cursuri st ec gest actuariat

8/6/2019 Cursuri St Ec Gest Actuariat

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THEME 3. TWO-DIMENSIONAL STATISTICAL DISTRIBUTIONS 50

X Y y1 . . . y j . . . ym Total

x1 f 11 . . . f 1 j . . . f 1m f 1...xi f i1 . . . f ij . . . f im f i ...

xn f n 1 . . . f nj . . . f nm f n Total f 1 . . . f j . . . f m

(relative frequencies).

Let the two-dimensional distribution of Z = ( X, Y ) as above.

The conditional distribution of Y given the event X = xi has thevalues y1, . . . , ym , the corresponding absolute frequencies f i1, . . . , f im ,and the corresponding relative frequencies

f i1f i

=f i1f i

, . . . ,f imf i

=f imf i

, suppose that f i > 0.

The conditional distribution of X given the event Y = y j has thevalues x1, . . . , x n , the corresponding absolute frequencies f 1 j , . . . , f nj ,and the corresponding relative frequencies

f 1 jf j

=f 1 jf j

, . . . ,f njf j

=f njf j

, suppose that f j > 0.

The (u, v)-th moment of Z = ( X, Y ) (or of the two-dimensionaldistribution of Z = ( X, Y )) is

muv =

n

i=1

m

j =1f ij xui yv j

n

i=1

m

j =1

f ij=

n

i=1

m

j =1

f ij xui y

v j .

We remark thatm10 = x and m01 = y,

where

x =

n

i=1f i x i

n

i=1f i

=n

i=1

f i xi


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and

y =

m

j =1f j y j

m

j =1f j

=m

j =1

f j y j

are the means of the marginal distributions (of X and Y , respec-tively).

The mean of Z = ( X, Y ) (or of the two-dimensional distribution of Z = ( X, Y )) is the pairz = ( x, y).

The conditional mean of Y given the event X = xi (the meaninside x i-group ) is the mean of the conditional distribution of Y givenX = xi , i.e.

i = mY/X = x i =

m

j =1f ij y j

f i =

m

j =1f ij y j

f i .

The functionB(xi) = i , i {1, . . . , n }

is called the regression function of the mean of Y with respect toX .The conditional mean of X given the event Y = y j (the meaninside y j -group ) is the mean of the conditional distribution of X given Y = y j , i.e.

j = mX/Y = yj =

n

i=1f ij xi

f j=

n

i=1f ij xi

f j.

The function A(y j ) = j , j {1, . . . , m }is called the regression function of the mean of X with respect toY .

We remark that the regression functions B(x) and A(y) can be approx-imated by Least Squares Method.


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3.3 Variation measures for two-dimensionalstatistical distributions

Let the two-dimensional distribution of Z = ( X, Y ) as above.

The (u, v)-th central moment of Z = ( X, Y ) (or of the two-dimensionaldistribution of Z = ( X, Y )) is

mcuv =

n

i=1

m

j =1f ij (xi x)u (y j y)v

n

i=1

m

j =1f ij

=n

i=1

m

j =1

f ij (xi x)u (y j y)v.

We remark thatmc20 =

2X and m

c02 =

2Y ,

where

2X =

n

i=1

m

j =1f ij (xi x)2

n

i=1

m

j =1f ij

=n

i=1

f i (xi x)2

and

2Y =

n

i=1

m

j =1

f ij (y j

y)2

n

i=1

m

j =1f ij

=m

j =1f j (y j y)2

are the variances of the marginal distributions (of X and Y ,respectively).

The covariance of X and Y (or of the two-dimensional distributionof Z = ( X, Y )) is

cov (X, Y ) = mc11

=

n

i=1

m

j =1f ij (xi x)(y j y)

n

i=1

m

j =1f ij

=n

i=1

m

j =1

f ij

(x i

x)(y j

y).

The conditional variance of Y given the event X = xi (the vari-ance inside x i-group ) is the variance of the conditional distributionof Y given X = xi , i.e.

2Y/X = x i =

m

j =1f ij (y j i)2

f i =

m

j =1f ij (y j i)2

f i .


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The conditional variance of X given the event Y = y j (the vari-ance inside y j -group ) is the variance of the conditional distributionof X given Y = y j , i.e.

2X/Y = yj =

n

i=1f ij (xi j )2

f j=

n

i=1f ij (x i j )2

f j.

The average of the variances within x-groups is

2Y =

n

i=1

m

j =1f ij (y j i)2

n

i=1

m

j =1 f ij

=n

i=1

f i 2Y/X = x i .

The average of the variances within y-groups is

2X =

n

i=1

m

j =1f ij (x i j )2

n

i=1

m

j =1f ij

=m

j =1

f j 2X/Y = yj .

The conditional variance of X given Y (the variance betweenx-groups ) is

2Y/X =

n

i=1

m

j =1f ij ( i y)2

n

i=1

m

j =1f ij

=n

i=1

f i ( i y)2.

The conditional variance of Y given X (the variance betweeny-groups ) is

2X/Y =

n

i=1

m

j =1f ij ( j x)2

n

i=1

m

j =1 f ij=

m

j =1

f j ( j x)2.

The rule of variances :

2Y = 2Y +

2Y/X ;

2X = 2X +

2X/Y

(The overall variation is the combination result between the random factorswithin each group and the essential factors determining the variation from agroup to another.)


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3.4 Correlation between variablesThe correlations (the dependence) that can be found between two variablesX and Y are classied as follows:

According to the way of change we can have: positive correlation (direct dependence ): if X is increasing

then Y will also increase and if X is decreasing then Y will alsodecrease.

negative correlation (opposite dependence ): if X is increas-ing then Y will decrease and if X is decreasing then Y will increase.

According to the intensity of the correlation we can have: high intensity (strong or tight); medium intensity ; low intensity .

According to the shape of the correlation we can have: linear correlation ;

nonlinear correlation , as exponential growth or logarithmic de-crease, for example.

Let the two-dimensional distribution of Z = ( X, Y ) as above. The degreeof correlation between the variables X and Y can be measured by using thefollowing indicators.

1. The covariance of X and Y (or of the two-dimensional distributionof Z = ( X, Y )), i.e.

cov (X, Y ) = mc11 =

n

i=1

m

j =1

f ij (xi

x)(y j

y)

n

i=1

m

j =1f ij

=n

i=1

m

j =1

f ij (x ix)(y j y).

It takes values between X Y and X Y . If X and Y are independent, then cov ( X, Y ) = 0. If cov (X, Y ) is close to zero, then there is no linear dependencebetween the variables X and Y .


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If cov(X, Y ) is positive, then we have a positive correlationbetween the variables X and Y .

cov (X, Y ) = X Y in the case of the perfect positive correlation(the linear increasing dependence). If cov (X, Y ) is negative, then we have a negative correlationbetween the variables X and Y . cov (X, Y ) = X Y in the case of the perfect negative correlation(the linear decreasing dependence).

2. The coefficient of correlation of X and Y (or of the two-dimensionaldistribution of Z = ( X, Y )) is

= (X, Y ) =cov(X, Y )

X Y .

It takes values between 1 and 1.The regression line (of Y with respect to X ) is

y y = Y X

(x x).

If X and Y are independent, then (X, Y ) = 0.

If (X, Y ) = 0, then there is no linear dependence between thevariables X and Y (the variables are independent or there is anonlinear dependence!).

If (X, Y ) = 1, then we have a direct linear dependence betweenthe variables X and Y , given by the regression liney y =

Y X

(x x).

If (X, Y ) =

1, then we have an opposite linear dependence

between the variables X and Y , given by the regression line

y y = Y X

(x x).

If 0 < (X, Y ) < 0.2, then we have a low positive correlationbetween the variables X and Y . If 0.2 < (X, Y ) < 0, then we have a low negative correlationbetween the variables X and Y .


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If 0.2

(X, Y )

0.5, then we have a weak positive correlation

between the variables X and Y , case needing a signicance testto be applied (like as the Student test).

If 0.5 (X, Y ) 0.2, then we have a weak negative correla-tion between the variables X and Y , case needing a signicancetest to be also applied.

If 0.5 < (X, Y ) 0.75, then we have a medium positive correla-tion between the variables X and Y . If 0.75 (X, Y ) < 0.5, then we have a medium negativecorrelation between the variables X and Y . If 0.75 < (X, Y ) 0.95, then we have a high positive correlationbetween the variables X and Y . If 0.95 (X, Y ) < 0.75, then we have a high negative corre-lation between the variables X and Y . If 0.95 < (X, Y ) < 1, then we have an extremely strong positivecorrelation between the variables X and Y , almost a direct linear

dependence.

If 1 < (X, Y ) < 0.95, then we have an extremely strongnegative correlation between the variables X and Y . almost anopposite linear dependence.

3. The coefficient of determination of Y with respect to X is

R2 =2Y/X2Y

,

and the coefficient of non-determination of Y with respect to X is

K 2 =2Y 2Y

.

By the rule of variances 2Y = 2Y + 2Y/X it follows that

R2 + K 2 = 1 .

The coefficient R2 shows the share of the variance between groups in theoverall variance, expressing the inuence of the classication factors.

If R2 = 1, then there is a strong functional relation between Y and X .


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If 0.7 < R 2 < 1, then the classication of the population accordingto X has a meaning, X variation inuencing Y variation.

If 0.5 < R 2 0.7, then the differences between the group meansare signicant. If R2 = 0 .5, then we cannot decide whether X variation has asignicant inuence over Y variation. If 0 < R 2 < 0.5, then X variation has no a signicant inuenceover Y variation. If R2 = 0, then X variation has no inuence over Y variation.

Exercise 3.1. Two hydrological stations make each a hundred measurementsof the level of a river during a year. The recorded data are given in thefollowing table.

