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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.

Each observation has uniform distribution density ${\theta }^{-1}{1}_{\{0<y_j<\theta \}}$. Independence makes the likelihood function their product. For positive observations the support reduces to $\theta >M$, givingFor data outside the positive orthant the likelihood is zero. Changing the convention at the endpoint does not change a continuous posterior distribution.

Multiplying the likelihood function by the Pareto distribution prior givesThe integral of this kernel is $m^{-(\alpha +n)}/(\alpha +n)$, so the normalized posterior distribution isThus it is $\mathrm{Pareto}(\alpha +n,m)$. This proves uniform-Pareto conjugacy: applying Bayes theorem preserves the family of prior distributions.

For independent identically distributed observations $Y_1,\dots ,Y_n$, the likelihood function and log-likelihood areA maximum-likelihood estimator is a measurable choice ${\hat{\theta }}_n\in {\mathrm{argmax}}_{\theta \in \Theta }{L}_n(\theta )$, when a maximizer exists. It need not be unique. The one-observation score function is $s_{\theta }(y)={\nabla }_{\theta }\log f(y,\theta )$, and the Fisher information matrix isUnder the regularity assumptions, differentiation beneath the integral gives ${\mathbb{E}}_{\theta }{s}_{\theta }={\nabla }_{\theta }\int f=0$. Differentiating again gives the second information identity. Independent observations have total Fisher information $nI(\theta )$.

Assume fixed parameter dimension, an interior true parameter ${\theta }_0$, a positive-definite matrix $I({\theta }_0)$, the standard differentiability and integrability conditions, and statistical consistency of ${\hat{\theta }}_n$. Then the asymptotic normality of a maximum likelihood estimator isHere $I$ is the information per observation. Thus the leading covariance matrix of the estimator itself is $I({\theta }_0{)}^{-1}/n$.

To prove this, statistical consistency places the estimator in an interior ball about ${\theta }_0$ with probability tending to one, where its score function vanishes. Write $d_n={\hat{\theta }}_n-{\theta }_0$ and use the integral first-order Taylor formula for a vector map:This integral matrix is needed in a vector problem; one does not have to assert a common scalar mean-value point for every score component. PutThe regular local uniform law of large numbers, continuity of the expected Hessian matrix, and statistical consistency give $J_n\stackrel{P}{\to }I({\theta }_0)$. The multivariate central limit theorem givesThe limiting information matrix is nonsingular, so $J_n^{-1}\stackrel{P}{\to }I({\theta }_0{)}^{-1}$. Solving the Taylor identity and applying the Slutsky theorem proves the displayed normal limit, since $I^{-1}I{I}^{-1}=I^{-1}$. Boundary parameters or singular information are outside this regular theorem.

Conditional on $\Theta =\theta$, summing the individual independent Poisson counts gives $Y_j\sim \mathrm{Poisson}(m_j\theta )$. The between-year conditional independence makes the likelihood functionMultiplying by the gamma distribution prior density, proportional to ${\theta }^{\alpha -1}{e}^{-\beta \theta }$, gives the Poisson-gamma conjugacy with unequal exposures:For an action $t$, the posterior squared-error loss decomposes asHence the Bayes estimator under squared error loss is the posterior mean, givingFuture counts are independent of the observed years conditional on $\Theta$, so the law of total expectation then gives the posterior predictive expected countThe Bayesian and credibility estimates coincide exactly. This exact Bühlmann–Straub credibility for Poisson-gamma counts occurs because the posterior mean is already affine in the exposure-weighted experience, and therefore belongs to the class over which the credibility estimate minimizes mean squared error.

The likelihood function for the ordered arrival times of a Poisson process, including the absence of further arrivals before $T$, isThe arrival-time constraint does not depend on $\lambda$. Equivalently, the Poisson distribution of the total count gives $L(\lambda )=e^{-\lambda T}(\lambda T{)}^n/n!$, which differs by a parameter-independent factor. Conditional on the count, the Poisson process conditional arrival times have density $n!/T^n$, independent of $\lambda$. Thus the total count contains all the information about the rate when the exposure is known; it is a sufficient statistic.

A conjugate prior is a family of prior distributions whose members remain in that family after updating by the likelihood function. Here multiplying a shape-rate gamma distribution density by the Poisson process likelihood changes its power of $\lambda$ and its exponential rate, leaving a gamma distribution. The parameters change with the observations; conjugate prior does not mean that the Bayesian posterior equals the prior distribution.

There is a notation problem here. The posterior predictive probability $p_0$ calculated above is already a fixed number conditional on $a,b,n,T$. Literally,No simulation is needed for that interpretation. The known exposure $T$ is also necessary, despite its omission from this part's list of inputs.

The natural uncertain quantity is instead $q(\lambda )=\mathbb{P}(M=0\mid \lambda )=e^{-\lambda }$. For its posterior probability of being below one half, draw the rate from its gamma distribution Bayesian posterior and average an indicator. Rough BUGS code isThe monitored average of belowHalf estimates $\mathbb{P}(\lambda >\log 2\mid \mathcal{D})$, equivalently $1-F_{\mathrm{Gamma}(A,B)}(\log 2)$. The equality boundary has zero probability. This code samples the already updated Bayesian posterior; adding the count likelihood function again would count the data twice.

For an Inhomogeneous Poisson process, the likelihood function is the product of the intensities at arrivals times the exponential of minus the integrated intensity. The integrated intensity is $\theta +2(T-\theta )=2T-\theta$. There are $j(\theta )$ arrivals at rate one and $n-j(\theta )$ at rate two, soSet $t_0=0$ and $t_{n+1}=T$ to include changes before the first or after the last arrival. An arrival exactly at the change has probability zero, so the convention for $j$ at that point does not affect the Bayesian posterior.

