Actuarial Science NO CLAIMS DISCOUNT NCD SYSTEMS This is an

Actuarial Science 2004/2005 13. NO CLAIMS DISCOUNT (NCD) SYSTEMS This is an example of an experience rating system whereby the premium paid by a policyholder is adjusted to reflect the past claims experience under the policy. NCD’s are most commonly applied to private motor insurance in the UK; they are also used in some household policies. A NCD system discourages small claims – it reduces claims costs to the insurer which offsets the decrease in premium income from NCD systems. Results/proofs marked *** will be completed in the lectures. 13.1 Ideas • Example 1 Suppose that there are 5 levels of discount 0, 30, 40, 50, 60%. At the end of each policy year, policyholders change levels according to the following rules. (1) A policyholder who has made no claims during a policy year moves to the next higher discount level (or remains at 60% if already at the highest level). (2) A policyholder who has made 1 claim during a policy year drops back 2 levels (or to 0% if that is reached). (3) Otherwise, policyholder drops back to 0% discount. • Transition matrix The rules of a NCD system can be summarised in a transition matrix showing probabilities of movements between each level – see Table 1 for the general notation. Table 1. 0 p00 p10 . . . pk0 Next discount level 1 2 ... k p01 p02 . . . p0k p11 p12 . . . p1k . . ... . . . ... . . . ... . pk1 pk2 . . . pkk Existing discount level 0 1 . . . k ´ D. Jimenez-Huerta 1 Actuarial Science 2004/2005 where the categories of discount level are labelled 0, 1, . . . , k, so the number of categories is k + 1, pij =probability that a policyholder moves from category i to category j, (i, j = 0, 1, ..., k), and pi+ = 1 (row sums = 1). • Example 1 continued Again consider the NCD systems in Example 1 but with the additional assumption that the number of claims per annum, N , by an individual policyholder has a Poisson distribution with mean 0.2. The transition matrix is given in Table 2 where ‘existing’ and ‘next’ refer to the existing and next discount levels expressed as percentages rather than categories 0,1,... as in Table 1. ´ D. Jimenez-Huerta 2 Actuarial Science 2004/2005 Table 2 Next (%) 0 0 30 Existing (%) 40 50 60 In the above transition matrix, α0 = Pr(N = 0) = 0.8187 so 1 − α0 = 0.1813, α1 = Pr(N = 1) = 0.1638, so α0 + α1 = 0.9825 giving 1 − α0 − α1 = 0.0175. 30 40 50 60 *** 13.2 Projection over a finite time horizon • Example 1 continued Suppose that there are 10000 policyholders at the start of year 1, all in the 0% category. What is the state of the NCD system at the start of each subsequent year? The movements from transitions between levels can be summarised in a matrix of expected numbers: see Tables 3a and 3b below. Table 3(a): end of year 1 (start of year 2). 0 1813 Next 30 40 8187 0 50 0 60 0 Total 10000 Existing 0 30 40 50 60 Total Proportion (1-α0 ) stay at 0% rest go to 30% (other elements 0) 1813 8187 0 0 0 10000 Table 3(b): end of year 2 (start of year 3). ´ D. Jimenez-Huerta 3 Actuarial Science 2004/2005 Next 30 40 1483.3 0 0 6702.7 (other elements 0) 1813 1484 6703 0 0 10000 Proportion (1-α0 ) stay at 0% rest go to 30% Proportion (1-α0 ) drop to 0% rest go to 40% Existing 0 30 40 50 60 Total 0 328.7 1484.3 50 0 0 60 0 0 Total 1813 8187 (and so on for further years). The state of the NCD at the start of each year is summarised by the column totals which are the expected numbers of policyholders in each category. ´ D. Jimenez-Huerta 4 Actuarial Science 2004/2005 • General results (These results, on Markov chains, were obtained in STATB087 and are repeated here for students who did not attend that course.) (1) In general, we require the distribution of number of policyholders in each category in year n. Let πi = probability policyholder is in category i at start of year n. If pij is the (i, j)th element of the transition matrix P then, by the law of total probability, πj (n+1) (n) = *** The above result can be expressed in ‘matrix form’ as follows. Let π (n) = (π0 , . . . , πk ) Then π (n+1) = (1) *** (2) Now find the expected number of policyholder for year n – denote this by m(n) . Similarly to the above, by conditioning on the previous year, we obtain m(n+1) = (2) *** (or multiply both sides of (1) by the total number of policyholders). • Example 1 continued Now proceed using the general notation, again with 10000 policyholders at the start of year 1, all in the 0% category. (n) (n) (a row vector). ´ D. Jimenez-Huerta 5 Actuarial Science 2004/2005 Initially: π (1) = (1, 0, 0, 0, 0) At start of year 2: π (2) = (0.1813, 0.8187, 0, 0, 0) At start of year 3: π (3) = (0.1813, 0.1484, 0.6703, 0, 0) and so on. • If we are only interested in the state of the NCD system at a particular value of n, repeated use of (1) and (2) gives π (n+1) = π (1) P n and m(n+1) = m(1) P n m(1) = (10000, 0, ..., 0) m(2) = (1813, 8187, 0, ..., 0) m(3) = (1813, 1484, 6703, 0, ..., 0) ´ D. Jimenez-Huerta 6 Actuarial Science 2004/2005 13.3 Projections over an infinite time horizon • This involves finding the equilibrium distribution π (n) as n → ∞ if it exists. Denote this limiting distribution by π = (π0 , . . . , πk ). Letting n → ∞ in (1) gives π= (3) *** The equilibrium distribution exists if the solution to this set of equations satisfies the properties of a