X Y 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4.03.2 1 13.3 1 1 23.4 1 2 2 13.5 3 5 2 1 23.6 1 3 5 4 1 1 13.7 1 1 10 3 1 13.8 1 1 11 2 13.9 1 1 12 1 1 14.0 2 1 14.1 2 2 1

a. Represent the data into a scatter diagram.b. Compute the means and the variances of X and Y and the covariance

of X and Y .c. Compute the linear regression function of the mean of Y with respect

to X .

d. Compute and interpret the coefficient of correlation of X and Y , theregression line of Y with respect to X , and the coefficient of determinationof Y with respect to X .

3.5 Nonparametric measures of correlationIf we do not have sufficient elements to identify the rule of distributions, thenwe can use nonparametric methods like as the coefficients of ranks correlationproposed by Kendall and Spearman.


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Let X and Y be two simple statistical variables for a statistical populationor for a statistical sample. Let x1, . . . , x n and y1, . . . , yn be the ungroupedvalues (variants) of the distributions of X and Y , respectively. The distribu-tions of X and Y are represented in a table of the following form:

Units u1 u2 . . . unX values x1 x2 . . . xnY values y1 y2 . . . yn

where n is the volume of population (the number of statistical units u i).Let a i be the rank of the variant xi inside the distribution of X , namely

the rank of x i in the increasing order of x1, . . . , x n . Let also bi be the rankof the variant yi inside the distribution of Y , namely the rank of yi in theincreasing order of y1, . . . , yn .

The Spearmans coefficient of correlation of the ranks is

S = 1 6

n

i=1d2i

n3 n,

wheredi = a i

bi ,

i

{1, . . . , n

}(the rank differences between variables).We remark that the Spearmans coefficient of correlation of the ranksS is even the coefficient of correlation (A, B ) of A and B, where Aand B are the statistical variables that represent the ranks of X andY , respectively. The distributions of A and B are represented in thefollowing table:

Units u1 u2 . . . unA values a1 a2 . . . an

B values b1 b2 . . . bn

If some ranks of X or Y are equal, then on can use the correctedSpearmans coefficient of correlation of the ranks , given by

S = 1 6

n

i=1d2i +

t3 t12

n3 n,

where t is the number of equal ranks.


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The Kendalls coefficient of correlation of the ranks is

K =2(P Q)

n2 n,

where

P =n

i=1

P i , Q =n

i=1

Q i ,

P i = |{ j = 1 , n /a j > a i and b j > b i}|,Qi = |{ j = 1 , n /a j > a i and b j < b i}|,

for all i {

1, . . . , n}.

We remark that the numbers P i are indicators of the concordance ,and the numbers Q i are indicators of the discordance between theranks.

The coefficients of correlation of the ranks take values between 1 and+1. They can interpreted similarly with the coefficient of correlation.These coefficients have the advantage that they can be used in the case of skewed distributions or a small number of units. Also, these coefficients areapplicable for studying the relation between qualitative variables that cannotbe expressed numerically, but can be classied by their ranks.

Exercise 3.2. A group of students obtained the following marks over twotests:

Students 1 2 3 4 5 6 7 8 9 10Test A marks 10 25 13 14 28 16 6 8 24 17Test B marks 17 23 15 12 26 18 8 13 20 22

Students 11 12 13 14 15 16 17 18 19 20Test A marks 30 15 23 4 26 12 21 19 29 18Test B marks 28 13 25 10 27 5 19 14 29 24

Compute and interpret the coefficient of correlation and the coefficients of correlation of the ranks between the results of these tests.


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Theme 4

Time series and forecasting

Usually, a time series Y = ( yi)i (i being the time) is inuenced by thefollowing factors (components) :

the trend (the tendency) ; the cyclical factor ; the seasonal factor ;

the random factor (the irregular factor) .

The main decomposition models for a time series Y = ( yi)i :

The additive model :yi = T i + C i + S i + R i ,

where T i , C i , S i , R i represent the trend, the cyclical, the seasonal andthe random components, respectively.

This model assumes that the components are independent and theyhave the same measurement unit.

The multiplicative model :

yi = T i C i S i R i .This model assumes that the components depend each other or theyhave different measurement units.

60


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THEME 4. TIME SERIES AND FORECASTING 61

4.1 The trend componentThis component can be determined by Least Squares Method.

4.2 The cyclical componentThe cyclical variation of a time series is the component that tends to oscillateabove and below the trend line for periods longer than 1 year (if the timeseries is composed by annual dates). This component explains most of thevariation of evolution that remains unexplained by the trend component.

The cyclical component can be expressed as:

The cyclical variation :C i = yi yi ,

where

yi is the value of time series Y at time i; yi = T i is the estimated trend value of time series Y at the same

time i.

The cycle : yiyi 100.

The relative cyclical residual :yi yi

yi 100.

Example 4.1. The sales of a company for the last nine years are as follows:

Yeaar 2002 2003 2004 2005 2006 2007 2008 2009 2010

Sales 5.7 5.9 6 6.2 6.3 6.3 6.4 6.4 6.6Determine the trend function of sales and evaluate the cyclical variation.

Solution. Using the Least Squares Method, we obtain that the trend line is

y = 5 .7 + 0.1y

(where y1 = 1 , . . . , y9 = 9).The estimated sales, the cyclical variations, the cycles and the relative

cyclical residuals are calculated in the following table:


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THEME 4. TIME SERIES AND FORECASTING 62

Year Sales (yi) Estimated sales ( yi) Cyclical variation Cycle Rel. cycl. residual2002 5.7 5.8 -0.1 98.28 -1.722003 5.9 5.9 0 100.00 0.002004 6 6 0 100.00 0.002005 6.2 6.1 0.1 101.64 1.642006 6.3 6.2 0.1 101.61 1.612007 6.3 6.3 0 100.00 0.002008 6.4 6.4 0 100.00 0.002009 6.4 6.5 -0.1 98.46 -1.542010 6.6 6.6 0 100.00 0.00

4.3 The seasonal componentThe seasonal variation of a time series is the repetitive and predictable move-ment around the trend line in 1 year or less. For detecting the seasonalvariation, the time intervals need to be measured in small periods such asquarters, months, weeks, ... .

Let Y = ( yi)i=1 ,n be a time series, and let k be the number of equalperiods per each year.The seasonal component can be expressed as:

The moving average value for each time interval :If k is odd, the moving average value corresponding to yi is

yi =1k

yi k 12 + + yi + + yi+ k 12 ,for all i {1 + k 12 , . . . , n k 12 }.If k is even, the moving average value corresponding to yi is

yi =1k

12

yi k2 + yi k2 +1 + + yi + + yi+ k2 1 +12

yi+ k2 ,

for all i {1 + k2 , . . . , n k2}. The percentage of actual value to the moving average value :

yiyi 100.


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Theme 5

The interest

5.1 A general model of interestDenition 5.1. The interest corresponding to the initial value (thepresent value, the principal) S 0 (expressed in units of currency (u.c.))over the time (the period of investment) t (usually expressed in years)is a function D : [0, ) [0, ) [0, ) that veries the following twoconditions:

1. D(S 0, 0) = 0 ,

S 0

0; D(0, t ) = 0 ,

t

0;

2. The function D(S 0, t ) increases in each of the two variables S 0 and t. Assuming that the function D(S 0, t ) has partial derivatives, thiscondition can be expressed as:

DS 0

(S 0, t ) > 0,Dt

(S 0, t ) > 0,S 0 > 0,t > 0.

Denition 5.2. The sum

S (S 0, t ) = S 0 + D(S 0, t )

is called the nal value (the future value, the amount) , and is alsodenoted by S t .

Remark 5.1. The nal value is a function S : [0, ) [0, ) [0, ) that veries the following two conditions:S (S 0, 0) = S 0,S 0 0; S (0, t ) = 0 ,t 0;

S S 0

(S 0, t ) > 1,S t

(S 0, t ) > 0,S 0 > 0,t > 0

64


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THEME 5. THE INTEREST 65

Denition 5.3. The annual interest rate , denoted by i, is the interest for 1 u.c. over 1 year, that is

i = D(1, 1).

The annual interest percentage , denoted by p, is the interest for 100 u.c.over 1 year, that is

p = D(100, 1).

Remark 5.2. Usually, p = 100i.

Denition 5.4. The function F : [0, ) [0, ) [0, ) given by F (S 0, t ) =

Dt

(S 0, t ),S 0 0,t 0is called the proportionality factor of the interest .

Remark 5.3.F (S 0, t ) =

S t

(S 0, t ),S 0 0,t 0.Proposition 5.1. We have

D(S 0, t ) = t

0 F (S 0, x)dx,S 0 0,t 0;

S t S (S 0, t ) = S 0 + t0 F (S 0, x)dx,S 0 0,t 0Corollary 5.1. We have

i = 10 F (1, x)dx; p = 10 F (100, x)dx.Denition 5.5. The function : [0,

)

[0,

) given by

(t) =S t (S 0, t )S (S 0, t )

,t 0is called the instantaneous interest rate .

Remark 5.4.(t) =

ln S t

(S 0, t ) =F (S 0, t )S (S 0, t )

,t 0.


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Proposition 5.2. We have

S t S (S 0, t ) = S 0e t

0(x)dx

,S 0 0,t 0;

D(S 0, t ) = S 0 e t0 (x)dx 1 ,S 0 0,t 0.Corollary 5.2. We have

i = e 1

0(x)dx

1; p = 100 e 1

0(x)dx

1 .

5.2 Equivalence of investmentsDenition 5.6. A multiple (nancial) investment consists in n ini-tial values S 01 , S 02 , . . . , S 0n invested over the times t1, t2, . . ., tn , with theannual interest rates i1, i2, . . ., in (or with the annual interest percentages p1, p2, . . ., pn ). Let D(S 01 , t1), D(S 02 , t2), . . ., D(S 0n , tn ) be the correspond-ing interests, and let S 1 = S (S 01 , t1), S 2 = S (S 02 , t2), . . ., S n = S (S 0n , tn )

be the corresponding nal values. The sumsn

k=1S 0k ,

n

k=1D(S 0k , tk) and

n

k=1S k

are called the total initial value , the total interest and the total -nal value of the given multiple investment, respectively. This multiple in-vestment can be expressed by a matrix of one of the following two forms

S 01 t1 i1S 02 t2 i2

......

