Bühlmann model

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:

${\displaystyle {\begin{aligned}\mathbb {E} \left[\left(a_{i0}+\sum _{j=1}^{m}a_{ij}X_{ij}-m(\vartheta )\right)^{2}\right]&=\mathbb {E} \left[\left(a_{i0}+\sum _{j=1}^{m}a_{ij}X_{ij}-\Pi \right)^{2}\right]+\mathbb {E} \left[\left(m(\vartheta )-\Pi \right)^{2}\right]-2\mathbb {E} \left[\left(a_{i0}+\sum _{j=1}^{m}a_{ij}X_{ij}-\Pi \right)\left(m(\vartheta )-\Pi \right)\right]\\&=\mathbb {E} \left[\left(a_{i0}+\sum _{j=1}^{m}a_{ij}X_{ij}-\Pi \right)^{2}\right]+\mathbb {E} \left[\left(m(\vartheta )-\Pi \right)^{2}\right]\end{aligned}}}$

The last equation follows from the fact that

${\displaystyle {\begin{aligned}\mathbb {E} \left[\left(a_{i0}+\sum _{j=1}^{m}a_{ij}X_{ij}-\Pi \right)\left(m(\vartheta )-\Pi \right)\right]&=\mathbb {E} _{\Theta }\left[\mathbb {E} _{X}\left.\left[\left(a_{i0}+\sum _{j=1}^{m}a_{ij}X_{ij}-\Pi \right)(m(\vartheta )-\Pi )\right|X_{i1},\ldots ,X_{im}\right]\right]\\&=\mathbb {E} _{\Theta }\left[\left(a_{i0}+\sum _{j=1}^{m}a_{ij}X_{ij}-\Pi \right)\left[\mathbb {E} _{X}\left[(m(\vartheta )-\Pi )|X_{i1},\ldots ,X_{im}\right]\right]\right]\\&=0\end{aligned}}}$

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

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

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

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

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

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

We can simplify derivative, noting that:

${\displaystyle {\begin{aligned}\mathbb {E} [X_{ij}X_{ik}]&=\mathbb {E} \left[\mathbb {E} [X_{ij}X_{ik}|\vartheta ]\right]=\mathbb {E} [{\text{cov}}(X_{ij}X_{ik}|\vartheta )+\mathbb {E} (X_{ij}|\vartheta )\mathbb {E} (X_{ik}|\vartheta )]=\mathbb {E} [(m(\vartheta ))^{2}]=v^{2}+\mu ^{2}\\\mathbb {E} [X_{ik}^{2}]&=\mathbb {E} \left[\mathbb {E} [X_{ik}^{2}|\vartheta ]\right]=\mathbb {E} [s^{2}(\vartheta )+(m(\vartheta ))^{2}]=\sigma ^{2}+v^{2}+\mu ^{2}\\\mathbb {E} [X_{ik}m(\vartheta )]&=\mathbb {E} [\mathbb {E} [X_{ik}m(\vartheta )|\Theta _{i}]=\mathbb {E} [(m(\vartheta ))^{2}]=v^{2}+\mu ^{2}\end{aligned}}}$

Taking above equations and inserting into derivative, we have:

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

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

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

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

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

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

Finally, the best estimator is

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