subject ct6 - acted corrections 2010.pdf · subject ct6 corrections to 2010 study material comment...

CT6: Corrections

The Actuarial Education Company © IFE: 2010 Examinations

Subject CT6

Corrections to 2010 study material

Comment This document contains details of any errors and ambiguities in the Subject CT6 study materials for the 2010 exams that have been brought to our attention. We will incorporate these changes in the study material each year. We are always happy to receive feedback from students, particularly details concerning any errors, contradictions or unclear statements in the courses. If you have any such comments on this course please email them to [email protected]. You may also find it useful to refer to the Subject CT6 Frequently Asked Questions thread on the Actuarial Discussion Forum (You can reach the forums by clicking on the “Discussion Forum” button in the bottom left-hand corner of ActEd’s website, or by going to http://www.acted.co.uk/forums/. This contains useful questions asked by students studying CT6, with answers written by ActEd’s tutors.

Important note This document was last revised significantly on 8 July 2010. The date on which any corrections have been added is noted at the start of each section. Course Notes Chapter 3 page 31 (updated on 18 November 2009) There is a typo in the exam-style question. The sum should read: 487,926=Â ix

CT6: Corrections

© IFE: 2010 Examinations The Actuarial Education Company

Chapter 6 page 28 (updated on 18 December 2009) The equation labelled (2.17) should read: ( ) [ ]+ -1 1 11 ( )Z X Z E m qqqq Replacement pages can be found at the end, Chapter 8 pages 7, 11, 12, 15 (updated on 27 November 2009) References to Chapter 7 Section 4 should read Section 3. Similarly, references to Chapter 7 equations (4.x) should read (3.x). Replacement pages can be found at the end. Chapter 9 page 67 (updated on 21 July 2010) The formula for the upper bound of the adjustment coefficient should read:

22[ ( )]

( )c E XrE Xl -<

Replacement pages can be found at the end. Q&A Bank 1 – solutions page 6 (updated on 8 July 2010) The is a mistake in the decision function table of Solution 1.6(ii). The values for 7d and 8d are the wrong way round. Replacement pages can be found at the end.

CT6: Corrections


Q&A Bank 2 – solutions page 26 (updated on 23 December 2009) There is a mistake in the solution to the variance, it should read:

2002 2 2

02003 2 3 3

2

0

1 1( )200 200

2,761.42600 200 600 200

M

MM

M

E Y x dx M dx

x M x M MM

= +

È ˘ È ˘= + = + - =Í ˙ Í ˙Í ˙ Í ˙Î ˚ Î ˚

Ú Ú

Hence: var( ) 20 2,761.42 55,228S = ¥ = (5 SF) Replacement pages can be found at the end. Q&A Bank 2 – solutions page 33 (updated on 5 July 2010) There is a mistake in the solution to the variance in Question 2.24, it should read:

{ }2

2

2 2

1

20,000 212

20,000 212

var( ) (1 )

100 (0.0004) (0.0004)(0.9996) 30,000

200 (0.001) (0.001)(0.999) 40,000

37.32m 326.35m 363.67m

n

i i i i ii

S q q qs m=

= + -

È ˘= ¥ + ¥Í ˙Î ˚È ˘+ ¥ + ¥Í ˙Î ˚

= + =

Â

Replacement pages can be found at the end. X2 assignment – solutions page 2 (updated on 19 February 2010) There is a typo in the estimators for 2[ ( )]E s q and var[ ( )]m q . The summations for i should go from 1 to 4 and the summations for j should go from 1 to 3.

CT6: Corrections


CT6 Mock Exam 2009 (updated 29 May 2009) The 2009 Mock exam was substantially revised for the summer session. Students should not use copies of the exam issued to them over the winter 2008 session.

CT6-06: Emperical Bayes Credibility theory Page 27


These four assumptions show that each risk in the collective satisfies the same assumptions as the particular risk that is of interest. The following two assumptions, which correspond to (1.9), (1.10) and (1.11) in Section 1.4, show the connection between different risks in the collective. The risk parameters 1 2, , , Nq q q… , regarded as random variables, are independent and identically distributed. (2.10) For πi k , the pairs ( , )i ijXq and ( , )k kmXq are independent. (2.11) Notice that, since the iq ’s are identically distributed, the values of [ ]( )iE m q ,

[ ]var ( )im q and È ˘Î ˚

2 ( )iE s q do not depend on i so that they can be denoted by

[ ]( )E m q , [ ]var ( )m q and È ˘Î ˚

2 ( )E s q , respectively, as in Section 2.2 and

Section 2.3. As in Section 1.4, more notation is needed. Denote:

=Â

1

n

ijj

P by iP (2.12)

=Â

1

N

ii

P by P (2.13)

( ) ( )-

=- -Â1

11 1 /

N

i ii

Nn P P P by *P (2.14)

=Â

1/

n

ij ij ij

P X P by iX (2.15)

= = =

=Â Â Â1 1 1

/ /N N n

i i ij iji i j

P X P P X P by X (2.16)

This notation is all given in the Tables (Page 30). Notice that what was denoted X in formula (2.5) is now denoted X1 and that X now has a different definition.

(This was exactly what happened in Section 1.4.) Notice also that Xi and X are weighted averages of the Xij ’s, the weights being the risk volumes Pij .

CT6-06: Emperical Bayes Credibility theory


With this new notation the credibility estimate of the pure premium, or number of claims, per unit of risk volume for the coming year for Risk Number 1 in the collective originally given by formula (2.5) can be reformulated as: ( ) [ ]+ -1 1 11 ( )Z X Z E m q (2.17) where:

= =

= Â Â1 1 1 11 1

n n

j j jj j

X P X P

and:

[ ]

=

=

=È ˘+ Î ˚

Â

Â

11

12

11

( ) var ( )

n

jj

n

jj

P

ZP E s mq q

Note that this credibility factor is specific to Risk 1, so is denoted by 1Z . For the other risks there will be different values of Z . It is important to realise that this is exactly the same as formula (2.5) but is written in the notation of this section rather than that of the previous section. Unbiased estimators for [ ]q( )E m , [ ]qvar ( )m and qÈ ˘

Î ˚2 ( )E s can be proposed

based on the observed values Y Pij ij jn

i

N,n s = =

RSTUVW1 1

.

CT6-08: Risk models (2)


To clarify which distributions are involved in this relationship, we could write:

f w f w MF MW

X

X( ) ( )

( )=

+−1

Question 8.3

What is the relationship between W and X ?

To specify the distribution for SR as given in formula (1.2) the distribution of NR is needed. In many contexts it will be obvious what this distribution is, but here is a general method for establishing the distribution. This is found as follows. Define:

NR = I1 + I2 + + IN where N denotes the number of claims from the risk (as usual). Ij is an indicator random variable which takes the value 1 if the reinsurer makes a (non-zero) payment on the j-th claim, and takes the value 0 otherwise. Thus NR gives the number of payments made by the reinsurer. Since Ij takes the value 1 only if

>jX M , P I P X Mj j( ) ( )= = > =1 π , say, and P I j( )= = −0 1 π In other words, I j has a B( , )1 π distribution. This means that NR has a compound Poisson distribution (as N is Poisson). Further, Ij has MGF:

M t tI ( ) exp{ }= + −π π1

and by formula (3.7) in Chapter 7 NR has MGF:

MNR(t) = MN (log MI (t))



Question 8.4

If N has a Poisson( )λ distribution, and half of the claims exceed the excess-of-loss retention limit, show that NR has a Poisson(½ )λ distribution.

Example Continuing the above example and using formula (1.2) as the model for SR, it can be seen that SR has a compound Poisson distribution with Poisson parameter 0 2 10 2. × = . Individual claims, Wi, have density function:

p(w) = ( )1 ( )f w M

F M+

- = 0.0005/0.2 = 0.0025, for 0 < w < 400

ie Wi is uniformly distributed on (0,400). [ ] 200iE W = , 2[ ] 53,333.33iE W = and

3[ ] 16,000,000iE W = , giving the same result as before. We haven’t worked out these precise quantities before, but if you multiply these figures by 2 (the Poisson parameter of SR ), you get ( ) 400RE S = , var( ) 106,667RS = and skew( ) , ,SR = 32 000 000 , which agree with the answers on Page 6. Thus, there are two ways to specify and evaluate the distribution of SR. The next question is a long question that uses a lot of the ideas we’ve covered so far. You will need to do some preliminary calculations in order to work out the answers.

CT6-08: Risk models (2) Page 11


Assumption (2.1) is very restrictive. It means that a maximum of one claim from each risk is allowed for in the model. This includes risks such as one-year term assurance, but excludes many types of general insurance policy. For example, there is no restriction on the number of claims that could be made in a policy year under household contents insurance. There are three important differences between this model and the collective risk model: (1) The number of risks in the portfolio has been specified. In the collective

risk model, there was no need to specify this number, nor to assume that it remained fixed over the period of insurance cover (not even when it was assumed that N ~ b(n, q)).