...S 0n tn in

(if the initial values are known), or

t1 i1 S 1t2 i2 S 2...

......

tn in S n

(if

the nal values are known).

Denition 5.7. We say that two multiple investments are equivalent by

interest and we denote

S 01 t1 i1S 02 t2 i2

......

...S 0n tn in

I

S 01 t1 i1S 02 t2 i2

......

...S 0m tm im

if the corre-

sponding total interest are equal, i.e.n

k=1D(S 0k , tk) =

m

k=1D(S 0k , tk).


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We say that two multiple investments are equivalent by present value

and we denote

t1 i1 S 1t2 i2 S 2...

......

tn in S n

P

t1 i1 S 1t2 i2 S 2...

......

tm im S m

if the corresponding

total initial values are equal, i.e.n

k=1S 0k =

m

k=1S 0k .

Denition 5.8. If

S 01 t1 i1S 02 t2 i2

... ... ...S 0n tn in

I

S (CI )0 , t , i

I

S 0, t

(CI )

, iI

S 0, t , i

(CI )

,

then the initial value S (CI )0 , the time of investment t (CI ) and the annual in-terest rate i(CI ) are called commonly replacements by interest .

Denition 5.9. If

S 01 t1 i1S 02 t2 i2

......

...S 0n tn in

I

S (MI )0 t1 i1S (MI )0 t2 i2

......

...S (MI )0 tn in

I

S 01 t (MI ) i1S 02 t (MI ) i2

......

...S 0n t (MI ) in

I

S 01 t1 i(MI )S 02 t2 i(MI )

......

...S 0n tn i(MI )

,

then the initial value S (MI )0 , the time of investment t (MI ) and the annual interest rate i(MI ) are called meanly replacements by interest .

Denition 5.10. If

t1 i1 S 1t2 i2 S 2...

......

tn in S n

P

t , i ,S (CP ) P

t(CP ) , i ,S P t, i

(CP ) , S .

then the nal value S (CP ) , the time of investment t (CP ) and the annual in-terest rate i(CP ) are called commonly replacements by present value .

Denition 5.11. If

t1 i1 S 1t2 i2 S 2...

......

tn in S n

P

t1 i1 S (MP )t2 i2 S (MP )...

......

tn in S (MP )

P

t (MP ) i1 S 1t (MP ) i2 S 2

......

...t (MP ) in S n

P

t1 i(MP ) S 1t2 i(MP ) S 2...

......

tn i(MP ) S n

,


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then the nal value S (MP ) , the time of investment t (MP ) and the annual interest rate i(MP ) are called meanly replacements by present value .

5.3 Simple interest

5.3.1 Basic formulasDenition 5.12. If the principal is not actualized over the time of invest-ment, then we say that we obtain a simple interest .

Proposition 5.3. For simple interest we have:

D D(S 0, t ) = S 0it =S 0 pt100

( the simple interest formula ),

S t S (S 0, t ) = S 0 + D = S 0(1 + it )( the compounding formula, the rule of interest ),

S 0 =S t

1 + it

( the discounting formula ),

i =D

S 0t=

S t S 0S 0t

, t =D

S 0i=

S t S 0S 0i

.

Remark 5.5. According to the above formulas, 1 + it is called the com-pounding factor , and 11+ it is called the discounting factor for simpleinterest.

Corollary 5.3. If t =hk

(i.e. k is the number of periods per year and h is

the number of such periods), then the simple interest is:

D S,hk

=S 0ih

k=

S 0 pt100k

.

Remark 5.6. If the time of investment t is given as the period from the initial date (d1, m 1, y1) to the nal date (d2, m 2, y2), (where di , m i , yi represents theday, the number of month and the year of the date), then we have threeconventions (procedures) to calculate the simple interest:


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1. The exact interest (actual/actual) :

D =S 0ih365

or D =S 0ih366

(for leap years) ,

where h is the number of calendar days from (d1, m 1, y1) to (d2, m 2, y2)(excluding either the rst or last day);

2. The bankers rule (actual/360) :

D =S 0ih360

,

where h is the number of calendar days from (d1, m 1, y1) to (d2, m 2, y2)(excluding either the rst or last day);

3. The ordinary interest (30/360) :

D =S 0ih360

,

whereh = 360( y2 y1) + 30( m2 m1) + d2 d1

(assumes that all months have 30 days, called the 30-day month convention ).

5.3.2 Simple interest with variable rateProposition 5.4. If the time of investment is t = t1 + t2 + ... + tm and theannual interest rate is i1 for the rst period t1, i2 for the second period t2,...,im for the last period tm , then we have:

D = S 0m

k=1

ik tk (the simple interest formula);

S t = S 0 1 +m

k=1

ik tk (the compounding formula);

S 0 =S t

1 +m

k=1ik tk

(the discounting formula).


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5.3.3 Equivalence by simple interestProposition 5.5. For simple interest we have:

S (CI )0 =

n

k=1S 0k ik tk

it, t(CI ) =

n

k=1S 0k ik tk

S 0i, i(CI ) =

n

k=1S 0k ik tk

S 0t,

S (MI )0 =

n

k=1S 0k ik tk

n

k=1iktk

, t(MI ) =

n

k=1S 0k ik tk

n

k=1S 0kik

, i(MI ) =

n

k=1S 0k ik tk

n

k=1S 0k tk

.

5.4 Compound interest

5.4.1 Basic formulasDenition 5.13. If the principal is actualized over each year of investment time (by adding the interest of the previous year), then we say that we obtain a compound interest .

Proposition 5.6. For compound interest we have:

D

D(S

0, t ) = S

0(1 + i)t

1

( the compound interest formula ),

S t S (S 0, t ) = S 0 + D = S 0(1 + i)t( the compounding formula, the rule of interest ),

S 0 =S t

(1 + i)t

( the discounting formula ).

Remark 5.7. According to the above formulas, (1 + i)t

is called the com-pounding factor , and

1(1 + i)t

is called the discounting factor for com-

pound interest. Denoting the annual compounding factor by

u = 1 + i

and the annual discounting factor by

v =1u

=1

1 + i,


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the above formulas can be written as

S t = S 0u t , S 0 = S t vt .

Remark 5.8. If

t = n +hk

( n being the integer part and hk

being the fractional part , i.e. the timeof investment cover only h periods from a total of k equal periods per the last year), then we have two conventions (procedures) to calculate the compound interest:

1. The rational procedure : we apply a compound interest for the inte-ger part and a simple interest for the fractional part, and hence

S t S n + hk = S 0(1 + i)n 1 + i hk

(the compounding formula);

D = S 0 (1 + i)n 1 + i hk 1 (the interest formula).

2. The commercial procedure : we extend the compound interest to the

fractional part, and henceS t S n+ hk = S 0(1+ i)n +

hk = S 0(1+ i)n k (1 + i)h (the compounding formula);

D = S 0 (1 + i)n +hk 1 = S 0 (1 + i)n k (1 + i)h 1 (the interest formula).

5.4.2 Nominal rate and effective rateDenition 5.14. For an initial value S 0 over n years at an annual rate j kcompounded k times per each year, the nal value is

S t = S 0 1 + jk

kkn

= S 0 (1 + i)n ,

where

1 + i = 1 +j kk

k

.

k is called the number of interest periods per year ; j k is called the nominal rate (annual interest rate);


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ik =

j k

kis called the interest rate per interest period (period

interest rate);

i is called the effective rate or the real rate (annual interest rate).

5.4.3 Compound interest with variable rateProposition 5.7. If the time of investment is t = t1 + t2 + ... + tm and the

annual interest rate is i1 for the rst period t1 = n1 +h1k1

, i2 for the second

period t2 = n2 +h2

k2,..., im for the last period tm = nm +

hm

km, then we have:

1. For the rational procedure:

D = S 0m

l=1

(1 + i l)n l 1 + il h lkl 1 (the compound interest formula);

S t = S 0m

l=1

(1 + i l)n l 1 + il h lkl

(the compounding formula);

2. For the commercial procedure:

D = S 0m

l=1

(1 + i l)t l 1 (the compound interest formula);

S t = S 0m

l=1

(1 + i l)t l (the compounding formula).

5.5 LoansThe amortization table for a loan of size (original balance ) V 0 u.c. pern years at an annual interest rate i has the following form:

Years start Years endYear Remaining Interest Principal Payment Remaining

principal part part (Rate) principal1 V 0 d1 = V 0 i Q1 T 1 = d1 + Q1 V 1 = V 0 Q12 V 1 d2 = V 1 i Q2 T 2 = d2 + Q2 V 2 = V 1 Q2

. . .k V k 1 dk = V k 1 i Qk T k = dk + Qk V k = V k 1 Qk

. . .n V n 1 dn = V n 1 i Qn T n = dn + Qn V n = V n 1 Qn = 0


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Obviously, we have:

V 0 = Q1 + Q2 + + Qn , V n 1 = Qn ,T n = Qn u, T k+1 T k = Qk+1 Qk u,

where u = 1 + i is the the annual compounding factor.We have two mainly procedures to calculate the payments of a loan:

1. The xed-principal amortization : Q1 = Q2 = = Qn = Q.In this case, we have:

Q =V 0

n;

T k+1 T k = Q i(arithmetic progression);

T k = Q[1 + (n k + 1) i].2. The xed-rate amortization : T 1 = T 2 = = T n = T.

In this case, we have:T = V 0

i1 vn

(the xed-rate formula );

Qk+1 = Qk u(geometric progression);

Qk = V 0 i

un 1 uk 1,

where u = 1 + i is the the annual compounding factor, and v =1u

=1

1 + i is the annual discounting factor.

Remark 5.9. The ination changes the purchasing power of money. After n years, the purchasing power of S n u.c. is reduced to

S 0 =S n

(1 + a1)(1 + a2) . . . (1 + an ),

where a1, a2, . . . , a n are the annual ination rates. S n is measured in futureunits of currency, and S 0 is measured in todays units of currency.


