= Solution
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>:
$$
\begin{aligned}
m&=\mathbb E\mu(\theta)=\frac\alpha{\lambda(k-1)},\\
v&=\operatorname{Var}(\mu(\theta))=\frac\alpha{\lambda^2(k-1)^2},\\
s^2&=\mathbb E[\operatorname{Var}(X_i\mid\theta)]
=\frac{\alpha(\alpha+1)}{\lambda^2(k-1)^2(k-2)}.
\end{aligned}
$$
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 imply
$$
\mathbb E X_i=m,\qquad
\operatorname{Cov}(\mu(\theta),X_i)=v,\qquad
\operatorname{Cov}(X_i,X_j)=v+s^2\mathbf1_{\{i=j\}}.
$$
All 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> gives
$$
\mathbb E\left[(Y-\sum_i c_iW_i)^2\right]
=v-2v\sum_i c_i+s^2\sum_i c_i^2+v\left(\sum_i c_i\right)^2.
$$
Its <normal equations> are $s^2c_i+v\sum_jc_j=v$ 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 gives
$$
\boxed{c_i=\frac v{s^2+nv}=\frac{k-2}{n(k-2)+\alpha+1}\quad(1\le i\le n),\qquad
c_0=\frac{\alpha(\alpha+1)}{\lambda(k-1)[n(k-2)+\alpha+1]}.}
$$
A <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> is
$$
\boxed{\widehat\mu_{\rm lin}=(1-Z)m+Z\overline X,\qquad
Z=\frac{nv}{s^2+nv}=\frac{n(k-2)}{n(k-2)+\alpha+1}.}
$$
Thus 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> update
$$
\theta\mid x_1,\ldots,x_n\sim\operatorname{Gamma}\left(\alpha+nk,\text{ rate }\lambda+\sum_i x_i^{-1}\right).
$$
Under <quadratic loss>, the conditional risk of an estimate $d$ is $\operatorname{Var}(\mu(\theta)\mid x)+(d-\mathbb E[\mu(\theta)\mid x])^2$. Its minimizer is therefore the <posterior mean>:
$$
\boxed{\widehat\mu_{\rm Bayes}=\frac{\alpha+nk}{(k-1)(\lambda+\sum_i x_i^{-1})}.}
$$
\b[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 $-2C\lambda/(1+\lambda x)^3\ne0$, 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 is
$$
\widehat\mu_{\rm Bayes}=(1-Z_B)m+Z_B\frac{kH}{k-1},\qquad Z_B=\frac A{\lambda+A}.
$$
This 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.
Back to article page
