Actuarial Science 1st chapter

8/12/2019 Actuarial Science 1st chapter

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Pengantar Individual Risk Models For a Short Term Exercise Penutup

Matematika Aktuaria I

Pertemuan ke-1, 12 Pebruari 2014

Departemen Matematika FMIPA IPB


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Penjelasan Umum

Deskripsi Singkat dan TIU

Deskripsi Singkat

Mata kuliah ini membahas terapan matematika yangberhubungan dengan aktuaria untuk pekerjaan di asuransi jiwa,dana pensiun, asuransi kesehatan, dan asuransi umum.Topik yang dibahas: model risiko individu jangka pendek,sebaran survival dan tabel hayat, asuransi hidup, anuitas hidup,premi, dan cadangan premi ( benet reserves ).


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Penjelasan Umum

Deskripsi Singkat dan TIU

Tujuan Instruksional UmumSetelah mengikuti perkuliahan ini, mahasiswa dapatmenjelaskan:

Model risiko individu jangka pendek dan aplikasinya dalambidang asuransi.Sebaran bertahan hidup dengan berbagai karakteristiknya,menjelaskan fungsi-fungsi yang terkait tabel hayat, danketerkaitan antar keduanya.

Jenis-jenis asuransi hidup, jenis-jenis anuitas hidup, baikdengan waktu kontinu dan waktu diskret.Penentuan besar premi dan cadangan premi untukbeberapa jenis asuransi, baik dengan waktu kontinu dan

waktu diskret.


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Penjelasan Umum

Pustaka

1 Bowers NL, Gerber HU, Hickman JC, Jones DA, NesbittCJ. 1997. Actuarial Mathematics. The Society ofActuaries. Schaumburg, Illinois.

2 Gerber HU. 1997. Life Insurance Mathematics. SwissAssociation of Actuaries Zurich. Springer-Verlag, NewYork.

3 Cunningham R, Herzog T, Richard L. 2006. Model forQuantifying Risk (Second Edition). London.


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Penjelasan Umum

Penentuan Nilai Akhir dan Huruf Mutu

Nilai Akhir (NA)UTS (40%)UAS (40%)Tugas + Kuis + Proyek (20%)

Huruf MutuA : NA75AB : 70 NA < 75B : 60 NA < 70BC : 50 NA < 60C : 40 NA < 50D : 25

NA < 40

E : NA < 25


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Introduction

For an insuring organization, let the random loss of a segmentof its risks be denoted by S . Then S is the random variable forwhich we seek a probability distribution. Historically, there havebeen two sets of postulates for distributions of S . The individual

risk model denesS = X 1 + X 2 + + X n

where X i is the loss on insured unit i and n is the number of risk

units insured. Usually the X s are postulated to be independentrandom variables, because the mathematics is easier and nohistorical data on the dependence relationship are needed.

d d l k d l h


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Models for Individual Claim Random Variables

First, we review basic concepts with a life insuranceproduct. In a one-year term life insurance the insurer

agrees to pay an amount b if the insured dies within a yearof policy issue and to pay nothing if the insured survivesthe year.The probability of a claim during the year is denoted by q .The claim random variable, X , has a distribution that canbe described by either its probability function, p.f., or itsdistribution function, d.f.

P I di id l Ri k M d l F Sh T E i P


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The p.f. is

f X (x ) = Pr (X = x ) =1 q , x = 0q , x = b 0 , elsewhere.

The d.f. is

F X (x ) = Pr (X x ) =0 , x < 01 q , 0 x < b 1 , x b .

E (X ) = bq , E (X 2) = b 2q , and Var (X ) = b 2q (1 q ).



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Random variable X can be noted by X = Ib ,b is the constant amount payable in the event of death,I is the random variable that is 1 for the event of death and0 otherwise.

Thus, Pr (I = 0) = 1 q , Pr (I = 1) = q , so that E (I ) = q and Var (I ) = q (1 q ).E (X ) = bE (I ) = bq ,Var (X ) = b 2Var (I ) = b 2q (1 q ).

Note: I is called an indicator , Bernoulli random variable , orbinomial random variable for a single trial.



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We now seek more general models in which the amount ofclaim is also a random variable and several claims can occur ina period. Health, automobile, and other property and liabilitycoverages provide immediate examples.

Random variable X can be noted by X = IB ,B is the random variable for the total claim amount incurredduring the period,I is the random variable for the event that at least one claimhas occurred.