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5.6 ProblemsExercise 5.1. A person deposits 1000 u.c. on 20 February 2011 at an annualinterest percent of 12%. Calculate the amount of this investment on 10November 2011 in each of the following cases:a) exact interest;b) bankers rule;c) ordinary interest.

Exercise 5.2. A person deposits 1000 u.c. for 3 years and seven months atan annual interest percent of 12%. Calculate the nal value of this investmentin each of the following cases:a) simple interest;b) compound interest, the rational procedure;c) compound interest, the commercial procedure;d) compounded monthly interest.

Exercise 5.3. Consider the following investments: 1000 u.c. for one year at12% per year, 800 u.c. for 9 months at 14% per year, and 1200 u.c. for 10months at 9% per year. Calculate the initial value, the time of investmentand the annual interest rate meanly replacements by simple interest.

Exercise 5.4. A person deposits 100 u.c. at the end of every month for 5years, at successive annual interest percents of 12%, 12%, 9%, 10%, 10%.Calculate the amount of this investment at the end of 5 years.

Exercise 5.5. Construct the amortization table for a loan of 2400 u.c. per4 years at an annual interest percent of 16%, in each of the following cases:a) xed-principal annually amortization;b) xed-rate annually amortization;c) xed-principal monthly amortization;d) xed-rate monthly amortization.Compare the obtained results when the annual successive ination rates are

4%, 6%, 5%, 6%.


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Theme 6

Introduction to Actuarial Math

6.1 A general model of insuranceIn an insurance model the insurer agrees to pay the insured one or moreamounts called claims (claim payments) , at xed times or when theinsured event occurs. In return of these claims, the insured pays one ormore amounts called premiums .

Usually the insure events are random events .For a mutually advantageous insurance, the present values (at the initial

moment of the insurance) of the premiums need to be equal to the presentvalue of the claims. These values are also called actuarial present values .

Denition 6.1. For a given insurance, the single premium payable at theinitial moment of the insurance is

P = E (X ),

where E (X ) denotes the mean of the random variable X that represents thepresent value of the claim.

Theorem 6.1. Let A be an insurance consisting in the partial insurances

A1, A2, . . . , A n ( n N

), and let P 1, P 2, . . . , P n be the single premiums cor-responding of these partial insurances. Then the single premium of the total insurance A is

P = P 1 + P 2 + + P n .Proof. Let X be the random variable that represents the present value of thetotal insurance A and let X 1, X 2, . . . , X n be the random variables represent-ing the present values of partial insurances A1, A2, . . . , A n , respectively. Wehave

X = X 1 + X 2 + + X n ,

75


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THEME 6. INTRODUCTION TO ACTUARIAL MATH 76

and hence

P = E (X ) = E (X 1 + X 2 + + X n ) = E (X 1) + e(X 2) + + E (X n )= P 1 + P 2 + + P n .

6.2 Biometric functionsThe mortality is the most important factor in the insurances of persons.The frequency of mortality for a population is measured by some statisticalfunction of age called biometric functions . We assume that the age ismeasured in years.

Denition 6.2. We denote by l0 the total number of persons of the analyzed population (the number of newborns).

Remark 6.1. Usually, l0 = 100000.

Remark 6.2. l0 represents the number of survivors to age 0 (from the ana-lyzed population).

6.2.1 Probabilities of life and deathDenition 6.3. Let x,n,m N. We denote

px = the probability that a person of age x will live at least one year; qx = the probability that a person of age x will die within one year, n px = the probability that a person of age x will attain age x + n; n qx = the probability that a person of age x will die until the age x + n;

m |n qx = the probability that a person of age x will attain age x + m but die until the age x + m + n. px is called probability of life for age x, and qx is called probability of death for age x.

Proposition 6.1. Let x,y,z N, x y z. We have:qx = 1 px ; (6.1)

n qx = 1 n px ; (6.2)


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0 px = 1; 0qx = 0; (6.3)1 px = px ; 1qx = qx ; (6.4)

0|n qx = n qx ; (6.5)n + m px = n px m px+ n ; (6.6)

n px = px px+1 . . . px+ n 1; (6.7)n qx = qx + px qx+1 + px px+1 qx+2 + . . . + px px+1 . . . px+ n 2 qx+ n 1;(6.8)

m |n qx = m px n qx+ m ; (6.9)m |n qx = m + n qx m qx = m px m + n px . (6.10)

Proof. Equalities (6.1), (6.2), (6.3), (6.4) and (6.5) are obvious.Denote by A(x, y) the event that a person of age x will attains age y.

ThenA(x, x + n + m) = A(x, x + n) A(x + n, x + n + m),

and the events A(x, x + n) and A(x + n, x + n + m) are independent. Hence

n + m px = P (A(x, x + n+ m)) = P (A(x, x + n))P (A(x+ n, x + n+ m)) = n pxm px+ n(where P (A) represents the probability of event A). Using (6.6) and (6.4)we have

n px = 1 px 1 px+1 . . . 1 px+ n 1 = px px+1 . . . px+ n 1.Also, we have

n qx = P (A(x, x + n))

= P A(x, x + 1) A(x, x + 1) A(x + 1 , x + 2) A(x, x + 2) A(x + 2 , x + 3) . . .A(x, x + n 1) A(x + n 1, x + n)

= qx + px qx+1 + px px+1 qx+2 + . . . + px px+1 . . . px+ n 2 qx+ n 1(where A represents the complementary event of A). We have

m |n qx = P (A(x, x + m) A(x + m, x + m + n))= P (A(x, x + m))P (A(x + m, x + m + n))= m px n qx+ m ;

m |n qx = P (A(x, x + m + n) A(x, x + m))


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= P (A(x, x + m + n)

\A(x, x + m))

= P (A(x, x + m + n)) P (A(x, x + m)) = m + n qx m qx= (1 m + n px ) (1 m px ) = m px m + n px .

6.2.2 The survival functionDenition 6.4. Let xN. We denote

lx = the expected number of survivors at age x (from the ana-lyzed population).

Proposition 6.2. For any xN we have

lx = l0 p(0, x). (6.11)

Proof. Obviously,lx = E (X ),

where X is the random variable that represents the number of survivors atage x. Let

X :0 . . . n . . . l0

x (0) . . . x (n) . . . x (l0)be the distribution of X , where, for any n {0, . . . , l 0}, x (n) denotes theprobability that the number of survivors at age x is equal to n. We have

x (n) = C nl0 (x p0)n (xq0)l0 n , n {0, . . . , l 0}.

Then X has a binomial distribution of parameters l0 and p(0, x). Therefore

lx = E (X ) = l0 x p0.

Denition 6.5. Let xN. We denote

s(x) = x p0 = the probability that a newborn will live to at least x.s(x) is called the survival function for age x.

Remark 6.3. xq0 = 1 x p0 = 1 s(x) represents the probability that a newborn will die until the age x.


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Proposition 6.3. Let x,n,m

N. We have:

lx = l0 p0 p1 . . . px 1; (6.12)n px =

lx+ nlx

; n qx =lx lx+ n

lx; (6.13)

m |n qx =lx+ m lx+ m + n

lx; (6.14)

px =lx+1lx

; qx =dxlx

, (6.15)

where

dx = lx lx+1 . (6.16)Proof. Equation (6.12) is an immediate consequence of (6.11) and (6.7). Us-ing (6.6) and (6.11) we have

n px =x+ n p0

x p0=

lx+ nl0

l0lx

=lx+ nlx

, and n qx = 1 n px =lx lx+ n

lx.

Using (6.9) and (6.12) we have

m |n qx = m px n qx+ m =lx+ m

lx lx+ m lx+ m + n

lx+ m=

lx+ m lx+ m + nlx

.

Taking n = 1 into (6.13) we obtain equalities (6.15).

Remark 6.4. dx represents the expected number of deaths at age x(i.e. at exactly age x or between ages x and x + 1 ).

Remark 6.5. There exists an age N such that

l > 0 and lx = 0 x > .

Denition 6.6. The value from the above remark is called the limiting

age .Remark 6.6. Usually, = 100.

6.2.3 The life expectancyDenition 6.7. Let xN, x . We denote

ex = the expected future lifetime for a person of age x (prior to death).


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ex is called the average remaining lifetime for age x, and x +

ex is called

the life expectancy for age x.Remark 6.7. We assume that the deaths are uniform distributed throughout the year.

Proposition 6.4. For any xN, x , we haveex =

12

+1lx

x

n =1

lx+ n . (6.17)

Proof. Obviously,lx = E (Y ),

where Y is the random variable that represents the future lifetime for aperson of age x. Let

Y :12

. . . n +12

. . . x +12

x (0) . . . x (n) . . . x ( x)be the distribution of Y , where, for any n {0, . . . , x}, x (n) representsthe probability that a person of age x will live only n (i.e. will die at agex + n). We have

x (n) = n |1qx = n px qx+ n , n {0, . . . , x},Using (6.13) and (6.16) it follows that

x(n) =lx+ nlx

dx+ nlx+ n

=dx+ n

lx=

lx+ n lx+ n +1lx

, n {0, . . . , x}. (6.18)Hence

ex = E (Y ) =

x

n =0

n +12

lx+ n lx+ n +1lx

=1lx

x

n =0

n +12

lx+ n n + 1 +12

lx+ n +1 + lx+ n +1

=1lx

12

lx x + 1 +12

l+1 + x

n =0

lx+ n +1 =12

+1lx

x

n =1

lx+ n ,

since l+1 = 0.

Remark 6.8. According to (6.17) and (6.13) we obtain:

ex =

12

+ x

n =1n px , xN, x . (6.19)


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6.2.4 Life tablesThe value of biometric functions are tabulated in life table (mortalitytable or actuarial table ) of the following form:

Age Nr. of survivors Nr. of deaths Probab. of death Average rem. lifetimex l x dx qx

ex

0 l0 = 1000001...

