Credibility. Chapters Stat Loss Models. Chapters (Stat 477) Credibility Brian Hartman - BYU 1 / 31

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1 Credibility Chapters Stat Loss Models Chapters (Stat 477) Credibility Brian Hartman - BYU 1 / 31

2 Why Credibility? You purchase an auto insurance policy and it costs $150. That price is mainly the expected cost of a policyholder with your characteristics (car make and model, age, driving record, etc.). After three years, you have no claims. You call the insurer to explain how you are better than the average driver with your characteristics, so should be charged less. Is your record a sign of your good driving, or just random chance? Chapters (Stat 477) Credibility Brian Hartman - BYU 2 / 31

3 Full Credibility If the sample mean X n is a stable estimator of, the true mean, then we should only use X n. We say the data is fully credible in this situation. More specically, P r( r X r) p which says that the dierence between the estimated mean and the true mean is proportionally small, with high probability (common choices of p and r are 0.9 and 0.05, respectively) Chapters (Stat 477) Credibility Brian Hartman - BYU 3 / 31

4 Full Credibility cont. We can restate that previous formula as Pr and nd the minimum value y p = inf y X = p n rp n Pr p X y p = p n If it is continuous then Pr X = p n y p = p Therefore, for full credibility, y p r p n= Chapters (Stat 477) Credibility Brian Hartman - BYU 4 / 31

5 Full Credibility cont. For full credibility, y p rp n In many cases, we can assume that! n X = p n follows a standard normal distribution. And yp 2 r p = Pr(jZj y p ) = 2 (y p ) 1 Therefore y p is the (1 + p)=2 percentile of the standard normal distribution. Chapters (Stat 477) Credibility Brian Hartman - BYU 5 / 31

6 Central Limit Example Suppose 10 past years' losses are available from a policyholder The sample mean is used to estimate = E(X j ). Determine the full credibility standard with r = 0:05 and p = 0:9. yp 2 n r 2 (1:645)(267:89) = (0:05)(184:6) = 2279:51 Note that we used the sample mean and standard deviation to estimate and and that 10 samples fall far short of the full credibility standard. Chapters (Stat 477) Credibility Brian Hartman - BYU 6 / 31

7 Poisson Example Assume that we are interested in estimating the number of claims and assume the follow a Poisson() distribution. Find the number of policies necessary for full credibility. We know that = E(N j ) = and 2 = V ar(n j ) = so the full credibility standard is n y p p r! 2 = y 2 p r 2 and will need to be estimated from the data. Chapters (Stat 477) Credibility Brian Hartman - BYU 7 / 31

8 Aggregate Payments Example Assume further that each claim size has a mean of Y and a variance of Y 2. What is the standard for full credibility in terms of the average aggregate claim amount? = E(S j ) = Y 2 = V ar(s j ) = ( 2 Y + 2 Y ) yp 2 (Y 2 n + 2 Y ) r 2 Y " 2 = y 2 p r Y Y 2 # Chapters (Stat 477) Credibility Brian Hartman - BYU 8 / 31

9 SOA Practice #2 You are given: The number of claims has a Poisson distribution. Claim sizes have a Pareto distribution with parameters = 0:5 and = 6. The number of claims and claim sizes are independent. The observed pure premium should be within 2% of the expected pure premium 90% of the time. Calculate the expected number of claims needed for full credibility. [16,913] Chapters (Stat 477) Credibility Brian Hartman - BYU 9 / 31

10 Partial Credibility Rather than either using either only the data or the manual rate, how about a weighted average? The partial credibility estimate (or credibility premium) is: P c = Z X + (1 Z)M where Z is the credibility factor. Chapters (Stat 477) Credibility Brian Hartman - BYU 10 / 31

11 Partial Credibility It can be shown that 2 r 2 y 2 p = V ar(p c ) = V ar[z X + (1 Z)M ] = Z 2 V ar( X) = Z 2 2 n ) Z = rp n y p It turns out that the credibility factor is the square root of the ratio of the actual count to the count needed for full credibility. Chapters (Stat 477) Credibility Brian Hartman - BYU 11 / 31

