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INSTITUTE AND FACULTY OF ACTUARIES

EXAMINATION 29 April 2015 (pm)

Subject CT6 – Statistical Methods Core Technical Time allowed: Three hours INSTRUCTIONS TO THE CANDIDATE 1.

Enter all the candidate and examination details as requested on the front of your answer booklet.

2.

You must not start writing your answers in the booklet until instructed to do so by the supervisor.

3.

Mark allocations are shown in brackets.

4.

Attempt all 10 questions, beginning your answer to each question on a new page.

5.

Candidates should show calculations where this is appropriate.

Graph paper is NOT required for this paper.

AT THE END OF THE EXAMINATION Hand in BOTH your answer booklet, with any additional sheets firmly attached, and this question paper. In addition to this paper you should have available the 2002 edition of the Formulae and Tables and your own electronic calculator from the approved list.

CT6 A2015

 Institute and Faculty of Actuaries

1

The matrix below shows the losses to Player A in a two player zero sum game. The strategies for Player A are denoted I, II, III and IV. I

Player B

2

1 2 3

10 8 3

Player A II III

IV

6 X 9

3 Y 4

4 6 7

(i)

Determine the values of X and Y for which there are dominated strategies for Player A. [4]

(ii)

Determine whether there exist values of X and Y which give rise to a saddle point. [3] [Total 7]

The table below shows cumulative claim amounts incurred on a portfolio of insurance policies. Accident Year 0 2011 2012 2013 2014

1,509 1,542 1,734 1,773

Development Year 1 2 1,969 2,186 1,924

2,106 2,985

3 2,207

Annual premiums written in 2014 were 4,013 and the ultimate loss ratio has been estimated as 93.5%. Claims can be assumed to be fully run off by the end of development year 3. Estimate the total claims arising from policies written in 2014 only, using the Bornhuetter-Ferguson method.

3

(i)

(a) (b)

[7]

Explain why an insurance company might purchase reinsurance. Describe two types of reinsurance. [3]

The claim amounts on a particular type of insurance policy follow a Pareto distribution with mean 270 and standard deviation 340. (ii)

CT6 A2015–2

Determine the lowest retention amount such that under excess of loss reinsurance the probability of a claim involving the reinsurer is 5%. [4] [Total 7]

4

Let X be a random variable with density f(x) = ex for x > 0. (i)

Construct an algorithm for generating random samples from X.

[2]

A sequence of simulated observations is required from the density function h( x )  2 xe

5

 x2

x  0.

(ii)

Construct a procedure using the Acceptance-Rejection method to obtain the required observations. [5]

(iii)

Calculate the expected number of pseudo-random numbers required to generate 10 observations from h using the algorithm in part (ii). [2] [Total 9]

An insurance company has for five years insured three different types of risk. The number of policies in the jth year for the ith type of risk is denoted by Pij for i = 1, 2, 3 and j = 1, 2, 3, 4, 5. The average claim size per policy over all five years for the ith type of risk is denoted by X i . The values of Pij and X i are tabulated below.

Risk type i

Year 1

1 2 3

17 42 43

Number of policies Year 2 Year 3 Year 4 23 51 31

21 60 62

29 55 98

Year 5

Mean claim size Xi

35 37 107

850 720 900

The insurance company will be insuring 30 policies of type 1 next year and has calculated the aggregate expected claims to be 25,200 using the assumptions of Empirical Bayes Credibility Theory Model 2. Calculate the expected annual claims next year for risks 2 and 3 assuming the number of policies will be 40 and 110 respectively. [9]

CT6 A2015–3

PLEASE TURN OVER

6

Annual numbers of claims on three different types of insurance policy follow a Poisson distribution with parameter µi for i = 1, 2, 3. Data for the last four years is given in the table below. Year

7

Type

1

2

3

4

Total

1 2 3

5 2 5

5 5 6

0 4 4

1 5 5

11 16 20

(i)

Derive the maximum likelihood estimate of µ1 and calculate the corresponding estimates of µ2 and µ3. [5]

(ii)

Test the hypothesis that µ1, µ2 and µ3 are equal using the scaled deviance. [5] [Total 10]

The following time series model is being used to model monthly data:

Yt  Yt 1  Yt 12  Yt 13  et  1et 1  12 et 12  1 12 et 13 where et is a white noise process with variance 2.