= 100

Usually, the values lx are derived by a census. The values dx , qx andex are

calculated according to (6.16), (6.15) and (6.17), respectively.The following actuarial table shows the life expectancy for the Romanian

population in 2008 (www.pensiileprivate.ro ).

x l x lx qx qx x +ex x +

ex

MALE FEMALE MALE FEMALE MALE FEMALE

18 100000 100000 0.0007 0.0004 66.7 7319 99930 99960 0.0007 0.0004 66.7 7320 99860 99920 0.001 0.0004 66.8 7321 99760 99880 0.0011 0.0004 66.8 7322 99654 99838 0.0011 0.0004 66.9 7323 99543 99794 0.0012 0.0005 66.9 73.124 99425 99748 0.0012 0.0005 67 73.125 99302 99700 0.0013 0.0005 67 73.126 99173 99651 0.0014 0.0005 67.1 73.127 99032 99597 0.0015 0.0006 67.1 73.228 98880 99539 0.0017 0.0006 67.2 73.229 98716 99477 0.0018 0.0007 67.3 73.230 98540 99412 0.0019 0.0007 67.3 73.231 98353 99342 0.0021 0.0008 67.4 73.332 98144 99261 0.0023 0.0009 67.5 73.333 97914 99167 0.0026 0.0011 67.6 73.334 97664 99062 0.0028 0.0012 67.7 73.435 97392 98945 0.003 0.0013 67.7 73.436 97100 98817 0.0036 0.0015 67.8 73.537 96754 98670 0.0041 0.0017 68 73.538 96356 98507 0.0047 0.0018 68.1 73.639 95905 98325 0.0052 0.002 68.2 73.7


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ex

MALE FEMALE MALE FEMALE MALE FEMALE40 95402 98127 0.0058 0.0022 68.4 73.741 94849 97911 0.0065 0.0025 68.5 73.842 94231 97670 0.0072 0.0027 68.7 73.943 93548 97404 0.008 0.003 68.9 7444 92804 97114 0.0087 0.0032 69.1 74.145 91998 96799 0.0094 0.0035 69.3 74.246 91133 96461 0.0101 0.0039 69.5 74.347 90209 96088 0.0109 0.0042 69.8 74.448 89228 95683 0.0116 0.0046 70 74.549 88191 95245 0.0124 0.0049 70.3 74.650 87101 94774 0.0131 0.0053 70.5 74.751 85960 94272 0.0143 0.0059 70.8 74.852 84731 93717 0.0155 0.0065 71.1 7553 83417 93112 0.0167 0.007 71.4 75.154 82024 92456 0.0179 0.0076 71.7 75.355 80556 91752 0.0191 0.0082 72 75.456 79017 91000 0.0208 0.009 72.3 75.657 77374 90182 0.0225 0.0098 72.7 75.858 75633 89302 0.0242 0.0105 73 7659 73803 88361 0.0259 0.0113 73.4 76.2

60 71891 87361 0.0276 0.0121 73.7 76.361 69907 86304 0.0299 0.0137 74.1 76.562 67814 85125 0.0323 0.0152 74.5 76.763 65625 83829 0.0346 0.0168 74.9 7764 63353 82423 0.037 0.0183 75.3 77.265 61011 80911 0.0393 0.0199 75.7 77.466 58614 79301 0.0427 0.0228 76.1 77.767 56111 77493 0.0461 0.0257 76.6 77.968 53524 75501 0.0495 0.0286 77 78.269 50875 73342 0.0529 0.0315 77.4 78.570 48183 71032 0.0563 0.0344 77.9 78.8

71 45471 68588 0.0682 0.0474 78.3 79.172 42371 65336 0.08 0.0604 78.8 79.573 38981 61387 0.0919 0.0735 79.4 79.974 35399 56877 0.1037 0.0865 80 80.475 31727 51959 0.1156 0.0995 80.6 8176 28059 46789 0.1275 0.1125 81.3 81.677 24483 41524 0.1393 0.1255 82 82.278 21072 36311 0.1512 0.1386 82.7 82.979 17886 31280 0.163 0.1516 83.4 83.6


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ex

MALE FEMALE MALE FEMALE MALE FEMALE80 14970 26538 0.1749 0.1646 84.2 84.381 12352 22170 0.1868 0.1776 85 85.182 10045 18232 0.1986 0.1906 85.8 85.983 8050 14757 0.2105 0.2037 86.6 86.784 6356 11751 0.2223 0.2167 87.4 87.585 4942 9205 0.2342 0.2297 88.3 88.386 3785 7091 0.2461 0.2427 89.1 89.287 2854 5370 0.2579 0.2557 90 9088 2118 3996 0.2698 0.2688 90.9 90.989 1546 2922 0.2816 0.2818 91.7 91.790 1111 2099 0.2935 0.2948 92.6 92.691 785 1480 0.3054 0.3078 93.5 93.592 545 1024 0.3172 0.3208 94.4 94.493 372 696 0.3291 0.3339 95.3 95.294 250 463 0.3409 0.3469 96.2 96.195 165 303 0.3528 0.3599 97 9796 107 194 0.3647 0.3729 97.8 97.897 68 122 0.3765 0.3859 98.6 98.698 42 75 0.3884 0.399 99.3 99.399 26 45 0.4002 0.412 99.8 99.8

100 15 26 1 1 101.8 101.9

6.3 ProblemsExercise 6.1. Calculate the probability that a 30 years old person will liveat least 35 years but at most 55 years.

Exercise 6.2. Consider a family of a 45 years old husband and a 43 yearsold wife.a) Calculate the probability that both spouses will die in the same year.

b) Calculate the probability that both spouses will die at the same age.Exercise 6.3. Calculate the average remaining lifetime and the life ex-pectancy for a 50 years old person.

Exercise 6.4. Calculate the probability that a 60 years old person will diebefore the integer number of years of his average remaining lifetime.

Exercise 6.5. For a 35 years old person, calculate the life expectancy andthe age of death having the maximum probability.


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Theme 7

Life annuities

7.1 A general model. ClassicationsIn a person insurance, the claims are payments while the insured survives.We have the following classications .

1. By period , the claims can be:

annuities ;

semiannual ;

quarterly ; monthly .

2. By amount , the claims can be:

constants ; variables .

3. By time of payment , the claims can be:

annuity-due , when the claims are payed at the beginning of eachperiod; annuity-immediate , when the claims are payed at the end of each period.

4. By time of rst payment , the claims can be:

immediate ; deferred .

84


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THEME 7. LIFE ANNUITIES 85

5. By number of payments , the claims can be:

single , when the claim is payed at a xed time, only if the insuredwill live at this time; temporary (limited) , when the claim is payed at xed times,while the insured survives; unlimited , when the claim is payed whole life.

7.2 Single claim

Denition 7.1. Let x, n N s.t. x + n . We denote

n E x = the single premium payable by a person of age x for a singleclaim of 1 u.c. over n years if the person survives.Remark 7.1. n E x is called the unitary premium .

Proposition 7.1. For any x, n N s.t. x + n , we haven E x =

Dx+ nDx

, (7.1)

where Dx = vx lx , (7.2)v =

11 + i

being the annual discounting factor, i being the annual interest rate.

Proof. For a mutually advantageous insurance, the single premium n E x needto be equal to the present value of the single claim, that is

n E x = E (X ),

where X is the random variable that represents the present value of the claim.We have

X =vn , if the insurer survives at least n years from the time of insurance issue ,0, otherwise .

Hence the distribution of X is

X : vn 0

n px n qx.


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By (6.13) and (7.2) we have

n E x = E (X ) = vn n px + 0 n qx = vn lx+ nlx

=vx+ n lx+ n

vx lx=

Dx+ nDx

.

Corollary 7.1. Let x, n N, x + n , T 0. The single premium payableby a person of age x for a single claim of T u.c. over n years if the person survives is

T n E x = T Dx+ nDx

.

Denition 7.2. Dx dened by (7.2) is called the commutation number .The single premium n E x dened by (7.1) is called the life discounting factor .

Remark 7.2. By (7.1) it follows that

y+ z E x = yE x zE x+ y , x,y,z N s.t. x + y . (7.3)

7.3 Life annuities-immediate

7.3.1 Whole life annuitiesDenition 7.3. Let xN, x . We denote

ax = the single premium payable by a person of age x for a whole lifeannuity-immediate of 1 u.c. per year.Proposition 7.2. For any xN, x , we have

ax =N x+1Dx

, (7.4)

whereN x = Dx + Dx+1 +

+ D. (7.5)

Proof. By Theorem 6.1 we have

ax = 1E x + 2E x + + xE x .Using (7.1) and (7.5) we obtain

ax =Dx+1Dx

+Dx+2Dx

+ +DDx

=N x+1Dx

.


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Corollary 7.2. Let x

N, x

, T

0. The single premium payable by a

person of age x for a whole life annuity-immediate of T u.c. per year is

T ax = T N x+1Dx

.

Denition 7.4. N x dened by (7.5) is called the cumulative commuta-tion number .

7.3.2 Deferred whole life annuitiesDenition 7.5. Let x, r

N s.t. x + r

. We denote

r | ax = the single premium payable by a person of age x for an r -year deferred whole life annuity-immediate of 1 u.c. per year (payable at theend of each year while the person survives from age x + r onward).

Proposition 7.3. For any x, r N s.t. x + r , we haver | ax =

N x+ r +1Dx

. (7.6)


r |ax = r +1 E x + r +2 E x + + xE x .Using (7.1) and (7.5) we obtain

r | ax =Dx+ r +1

Dx+

Dx+ r +2Dx

+ +DDx

=N x+ r +1

Dx.

Corollary 7.3. Let x, r N, x + r , T 0. The single premium payableby a person of age x for an r -year deferred whole life annuity-immediate of T u.c. per year is

T r | ax = T N x+ r +1

Dx.

Remark 7.3. By (7.6), (7.1) and (7.4) it follows that

r |ax = r E x ax+ r , x, r N s.t. x + r , (7.7)0|ax = ax , xN, x ,

x |ax = 0 , xN, x .


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7.4 Temporary life annuitiesDenition 7.6. Let x, r N s.t. x + r . We denote

ax: r = the single premium payable by a person of age x for an r -year temporary life annuity-immediate of 1 u.c. per year (payable at the end of each year while the person survives during the next r years).

