Bühlmann model explained

In credibility theory, a branch of study in actuarial science, the Bühlmann model is a random effects model (or "variance components model" or hierarchical linear model) used to determine the appropriate premium for a group of insurance contracts. The model is named after Hans Bühlmann who first published a description in 1967.[1]

Model description

Consider i risks which generate random losses for which historical data of m recent claims are available (indexed by j). A premium for the ith risk is to be determined based on the expected value of claims. A linear estimator which minimizes the mean square error is sought. Write

\scriptstyle=

1
m
m
\sum
j=1

Xij

for the average value.

\Thetai

- the parameter for the distribution of the i-th risk

m(\vartheta)=\operatornameE\left[Xij|\Thetai=\vartheta\right]

\Pi=\operatornameE(m(\vartheta)|Xi1,Xi2,...Xim)

- premium for the i-th risk

\mu=\operatornameE(m(\vartheta))

s2(\vartheta)=\operatorname{Var}\left[Xij|\Thetai=\vartheta\right]

\sigma2=\operatornameE\left[s2(\vartheta)\right]

v2=\operatorname{Var}\left[m(\vartheta)\right]

Note:

m(\vartheta)

and

s2(\vartheta)

are functions of random parameter

\vartheta

The Bühlmann model is the solution for the problem:

\underset{ai0,ai1,...,aim

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

where

ai0

m
+\sum
j=1

aijXij

is the estimator of premium

\Pi

and arg min represents the parameter values which minimize the expression.

Model solution

The solution for the problem is:

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

where:

Z=1
1+\sigma2
v2m

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.[2]

Lemma. The problem can be stated alternatively as:

f=E\left[\left(ai0

m
+\sum
j=1

aijXij-m(\vartheta)\right)2\right]\tomin

Proof:

\begin{align} E\left[\left(ai0

m
+\sum
j=1

aijXij-m(\vartheta)\right)2\right]&=E\left[\left(ai0

m
+\sum
j=1

aijXij-\Pi\right)2\right]+E\left[\left(m(\vartheta)-\Pi\right)2\right]-2E\left[\left(ai0

m
+\sum
j=1

aijXij-\Pi\right)\left(m(\vartheta)-\Pi\right)\right]\\ &=E\left[\left(ai0

m
+\sum
j=1

aijXij-\Pi\right)2\right]+E\left[\left(m(\vartheta)-\Pi\right)2\right] \end{align}

The last equation follows from the fact that

\begin{align} E\left[\left(ai0

m
+\sum
j=1

aijXij-\Pi\right)\left(m(\vartheta)-\Pi\right)\right]&=E\Theta\left[EX\left.\left[\left(ai0

m
+\sum
j=1

aijXij-\Pi\right)(m(\vartheta)-\Pi)\right|Xi1,\ldots,Xim\right]\right]\\ &=E\Theta\left[\left(ai0

m
+\sum
j=1

aijXij-\Pi\right)\left[EX\left[(m(\vartheta)-\Pi)|Xi1,\ldots,Xim\right]\right]\right]\\ &=0 \end{align}

We are using here the law of total expectation and the fact, that

\Pi=E[m(\vartheta)|Xi1,\ldots,Xim].

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

1
2
\partialf
\partialai0

=E\left[ai0

m
+\sum
j=1

aijXij-m(\vartheta)\right]=ai0

m
+\sum
j=1

aijE(Xij)-E(m(\vartheta))=ai0+\left

m
(\sum
j=1

aij-1\right)\mu

ai0=\left(1-

m
\sum
j=1

aij\right)\mu

For

k0

we have:
1
2
\partialf
\partialaik

=E\left[Xik\left(ai0

m
+\sum
j=1

aijXij-m(\vartheta)\right)\right]=E\left[Xik\right]ai0

m
+\sum
j=1,jk

aijE[XikXij]+aik

2
E[X
ik

]-E[Xikm(\vartheta)]=0

We can simplify derivative, noting that:

\begin{align} E[XijXik]&=E\left[E[XijXik|\vartheta]\right]=E[cov(XijXik|\vartheta)+E(Xij|\vartheta)E(Xik|\vartheta)]=E[(m(\vartheta))2]=v2+\mu2

2
\\ E[X
ik

]&=E\left

2
[E[X
ik

|\vartheta]\right]=E[s2(\vartheta)+(m(\vartheta))2]=\sigma2+v2+\mu2\\ E[Xikm(\vartheta)]&=E[E[Xik

2]=v
m(\vartheta)|\Theta
i]=E[(m(\vartheta))

2+\mu2 \end{align}

Taking above equations and inserting into derivative, we have:

1
2
\partialf
\partialaik

=\left(

m
1-\sum
j=1

aij\right

m
)\mu
j=1,jk

aij(v2+\mu

2)+a
ik

(\sigma2+v2+\mu2)-(v2+\mu

2)=a
ik

\sigma2-\left(

m
1-\sum
j=1

aij\right)v2=0

2a
\sigma
ik

=v2\left

m
(1-\sum
j=1

aij\right)

Right side doesn't depend on k. Therefore, all

aik

are constant

ai1==aim=

v2
\sigma2+mv2

From the solution for

ai0

we have

ai0=(1-maik)\mu=\left(1-

mv2
\sigma2+mv2

\right)\mu

Finally, the best estimator is

ai0

m
+\sum
j=1

aijXij=

mv2
\sigma2+mv2

\bar{Xi}+\left(1-

mv2
\sigma2+mv2

\right)\mu=Z\bar{Xi}+(1-Z)\mu

References

Sources

Notes and References

  1. Bühlmann . Hans . 1967 . Experience rating and credibility . ASTIN Bulletin . 4 . 3 . 199 - 207.
  2. Proof can be found on this site: Web site: Hanspeter . Schmidli . Lecture notes on Risk Theory . Institute of Mathematics, University of Cologne . August 11, 2013 . https://web.archive.org/web/20130811041617/http://www.math.ku.dk/~schmidli/rt.pdf .