12 SOA Example #65 You are given the following information about a general liability book of business comprised of 2500 insureds: X i = P N i j=1 Y ij is a random variable representing the annual loss of the ith insured. N 1 ; N 2 ; : : : ; N 2500 are independent and identically distributed random variables following a negative binomial distribution with parameters r = 2 and = 0:2. Y i1 ; Y i2 ; : : : ; Y ini are independent and identically distributed random variables following a Pareto distribution with = 3 and = The full credibility standard is to be within 5% of the expected aggregate losses 90% of the time. Using limited uctuation credibility theory, calculate the partial credibility of the annual loss experience for this book of business. [0.47] Chapters (Stat 477) Credibility Brian Hartman - BYU 12 / 31

13 Greatest Accuracy Credibility Example 1 Assume there are two dierent types of drivers, good and bad drivers. The variable x is the number of claims in any one year. x P r(xjg) P r(xjb) P r(g) = 0: P r(b) = 0: Chapters (Stat 477) Credibility Brian Hartman - BYU 13 / 31

14 Example 1 cont. Suppose a policyholder had 0 claims the rst year and 1 claim the second year. Determine the predictive distribution of his claims in the third year [0.65, 0.23, 0.13] the posterior probability of him being a good driver. [0.737] the Bayesian premium [0.479] Chapters (Stat 477) Credibility Brian Hartman - BYU 14 / 31

15 Greatest Accuracy Credibility Example 2 Assume that the amount of a claim has an exponential distribution with mean 1=. The parameter follows a gamma distribution with = 4 and = 1=1000. f (xj) =e x f () = e 1000 Chapters (Stat 477) Credibility Brian Hartman - BYU 15 / 31

16 Example 2 cont. Suppose a policyholder had claims of 100, 950, and 450. Determine the predictive distribution of the fourth claim. the posterior distribution of. the Bayesian premium [416.67] Chapters (Stat 477) Credibility Brian Hartman - BYU 16 / 31

17 Buhlmann Model The Buhlmann model is the simplest credibility model. Under this model, past losses X 1 ; : : : ; X n have the same mean and variance and are i.i.d. conditional on. We dene () = E(X j j = ) () = V ar(x j j = ) [hypothetical mean] [process variance] We further dene = E[( )] = E[( )] a = V ar[( )] Chapters (Stat 477) Credibility Brian Hartman - BYU 17 / 31

18 Buhlmann Model continued With the denitions on the previous slide, we can nd the mean, variance, and covariance of the X j s. E(X j ) = V ar(x j ) = + a Cov(X i ; X j ) = a for i 6= j Chapters (Stat 477) Credibility Brian Hartman - BYU 18 / 31

19 Buhlmann Model continued The credibility premium follows a familiar form Z X + (1 Z) where and Z = n n + k k = a = E[V ar(x jj )] V ar[e(x j j )] Chapters (Stat 477) Credibility Brian Hartman - BYU 19 / 31

20 Buhlmann Example Recall example 1, x P r(xjg) P r(xjb) P r(g) = 0: P r(b) = 0: Find the Buhlmann estimate of E(X 3 j0; 1). = 0:475; a = 0:016875; = 0:4825 k = 28:5926; Z = 0:0654; E(X 3 j0; 1) = 0:4766 Chapters (Stat 477) Credibility Brian Hartman - BYU 20 / 31

21 SOA Example #18 Two risks have the following severity distributions: Probability of Claim Probability of Claim Amount of Claim Amount for Risk 1 Amount for Risk , , Risk 1 is twice as likely to be observed as Risk 2. A claim of 250 is observed. Calculate the Bhlmann credibility estimate of the second claim amount from the same risk. [10,622] Chapters (Stat 477) Credibility Brian Hartman - BYU 21 / 31