Proposition 7.4. For any x, r N s.t. x + r , we haveax: r =

N x+1 N x+ r +1Dx

. (7.8)


ax: r = 1E x + 2E x + + r E x .By (7.1) and (7.5) we obtain

ax: r =Dx+1Dx

+Dx+2Dx

+ +Dx+ rDx

=N x+1 N x+ r +1

Dx.

Corollary 7.4. Let x, r N, x + r , T 0. The single premium payableby a person of age x for an r -year temporary life annuity-immediate of T

u.c. per year is

T ax: r = T N x+1 N x+ r +1

Dx.

Remark 7.4. By (7.4), (7.8), (7.6) and (7.7) it follows that

ax = ax: r + r | ax , x, r N s.t. x + r , (7.9)ax: r = ax r E x ax+ r , x, r N s.t. x + r ,ax: 0 = 0 ,

x

N, x

,

ax: x = ax , xN, x .

7.5 Life annuities-immediate with k-thly pay-ments

In this case the claims are payable at the end of each k-th period of the year.


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7.5.1 Whole life annuities with k-thly paymentsDenition 7.7. Let xN, x and kN. We denote

a(k)x = the single premium payable by a person of age x for a whole life

annuity-immediate of 1k

u.c. per each k-th period of the year (i.e. 1u.c. per each year).

Denition 7.8. For any xN, x and kN, k 2, we dene theintermediate commutation numbers Dx+ 1k , D x+ 2k , . . . , D x+ k 1k such that Dx , D x+ 1k , D x+ 2k , . . . , D x+ k 1k , D x+1 is a arithmetic progression.

Lemma 7.1. For any xN, x , kN and h {0, 1, . . . , k}we have

Dx+ hk =k h

k Dx +hk Dx+1 . (7.10)

Proof. The arithmetic progression Dx , D x+ 1k , D x+ 2k , . . . , D x+ k 1k , D x+1 has k+

1 terms, so its ratio isDx+1 Dx

kand hence its ( h + 1)-th term is

Dx+ hk = Dx + h Dx+1 Dx

k=

k hk Dx +

hk Dx+1 .

Proposition 7.5. For any xN, x and kNwe havea (k)x =

N x+1Dx

+k 1

2k. (7.11)


a(k)x =1k

x

n =0

k

h=1

n + hkE x . (7.12)

By (7.1), (7.10) and (7.5) we obtain

a (k)x =1k

x

n =0

k

h=1

Dx+ n + hkDx

=1

k Dxk

h=1

x

n =0

k hk Dx+ n +

hk Dx+ n+1

=1

k Dxk

h=1

k hk

x

n =0

Dx+ n +hk

x

n =0

Dx+ n +1


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=1

k Dxk

h=1

k

h

k (Dx + N x+1 ) +hk N x+1

=1

k Dxk

h=1

k hk Dx + N x+1 =

1k Dx

k k(k + 1)

2k Dx + kN x+1

=N x+1Dx

+k 1

2k.

Corollary 7.5. Let xN, x , kN and T 0. The single premium payable by a person of age x for a whole life annuity-immediate of T u.c. per each k-th period of the year is

T k a(k)x = T k N x+1Dx

+k 1

2.

Remark 7.5. By (7.11) and (7.4) it follows that

a (k)x = ax +k 1

2k, xN, x , kN,

a(1)x = ax , xN, x .

7.5.2 Deferred whole life annuities with k-thly pay-ments

Denition 7.9. Let x, r N s.t. x + r and let kN. We denote r | a

(k)x = the single premium payable by a person of age x for an r -year

deferred whole life annuity-immediate of 1k

u.c. per each k-th period of the year (payable at the end of each k-th period of the year while theperson survives from age x + r onward).

Proposition 7.6. For any x, r N s.t. x + r and any kNwe have

r | a (k)x =N x+ r +1

Dx+

k 12k

Dx+ rDx

. (7.13)


r |a (k)x =1k

r x

n =0

k

h=1r + n + hk

E x . (7.14)


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By (7.3) and (7.12) we obtain

r | a(k)x =1k

r x

n =0

k

h=1r E x n + hk E x+ r = r E x a

(k)x+ r ,

and using (7.1) and (7.11) we obtain

r | a(k)x =Dx+ rDx

N x+ r +1Dx+ r

+k 1

2k.

Corollary 7.6. Let x, r N, x + r , k N and T 0. The sin-gle premium payable by a person of age x for an r -year deferred whole life

annuity-immediate of T u.c. per each k-th period of the year is

T k r | a(k)x = T k N x+ r +1

Dx+

k 12

Dx+ rDx

.


r |a (k)x = r E x a(k)x+ r , x, r N s.t. x + r , kN, (7.15)

0|a(k)x = a

(k)x , x

N, x , kN

, x |a (k)x = 0 , xN, x , kN,

r |a (1)x = r |ax , x, r N, x .

7.5.3 Temporary life annuities with k-thly paymentsDenition 7.10. Let x, r N s.t. x + r and let kN. We denote

a(k)x: r = the single premium payable by a person of age x for an r -year

temporary life annuity-immediate of 1

ku.c. per each k-th period of the

year (payable at the end of each each k-th period of the year while theperson survives during the next r years).

Proposition 7.7. For any x, r N s.t. x + r and any kNwe havea (k)x: r =

N x+1 N x+ r +1Dx

+k 1

2k1

Dx+ rDx

. (7.16)


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Proof. By Theorem 6.1, (7.12) and (7.14) we have

a (k)x: r =1k

r 1

n=0

k

h=1n + hk

E x =1k

x

n =0

k

h=1n + hk

E x 1k

x

n = r

k

h=1n + hk

E x

=1k

x

n=0

k

h=1n + hk

E x 1k

r x

n=0

k

h=1r + n + hk

E x = a (k)x r | a(k)x ,

and using (7.11) and (7.13) we obtain the equality from enounce.

Corollary 7.7. Let x, r N, x + r , k N and T 0. The singlepremium payable by a person of age x for an r -year temporary life annuity-immediate of T u.c. per each k-th period of the year is

T k a(k)x: r = T k

N x+1 N x+ r +1Dx

+k 1

21

Dx+ rDx

.

Remark 7.7. By (7.11), (7.16), (7.13), (7.15) and (7.8) it follows that

a(k)x = a(k)x: r + r |a

(k)x , x, r N s.t. x + r , kN,

a(k)x: r = a(k)x r E x a

(k)x+ r , x, r N s.t. x + r , kN,

a (k)x: 0 = 0 , xN, x , kN,

a(k)

x: x = a(k)

x , xN

, x , kN

,a(1)x: r = ax: r , x, r N, x .

7.6 Pension

7.6.1 Annually pensionWe denote by r the number of years until the time of retirement.

Denition 7.11. Let x, r N s.t. x + r . We denote

P x: r (r | ax ) = the r -year temporary life premium payable by a person of age x (at the end of each year while the person survives during the next r years) for an r -year deferred whole life annually pension of 1 u.c. per each year (payable at the end of each year while the person survives from age x + r onward).

Proposition 7.8. For any x, r N s.t. x + r , we haveP x: r (r | ax ) =

r | axax: r

=N x+ r +1

N x+1 N x+ r +1. (7.17)


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Proof. For a mutually advantageous insurance, the present values (at theinitial moment of the insurance) of the premiums need to be equal to thepresent value of the pensions. By Denition 7.6, Corollary 7.4, Denition 7.5and Proposition 7.3 we have

P x: r (r |ax ) ax: r = r | ax , so P x: r (r |ax ) N x+1 N x+ r +1

Dx=

N x+ r +1Dx

,

and hence we obtain the equality from enounce.

Corollary 7.8. Let x, r N, x + r and T 0. The r -year temporary life premium payable by a person of age x for an r -year deferred whole lifeannually pension of T u.c. per each year is

T P x: r (r |ax ) = T N x+ r +1

N x+1 N x+ r +1.

7.6.2 Monthly pensionWe denote by r the number of years until the time of retirement.

Denition 7.12. Let x, r N s.t. x + r . We denote P

(12)x: r (r | a

(12)x ) = the r -year temporary life premium payable by a person

of age x at the end of each month (while the person survives during the next r years) for an r -year deferred whole life monthly pension of 1u.c. per each month (payable at the end of each month while the person survives from age x + r onward).

Proposition 7.9. For any x, r N s.t. x + r , we haveP (12)x: r (r | a

(12)x ) =

r | a(12)x

a (12)x: r=

24N x+ r +1 + 11 Dx+ r24(N x+1 N x+ r +1 ) + 11( Dx Dx+ r )

. (7.18)

Proof. For a mutually advantageous insurance, the present values (at theinitial moment of the insurance) of the premiums need to be equal to thepresent value of the pensions. By Denition 7.10, Corollary 7.7, Denition7.9 and Proposition 7.6 we have

P (12)x: r (r |a(12)x ) 12 a

(12)x: r = 12 r | a(12)x ,

so

P (12)x: r (r |a(12)x ) 12

N x+1 N x+ r +1Dx

+112

1 Dx+ rDx

= 12N x+ r +1

Dx+

112

Dx+ rDx

,

and hence we obtain the equality from enounce.


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Corollary 7.9. Let x, r

N, x + r

and T

0. The r -year temporary

life premium payable by a person of age x at the end of each month for an r -year deferred whole life monthly pension of T u.c. per each month is

T P (12)x: r (r |a

(12)x ) = T

24N x+ r +1 + 11 Dx+ r24(N x+1 N x+ r +1 ) + 11( Dx Dx+ r )

.

7.7 ProblemsExercise 7.1. Calculate the single premium payable by a 30 years old personfor a single claim of 10000$ over 35 years if the person survives. The annualinterest percent is 8%.

Exercise 7.2. Calculate the single premium payable by a 30 years old per-son for a whole life annuity-immediate of 12000RON per year. The annualinterest percent is 14%.

Exercise 7.3. Calculate the single premium payable by a 30 years old personfor a 35-year deferred whole life annuity-immediate of 12000RON per year.The annual interest percent is 14%.

