Answer to Question #126844 in Statistics and Probability for Ruth Lucy

Question

A No Claims Discount system is operated by a car insurer. There are four levels of discount: 0%, 10%, 25% and 40%. After a claim-free year a policy holder moves up one level (or remains at the 40% level if they are already there). If a policy holder makes one claim in a year he or she moves down one level (or remains at the 0% level if already there). A policyholder who makes more than one claim in a year moves down two levels (or moves to or remains at the 0% level). Changes in level can only happen at the end of each year. The probability of a claim in any given month is assumed to be constant at 0.04 (i.e. 4%). At most one claim can be made per month and claims are assumed independent. Calculate the proportion of policyholders in the long run who are at the 25% level.

Answer 1

Solution:

Let A₀ be the event of zero claim in a year.

Then p₀=P(A₀)=(1-0.04)¹²=(24/25)¹²=0.6127...

Let A₁ be the event of one claim in a year.

Then p₁=P(A₁)=12*0.04*0.96¹¹= 0.3063...

Let A_m be the event of more than one claim in a year.

Then p_m=P(A_m)=1-P(A₀)-P(A₁)=0.0809...

Let A_0,n be the event of level of discount 0% after n years,

A_10,n be the event of level of discount 10% after n years,

A_25,n be the event of level of discount 25% after n years,

and A_40,n be the event of level of discount 0% after n years.

Then for any n we obtain

P(A_0,n)=p₁(P(A_0,n-1)+P(A_10,n-1))+p_m(P(A_0,n-1)+P(A_10,n-1)+P(A_25,n-1))

P(A_10,n)=p₀P(A_0,n-1)+p₁P(A_25,n-1)+p_mP(A_40,n-1)

P(A_25,n)=p₀P(A_10,n-1)+p₁P(A_40,n-1)

P(A_40,n)=p₀P(A_25,n-1)+p₀P(A_40,n-1)

P(A_0,n)+P(A_10,n)+P(A_25,n)+P(A_40,n)=1

Really, for example, one can be have 25%-level discount after n year if after (n-1) year one had 10%-level discount and zero claim in a year or if he had 40%-level discount and one claim in the last year.

Let

P(A_i,n)$\rarr$ T_i , $n \rarr \infty$, i=0,10,25,40

Then we obtain system

T₀=p₁T₀+p₁T₁₀+p_mT₀+p_mT₁₀+p_mT₂₅

T₁₀=p₀T₀+p₁T₂₅+p_mT₄₀

T₂₅=p₀T₁₀+p₁T₄₀

T₄₀=p₀T₂₅+p₀T₄₀

T₀+T₁₀+T₂₅+T₄₀=1

T₂₅=1/p₀(T₄₀-p₀T₄₀),

T₁₀=1/p₀(T₂₅-p₁T₄₀)=...

After solving this system we obtain

T₀=0.1624...

T₁₀=0.2057...

T₂₅=0.2446...

T₄₀=0.3871...

Answer:

The proportion of policyholders in the long run who are at the 25% level is equal 24,5%.