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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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Pengantar Individual Risk Models For a Short Term Exercise Penutup
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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Pengantar Individual Risk Models For a Short Term Exercise Penutup
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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Pengantar Individual Risk Models For a Short Term Exercise Penutup
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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Pengantar Individual Risk Models For a Short Term Exercise Penutup
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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Pengantar Individual Risk Models For a Short Term Exercise Penutup
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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Pengantar Individual Risk Models For a Short Term Exercise Penutup
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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Pengantar Individual Risk Models For a Short Term Exercise Penutup
Models for Individual Claim Random Variables
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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Pengantar Individual Risk Models For a Short Term Exercise Penutup
Models for Individual Claim Random Variables
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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Pengantar Individual Risk Models For a Short Term Exercise Penutup
Models for Individual Claim Random Variables
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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Pengantar Individual Risk Models For a Short Term Exercise Penutup
Models for Individual Claim Random Variables
Prove: E (X ) = q Use formula E (X ) = E (E (X
|I )) .
X = IB .If I = 0, X = 0, so that E (X |I = 0) = E (0) = 0.If I = 1, X = B , so that E (X |I = 1) = E (B |I = 1) = .E (X
|I ) = I
E (X ) = E (E (X |I )) = E (I ) = E (I ) = q .
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Pengantar Individual Risk Models For a Short Term Exercise Penutup
Models for Individual Claim Random Variables
Prove: Var (X ) = 2q (1 q ) + 2q Use formula Var (X ) = Var (E (X |I )) + E (Var (X |I )) .X
= IB .If I = 0, X = 0, so that Var (X |I = 0) = Var (0) = 0.If I = 1, X = B , so that Var (X |I = 1) = Var (B |I = 1) = 2 .Var (X |I ) = 2 I E (Var (X |I )) = E (2I ) = 2q .E (X |I ) = I Var (E (X |I )) = Var (I ) = 2Var (I ) = 2q (1 q ).Var (X ) = Var (E (X |I )) + E (Var (X |I )) = 2q (1 q ) + 2q
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g p
Models for Individual Claim Random Variables
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
Models for Individual Claim Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
Solution 1.4 (Continued)
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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Sums of Independent Random Variables
Solution 1.4 (Continued)
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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Sums of Independent Random Variables
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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SELAMAT BELAJAR