Exercise 7.4. Calculate the single premium payable by a 30 years old personfor a 35-year temporary life annuity-immediate of 12000RON per year. The

annual interest percent is 14%.Exercise 7.5. Calculate the single premium payable by a 30 years old personfor a whole life annuity-immediate of 1000RON per month. The annualinterest percent is 14%.

Exercise 7.6. Calculate the single premium payable by a 30 years old personfor a 35-year deferred whole life annuity-immediate of 1000RON per month.The annual interest percent is 14%.

Exercise 7.7. Calculate the single premium payable by a 30 years old personfor a 35-year temporary life annuity-immediate of 1000RON per month. Theannual interest percent is 14%.

Exercise 7.8. Calculate the annuity-immediate premium payable by a 30years old person for an annuity-immediate pension of 12000RON per year.The annual interest percent is 14% and the age of retirement is 65 years.

Exercise 7.9. Calculate the monthly-immediate premium payable by a 30years old person for a monthly pension of 1000RON per month. The annualinterest percent is 14% and the age of retirement is 65 years.


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Theme 8

Life insurances

8.1 A general model. ClassicationIn a life insurance, the single claim is payable at the moment of death, if thedeath occurs in the period covered by the insurance. The life insurance canbe:

immediate and unlimited , when the claim is payed at the momentof death, whenever this occurs.

deferred , when the claim is payed only if the insured dies after a xedterm from the time of insurance issue. temporary (limited) , when the claim is payed only if the insureddies within a xed term from the time of insurance issue.

8.2 Whole life insuranceDenition 8.1. Let xN, x . We denote

Ax = the single premium payable by a person of age x for a whole lifeinsurance of 1 u.c. (payable at the moment of death, whenever thisoccurs).

Proposition 8.1. For any xN, x , we have

Ax =M xDx

, (8.1)

where

M x = C x + C x+1 + + C , with C x = dx vx+12 = ( lx lx+1 )vx+

12 , (8.2)

95


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THEME 8. LIFE INSURANCES 96

v =1

1 + ibeing the annual discounting factor, i being the annual interest

rate.

Proof. For a mutually advantageous insurance, the single premium A(x) needto be equal to the present value of the single claim, that is

Ax = E (X ),

where X is the random variable that represents the present value of the claim.Assuming that the deaths are uniform distributed throughout the year, wehave

X = vn +12 , if n is the number of complete years lived by the insured since issue ,

for any n {0, . . . , x}. Hence the distribution of X isX : v

12 . . . vn +

12 . . . v x+

12

x (0) . . . x (n) . . . x ( x),

where, for any n {0, . . . , x}, x (n) represents the probability that aperson of age x will live only n (i.e. will die at age x + n). By (6.18) we have x(n) =

dx+ n

lx,

n

{0, . . . ,

x

},

where dx+ n = lx+ n lx+ n +1 represents the number of deaths at age x + n.Using (8.2) it follows that

Ax = E (X ) = x

n =0

x (n)vn +12 =

x

n =0

dx+ nlx v

n + 12 = x

n =0

dx+ n vx+ n +12

lx vx

= x

n =0

C x+ nDx

=M xDx

.

Corollary 8.1. Let x N, x and let T 0. The single premium payable by a person of age x for a whole life insurance of T u.c. isT

Ax = T

M xDx

.

Corollary 8.2. For any xN, x , we have

Ax = v (1 i ax ) . (8.3)


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8.3 Deferred life insuranceDenition 8.2. Fie x, r N s.t. x + r . We denote

r |

Ax = the single premium payable by a person of age x for a r -year deferred life insurance of 1 u.c. (payable at the moment of death only if the insured die at least r years following insurance issue).

Proposition 8.2. For any x, r N s.t. x + r , we haver |

Ax =

M x+ rDx

. (8.4)

Proof. Similar to Proposition 8.1 we have

r |

Ax = E (r | X ),

where r X is the random variable having the distribution

r | X :0 0 . . . 0 vr +

12 vr +1+

12 . . . v x+

12

x (0) x(1) . . . x (r 1) x (r ) x (r + 1) . . . x ( x).

Using (6.18) and (8.2) we obtain that

r |

Ax = E (r |X ) =

x

n = r x (n)vn + 12

=

x

n = r

dx+ nlx v

n + 12

=

x

n = r

dx+ n

vx+ n +

12

lx vx=

x

n = r

C x+ nDx

=M x+ r

Dx.

Corollary 8.3. Let x, r N, x + r , T 0. The single premium payableby a person of age x for a r -year deferred life insurance of T u.c. isT r |

Ax = T

M x+ rDx

.


r |

Ax = r E x

Ax+ r , x, r N s.t. x + r , (8.5)0|

Ax =

Ax , xN, x .


r |

Ax = v r E x i r | ax , x, r N s.t. x + r . (8.6)


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8.4 Temporary life insuranceDenition 8.3. Let x, r N s.t. x + r . We denote

A1x: r

= the single premium payable by a person of age x for a r -year term life insurance of 1 u.c. (payable at the moment of death only if the insured die within r years following insurance issue).

Proposition 8.3. For any x, r N s.t. x + r , we have

A1x: r =M x M x+ r

Dx. (8.7)

Proof. Similar to Proposition 8.1 we have

A1x: r

= E (X r ),

where X r is the random variable having the distribution

X r :v

12 v1+

12 . . . vr 1+

12 0 0 . . . 0

x (0) x (1) . . . x (r 1) x (r ) x (r + 1) . . . x ( x).

Using(6.18) and (8.2) we obtain

A1

x: r= E (X r ) =

r 1

n=0

x(n)vn +12 =

r 1

n =0

dx+ nlx v

n+ 12 =r 1

n =0

dx+ n vx+ n +12

lx vx

=r 1

n =0

C x+ nDx

=M x M x+ r

Dx.

Corollary 8.4. Let x, r N, x + r , T 0. The single premium payableby a person of age x for a r -year term life insurance of T u.c. isT

A1

x: r= T

M x M x+ rDx

.


Ax = A1x: r + r |Ax , x, r N s.t. x + r , (8.8)

A1

x: r=

Ax r E x

Ax+ r , x, r N s.t. x + r ,

A1

x: 0= 0 , xN, x .


A1x: r

= v 1 r E x i ax: r , x, r N s.t. x + r . (8.9)


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8.5 ProblemsExercise 8.1. Calculate the single premium payable by a 30 years old personfor a whole life insurance of 1000RON. The annual interest percent is 14%.

Exercise 8.2. Calculate the single premium payable by a 30 years old personfor a 35-year deferred life insurance of 1000RON. The annual interest percentis 14%.

Exercise 8.3. Calculate the single premium payable by a 30 years old personfor a 35-year term life insurance of 1000RON. The annual interest percent is14%.


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Theme 9

Collective annuities and

insurancesNext, we consider an insured group of m persons having the ages x1, x2, . . . , x m(mN

, x j N, x j j {1, . . . , m }).

9.1 Multiple life probabilitiesDenition 9.1. Let a group of m persons having the ages x1, x2, . . . , x m ,

where mN

, x j N, x j j {1, . . . , m }. Let n, k N s.t. k m.We denote

n px1 x2 ...x m = the probability that all members of the group will surviven years; n p [k ]x1 x2 ...x m = the probability that exactly k of the group members will

survive n years;

n p kx1 x2 ...x m = the probability that at least k of the group members will survive n years;

n px1 x2 ...x m is called the probability of joint survival ( probability of the joint-life ) for the group, and n p [k ]

x1 x2 ...x mand n p k

x1 x2 ...x mare called probabil-

ities of partial survival for the group.

Remark 9.1. We assume that the deaths of the group members are indepen-dent.

Denition 9.2. We denote by x the maximum age of the group , i.e.

x = max {x1, x2, . . . , x m}.

100


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THEME 9. COLLECTIVE ANNUITIES AND INSURANCES 101

Also, we denote by x the minimum age of the group , i.e.

x = min {x1, x2, . . . , x m}.Remark 9.2. Obviously, if n > x then n px1 x2 ...x m = 0 .Proposition 9.1. Let n, k N s.t. k m. We have:

n px1 x2 ...x m = n px1 n px2 . . . n pxm =lx1 + nlx1

lx2 + nlx2 . . .

lxm + nlxm

; (9.1)

n p [k ]x1 x2 ...x m

=m k

s=0

(1)s C sk+ s1 i1


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9.3 Single claims for partial survivalDenition 9.4. Let a group of m persons having the ages x1, x2, . . . , x m ,where mN

, x j N, x j j {1, . . . , m }. Let n, k N s.t. k m.We denote n E [k ]x1 ,x 2 ,...,x m = the single premium payable by the group for a single

claim of 1 u.c. over n years if exactly k of the members survive;

n E kx1 ,x 2 ,...,x m = the single premium payable by the group for a singleclaim of 1 u.c. over n years if at least k of the members survive.


n E [k ]x1 ,x 2 ,...,x m

=m k

s=0

(1)s C sk+ s1 i1


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ax1 ,x 2 ,...,x m = the single premium payable by the group for a whole joint-life annuity-immediate of 1 u.c. per year (payable at the end of each year while all of the members survive).


ax1 ,x 2 ,...,x m =N x1 +1 ,x 2 +1 ,...,x m +1

Dx1 ,x 2 ,...,x m, (9.8)

where

N x1 ,x 2 ,...,x m = x

n =0

Dx1 + n,x 2 + n,...,x m + n , (9.9)

x = max {x1, x2, . . . , x m}being the maximum age of the group.Corollary 9.3. Let a group of m persons having the ages x1, x2, . . . , x m ,where m N

, x j N, x j j {1, . . . , m }. Let T 0. The singlepremium payable by the group for a whole joint-life annuity-immediate of T u.c. per year is

T ax1 ,x 2 ,...,x m = T N x1 +1 ,x 2 +1 ,...,x m +1

Dx1 ,x 2 ,...,x m.