On the other hand you could argue that there was an implicit assumption of a constant number of risks in the very fact that you are using a binomial distribution to model the number of claims. (2) The number of claims from each individual risk has been restricted. There

was no such restriction in the collective risk model. (3) It is assumed that individual risks are independent. In the collective risk

model it was individual claim amounts that were independent. The contrast here is between the occurrence of claims and the size of claims. Assumptions (2.1) and (2.2) say that N b qj j~ ( , )1 . Thus, the distribution of Yj is compound binomial, with individual claims distributed as Xj. From formulae (3.13) and (3.14) in Chapter 7 it follows immediately that: [ ]j j jE Y q m= (2.3) var[Yj] = qj σ j

2 + qj (1 − qj) μ j2 (2.4)

S is the sum of n independent compound binomial random variables. It was seen in Section 3.5 of Chapter 7 that there is no general result about the distribution of such a sum. This distribution can be stated only when the compound binomial variables are identically distributed, as well as independent. It is possible, but complicated, to compute the distribution function of S under certain conditions.



However, it is easy to find the mean and variance of S:

1 1 1

[ ] [ ]n n n

j j j jj j j

E S E Y E Y q m= = =

È ˘Í ˙= = =Í ˙Î ˚Â Â Â (2.5)

The assumption that individual risks are independent is needed to write:

s m= = =

È ˘Í ˙= = = + -Í ˙Î ˚Â Â Â 2 2

1 1 1var [ ] var var[ ] ( (1 ) )

n n n

j j j j j j jj j j

S Y Y q q q (2.6)

In the special case when { }Y j j

n=1 is a sequence of identically distributed, as well

as independent, random variables, then for each policy the values of q j , μ j and

σ j2 are identical, say q , μ and σ 2 . Also, F xj ( ) is independent of j, say F(x).

Hence, S is compound binomial, with binomial parameters n and q, and individual claims have distribution function F(x). In this special case, it reduces to the collective risk model, and it can be seen from (2.5) and (2.6) that: [ ]E S nqm= s m= + -2 2var [ ] (1 )S nq nq q which correspond to (3.13) and (3.14) respectively in Chapter 7.

Question 8.6

The probability of a claim arising on any given policy in a portfolio of 1,000 one year term assurance policies is 0.004. Claim amounts have a Gamma( , . )5 0 002 distribution. Find the mean and variance of the aggregate claim amount.

The following section forms part of the Core Reading, but does not address any specific syllabus objectives. You can treat the material here as useful practice in applying the models we have studied.

CT6-08: Risk models (2) Page 15


(i) The moments of Si can be calculated by conditioning on the value of λ i . Since Si i|λ has a straightforward compound Poisson distribution, formulae (3.8) and (3.9) in Chapter 7 can be used to write:

1 1[ ] [ [ | ] ] [ ] 0.2i i i iE S E E S E m ml l= = =

l l

l l= += +

= +2 1

22 1

var [ ] [ var [ | ] ] var [ [ | ] ][ ] var[ ]

0.2 0.01

i i i i i

i i

S E S E SE m m

m m

(ii) The random variables { }Si i

n=1are independent and identically distributed,

each with the distribution of Si given in part (i). Hence, the result in (i) above can be used to write:

11

[ ] 0.2n

i ii

E S n E S n m=

È ˘Í ˙ = =Í ˙Î ˚Â

=

È ˘Í ˙ = = +Í ˙Î ˚Â 2

2 11

var var [ ] 0.2 0.01n

i ii

S n S n m n m

Example Suppose the Poisson parameters for individual policies are drawn from a gamma distribution with parameters α and δ . Find the distribution of the number of claims from a policy chosen at random from the portfolio. Solution Let Ni denote the number of claims from the i-th policy in the portfolio and let λ i be its Poisson parameter. Then Ni has a Poisson distribution with parameter λ i but the problem is that (by assumption) the value of λ i is not known. What is known is the distribution from which λ i has been chosen.