8

(i)

Perform two differencing transformations and show that the result is a moving average process which you may assume to be stationary. [3]

(ii)

Explain why this transformation is called seasonal differencing.

(iii)

Derive the auto-correlation function of the model generated in part (i). [8] [Total 12]

[1]

The number of claims, N, in a given year on a particular type of insurance policy is given by: P(N = n) = 0.8  0.2n

n = 0, 1, 2, …

Individual claim amounts are independent from claim to claim and follow a Pareto distribution with parameters  = 5 and  = 1,000. (i)

Calculate the mean and variance of the aggregate annual claims per policy. [4]

(ii)

Calculate the probability that aggregate annual claims exceed 400 using: (a) (b)

a Normal approximation. a Lognormal approximation. [6]

(iii)

CT6 A2015–4

Explain which approximation in part (ii) you believe is more reliable. [2] [Total 12]

9

Let p be an unknown parameter and let f(px) be the probability density of the posterior distribution of p given information x. (i)

Show that under all-or-nothing loss the Bayes estimate of p is the mode of f(px). [2]

John is setting up an insurance company to insure luxury yachts. In year 1 he will insure 100 yachts and in year 2 he will insure 100 + g yachts where g is an integer. If there is a claim the insurance company pays a fixed sum of $1m per claim. The probability of a claim on a policy in a given year is p. You may assume that the probability of more than one claim on a policy in any given year is zero. Prior beliefs about p are described by a Beta distribution with parameters  = 2 and  = 8. In year 1 total claims are $13m and in year 2 they are $20m.

10

(ii)

Derive the posterior distribution of p in terms of g.

[4]

(iii)

Show that it is not possible in this case for the Bayes estimate of p to be the same under quadratic loss and all-or-nothing loss. [6] [Total 12]

Claims on a certain portfolio of insurance policies arise as a Poisson process with annual rate . Individual claim amounts are independent from claim to claim and follow an exponential distribution with mean . The insurance company has purchased excess of loss reinsurance with retention M from a reinsurer who calculates premiums using a premium loading of . Denote by Xi the amount paid by the reinsurer on the ith claim (so that Xi = 0 if the ith claim amount is below M). (i)

Explain why the claims arrival process for the reinsurer is also a Poisson process and specify its parameter. [3]

(ii)

Show that M X i (t )  1  e

(iii)

(iv)

M





t . 1  t

(a)

Determine E(Xi).

(b)

Write down and simplify the equation for the reinsurer’s adjustment coefficient. [6]

Comment on your results to part (iii).

END OF PAPER

CT6 A2015–5

[4]

[2] [Total 15]

INSTITUTE AND FACULTY OF ACTUARIES

EXAMINERS’ REPORT April 2015 examinations

Subject CT6 – Statistical Methods Core Technical

Introduction The Examiners’ Report is written by the Principal Examiner with the aim of helping candidates, both those who are sitting the examination for the first time and using past papers as a revision aid and also those who have previously failed the subject. The Examiners are charged by Council with examining the published syllabus. The Examiners have access to the Core Reading, which is designed to interpret the syllabus, and will generally base questions around it but are not required to examine the content of Core Reading specifically or exclusively. For numerical questions the Examiners’ preferred approach to the solution is reproduced in this report; other valid approaches are given appropriate credit. For essay-style questions, particularly the open-ended questions in the later subjects, the report may contain more points than the Examiners will expect from a solution that scores full marks. The report is written based on the legislative and regulatory context at the date the examination was set. Candidates should take into account the possibility that circumstances may have changed if using these reports for revision. F Layton Chairman of the Board of Examiners June 2015

 Institute and Faculty of Actuaries

Subject CT6 (Statistical Methods Core Technical) – April 2015 – Examiners’ Report

General comments on Subject CT6 The examiners for CT6 expect candidates to be familiar with basic statistical concepts from CT3 and so to be comfortable computing probabilities, means, variances etc. for the standard statistical distributions. Candidates are also expected to be familiar with Bayes’ Theorem, common types of reinsurance, and risk models, and to be able to apply it to given situations. Many of the weaker candidates are not familiar with this material. The examiners will accept valid approaches that are different from those shown in this report. In general, slightly different numerical answers can be obtained depending on the rounding of intermediate results, and these will still receive full credit. Numerically incorrect answers will usually still score some marks for method, providing candidates set their working out clearly. Comments on the April 2015 paper The examiners felt that this paper was generally not as well answered as recent papers. A disappointing number of candidates were seemingly unfamiliar with core concepts such as Bayes’ Theorem or EBCT Model 2. Well prepared candidates were typically able to score well, although some struggled with unfamiliar applications of theory.