9.5 Whole life annuities for partial survivalDenition 9.6. Let a group of m persons having the ages x1, x2, . . . , x m ,where mN

, x j N, x j j {1, . . . , m }. Let kN s.t. k m. Wedenote a [k ]x1 ,x 2 ,...,x m = the single premium payable by the group for a whole life

annuity-immediate of 1 u.c. per year payable (at the end of each year)while exactly k of the members survive;

a kx1 ,x 2 ,...,x m = the single premium payable by the group for a whole lifeannuity-immediate of 1 u.c. per year payable (at the end of each year)while at least k of the members survive.


a [k ]x1 ,x 2 ,...,x m

=m k

s=0

(1)s C sk+ s1 i1


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Corollary 9.4. Let a group of m persons having the ages x1, x2, . . . , x m ,where m

N, x j N, x j j {1, . . . , m }. Let kN s.t. k m and let T 0.

1. The single premium payable by the group for a whole life annuity-immediate of T u.c. per year payable while exactly k of the memberssurvive is

T a [k ]x1 ,x 2 ,...,x m = T m k

s=0

(1)s C sk+ s1 i1


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9.7 Deferred whole life annuities for partialsurvival

Denition 9.8. Let a group of m persons having the ages x1, x2, . . . , x m ,where mN

, x j N, x j j {1, . . . , m }. Let r N s.t. x + r and let kN s.t. k m. We denote r | a [k ]x1 ,x 2 ,...,x m = the single premium payable by the group for an r -year

deferred whole life annuity-immediate of 1 u.c. per year payable (after r -years, at the end of each year) while exactly k of the members survive;

r | a kx1 ,x 2 ,...,x m = the single premium payable by the group for an r -year deferred whole life annuity-immediate of 1 u.c. per year payable (after r -years, at the end of each year) while at least k of the members survive.


r |a [k ]x1 ,x 2 ,...,x m

=m k

s=0

(1)s C sk+ s1 i1


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9.8 Temporary life annuities for joint survivalDenition 9.9. Let a group of m persons having the ages x1, x2, . . . , x m ,where mN

, x j N, x j j {1, . . . , m }. Let r N s.t. x + r .We denote ax1 ,x 2 ,...,x m : r = the single premium payable by the group for an r -year temporary joint-life annuity-immediate of 1 u.c. per year (payable at

the end of each year while all of the members survive during the next ryears).


ax1 ,x 2 ,...,x m : r =N x1 +1 ,x 2 +1 ,...,x m +1 N x1 + r +1 ,x 2 + r +1 ,...,x m + r +1

Dx1 ,x 2 ,...,x m. (9.15)

Corollary 9.7. Let a group of m persons having the ages x1, x2, . . . , x m ,where mN

, x j N, x j j {1, . . . , m }. Let r N s.t. x + r and let T 0. The single premium payable by the group for an r -year temporary joint-life annuity-immediate of T u.c. per year isT ax1 ,x 2 ,...,x m : r = T

N x1 +1 ,x 2 +1 ,...,x m +1 N x1 + r +1 ,x 2 + r +1 ,...,x m + r +1Dx1 ,x 2 ,...,x m

.

9.9 Temporary life annuities for partial sur-vival

Denition 9.10. Let a group of m persons having the ages x1, x2, . . . , x m ,where m N

, x j N, x j j {1, . . . , m }. Let r N s.t. x + r and let kN s.t. k m. We denote a [k ]x1 ,x 2 ,...,x m : r = the single premium payable by the group for an r -year

temporary life annuity-immediate of 1 u.c. per year payable (at the end of each year) while exactly k of the members survive (during the next ryears);

a kx1 ,x 2 ,...,x m : r = the single premium payable by the group for an r -year temporary life annuity-immediate of 1 u.c. per year payable (at the end of each year) while at least k of the members survive (during the next r years).


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a [k ]x1 ,x 2 ,...,x m : r

=m k

s=0

(1)s C sk+ s1 i1


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x = max

{x1, x2, . . . , x m

}being the maximum age of the group, with

C x1 ,x 2 ,...,x m = ( lx1 lx2 . . . lxm lx1 +1 lx2 +1 . . . lxm +1 ) vx 1 + x 2 + ... + x m

m +12 ,

(9.20)

v =1

1 + ibeing the annual discounting factor, i being the annual interest

rate.

Corollary 9.9. Let a group of m persons having the ages x1, x2, . . . , x m ,where m N

, x j N, x j j {1, . . . , m }. Let T 0. The singlepremium payable by the group for an insurance of T u.c. payable at themoment of the rst death, whenever this occurs, is

T

Ax1 ,x 2 ,...,x m = T M x1 ,x 2 ,...,x mDx1 ,x 2 ,...,x m

.

9.11 Group insurance payable at the k-th deathDenition 9.12. Let a group of m persons having the ages x1, x2, . . . , x m ,where mN

, x j N, x j j {1, . . . , m }. Let kN s.t. k m. Wedenote

A [k ]x1 ,x 2 ,...,x m

= the single premium payable by the group for an insurance

of 1 u.c. payable at the moment of the k-th death, whenever this occurs.


A [k ]

x1 ,x 2 ,...,x m=

k 1

s=0

(1)s C sm k+ s1 i1


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9.12 ProblemsExercise 9.1. Consider a group of 4 members of 55, 53, 30, and 28 yearsold.a) Calculate the probability that all of the members survive 15 years.b) Calculate the probability that exactly 2 of the members survive 25 years.c) Calculate the probability that at least 3 of the members survive 20 years.d) Calculate the probability that at most 3 of the members survive 10 years.

Exercise 9.2. Calculate the single premium payable by a family of twopersons of 32 and 30 years old for a single claim of 20000$ over 35 years if both members will be alive. The annual interest percent is 12%.

Exercise 9.3. Calculate the single premium payable by a family of twopersons of 32 and 30 years old for a single claim of 20000$ over 35 years if just one member will be alive. The annual interest percent is 12%.

Exercise 9.4. Calculate the single premium payable by a family of twopersons of 32 and 30 years old for a single claim of 20000$ over 35 years if atleast one member will be alive. The annual interest percent is 12%.

Exercise 9.5. Calculate the single premium payable by a family of threepersons of 46, 44 and 22 years old for a life annuity-immediate of 10000$ per

year while all of the members survive. The annual interest percent is 12%.Exercise 9.6. Calculate the single premium payable by a family of threepersons of 46, 44 and 22 years old for a life annuity-immediate of 10000$ peryear while exactly two of the members survive. The annual interest percentis 12%.

Exercise 9.7. Calculate the single premium payable by a family of threepersons of 46, 44 and 22 years old for a life annuity-immediate of 10000$ peryear while at least two of the members survive. The annual interest percentis 12%.

Exercise 9.8. Calculate the single premium payable by a family of twopersons of 42 and 37 years old for a 10-year deferred life annuity-immediateof 10000$ per year while all of the members survive. The annual interestpercent is 12%.

Exercise 9.9. Calculate the single premium payable by a family of twopersons of 42 and 37 years old for a 10-year deferred life annuity-immediateof 10000$ per year while just one member survives. The annual interestpercent is 12%.


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Exercise 9.10. Calculate the single premium payable by a family of twopersons of 42 and 37 years old for a 10-year deferred life annuity-immediateof 10000$ per year while at least one member survives. The annual interestpercent is 12%.

Exercise 9.11. Calculate the single premium payable by a family of threepersons of 60, 54 and 35 years old for a 30-year temporary life annuity-immediate of 10000$ per year while all of the members survive. The annualinterest percent is 12%.

Exercise 9.12. Calculate the single premium payable by a family of threepersons of 60, 54 and 35 years old for a 30-year temporary life annuity-immediate of 10000$ per year while just one member survives. The annualinterest percent is 12%.

Exercise 9.13. Calculate the single premium payable by a family of threepersons of 60, 54 and 35 years old for a 30-year temporary life annuity-immediate of 10000$ per year while at least one member survives. Theannual interest percent is 12%.

Exercise 9.14. Calculate the single premium payable by a family of twopersons of 28 and 25 years old for an insurance of 50000$ payable at themoment of the rst death. The annual interest percent is 12%.

Exercise 9.15. Calculate the single premium payable by a family of twopersons of 28 and 25 years old for an insurance of 50000$ payable at themoment of the last death. The annual interest percent is 12%.

Exercise 9.16. Calculate the single premium payable by a family of threepersons of 50, 49 and 25 years old for an insurance of 50000$ payable at themoment of the rst death. The annual interest percent is 12%.

Exercise 9.17. Calculate the single premium payable by a family of threepersons of 50, 49 and 25 years old for an insurance of 50000$ payable at the

moment of the second death. The annual interest percent is 12%.

Exercise 9.18. Calculate the single premium payable by a family of threepersons of 50, 49 and 25 years old for an insurance of 50000$ payable at themoment of the last death. The annual interest percent is 12%.


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Theme 10

Bonus-Malus system in

automobile insurance

10.1 A general modelThe Bonus-Malus system is the most well known system of goods insurance,especially car insurance. In this type of insurance, policies are categorizedbased on characteristics of the insured vehicle (the insured good), and onBonus-Malus level , given by the previous number of claims. The insurance

period for goods is usually one year. In this case, one policy remains in acertain payment class for one year and then it can be transferred to anotherpayment class, based on the number of accidents from the previous year. If the insured vehicle didnt have any accident, then the new payment classwill be better, so the premium will be reduced ( bonus ). As the numberof accidents grows, the new class will be worst, so the premium will beincreased (malus ).

Denition 10.1. A Bonus-Malus insurance system can be represented as S = ( C,D,T, ), where:

C =

{1, . . . , c

}represents the set of payment classes ( c

N). If

i > j , i, j C , we say thai i is a better class than j .

D = {0, . . . , r }represents the set of annual number of accidentspossible for an insurance policy ( rN). T : C D C is a function called the rule of passing of the system; for any iC and j D, T (i, j ) represents the payment class in which it will be transferred the next year every insurance policy from class

i that had j accidents during the current year. The function T (i, j )increases in i (for any xed j ) and decreases in j (for any xed i).

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