distribution, i.e. 0 ≤ πi ≤ 1 (for all i = 0, 1, . . . , k) and • Example 1 continued (a) Finding the equilibrium distribution. Using (3): π0 = (π0 + π1 + π2 )(1 − α0 ) + (π3 + π4 )(1 − α0 − α1 ) = (1 − α0 ) πi − α1 (π3 + π4 ) π1 = π0 α0 + π3 α1 π2 = π1 α0 + π4 α1 π 3 = π 2 α0 π4 = π3 α0 + π4 α0 Adding the last four equations gives π1 + . . . + π4 = ( πi )α0 + (π3 + π4 )α1 Adding π0 to each side gives the first equation so we don’t have five linearly independent equations, which is a consequence of row sums = 1 in P . So in addition to the above equations, we use πi = 1. α0 Let β = 1−α0 . Then the solution is: π4 = βπ3 , π2 = π1 α0 + π2 βα0 α1 π1 = π0 α0 +π1 2 α0 α1 1 − βα0 α1 k i=0 πi = 1. π3 = π2 α0 , π2 (1 − βα0 α1 ) = π1 α0 , π1 1 − 2 α0 α1 1 − βα0 α1 giving giving = π0 α0 Hence π1 = 1.1345π0 , π2 = 2.3548π0 , π3 = 1.9279π0 , π4 = 8.7058π0 . πi = 1 gives π0 (1 + 1.1345 + ... + 8.7058) = 1, hence π0 = 1 15.1230 = 0.06612, π1 = 0.07502, π2 = 0.1557, π3 = 0.1275 and π4 = 0.5757. ´ D. Jimenez-Huerta 7 Actuarial Science 2004/2005 (b) Finding the mean long term premium as a percentage of the full premium. If the full premium is c, then the average premium paid is *** i.e. about 50% of the full premium. 13.4 Heterogeneity in the portfolio NCD systems may not work as well as you hope – the premium may not differentiate too well between those who claim frequently and those who do not. Example 2. This is a simple example to illustrate the above point. Suppose there are two categories of claimant: ‘good driver’ and ‘bad driver’, and the corresponding probabilities of a claim are 0.1 and 0.2, respectively. Suppose that there are three categories of discount: 0%, 25%, 50% and the NCD system is: • move up 1 category if no claim made, • move down 1 category if ≥ 1 claim made. Let α0 denote the probability of no claim made. Let 0, 1, 2 denote the three categoies 0%, 25%, 50% of discount, respectively. The transition matrix P is Next level 0 1 2 1-α0 α0 0 1-α0 0 α0 0 1-α0 α0 Existing level 0 1 2 For a good driver, α0 = 0.9 and, by using (3), the equilibrium distribution 1 9 is π = 91 , 91 , 81 . 91 1 4 16 For a bad driver, α0 = 0.8 and the equilibrium distribution is π = 21 , 21 , 21 . The average premiums paid in the long term are as follows: for a good driver: for a bad driver: 1 91 1 21 ×c+ 9 91 × 0.75c + 81 91 × 0.5c = 0.530c, ×c+ 4 21 × 0.75c + 16 21 × 0.5c = 0.571c. 8 ´ D. Jimenez-Huerta Actuarial Science 2004/2005 So we see that bad drivers are twice as likely to claim as good drivers but the premium is only, on average marginally higher. We could adjust discounts so that bad drivers pay a premium double that for good drivers – but the discount in category 2 has to become very large to do this – this is not in the insurer’s interest! Alternatively, we could increase the number of categories of discount but that might not help enough without the NCD system getting too complex. Note: NCD system’s are usually more complex than used in these examples. There are exemptions to some claims counting against the NCD (eg windscreen), also protected policies and introductory offers. 13.5 Effect of NCD system on likelihood to claim A policyholder needs to assess the costs of not claiming and hence getting more discount next year, against the costs of claiming and losing or getting less discount next year. Example 2 continued (a) When should a policyholder claim? With the NCD system as in the previous example, we can calculate the extra cost of the premium by claiming as follows. Suppose the full premium £500, there is no policy excess and the horizon is 3 years (i.e. the policyholder is assuming no more than one claim over the next 3 years). Current Discount level 0% 25% 50% *** So, for example, a policyholder currently in the 0% category will claim if the cost of repair is > 250. (The same results apply if there is an infinite horizon.) (b) Probability of claiming if policyholder has had an accident. We now need to know the loss distribution: suppose this is lognormal with parameters µ = 5, σ = 2 (using the notation of section 8). We require Pr(X > x) where x =cost in last column of above table. ´ D. Jimenez-Huerta 9 Premium in next 3 years No claim Claim 375, 250, 250 250, 250, 250 250, 250, 250 Extra cost of premium by claiming Actuarial Science 2004/2005 Current discount level 0 25 50 x 250 375 125 Pr(X > x) 0.397 0.322 0.534 Calculate probabilities as in section 10 of notes, i.e. *** (c) Marginal probability of a claim. Suppose Pr(accident) is 0.1 for good drivers and 0.2 for bad drivers. (These were the claim probabilities in the previous example but these have changed here because a claim is not automatically made if policyholder has an accident.) Using Pr(claim) = *** where Pr(claim|accident) was calculated in (b), we obtain Category 0 25 50 Pr(claim) Good Bad 0.0397 0.0794 0.0322 0.0644 0.0534 0.1068 (d) Mean long term premium. The transition matrices, P , for each category of driver are then: Good 0.0397 0.0322 0 0.9603 0 0.0534 0 0.9678 0.9466 0.0794 0.0644 0 Bad 0.9206 0 0.1068 0 0.9356 0.8932 The corresponding equilibrium distributions are: (0.0018 0.0522 0.9461) and the long term average premiums are: 0.5139c 0.5290c (0.0071 0.1017 0.8912) (e) Including a policy excess: think about this as part of the problem set in Exercises 7. ´ D. Jimenez-Huerta 10