If we dene = E (B |I = 1) and 2 = Var (B |I = 1),then we haveE (X ) = q Var (X ) = 2q (1 q ) + 2 q



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g p


Example 1.1Suppose we have the p.d.f. for B given I = 1 by

f B |I (x |1) = 0.0009 1 x 2000 , 0 < x < 2000 ,0 , elsewhere,

with Pr (B = 2000

|I = 1) = 0.1 and Pr (I = 1) = q = 0.15.

Calculate E (X ) and Var (X ).



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g p


Solution:

= E (X |I = 1) = E (B |I = 1) =

2000

0 0.0009 x 1

x 2000 dx + ( 0.1)(2000 ) = 800.

E (X 2|I = 1) = E (B 2 |I = 1) =

2000

0 0.0009 x 2 1 x 2000 dx + ( 0.1)(2000 )2 = 1000000 ,

2 = E (X 2|I = 1) [E (X |I = 1)]2 = 1000000 (800 )2 =360000.

E (X ) = q = ( 800 )(0.15 ) = 120.Var (X ) = 2q (1 q ) + 2q =(800 )2(0.15 )(0.85 ) + ( 360000 )(0.15 ) = 135600.



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Sums of Independent Random Variables

S = X + Y , where X and Y are discrete random variables.The d.f. of S is

F S (s ) = Pr (S s ) = Pr (X + Y s )

F S (s ) =y s

Pr (X + Y s |Y = y )Pr (Y = y )

=y s

Pr (X s y |Y = y )Pr (Y = y )



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When X and Y are independent, this last sum can bewritten

F S (s ) =y s

F X (s y )f Y (y ).

The p.f. corresponding to this d.f. can be calculated by

f S (s ) =y s

f X (s y )f Y (y ).



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S = X + Y , where X and Y are kontinu random variables.When X and Y are independent, the d.f. of S can bewritten

F S (s ) = s 0 F X (s y )f Y (y ) dy .The p.f. corresponding to this d.f. can be calculated by

f S (s ) = s 0 f X (s y )f Y (y ) dy .Note: This operation is called convolution process .



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To determine the distribution of the sum of more than tworandom variables, we can use the convolution processiteratively. For S = X l + X 2 + + X n where X i are independentrandom variables, F i is the d.f. of X i and F (k ) is the d.f. of

X l + X 2 + + X k , we proceed thus:F (2) = F 2F

(1) = F 2F 1F (3) = F 3F

(2)

F (4) = F 4

F (3)

...F (n ) = F n F

(n 1)



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Example 1.2The random variables X 1 , X 2 , and X 3 are independent withdistributions dened by

x f 1(x ) f 2 (x ) f 3(x )0 0.4 0.5 0.61 0.3 0.2 0.02 0.2 0.1 0.13 0.1 0.1 0.14 0.1 0.15 0.1

Derive the p.f. and d.f. of S = X 1 + X 2 + X 3 .



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Solution 1.2

S 2 = X 1 + X 2 f (2)(s ), s = 0, . . . , 7

f (2)(s ) =s

x = 0f 1 (s x )f 2(x )

f (2) (0) = f 1(0)f 2(0) = ( 0 .4)(0 .5) = 0.20f (2) (1) = f

1(1)f

2(0) + f

1(0)f

2(1) = 0.23

f (2) (2) = f 1(2)f 2(0) + f 1 (1)f 2(1) + f 1(0)f 2(2) = 0.20f (2) (3) = 0.16 ; f (2)(4) = 0.11 ; f (2)(5) = 0.06 ;f (2) (6) = 0.03 ; f (2)(7) = 0.01 .



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Solution 1.2 (Continued)

S 3 = X 1 + X 2 + X 3 = S 2 + X 3 f (3)(s ), s = 0, . . . , 12

f (3)(s ) =s

x = 0

f (2)(s x )f 3(x )

f (3) (0) = f (2)(0)f 3 (0) = ( 0.4)(0.5) = 0.12

f (3)

(1) = f (2)

(1)f 3 (0) + f (2)

(0)f 3 (1) = 0.138f (3) (2) = f (2)(2)f 3 (0) + f (2)(1)f 3 (1) + f (2)(0)f 3 (2) = 0.14f (3) (3) = 0.139 ; f (3)(4) = 0.129 ; f (3)(5) = 0.115 ; f (3)(6) =0.088 ; f (3)(7) = 0.059 ; f (3)(8) = 0.036 ; f (3)(9) =0.021 ; f (3)(10 ) = 0.01 ; f (3)(11 ) = 0.004 ; f (3)(12 ) = 0.001.



