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The Bühlmann model (named after Hans Bühlmann) is a random effects model (or "variance components model" or hierarchical linear model) used in credibility theory in actuarial science to determine the appropriate premium for a group of insurance contracts.

Bühlmann model was introduced in the paper "Experience rating and credibility" by Hans Bühlmann (1967). It answers the following question:

We have i risks that generate random losses, and possess historical data on the claims produced by that risks. Especially, we have j samples for the i-th risk. We want to set a premium for the i-th risk based on the expected value of claims. The model answers the question, what is the best estimator of the expected value, considering information of all claims. Precisely, we look for the estimator that is best of all linear estimators in terms of minimizing the mean square error.

Let us have:

$$X_{ij}$$ - the j-th claim for the i-th risk. We assume that all claims for i-th risk are independent and identically distributed

$$\bar{X}_i=\frac{1}{m}\sum_{j=1}^{m}X_{ij}$$

$$\Theta_i$$ - the parameter for the distribution of the i-th risk

$$m(\vartheta)= E\left [ X_{ij} |\Theta_i = \vartheta\right ]$$

$$\Pi=E(m(\vartheta)|X_{i1},X_{i2},...X_{im})$$ - premium for the i-th risk

$$\mu = E(m(\vartheta))$$

$$s^2(\vartheta)=Var\left [ X_{ij} |\Theta_i = \vartheta\right ]$$

$$\sigma^2=E\left [ s^2(\vartheta) \right ]$$

$$v^2=Var\left [ m(\vartheta) \right ]$$

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

The Bühlmann model is the solution for the problem:
 * $$E\left [ \left ( a_{i0}+\sum_{j=1}^{m}a_{ij}X_{ij}-\Pi \right)^2\right ]\rightarrow min $$

The solution for the problem is:

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

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

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.

Proof
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. Proof can be found on this site.

Lemma: The problem can be stated alternatively as:

$$f=E\left [ \left ( a_{i0}+\sum_{j=1}^{m}a_{ij}X_{ij}-m(\vartheta)\right )^2\right ]\rightarrow max$$

Proof:

$$E\left [ \left ( a_{i0}+\sum_{j=1}^{m}a_{ij}X_{ij}-m(\vartheta)\right )^2\right ]=$$ $$=E\left [ \left ( a_{i0}+\sum_{j=1}^{m}a_{ij}X_{ij}-\Pi\right )^2\right ]+E\left [  \left ( m(\vartheta)-\Pi\right )^2\right ]+2E\left [  \left ( a_{i0}+\sum_{j=1}^{m}a_{ij}X_{ij}-\Pi\right ) \left ( m(\vartheta)-\Pi\right )\right ]=$$ $$=E\left [ \left ( a_{i0}+\sum_{j=1}^{m}a_{ij}X_{ij}-\Pi\right )^2\right ]+E\left [  \left ( m(\vartheta)-\Pi\right )^2\right ]$$

The last equation follows from the fact that $$E\left [ \left ( a_{i0}+\sum_{j=1}^{m}a_{ij}X_{ij}-\Pi\right ) \left ( m(\vartheta)-\Pi\right )\right ]=E_{\Theta}\left \{  E_X\left [  \left ( a_{i0}+\sum_{j=1}^{m}a_{ij}X_{ij}-\Pi\right ) \left ( m(\vartheta)-\Pi\right ) |  X_{i1},X_{i2},X_{im}\right ]\right \}=$$ $$=\left ( a_{i0}+\sum_{j=1}^{m}a_{ij}X_{ij}-\Pi\right )E_{\Theta}\left \{ E_X\left [   \left ( m(\vartheta)-\Pi\right ) |  X_{i1},X_{i2},X_{im}\right ]\right \}=0$$

We are using here the law of total expectation and the fact, that $$\Pi=E(m(\vartheta)|X_{i1},X_{i2},...X_{im})$$

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

$$\frac{1}{2}\frac{\partial f}{\partial a_{01}}=$$

$$E\left [  a_{i0}+\sum_{j=1}^{m}a_{ij}X_{ij}-m(\vartheta)\right ]=a_{i0}+\sum_{j=1}^{m}a_{ij}E(X_{ij})-E(m(\vartheta))=a_{i0}-\left (\sum_{j=1}^{m}a_{ij}-1  \right )\mu$$

$$a_{i0}=\left (\sum_{j=1}^{m}a_{ij}-1 \right )\mu$$

For $$k\neq 0$$ we have:

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

We can simplify derivative, noting that:

$$E[X_{ij}X_{ik}]=E[E[X_{ij}X_{ik}|\vartheta]]=E[cov(X_{ij}X_{ik}|\vartheta)+E(X_{ij}|\vartheta)E(X_{ik}|\vartheta)]=E[(m(\vartheta))^2]=v^2+\mu^2$$

and

$$E[X^2_{ik}]=E[E[X^2_{ik}|\vartheta]]=E[s^2(\vartheta)+(m(\vartheta))^2]=\sigma^2+v^2+\mu^2$$

and

$$E[X_{ik}m(\vartheta)]=E[E[X_{ik}m(\vartheta)|\Theta_i]=E[(m(\vartheta))^2]=v^2+\mu^2$$

Taking above equations and inserting into derivative, we have:

$$\frac{1}{2}\frac{\partial f}{\partial a_{ik}}=\left ( 1-\sum_{j=1}^{m}a_{ij} \right )\mu^2+\sum_{j=1,j\neq k}^{m}a_{ij}(v^2+\mu^2)+a_{ik}(\sigma^2+v^2+\mu^2)-(v^2+\mu^2)=a_{ik}\sigma^2-\left ( 1-\sum_{j=1}^{m}a_{ij} \right )v^2=0$$

$$\sigma^2a_{ik}=v^2\left (1- \sum_{j=1}^{m} a_{ij}\right)$$

Right side doesn't depend on k. Therefore all $$a_{ik}$$ are constant

$$a_{i1}=a_{i2}=...=a_{im}=\frac{v^2}{\sigma^2+mv^2}$$

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

$$a_{i0}=(1-ma_{ik})\mu=\left ( 1-\frac{mv^2}{\sigma^2+mv^2} \right )\mu$$

Finally, the best estimator is

$$a_{i0}+\sum_{j=1}^{m}a_{ij}X_{ij}=\frac{mv^2}{\sigma^2+mv^2}\bar{X_i}+\left ( 1-\frac{mv^2}{\sigma^2+mv^2} \right )\mu=Z\bar{X_i}+(1-Z)\mu$$