Page 2

Subject CT6 (Statistical Methods Core Technical) – April 2015 – Examiners’ Report

1

(i)

III will dominate II if X  6. III will dominate IV if X  Y. I will dominate IV if Y  8 I does not dominate II and vice versa (since I is better if Player B chooses 1 and II is better if Player B chooses 2). III cannot dominate I (since I is better if Player B chooses 1). IV cannot dominate any strategy since it gives the worst outcome if Player B chooses 3. Similarly II cannot dominate any strategy since it gives the worst result if Player B chooses 1. So there will exist dominated strategies if X  6 or X  Y or Y  8.

(ii)

A saddle point exists if an entry is both the largest in its column and the smallest in its row. This can only occur in row 2 (the smallest values in rows 1 and 3 are not the largest in their columns). X cannot give a saddle since this would require X  8 (to be the smallest in row 2) but then X would not be the largest in its column. Equally Y cannot give a saddle point as this would require that Y  8 in which case Y would not be the largest in its column. So there are no values of X and Y which give a saddle point.

This straightforward question was relatively poorly answered, with many candidates seemingly put off by the unfamiliar nature of the question.

2

The development factors are: DF2,3 =

2207 = 1.0480 2106

DF1,2 =

(2106  2985) = 1.2253 (1969  2186)

DF0,1 =

(1969  2186  1924) = 1.2704 (1509  1542  1734)

Page 3

Subject CT6 (Statistical Methods Core Technical) – April 2015 – Examiners’ Report

The initial ultimate loss for 2014 is 0.935 × 4013 = 3752.16.  1  Total emerging liability = initial UL ×  1   f  1   = 3752.16 ×  1    1.0480 1.2253  1.2704 

= 1452.01 Total claims = 1452.01 + 1773 = 3225.01 This standard chain ladder question was very well answered by the majority of candidates.

3

(i)

(a)

To protect itself from the risk of large claims.

(b)

(ii)



Excess of loss reinsurance where the reinsurer pays any amount of a claim above the retention.



Proportional reinsurance where the reinsurer pays a fixed proportion of any claim.

We must first find the parameters  and  of the Pareto distribution.

 2  = 270 and = 3402 2  1 (  1) (  2)  2  = 3402   2 (  1)2

so

3402  = 1.585733882 = 2 2702

so  =

2 1.585733882 1.585733882 1

= 5.4145 and  = 270 × 4.4145 = 1191.920375

Page 4

Subject CT6 (Statistical Methods Core Technical) – April 2015 – Examiners’ Report

We need to find M such that P(X > M) = 0.05 

i.e.

     = 0.05   M    M



M =

=

1  0.05

=

1  0.05

(  M )

1    1  0.05       1

0.05 

=

1   1191.920375  1  0.05 5.4145      1 5.4145 0.05

= 880.8 This question was the best answered on the paper with most candidates scoring well. Some candidates were unable to manipulate the Pareto distribution.

4

(i)

X has an exponential distribution with parameter 1. Let u be a sample from a U(0,1) distribution. Then using the inverse transform method we set u = F(x) = 1  ex i.e. 1  u = ex i.e. x = log(1  u) so the algorithm is Step 1 Step 2

Generate u from U(0,1). Set x = log(1  u).

Page 5

Subject CT6 (Statistical Methods Core Technical) – April 2015 – Examiners’ Report

(ii)

We first find C = max x 0

h( x) f ( x) 

2

2 xe x = max  x x0 e

= max 2 xe x x

2

x0

To find the maximum consider g(x) = 2 xex x

2

then

log g(x)

= log2 + logx + x  x2

d log g ( x) dx

= 1  1  2x x

setting this equal to zero we have 1 1  2 x = 0 x

i.e.

