INSTITUTE AND FACULTY OF ACTUARIES CURRICULUM 2019 SPECIMEN SOLUTIONS. Subject CS1B Actuarial Statistics

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1 INSTITUTE AND FACULTY OF ACTUARIES CURRICULUM 2019 SPECIMEN SOLUTIONS Subject CS1B Actuarial Statistics

2 Question 1 (i) # Data entry before <- c(155, 152, 146, 153, 146, 160, 139, 148) after <- c(145, 147, 123, 137, 141, 142, 140, 138) # define x as the pair-wise differences of the study's results x <- before - after # define and calculate intermediate variables mx <- mean(x) sdx <- sd(x) nx <- length(x) alpha <- 0.1 # (the confidence level) t_quantile <- qt(p = alpha / 2, # two-sided interval need to divide alpha by 2 df = nx - 1, lower.tail = FALSE) # gives upper tail i.e. P(X > x) # assuming x follows a normal distribution with unknown variance: c_int <- c(mx - t_quantile * sdx / sqrt(nx), mx + t_quantile * sdx / sqrt(nx)) c_int [1] ALTERNATIVE SOLUTIONS Using the t.test function t.test(x = before - after, conf.level = 0.9) One Sample t-test data: before - after t = , df = 7, p-value = alternative hypothesis: true mean is not equal to 0 90 percent confidence interval: sample estimates: mean of x Restricting function output to only return the required confidence interval t.test(x = before - after, conf.level = 0.9)$conf.int [1] attr(,"conf.level") [1] 0.9 Using the Paired t-test functionality of the t.test function t.test(x = before, y = after, paired = TRUE, conf.level = 0.9)

3 Paired t-test data: before and after t = , df = 7, p-value = alternative hypothesis: true difference in means is not equal to 0 90 percent confidence interval: sample estimates: mean of the differences [8] (ii) # define and calculate intermediate values mu <- 10 t_stat <- (mx - mu) / (sdx / sqrt(nx)) pval <- pt(t_stat, df = nx - 1, lower.tail = FALSE) pval [1] With p-value of we do not reject the null hypothesis at 0.01 significance level ALTERNATIVE SOLUTIONS Using the t.test function which also returns the p-value t.test(x = before - after, alternative = "greater", mu = 10, conf.level = 0.99) One Sample t-test data: before - after t = , df = 7, p-value = alternative hypothesis: true mean is greater than percent confidence interval: Inf sample estimates: mean of x Restrict the t.test function ouptut to only include the p-value t.test(x = before - after, alternative = "greater", mu = 10, conf.level = 0.99)$p.value [1] Using the Paired t-test functionality of the t.test function t.test(x = before, y = after, paired = TRUE, alternative = "greater", mu = 10,

4 conf.level = 0.99) Paired t-test data: before and after t = , df = 7, p-value = alternative hypothesis: true difference in means is greater than percent confidence interval: Inf sample estimates: mean of the differences [7] [Total 15] Question 2 (i) (a) Posterior mean given as Z*x + (1-Z)*alpha/beta where alpha/beta is the prior (gamma) mean of λ. {2} M = 1000 alpha = 100 beta = 1 Z = 1/(beta+1) pm = X = numeric(m) for(m in 1:M){ lam = rgamma(1,shape=alpha,rate=beta) x = rpois(1,lam) X[m] = x pm[m] = Z*x + (1-Z)*alpha/beta } {10} (b) hist(pm,main="histogram of posterior means", xlab="posterior mean",ylab="frequency")

5 {3} [15] (ii) (a) round(mean(pm),3) round(var(pm),3) (b) MC mean and variance of posterior mean estimates: , {2} mg = numeric(m) for(m in 1:M){ lam = rgamma(1,shape=alpha,rate=beta) x = rpois(1,lam) y = rgamma(1000,shape=alpha+x, rate=beta+1) mg[m] = mean(y) } round(mean(mg),3) round(var(mg),3) MC mean and variance of posterior from Gamma samples: , {10} [12] Note that this solution to part (ii) uses a new set of Monte Carlo repetitions. This is not necessary, and full credit can be given for combining parts (i) and (ii) in a single exercise. Clearly the precise numerical values for the means and variances will differ from implementation to implementation.