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Example 1.3Let X i for i = 1, 2 be independent and identically distributed

with the d.f.

F (x ) =0 , x < 0x , 0 x < 11 , x 1.

Let S = X 1 + X 2 . Find the p.f. and d.f. of S .



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Solution 1.3The d.f. for S can be nd by formula

F S (x ) =

x

0

F 1(x

y )f 2(y ) dy

,For x < 0, F S (x ) = 0.For 0 x < 1,

F S (x ) = x

0F 1 (x y )f 2 (y ) dy =

x

0(x y )(1) dy

= x 2

2 .



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Solution 1.3 (Continued)For 1 x < 2,

F S (x ) = x 1

0F 1(x y )f 2 (y ) dy +

1

x 1F 1 (x y )f 2(y ) dy

= x 1

0(1)(1) dy +

1

x 1(x y )(1) dy

= 1 + 2x x 2

2 .

For x 2 , F S (x ) = 1.



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Solution 1.3 (Continued)Finally, the d.f. and p.f. of S are

F S (x ) =

0 , x < 0x

2

2 , 0 x < 11 + 2x x

2

2 , 1 x < 21 , x 2

f S (x ) =

x , 0 < x < 1

2 x , 1 x < 20 , otherwise .

Note: f S (x ) = F S (x )

x .



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Example 1.4Consider three independent random variables X 1 , X 2 , X 3 . Fori = 1, 2, 3, X i has an exponential distribution and E [X i ] = 1/ i .Derive the p.f. of S = X 1 + X 2 + X 3 by the convolution process.



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Solution 1.4We have p.f. of X 1 , X 2 , and X 3 are

f 1 (x ) = e x , x > 0

0 , otherwise

f 2 (x ) = 2e 2x , x > 00 , otherwise

f 3 (x ) = 3e 3x , x > 0

0 , otherwise.



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S 2 = X 1 + X 2 note that f (2)(x ) is p.f. of S 2

We have

f (2)(x ) = x

0f 1(x y )f 2 (y ) dy

= x

0 e (x y )

2e 2y

dy

= 2e x x

0e y dy

= 2e x 2e 2x , x > 0.



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S = X 1 + X 2 + X 3 = S 2 + X 3 note that f (3)(x ) is p.f. of S

We have

f S (x ) = f (3)(x ) = x

0f (2) (x y )f 3 (y ) dy

= x

0(2e (x y ) 2e 2(x y ))3e 3y dy

= 6e x x

0e 2y dy 6e 2x

x

0e y dy

= 3e x

6e 2x

+ 3e 3x

, x > 0.



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Example 1.5Let X have a uniform distribution on (0 , 2) and let Y beindependent of X with a uniform distribution over (0, 3).Determine the d.f. of S = X + Y .



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Solution 1.5We have p.f. of X and Y are

f 1(x ) = 12 , 0 < x < 20 , otherwise

; f 2(y ) =13 , 0 < y < 30 , otherwise

and the d.f. of X is

F 1(x ) =0 , x < 0x 2

, 0 x < 21 , x 2.



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Solution 1.5 (Continued)Case A: s < 0, F S (s ) = 0.Case B: 0

s < 2,

F S (s ) = s

0F 1 (s y )f 2(y ) dy

=

s

0

s y 2

1

3

dy

= s 2

12.



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Solution 1.5 (Continued)Case C: 2 s < 3,

F S (s ) = s

0 F 1(s y )f 2(y ) dy =

s 2

0F 1(s y )f 2(y ) dy +

s

s 2F 1(s y )f 2(y ) dy

= s 2

0 (1)13 dy +

s

s 2

s

y

213 dy

= s 1

3 .



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Solution 1.5 (Continued)Case D: 3 s < 5,F S (s ) =

s

0

F 1(s y )f 2(y ) dy

= s 2

0F 1(s y )f 2(y ) dy +

3

s 2F 1(s y )f 2(y ) dy

=

s 2

0

(1)13

dy +

3

s 2

s y 2

13

dy

= s 2 + 10s 1312

.

Case E: s

5, F S (s ) = 1.



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Solution 1.5 (Continued)Finally we have

F S (s ) =

0 , s < 0s 2

12 , 0 s < 2

s 13

, 2 s < 3

s 2

+ 10s 1312 , 3 s < 51 , s 5.