1 + x  2x2 = 0 2x2  x  1 = 0 (2x + 1)(x 1) = 0 x = 1 or x = ½

but x > 0 so C = 2 × 1 × e11 = 2 so now set w(x) =

2 h( x ) = xex  x 2 f (x )

and our algorithm is Step 1 Step 2 Step 3 Step 4

Page 6

Generate u from U(0,1) distribution. Generate x = log(1  u). Generate v from U(0,1). If v > w(x) return to step 1 else return x.

Subject CT6 (Statistical Methods Core Technical) – April 2015 – Examiners’ Report

(iii)

1 1 so on average 2 simulations are needed to = 2 c return 1 value, so in this case we need 20 simulations. Each simulation requires 2 pseudo-random numbers so on average we will need 40 pseudorandom numbers.

We accept a proportion

Most candidates were able to produce good quality answers to part (i), although the majority struggled with part (ii). Candidates with the confidence to apply the acceptance-rejection method scored well.

5

P1 = 17 + 23 + 21 + 29 + 35 = 125

P2 = 42 + 51 + 60 + 55 + 37 = 245 P3 = 43 + 31 + 62 + 98 + 107 = 341 P = 125 + 245 + 341 = 711 X =

(850 125  720  245  900  341) = 829.18 711

expected claims per policy for risk 1 next year = so

840 = Z1 × 850 + (1  Z1) × 829.18

Z1 =

so

25, 200 = 840 30

840  829.18 = 0.519594 850  829.18

125 = 0.51969 E( s2 ( )) 125  Var[ m( )]

so

125  0.51969 125 E ( s 2( )) = 115.57217 = Var[ m( )] 0.51969

so

Z2 =

245 = 0.6794756 245  115.528

Z3 =

341 = 0.74686 341  115.528

Page 7

Subject CT6 (Statistical Methods Core Technical) – April 2015 – Examiners’ Report

So credibility premium per policy are Type 2: 0.67956 × 720 + (1  0.67956) × 829.18 = 755.0 Type 3: 0.74694 × 900 + (1  0.74694) × 829.18 = 882.1 so overall expected claims Type 2: 754.98 × 40 = 30,200 Type 3: 882.08 × 110 = 97,028

Candidates with good knowledge of EBCT Model 2 scored well here, however a disappointing number of candidates were apparently unfamiliar with this method.

6

(i)

The likelihood for type 1 policies is given by

l=

41 11 1 4

e

 y ij !

j 1

Taking logarithms gives 4

L = log l =  41  11log 1   log yij ! j 1

so

L 11  4  1 

and setting

i.e.

4 

11 =0 ˆ1 

ˆ1 =

11 = 2.75 4

similarly

ˆ 3 =

Page 8

L = 0 we have 1

ˆ 2 =

16 =4 4

20 =5 4

Subject CT6 (Statistical Methods Core Technical) – April 2015 – Examiners’ Report

(ii)

Assuming  = 1 = 2 = 3 we have ˆ =

11 16  20 = 3.916667 12

The difference in scaled deviance is given by  = 2(log L1 log L2  log L3 log L)

= 2( 4ˆ 1 11 log ˆ 1  4ˆ 2  16 log ˆ 2  4ˆ 3  20 log ˆ 3 ˆ  47 log ˆ ) 12 [The logarithms of factorials cancel] = 2(47 + 11 log2.75 + 16 log4 + 20 log5 + 47  47 log3.91667) = 2.6615 2 Under H0: 1 = 2 = 3 we have that  comes from a  distribution with 3  1 = 2 degrees of freedom.

The upper 5% point of the 22 distribution is 5.991. The observed value is below this and so there is no evidence to suggest that the underlying parameters are different for each risk. Most candidates were able to score well on part (i), although only the better prepared candidates were able to complete part (ii).

7

(i)

Set Xt = (1  B12)(1  B) Yt where B is the background shift operator i.e. Xt = Yt  Yt1  Yt12 + Yt13 then we have Xt = et + 1 et1 + 12 et12 + 112et13 = (1 + 1B)(1 + 12B12)et which is a moving average process [of order 13].

(ii)

This is called seasonal differencing because it compares the monthly change in Yt with the corresponding monthly change at the same time last year.

Page 9

Subject CT6 (Statistical Methods Core Te...