6 (iii) The similarity of the Monte Carlo estimates of the mean and variance and those from the Gamma(α + xx, β + 1) sample demonstrates that the posterior Distribution for the Poisson/Gamma credibility model is the Gamma(α + xx, β + 1). [3] [Total 30] Question 3 (i) All values are positive integer with some values more than 1, so use Poisson distribution as error structure. [5] (ii) model <- glm(formula = Claim.number ~ Age + factor(car.group) + Area + factor(ncd) + Gender, data = datatrain, family = poisson()) summary(model) Deviance Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error z value Pr(> z ) (Intercept) e-14 *** Age e-11 *** factor(car.group) factor(car.group) factor(car.group) factor(car.group) * factor(car.group) *** factor(car.group) factor(car.group) *** factor(car.group) *** factor(car.group) *** factor(car.group) ** factor(car.group) e-05 *** factor(car.group) e-05 *** factor(car.group) e-06 *** factor(car.group) e-06 *** factor(car.group) e-05 *** factor(car.group) e-06 *** factor(car.group) e-07 *** factor(car.group) e-09 *** factor(car.group) e-08 *** AreaEast Midlands AreaLondon * AreaNI ** AreaNorth East AreaNorth West AreaSouth East * AreaSouth West AreaWales * AreaWest Midlands AreaYorkshire and the Humber factor(ncd) e-07 *** factor(ncd) e-14 *** factor(ncd) < 2e-16 ***

7 factor(ncd) < 2e-16 *** factor(ncd) < 2e-16 *** GenderMale e-05 *** --- Signif. codes: 0 *** ** 0.01 * (Dispersion parameter for poisson family taken to be 1) Null deviance: on 7999 degrees of freedom Residual deviance: on 7963 degrees of freedom AIC: [10] (iii) Male policyholders have higher mean of reported claims (by exp( ) 1 = 29.7%) than female policyholders. The difference is significant (p-value = 6.14e-5). [10] (iv) Compare to Null model; the deviance is reduced by while the degrees of freedom reduce by 36. The observed difference in deviance (484.3) is very high compared to the values of the χχ 2 36 distribution, so the fitted model is significant/good (alternatively, 2 compare the deviance of the fitted model (4085.6) to the χχ 7963 distribution.) [10] (v) (a) datatrain$age2= datatrain$age^2 {2} (b) model1 <- glm(formula = Claim.number ~ Age + Age2 + factor(car.group) + Area + factor(ncd) + Gender, data = datatrain, family = poisson()) summary(model1) Deviance Residuals: Min 1Q Median 3Q Max Coefficients: Estimate Std. Error z value Pr(> z ) (Intercept) e e Age e e < 2e-16 *** Age e e < 2e-16 *** factor(car.group) e e factor(car.group) e e factor(car.group) e e factor(car.group) e e * factor(car.group) e e *** factor(car.group) e e factor(car.group) e e *** factor(car.group) e e *** factor(car.group) e e *** factor(car.group) e e ** factor(car.group) e e e-05 *** factor(car.group) e e *** factor(car.group) e e e-06 *** factor(car.group) e e e-06 *** factor(car.group) e e e-05 ***

8 factor(car.group) e e e-06 *** factor(car.group) e e e-07 *** factor(car.group) e e e-09 *** factor(car.group) e e e-08 *** AreaEast Midlands 9.654e e AreaLondon 3.190e e * AreaNI 3.837e e ** AreaNorth East e e AreaNorth West e e AreaSouth East e e * AreaSouth West e e AreaWales e e * AreaWest Midlands e e AreaYorkshire and the Humber 1.060e e factor(ncd) e e e-07 *** factor(ncd) e e e-13 *** factor(ncd) e e < 2e-16 *** factor(ncd) e e < 2e-16 *** factor(ncd) e e < 2e-16 *** GenderMale 2.657e e e-05 *** --- Signif. codes: 0 *** ** 0.01 * (Dispersion parameter for poisson family taken to be 1) Null deviance: on 7999 degrees of freedom Residual deviance: on 7962 degrees of freedom AIC: {12} (c) The p-value of the age squared coefficient shows that it is significant. Also, the deviance is reduced more than twice the change in degrees of freedom. So the variable is significantly associated with the number of reported claims. {6} [20] [Total 55] END OF MARKING SCHEDULE