MAT00061M Statistics for Insurance 2018/19
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MAT00061M
MMath and MSc Examinations 2018/19
Statistics for Insurance
1 (of 3) . (a) In the following table, we have the cumulative payments made from the
home insurance claims by origin year and development year . Assume that all claims are settled by the end of development year 3 and the figures in the table have been inflation adjusted .
Development year
Origin year 0 1 2 3
2005 2062 2478 2723 3036
2006 2115 2492 2756
2007 2140 2518
2008 2176
(i) Use the basic chain ladder method with the weighted development factors to estimate the amount of reserves which should be set aside for the future claims at the end of 2008 . [10]
(ii) In addition to the table on the cumulative payments given above, we also have the following table on annual premiums written in 2005– 2008 as well as the loss ratios .
Origin year |
2005 |
2006 |
2007 |
2008 |
Premium |
3725 |
3810 |
3925 |
3958 |
Loss ratio |
86% |
86% |
87% |
87% |
Use the Bornhuetter-Ferguson method and the weighted develop- ment factors to estimate the amount of reserves which should be set aside for the future claims at the end of 2008 . [10]
(b) In an insurance portfolio, a typical claim variable X follows an exponential distribution with mean 400 . The insurance company handling the claims has made an excess of loss treaty with a reinsurer with the excess level M = 500 .
(i) Letting Y be the part paid by the baseline insurance company for the claim X, calculate E(Y) . [7]
(ii) If claims inflation of 3% is expected for the next year and the excess level remains at M = 500, what will be the expected cost of a claim to the reinsurance company? [7]
2 (of 3) . (a) In an insurance company, three separate Poisson surplus processes are
being managed, where the respective parameters are given in the following table . Here for any process, U denotes the initial reserves, λ denotes the Poisson parameter, θ is the security loading factor and X is the typical claim size .
Process |
U |
λ |
θ |
|
Claim size X |
A |
500 |
10 |
0 .25 |
~ |
exponential with mean 20 |
B |
500 |
10 |
0 .3 |
|
X = 20 |
C |
500 |
10 |
0 .2 |
|
X ~ Γ(2, 0.1) |
(i) For Process A, write down the adjustment equation and calculate the adjustment coefficient . [5]
(ii) For Process B, find an upper bound R0 for the adjustment coefficient of the form R0 = 2θE(X)/E(X2 ) and write down the adjustment equation . [5]
(iii) For Process C, write down the adjustment equation and calculate the adjustment coefficient . [5]
(iv) Using the Newton-Raphson method, the adjustment coefficient for process B is R = 0.0252 . Calculate the Lundberg’s upper bound as an approximation for the probability of ruin for the three processes
A, B and C, and rank these three processes . [5]
(b) The process used to model a risk is a Poisson surplus process with Pois- son parameter λ and individual claim size X . The distribution for X is exponential with mean 100 . The baseline insurance company uses a se- curity loading of θ = 0.3 . A proportional reinsurance is available whereby a security loading of ξ = 0.4 (for reinsurance) is used and the baseline insurance company pays the proportion α of each individual claim X .
(i) Give the minimum value of α if the baseline insurance company wants its expected net profit to be non-negative . [2]
(ii) If the insurance company decides to go ahead with a proportional reinsurance with α = 0.6, determine the adjustment coefficient . [4]
(c) In a study undertaken to model the five-year survival rates for the patients after taking a surgery, a binomial generalised linear model with the logistic link function was used with a linear predictor η = α6 + β6 x, where α6 and β6 are the parameters for the factor gender and x is the variate age (in years) when the patient takes the surgery. For the female patients, let α6 = αF and β6 = βF , and for the male patients, let α6 = αM and β6 = βM . Using the maximum likelihood method, for the female patients, the estimates of the two parameters were F = 2.2 and F = _0.07; and for the male patients, the estimates of the two parameters were M = 3.1 and M = _0.08 .
(i) Using the model and the estimation results, what is the five-year survival rate for a 52-year-old female patient after taking this surgery?
[3]
(ii) What is the five-year survival rate for a 65-year old male patient after taking this surgery? [3]
3 (of 3) . (a) Let S = X1 + . . . + XN be a compound Poisson distribution where N
is Poisson with parameter 20 and Xi are independent and identically dis- tributed random variables . If the distribution Xi is uniform on [0, 150] ,
calculate the mean, variance and skewness of S . [9]
(b) There are 1600 houses insured against fire by a company. The houses are classified into two types for purposes of this fire insurance . It is assumed that fires occur independently of one another, and the chance of more than one claim on any individual house in a given year is negligible . The number of houses in each class and the corresponding claim probability are given in the following table . The claim size distribution in type 1 is identical and uniform on [0, 400], and the claim size distribution in type 2 is identical and exponential with mean 300 .
class type k |
number of houses in class |
claim probability |
k = 1 |
600 |
0.02 |
k = 2 |
1000 |
0.01 |
We use a security loading of 1.5θ for houses in type 1, and θ for houses in type 2 . Use the expected value principle and the normal approximation to find the value of θ which gives us a 99% probability that premiums exceed claims . You may have to use the 0.99-quantile of the standard
(ii) A claim data set of 100 observations was collected from a portfolio of home insurance policies . The following R codes were used to implement the Kolmogorov-Smirnoff test for this claim data set .
¿ mu=mean(log(home))
¿ sigma2=mean((log(home)-mu)ˆ2)
¿ ks .test(home,”plnorm”,mu,sqrt(sigma2))
One-sample Kolmogorov-Smirnov test
data: home
D = 0 .063485, p-value = 0 .8151
alternative hypothesis: two-sided
Based on the above R codes and output, write out the null hypothesis in the test . Should this null hypothesis be rejected according to the R output? [5]
2022-08-11