The problem can be summarised as follows: Given that: N Pi i i| ~ ( )λ λ and λ α δi G~ ( , ) find the marginal distribution of Ni. This problem can be solved by removing the conditioning in the usual way. We will use the techniques that we saw when we looked at mixture distributions in Chapter 3. For x = 0, 1, 2, ...

aa

aa

l dl l d l la

d l d l la

• -

• + -

= = - -G

= - +G

Ú

Ú

10

10

( ) exp{ } exp { }! ( )

exp { ( 1) }( ) !

xi

x

P N x dx

dx

Evaluate the integral by comparing the integrand with a gamma density, so that:

P N xx

xi x( )

( ) !( )

( )= =

+

+ +δα

α

δ

α

αΓΓ

1

which shows that the marginal distribution of Ni is negative binomial with

parameters α and δδ + 1

.

What we have done here is to make the integral look like the density function of a

( , 1)gamma x a d+ + distribution. This will involve adding a constant of ( 1)( )

x

x

ada

++G +

inside the integral. So we need an additional constant of ( )( 1)x

xa

ad +G ++

on the outside of

the integral. Note that the integral is with respect to l , so we can take terms involving any other parameters outside.

CT6-09: Ruin theory Page 67

The Actuarial Education Company IFE: 2010 Examinations

Adjustment coefficient For a compound Poisson process, the adjustment coefficient r is the unique positive root of the equation:

λ λ+ =cr M rX ( ) Upper and lower bounds for r :

22[ / ( )]

( )c E Xr

E Xl -< r

Mc m> 1

1log( / )λ

If the insurer uses a security loading of θ , the equation for the adjustment coefficient is: 1 1 1+ + =( ) ( )θ m r M rX If reinsurance is effected this equation becomes:

1 1 1+ + − + =( ) ( ) ( ) ( ) ( )θ ξE X E Z r M rY where ξ denotes the reinsurer’s security loading, Y represents the amount of an individual claim paid by the insurer, net of reinsurance, and Z denotes the amount of an individual claim paid by the reinsurer. For a general aggregate claims process, the adjustment coefficient r is the unique positive root of the equation: E er S ci[ ]( )− = 1, where Si denotes the aggregate claims from a risk in time period i and c is the constant premium charged to insure this risk. Lundberg’s inequality

ψ ( )u e ru≤ −

CT6-09: Ruin theory

IFE: 2010 Examinations The Actuarial Education Company

This page has been left blank so that you can keep the chapter summaries together as a revision tool.

CT6: Q&A Bank Part 1 – Solutions Page 5


0

5

10

15

20

25

30

35

0 0.2 0.4 0.6 0.8 1

p

expe

cted

pro

fitumbrellashot foodice cream

Finding first where the line for 3d cuts 1d : 1

1816 3 17 15p p p+ = - fi = Now we find where 1d cuts 2d : 7

1816 3 9 21p p p+ = + fi = So the revised Bayesian criterion solution is as follows: If 1

180 p£ £ , 3d is optimal.

If 7118 18p£ £ , 1d is optimal.

If 718p ≥ , 2d is optimal. [4]

CT6: Q&A Bank Part 1 – Solutions


Solution 1.6

(i) Loss function The table of the nine values of the loss function is as follows:

Guessed parameter value "θ

a1 = ¼ a2 = ½ a3 = ¾

θ1 = ¼ 0 50 100

θ2 = ½ 50 0 50 Actual

parameter value θ θ3 = ¾ 100 50 0

[2] (ii) Decision and risk functions A single observation from a B( , )1 θ distribution will take the value 0 or 1. For each of these two values we can choose either θ = ¼ , θ = ½ or θ = ¾ . So there will be 3² 9= decision functions altogether. If x is the single observation from our distribution, we define the 9 different functions as follows:

Observation 0x = 1x =

1( )d x ¼ ¼

2( )d x ½ ½

3( )d x ¾ ¾

4( )d x ¼ ½

5( )d x ¼ ¾

6( )d x ½ ¾

7 ( )d x ½ ¼

8( )d x ¾ ¼

9( )d x ¾ ½ [2]



So the mean and variance of the net claims for the direct insurer and the reinsurer are:

3( ) 300 2254IE X = ¥ = 2var( ) (3 / 4) 180,000 101,250IX = ¥ = [1]

1( ) 300 754RE X = ¥ = 2var( ) (1/ 4) 180,000 11, 250RX = ¥ = [1]

Using the formula for the variance of a compound Poisson distribution: 2 2var( ) [ ] 250[101,250 225 ] 37,968,750I IS E Xl= = + = [1] 2 2var( ) [ ] 250[11,250 75 ] 4,218,750R RS E Sl= = + = [1] where SD and SR are the aggregate claim amounts. Alternatively we could use the formula for ( )rE X given in the Tables. Solution 2.20

