variance decreases credibility, because a fixed amount of observed experience becomes noisier. The future exposure supplies no new observation, so it does not enter $Z_W$. It only scales the total premium:The independent-life variance scaling is part of the usual exposure interpretation. Without it, a known variance $v$ for one life does not determine the variance of a total: a common pairwise conditional covariance $c$ would add $m_j(m_j-1)c$ to $m_{j}v$. Similarly, cross-year conditional dependence would change the joint likelihood. Those assumptions are implicit in applying the standard credibility model to the final paragraph.

Given $\Theta =\theta$, the sum of the independent yearly counts is $S_n\sim \mathrm{Poisson}(n\theta )$: their probability generating functions multiply to $\exp (n\theta (z-1))$. Averaging this conditional law over the prior gives the mixed Poisson distributionThe likelihood for the full vector depends on $\theta$ only through $s$, so $S_n$ is a sufficient statistic; conditioning on the total or on the full observed vector gives the same posterior for $\Theta$.

For an event of positive probability $p_n(s)>0$, the posterior density is proportional to $e^{-n\theta }{\theta }^s\pi (\theta )$. Put $I_s={\int }_0^{\infty }{e}^{-n\theta }{\theta }^s\pi (\theta )d\theta$. Its mean is $I_{s+1}/I_s$. The future count has conditional mean $\Theta$, soThis is the posterior mean from adjacent mixed Poisson probabilities. Both probabilities refer to the same $n$-year exposure; the numerator is not a probability for $S_{n+1}$. For a proper prior and $n>0$, the integrals defining this posterior mean are finite because ${\theta }^{s+1}{e}^{-n\theta }$ is bounded on $[0,\infty )$.

For the independent normal distribution observations, the log likelihood isIts derivative is ${\sum }_i(X_i-\theta )$ and its second derivative is $-n<0$, so the unique maximum-likelihood estimate isFor the prior $\theta \sim N(\mu ,{\tau }^{-1})$ with $\tau >0$, multiplying prior density by likelihood and completing the square givesThus normal-normal conjugacy gives the posterior distributionFor quadratic loss, the conditional expected loss of action $a$ is the posterior variance plus $(a-\mathbb{E}[\theta \mid X]{)}^2$. Its unique minimizer is the posterior mean, so the Bayes estimator isThis weights the sample and prior means by their respective precisions.

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.

For a prior $\Pi$ and loss $L$, the Bayes risk of a rule $d$ is $r(\Pi ,d)=\int {\mathbb{E}}_{\theta }L(\theta ,d(X))\Pi (d\theta )$. A Bayes decision rule minimizes this integrated risk. An Extended Bayes rule is a rule whose excess over the optimal Bayes risk can be made arbitrarily small by a suitable proper prior: for every $\epsilon >0$ some $\Pi$ has $r(\Pi ,d)\le {\inf }_{d^{\prime }}r(\Pi ,d^{\prime })+\epsilon$, with finite risks.

Condition on the observation and use iterated expectation. A rule is Bayes precisely when it chooses a minimizer of the posterior expected loss almost surely, up to the usual measurability requirement. Under squared Euclidean loss, with a finite posterior second moment, write $m=\mathbb{E}[\theta \mid X]$. The identityshows that the Bayes rule is the posterior mean.

By normal-normal conjugacy, the posterior distribution isThe posterior mean is a precision-weighted average of the population mean and the observed personal measurement.

A constant improper prior would give the formal posterior density kernel $N^{-1}{1}_{\{N\ge y\}}$. However,There is no normalized posterior distribution under this prior. The failure is not simply that the prior is improper: some improper priors give proper posteriors. Here posterior propriety itself fails after this single observation, so probabilities, quantiles and a posterior mean cannot be assigned to that kernel.

Write $H_k={\sum }_{j=1}^{k}1/j$, with $H_0=0$. The truncated uniform prior gives the posterior distributionIts posterior mean isEach term in the numerator of the expectation is $N/N=1$, which explains the exact count $M-y+1$.

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.