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Incredibility theory, a branch of study inactuarial science, theBühlmann model is arandom effects model (or "variance components model" orhierarchical linear model) used to determine the appropriatepremium for a group of insurance contracts. The model is named after Hans Bühlmann who first published a description in 1967.^[1]

Consideri risks which generate random losses for which historical data ofm recent claims are available (indexed byj). A premium for theith risk is to be determined based on the expected value of claims. A linear estimator which minimizes the mean square error is sought. Write

• X_ij for thej-th claim on thei-th risk (we assume that all claims fori-th risk areindependent and identically distributed)
• ${\displaystyle {\overline{X}}_i=\frac{1}{m}\sum _{j=1}^m{X}_{ij}}$ for the average value.
• ${\displaystyle {\mathrm{\Theta }}_i}$ - the parameter for the distribution of the i-th risk
• ${\displaystyle m(\vartheta )=\mathrm{E}\left[X_{ij}|{\mathrm{\Theta }}_i=\vartheta \right]}$
• ${\displaystyle \mathrm{\Pi }=\mathrm{E}(m(\vartheta )|X_{i1},X_{i2},...X_{im})}$ - premium for the i-th risk
• ${\displaystyle \mu =\mathrm{E}(m(\vartheta ))}$
• ${\displaystyle s^2(\vartheta )=\mathrm{Var}\left[X_{ij}|{\mathrm{\Theta }}_i=\vartheta \right]}$
• ${\displaystyle {\sigma }^2=\mathrm{E}\left[s^2(\vartheta )\right]}$
• ${\displaystyle v^2=\mathrm{Var}\left[m(\vartheta )\right]}$

Note:${\displaystyle m(\vartheta )}$ and${\displaystyle s^2(\vartheta )}$ are functions of random parameter${\displaystyle \vartheta }$

The Bühlmann model is the solution for the problem:

${\displaystyle \underset{a_{i0},a_{i1},...,a_{im}}{\mathrm{a}\mathrm{r}\mathrm{g}\mathrm{m}\mathrm{i}\mathrm{n}}\mathrm{E}\left[{\left(a_{i0}+\sum _{j=1}^m{a}_{ij}{X}_{ij}-\mathrm{\Pi }\right)}^2\right]}$

where${\displaystyle a_{i0}+\sum _{j=1}^m{a}_{ij}{X}_{ij}}$ is the estimator of premium${\displaystyle \mathrm{\Pi }}$ andarg min represents the parameter values which minimize the expression.

The solution for the problem is:

${\displaystyle Z{\overline{X}}_i+(1-Z)\mu }$

where:

${\displaystyle Z=\frac{1}{1+\frac{{\sigma }^2}{v^{2}m}}}$

We can give this result the interpretation, that Z part of the premium is based on the information that we have about the specific risk, and (1-Z) part is based on the information that we have about the whole population.

The following proof is slightly different from the one in the original paper. It is also more general, because it considers all linear estimators, while original proof considers only estimators based on average claim.^[2]

Lemma. The problem can be stated alternatively as:

${\displaystyle f=\mathbb{E}\left[{\left(a_{i0}+\sum _{j=1}^m{a}_{ij}{X}_{ij}-m(\vartheta )\right)}^2\right]\to min}$

Proof:

The last equation follows from the fact that

We are using here the law of total expectation and the fact, that${\displaystyle \mathrm{\Pi }=\mathbb{E}[m(\vartheta )|X_{i1},\dots ,X_{im}].}$

In our previous equation, we decompose minimized function in the sum of two expressions. The second expression does not depend on parameters used in minimization. Therefore, minimizing the function is the same as minimizing the first part of the sum.

Let us find critical points of the function

${\displaystyle \frac{1}{2}\frac{\mathrm{\partial }f}{\mathrm{\partial }{a}_{i0}}=\mathbb{E}\left[a_{i0}+\sum _{j=1}^m{a}_{ij}{X}_{ij}-m(\vartheta )\right]=a_{i0}+\sum _{j=1}^m{a}_{ij}\mathbb{E}(X_{ij})-\mathbb{E}(m(\vartheta ))=a_{i0}+\left(\sum _{j=1}^m{a}_{ij}-1\right)\mu }$

${\displaystyle a_{i0}=\left(1-\sum _{j=1}^m{a}_{ij}\right)\mu }$

For${\displaystyle k\ne 0}$ we have:

${\displaystyle \frac{1}{2}\frac{\mathrm{\partial }f}{\mathrm{\partial }{a}_{ik}}=\mathbb{E}\left[X_{ik}\left(a_{i0}+\sum _{j=1}^m{a}_{ij}{X}_{ij}-m(\vartheta )\right)\right]=\mathbb{E}\left[X_{ik}\right]a_{i0}+\sum _{j=1,j\ne k}^m{a}_{ij}\mathbb{E}[X_{ik}{X}_{ij}]+a_{ik}\mathbb{E}[X_{ik}^2]-\mathbb{E}[X_{ik}m(\vartheta )]=0}$

We can simplify derivative, noting that:

Taking above equations and inserting into derivative, we have:

${\displaystyle \frac{1}{2}\frac{\mathrm{\partial }f}{\mathrm{\partial }{a}_{ik}}=\left(1-\sum _{j=1}^m{a}_{ij}\right){\mu }^2+\sum _{j=1,j\ne k}^m{a}_{ij}(v^2+{\mu }^2)+a_{ik}({\sigma }^2+v^2+{\mu }^2)-(v^2+{\mu }^2)=a_{ik}{\sigma }^2-\left(1-\sum _{j=1}^m{a}_{ij}\right)v^2=0}$

${\displaystyle {\sigma }^2{a}_{ik}=v^2\left(1-\sum _{j=1}^m{a}_{ij}\right)}$

Right side doesn't depend onk. Therefore, all${\displaystyle a_{ik}}$ are constant

${\displaystyle a_{i1}=\cdots =a_{im}=\frac{v^2}{{\sigma }^2+m{v}^2}}$

From the solution for${\displaystyle a_{i0}}$ we have

${\displaystyle a_{i0}=(1-m{a}_{ik})\mu =\left(1-\frac{m{v}^2}{{\sigma }^2+m{v}^2}\right)\mu }$

Finally, the best estimator is

${\displaystyle a_{i0}+\sum _{j=1}^m{a}_{ij}{X}_{ij}=\frac{m{v}^2}{{\sigma }^2+m{v}^2}\overline{X_i}+\left(1-\frac{m{v}^2}{{\sigma }^2+m{v}^2}\right)\mu =Z\overline{X_i}+(1-Z)\mu }$

• Actuarial science
• Analysis of variance

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