This is Subject C2, September 1995, Question 11. We have ~ (0,200)X U and:

0X X M X M

Y ZM X M X M X M

< <Ï Ï= =Ì Ì> - >Ó Ó

The expected amount paid by the insurer on any claim is:

200

0

1 1( ) 50200 200

M

M

E Y x dx M dx= + =Ú Ú [1]

Solving this gives:

2002

0

50400 200

M

M

x MxÈ ˘ È ˘+ =Í ˙ Í ˙Î ˚Í ˙Î ˚

2 250

400 200M MMfi + - = [1]



2 400 20,000 0 58.579, 341.42M M Mfi - + = fi = [1] Since claims are a maximum of 200, it must be the first value of M . The variance of the aggregate annual claims, S , paid by the insurer is given by: 2 2var( ) ( ) 20 ( )S E Y E Yl= = where:

2002 2 2

02003 2 3 3

2

0

1 1( )200 200

2,761.42600 200 600 200

M

MM

M

E Y x dx M dx

x M x M MM

= +

È ˘ È ˘= + = + - =Í ˙ Í ˙

Í ˙ Í ˙Î ˚ Î ˚

Ú Ú

[2]

Hence: var( ) 20 2,761.42 55,228S = ¥ = (5 SF) [1] Solution 2.21

This is Subject C2, September 1998, Question 12 (part) Let Z be the amount paid on each claim by the reinsurer. We have:

0 100

100 100 200100 200

XZ X X

X

<ÏÔ= - £ <ÌÔ =Ó

[½]

We first want ( )E Z , 2( )E Z and 3( )E Z . So we have:

200

0.01 2

100

( ) ( 100)0.01 100xE Y x e dx e- -= - +Ú [1]



We can find the second moment of Y + 2 in a similar way:

2

2

2

[( 2) ] 4 ( 0) 4 ( 1) 4 ( 2) 9 ( 3)

( ) 4 ( 0) 3 ( 1)

1 9var( ) [ ( )] 4 3 40.08854 64

E Y P S P S P S P S

E S P S P S

S E S

+ = = + = + = + = +

= + = + =

= + + ¥ + ¥ =

"

[2]

So: 2var( 2) 40.0885 4.6406 18.5533Y + = - = So the variance of Y is also equal to 18.55. [1] Solution 2.24

For each age group, the individual claim amounts have a uniform distribution. So the mean and variance of the individual claim distributions are:

12( ) ( ) 1,000E X b a x= + =

220,0002112 12var( ) ( )X b a= - = [2]

This is the individual risk model so using the formulae from Chapter 7, the variance of the aggregate claims, S , will be:

{ }2

2

2 2

1

20,000 212

20,000 212

var( ) (1 )

100 (0.0004) (0.0004)(0.9996) 30,000

200 (0.001) (0.001)(0.999) 40,000

37.32m 326.35m 363.67m

n

i i i i ii

S q q qs m=

= + -

È ˘= ¥ + ¥Í ˙Î ˚È ˘+ ¥ + ¥Í ˙Î ˚

= + =

Â

[3] Alternatively, we could treat each group as a compound binomial and use the collective risk model variance given on page 16 of the Tables. For example,

~ (100,0.0004)N Bin for the 100 lives aged exactly 30.



Solution 2.25

Using the formulae for the lognormal distribution, the mean and variance of the amounts B are:

2 2½ 3 ½ 1.2( ) 41.264E B e em s+ + ¥= = =

and: 2 22var( ) ( 1) 5, 484.04B e em s s+= - = [1]

The indicator variable I has a B( , . )1 0 002 distribution, which has mean and variance: ( ) 1 0.002 0.002E I = ¥ = and: var( ) 1 0.002 0.998 0.001996I = ¥ ¥ = [1] Assuming that the probability of a claim is independent of the benefit amount, the mean and variance of X are: ( ) ( ) ( ) ( ) 0.002 41.264 0.08253E X E IB E I E B= = = ¥ = [1] and:

2 2 2

2 2 2 2

2 2 2

var( ) var( ) ( ) [ ( ) ( )]

( ) ( ) [ ( )] [ ( )]

0.002 (5484.04 41.264 ) 0.002 41.264 14.367

X IB E I B E I E B

E I E B E I E B

= = -

= -

= ¥ + - ¥ = [1] Alternatively, you can work these out using conditional means and variances.

subject ct6 - acted corrections 2010.pdf · subject ct6 corrections to 2010 study material comment...

Documents