22 Buhlmann-Straub Model The Buhlmann model assumes that all the past claims are i.i.d. but what if they are not. The Buhlmann-Straub model generalizes the Buhlmann model to allow each of the X i to be eected by its own exposure, m i. The X i are independent, conditional on, with common mean but with conditional variances () = E(X j j = ) V ar(x j j = ) = () m j Chapters (Stat 477) Credibility Brian Hartman - BYU 22 / 31

23 Buhlmann-Straub Model Similar to the Buhlmann model, dene = E[( )] = E[( )] a = V ar[( )] Then we can nd the mean, variance, and covariance of the X j s. E(X j ) = V ar(x j ) = m j + a Cov(X i ; X j ) = a for i 6= j Chapters (Stat 477) Credibility Brian Hartman - BYU 23 / 31

24 Buhlmann-Straub Model continued For convenience, let m = m 1 + m m n The credibility premium again follows a familiar form Z X + (1 Z); X = nx j=1 m j m X j where Z = m m + =a Chapters (Stat 477) Credibility Brian Hartman - BYU 24 / 31

25 SOA Example #21 You are given: (i) The number of claims incurred in a month by any insured has a Poisson distribution with mean. (ii) The claim frequencies of dierent insureds are independent. (iii) The prior distribution of is gamma with probability density function: (iv) f () = (100)6 e Month Number of Insureds Number of Claims ? Calculate the Buhlmann-Straub credibility estimate of the number of claims in Month 4. [16.9] Chapters (Stat 477) Credibility Brian Hartman - BYU 25 / 31

26 Nonparametric Estimation We may want to make as few distributional assumptions as possible. We can then simply use nonparametric, unbiased estimators of the Buhlmann quantities as follows: ^ = X 1 ^ = r(n 1) ^a = 1 r 1 rx i=1 rx nx i=1 j=1 (X ij X i ) 2 ( X i X) 2 1 rn(n 1) rx nx i=1 j=1 (X ij X i ) 2 Chapters (Stat 477) Credibility Brian Hartman - BYU 26 / 31

27 SOA Example #12 You are given total claims for two policyholders: Year Policyholder X Y Using the nonparametric empirical Bayes method, calculate the Buhlmann credibility premium for Policyholder Y. [687.4] Chapters (Stat 477) Credibility Brian Hartman - BYU 27 / 31

28 Nonparametric Estimation Cont. We can similarly nd nonparametric, unbiased estimators of the Buhlmann-Straub quantities as follows: ^ = X ^ = P r i=1 P ni m j=1 ij(x ij X i ) P 2 r (n i=1 i 1) ^a = (P r i=1 m i( X i X) 2 ^(r 1) P r i=1 m m 1 P r i=1 m2 i m i m ( X i ) 2 r m ^ if is unknown if is known Chapters (Stat 477) Credibility Brian Hartman - BYU 28 / 31

29 Example For a group policyholder, we have the data in the table below. Year 1 Year 2 Year 3 Total Claims 60,000 70,000 - Number in Group If the manual rate is 500 per year per member, estimate the total credibility premium for year 3. [94,874] Chapters (Stat 477) Credibility Brian Hartman - BYU 29 / 31

30 Semiparametric Estimation Sometimes it makes sense to assume a parametric distribution. For example, if claim counts (m ij X ij ) are assumed to follow a Poisson(m ij i ) distribution. In that case, E(m ij X ij j i ) = V ar(m ij X ij j i ) = m ij i Implying that ( i ) = ( i ) = i and = which means that we could use X to estimate. Chapters (Stat 477) Credibility Brian Hartman - BYU 30 / 31

31 Example In the past year, the distribution of automobile insurance policyholders by number of claims is Claims Insureds For each policyholder, nd a credibility estimate of the number of claims next year based on the past year's experience, assuming a (conditional) Poisson distribution on the number of claims. [(0:14)X i + 0:86(0:194)] Chapters (Stat 477) Credibility Brian Hartman - BYU 31 / 31