A uniform prior on $(0,T)$ makes the posterior density proportional to the Poisson change-point posterior likelihood above. The zeros trick introduces an observed zero with Poisson distribution mean $K-\log L(\theta )$, making its likelihood function equal to $e^{-K}L(\theta )$. Choose $K=T+n\log 2+1$; since $\log L(\theta )\le n\log 2$, this mean is strictly positive. Rough BUGS code, with zero=0 supplied as data, isFor $n=0$, omit the array and set j <- 0. Monitor the sampled theta to obtain its posterior mean, credible interval and interval probabilities.

There is also an exact sampling method. On $(t_j,t_{j+1})$ the posterior density is proportional to $2^{-j}{e}^{\theta }$, so choose the interval with weightsThen draw $U$ from a uniform distribution on $(0,1)$ and set $\theta =\log (e^{t_j}+U(e^{t_{j+1}}-e^{t_j}))$. These weighted interval draws sample the posterior directly, without asking a local Markov chain Monte Carlo update to cross its discontinuities.

The overall mean $\mu$ and the between-study heterogeneity $\tau$ encode different information. A proper broad normal distribution prior such as $\mu \sim N(0,2^2)$ is one possible weak prior for the mean log odds ratio; information about the spread of trials alone does not determine its center.

A concrete prior calibration for normal random-effect range can make the factor-of-50 statement simultaneous across all six trials. Conditional on $\tau$, each pair difference has normal distribution ${\beta }_j-{\beta }_k\sim N(0,2{\tau }^2)$. Set $R=\log 50$, $m=\left(\genfrac{}{}{0ex}{}{6}{2}\right)=15$, $z={\Phi }^{-1}(1-0.05/(2m))$ andFor every $\tau \le A$, each pair exceeds $R$ in absolute value with probability at most $0.05/m$. The union bound therefore gives $\mathbb{P}({\max }_j{\beta }_j-{\min }_j{\beta }_j>R)\le 0.05$, and integrating over the uniform prior preserves that bound. This is one explicit interpretation of “very unlikely”; a different elicited probability would change the bound. A smoother proper scale prior could be calibrated similarly.

The proposed $1/\tau$ improper prior is unsuitable. The observed-data likelihood function approaches the positive common-effect likelihood as $\tau \downarrow 0$. After restricting $\mu$ and the intercepts to a compact interior region, it is bounded below there by a positive constant. HenceThis is an improper posterior from a log-uniform random-effect scale prior. Proper conditional sampling distributions do not repair the improper joint posterior, and an arbitrary tiny cutoff would make inference depend on that cutoff.

For the common Bayesian deviance convention $D=-2\log L$ used in all three models,Here Dhat is $D(\overline{\theta })$, an at-posterior-mean fit measure, and $p_D$ is an effective parameter count. The independent model's Dhat of 53.1 is almost identical to the exchangeable model's 53.2; both improve on the common model's 57.8. The common model's $p_D=7$ corresponds to six intercepts plus one shared effect. Independence uses roughly twelve effective parameters. Partial pooling reduces the exchangeable model's effective complexity to about 8.7 while retaining nearly the same fitted likelihood function as independence.

The exchangeable model has the lowest reported DIC, but the common model is competitive. Their difference is only about 1.3, whereas independence is worse by about 6.3. The deviance information criterion measures penalized fit for a predictive comparison, not model posterior probabilities, and these numbers do not establish overwhelming evidence for heterogeneity. The displayed exchangeable $\overline{D}+p_D$ is $61.9+8.7=70.6$, rather than the printed 70.5; rounding of the underlying values can account for a tenth and does not change this interpretation.

Among the supplied candidates, choose ARMA(2,1). It has the smallest Akaike information criterion. Relative to ARMA(2,2), its AIC improvement is $1.71$; the additional second moving-average estimate is only half a standard error from zero and increases the log likelihood function by only about $0.14$. Dropping it is a reasonable parsimony choice, although that AIC gap is small. ARMA(1,1) has AIC larger by $17.55$, a much clearer loss of fit. In the chosen model the second autoregressive term is about seven standard errors from zero, so it should not be dropped merely to obtain order one. The first autoregressive term is less precisely estimated; this does not justify automatically deleting it without fitting and comparing the reduced candidate.

Using the usual positive-sign moving-average convention, the fitted autoregressive moving-average model isThese are plug-in values rounded as given, not exact population parameters. The model has zero mean. Its polynomials are $\varphi (z)=1+0.30z-0.28z^2=(1-0.4z)(1+0.7z)$ and $\theta (z)=1+0.5z$.

With angular frequency $\omega \in [-\pi ,\pi ]$, so that the autocovariance is $\gamma (h)={\int }_{-\pi }^{\pi }{e}^{ih\omega }{f}_X(\omega )d\omega$, the spectral density of a stationary process isThe frequency convention makes the normalization unambiguous.

Use the chain rule for conditional densities. With $x_0=x_1=0$, the likelihood function of the nondegenerate observations isThe first residual is $x_2$ and carries no parameter information. There is no ordinary joint Lebesgue density for $X_1,\dots ,X_n$ because $X_1=0$ deterministically; this expression is the conditional likelihood function given the fixed initial values, or the likelihood function on $x_2,\dots ,x_n$. The initial point mass is parameter independent and has no effect on inference. This is the conditional likelihood of an initialized Gaussian AR(2) process.