GLMs: Gompertz s Law. GLMs in R. Gompertz s famous graduation formula is. or log µ x is linear in age, x,

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1 Computing: an indispensable tool or an insurmountable hurdle? Iain Currie Heriot Watt University, Scotland ATRC, University College Dublin July 2006 Plan of talk General remarks The professional syllabus The professional examinations The R program Some examples GLMs NCD systems Simulation Graduation Markov models 1 2 GLMs: CT6 syllabus exponential families, mean, variance, etc variables, factors, interactions link functions, linear predictors deviance and model selection Pearson and deviance residuals GLMs: CT6 examination question September, 2005: Q2 Y 1, Y 2,..., Y n are independent variables, and Y i P(µ i ). The fitted values are denoted by ˆµ i. Derive the form of the scaled deviance. 3 4

2 mortality Mortality Log(Mortality) Number of claims Number of claims Central exposure Central exposure 5 6 GLMs: Gompertz s Law Gompertz s famous graduation formula is µ x = AB x or log µ x is linear in age, x, log µ x = log A + log B x = a + bx This is a GLM with Data: {d x, Ex} c Random: d x P(Exµ c x ) Linear: log Exµ c x = log Ex c + a + bx GLMs in R Gompertz = glm(dth, offset = log(exp), plot(, log(dth/exp)) lines(, log(gompertz$fit/exp)) plot(, residuals(gompertz, type = "dev")) abline(h = 0) 7 8

3 Models with factors and interactions Log(Mortality) Deviance residuals Example: Mortality by age and sex deviance residuals = 71 Data: Deaths: d i, Exposures: e i, Sex: f i, : x i Models glm(dth 1, offset = log(exp), glm(dth Sex, offset = log(exp), glm(dth, offset = log(exp), glm(dth + Sex, offset = log(exp), glm(dth *Sex, offset = log(exp), Model: Sex Model: Common regression GLMs: Heriot-Watt examination question September, 2005: Q8 (b) Table (omitted) gives the number of deaths, d x, and the central exposed to risk, E c x, x = 70,..., 80. Suppose the number of deaths is modelled by a GLM with Poisson error and log link where the linear predictor is Model: Sex + Model: Sex * log η x = log E c x + a + bx The fitted values are â = and ˆb = Calculate the observed and fitted force of mortality when x = 70. Calculate the fitted number of deaths when x = Plot the logarithm of the observed force of mortality at x = 70,..., 80 and the logarithm of the fitted force of mortality for 70 x

4 Simulation: CT6 syllabus concept of Monte Carlo simulation random and pseudo-random numbers generating pseudo-random numbers from specified distributions deciding the number of simulations needed GLMs: CT6 examination question March, 2006: Q5 (a) X 1, X 2,..., X n are iid E(λ). Show that X X n is G(n, λ). (b) Use (b) to generate a random sample (of size one!) from a gamma distribution with mean 30 and variance 300 using the 5 digit pseudo-random numbers September, 2006: No question! The R simulation toolkit Simulation: HW syllabus basic theory, pseudo random numbers, PIT the R simulation toolkit estimation compound distributions and aggregate claims bootstrap rbinom(sample.size, n, p) runif(sample.size, min, max) rnorm(sample.size, mean, sd) rpois(sample.size, mean) rlnorm(sample.size, meanlog, sdlog) rgamma(sample.size, shape, rate) dnorm(x, mean, sd) pnorm(x, mean, sd) qnorm(x, mean, sd) Bootstrap.sample = sample(x, replace = T) 15 16

5 Histogram of Median Distribution of median in samples of size 10 from in LN(1,1) N.simulations = Median = numeric(n.simulations) for(i in 1:N.simulations) Median[i] = median(rlnorm(10, 1, 1)) Density Normal Gamma Median Aggregate claims N.years = Aggregate claims process CP (10, LN(1, 1)) N.claims = rpois(n.years, 10) Agg.claims = numeric(n.years) for (i in 1:N.years) Agg.claims[i] = sum(rlnorm(n.claims[i], 1, 1)) 19 Density Normal Lognormal Gamma Aggregate claims

6 Bootstrap sampling distribution of the mean Bootstrap methods Data = c(0.15,0.24,...,5.15,8.34) Compute 95% CI from 10 skew values Normal methods: t.test(data)$conf.int Bootstrap: for (i in 1:10000) Mean[i] = mean(sample(data, replace = T)) 21 Density Bootstrap 95% CI = ( 0.61, 3.82 ) Students t 95% CI = ( 0.06, 3.99 ) Students t, 9df Sample mean Simulation: Heriot-Watt examination question September, 2005: Q10 (c) The following eleven values are taken at random from the mark list from a Risk Theory paper: 40, 43, 45, 60, 62, 67, 70, 70, 72, 73, 85. The values are held in the variable marks in the program R. Explain how the following R code enables a 95% confidence interval for the median mark to be estimated. Iter = 1000 Bootstrap = numeric(iter) for (i in 1:1000) Bootstrap[i] = median(sample(marks, replace = T)) quantile(bootstrap, probs = c(0.025,0.975)) Experience rating: CT6 syllabus simple experience rating systems claim thresholds stationary distributions 23 24

7 Experience rating: CT6 examination question March, 2006: Q2 An NCD system has 3 levels of discount: Level 0 discount = 0 Level 1 Level 2 where 0 < p < 0.5. discount = p discount = 2p Probability of no claim = 0.9. Claim: go to level 0; no claim: up one level. Find the proportion of the full premium paid over the portfolio once the steady state is reached. Transition matrix is We seek repeated evaluation of P = p n+1 = p n P where p n gives the distribution over states at time n Proportion of premium over time NCD over time P = matrix(c(0.1,0.1,0.1, 0.9,0,0, 0,0.9,0.9), 3, 3) N.years = 30 Proportion = matrix(0, N.years, 3) Proportion[1, ] = c(1, 0, 0) for (i in 1:(N.years-1)) Proportion[i+1, ] = Proportion[i, ] %*% P Premium = Proportion %*% c(1, 0.8, 0.6) Proportion of premium Number of years 27 28

8 NCD: Heriot-Watt examination question 2005: Q11 An insurer s NCD scale has 7 levels of discount: 0% to 60% in 10% bands, etc. Conclusions R opens up the possibility of realistic practical applications. Similar arguments apply to Excel and Maple. Accreditation would allow freedom to pursue these ideas. Something needs to be done!! 29 30