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Moment Generating FunctionAnother method to determine the distribution of the sum ofrandom variables is based on the uniqueness of themoment generating function (m.g.f.), which, for the random

variable X , is dened by M X (t ) = E [e tX

].If this expectation is nite for all t in an open interval aboutthe origin, then M X (t ) is the only m.g.f. of the distribution ofX , and it is not the m.g.f. of any other distribution.

This uniqueness can be used as follows. For the sumS = X 1 + X 2 + + X n ,M S (t ) = E (e tS ) = E (e t (X 1 + X 2 + + X n ))

= E (e tX 1 e tX 2 e tX n ).



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Moment Generating FunctionIf X 1 , X 2 , . . . , X n are independent, then the expectation ofthe product is equal to

M S (t ) = E (e tX 1 )E (e tX 2 ) E (e tX n )= M X 1 (t )M X 2 (t ) M X n (t ).

Recognition of the unique distribution corresponding thisformula would complete the determination of S sdistribution.



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Example 1.6

Consider three independent random variables X 1 , X 2 , X 3 . Fori = 1, 2, 3, X i has an exponential distribution and E [X i ] = 1/ i .Derive the p.d.f. of S = X 1 + X 2 + X 3 by recognition of the m.g.f.of S .



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Solution 1.6M.g.f. of X 1 is

M X 1 (t ) = E [e tX 1 ] =

0e tx e x dx =

11 t

, t < 1.

M.g.f. of X 2 and X 3 are M X 2 (t ) = 22 t and

M X 3 (t ) = 33 t

, t < 1.

M.g.f. of S is

M S (t ) =3

i = 1M X i (t ) = 11 t

22 t 33 t

= 31 t

62 t

+ 33 t

(by method of partial fraction)

P.d.f of S is f S (x ) = 3e x

3(2e 2x ) + ( 3e 3x ), x > 0.



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Example 1.7

Consider n independent random variables X 1 , X 2 , . . . , X n . Fori = 1, 2, . . . , n , X i has a normal distribution with mean andvariance 2 . Derive the p.d.f. of S = X 1 + X 2 + + X n byrecognition of the m.g.f. of S .



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Solution 1.7

We have X i normal (, 2 ). The p.d.f. of normaldistribution is

f X i (x , ,2) =

1 2 e

(x )2

2 2 , < x < .

M.g.f. of X i is M X i (t ) = e t +

22 t

2.

M.g.f. of S is M S (t ) =n i = 1 M X i (t ) = e

n t + n 22 t 2

Finally we have S normal (n , n 2 ).



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FINISH



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Exercises

1 Obtain the mean and variance of the claim randomvariable X where q = 0.05 and the claim amount randomvariable B is uniformly distributed between 0 and 20.

2

Let X be the number showing when one true die is thrown.Let Y be the number of heads obtained when X true coinsare then tossed. Calculate E [Y ] and Var (Y ).

3 The probability of a re in a certain structure in a giventime period is 0.02. If a re occurs, the damage to thestructure is uniformly distributed over the interval (0, a )where a is its total value. Calculate the mean and varianceof re damage to the structure within the time period.



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Exercises

4 Independent random variables X k for three lives have thediscrete probability functions given below.

x Pr (X 1 = x ) Pr (X 2 = x ) Pr (X 3 = x )

0 0.6 0.7 0.61 0.0 0.2 0.02 0.3 0.1 0.03 0.0 0.0 0.44 0.1 0.0 0.0

Use a convolution process on the non-negative integervalues of x to obtain F S (x ) for x = 0, 1, . . . , 9 whereS = X 1 + X 2 + X 3 .



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Exercises

5 Let X i for i = 1, 2, 3 be independent and identicallydistributed with the d.f.

F (x ) =0 , x < 0x , 0 x < 11 , x 1 .

Let S = X 1 + X 2 + X 3 . Find p.f. and d.f. for S .



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Exercises

Show that F S (x ) is given by

F S (x ) =

0 , x < 0x 3

6

, 0

x < 1

x 3 3(x 1)36

, 1 x < 2x 3 3(x 1)3 + 3(x 2)3

6 , 2 x < 3

1 , x

3.

Show that E [S ] = 1.5 and Var (S ) = 0.25.Evaluate the following probabilities:(a) Pr (S 0.5), (b) Pr (S 1.0), (c) Pr (S 1 .5).



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Actuarial Science 1st chapter