stt455_Actuarial Models.pdf

8/11/2019 stt455_Actuarial Models.pdf

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STT 455-6: Actuarial Models

Albert Cohen

Actuarial Sciences ProgramDepartment of Mathematics

Department of Statistics and ProbabilityA336 Wells Hall

Michigan State UniversityEast Lansing MI

[email protected]@stt.msu.edu

Albert Cohen (MSU) STT 455-6: Actuarial Models MSU 2011-12 1 / 283


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Copyright Acknowledgement

Many examples and theorem proofs in these slides, and on in class exampreparation slides, are taken from our textbook Actuarial Mathematics forLife Contingent Risks by Dickson,Hardy, and Waters.

Please note that Cambridge owns the copyright for that material.No portion of the Cambridge textbook material may be reproduced

in any part or by any means without the permission of thepublisher. We are very thankful to the publisher for allowing postingof these notes on our class website.

Also, we will from time-to-time look at problems from released

previous Exams MLC by the SOA. All such questions belong incopyright to the Society of Actuaries, and we make no claim onthem. It is of course an honor to be able to present analysis of suchexamples here.



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Survival Models

An insurance policy is a contract where the policyholder pays apremiumto the insurer in return for a benefitor payment later.

The contract specifies what event the payment is contingenton. This

event may be random in nature

Assume that interest rates are deterministic, for now

Consider the case where an insurance company provides a benefitupon death of the policyholder. This time is unknown, and so the

issuer requires, at least, a model of of human mortality



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Survival Models

Define (x) as a human at age x. Also, define that persons future lifetimeas the continuous random variable Tx. This means that x+Tx representsthat persons age at death.

Define the lifetime distribution

Fx(t) = P[Tx t] (1)the probabiliity that (x) does not survive beyond age x+tyears, and its

complement, the survival function Sx(t) = 1 Fx(t).



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Conditional Equivalence

We have an important conditional relationship

P[Tx t] = P[T0 x+t| T0 >x]



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Conditional Equivalence

We have an important conditional relationship

P[Tx t] = P[T0 x+t| T0 >x]

= P[x x]

(2)

and so

Fx(t) =F0(x+t) F0(x)

1 F0(x)Sx(t) =

S0(x+t)

S0(x)

(3)

In general we can extend this to

Sx(t+u) =Sx(t)Sx+t(u) (4)



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Conditions and Assumptions

Conditions on Sx(t)

Sx(0) = 1

limtSx(t) = 0 for all x 0Sx(t1)

Sx(t2) for all t1

t2 and x

0

Assumptions on Sx(t)

ddt

Sx(t) existst R+limtt Sx(t) = 0 for all x 0limtt2 Sx(t) = 0 for all x 0

The last two conditions ensure that E[Tx] and E[T2x] exist, respectively.



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Example 2.1

Assume that F0(t) = 1

1 t120

16 for 0 t 120. Calculate the

probability that

(0) survives beyond age 30

(30) dies before age 50




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Example 2.1


1 t120

16 for 0 t 120. Calculate the

probability that


(30) dies before age 50


P[(0) survives beyond age 30] =S0(30) = 1 F0(30)

=

1 30

120

16

= 0.9532

P

[(30) dies before age 50] = F30(20)=

F0(50) F0(30)1 F0(30) = 0.0410

P[(40) survives beyond age 65] = S40(25) =S0(65)

S0(40)= 0.9395

(5)



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The Force of Mortality

Recall from basic probability that the density ofFx(t) is defined asfx(t) :=

ddt

Fx(t).



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Recall from basic probability that the density ofFx(t) is defined asfx(t) :=

ddt

Fx(t).

It follows that

f0(x) := d

dxF0(x) = lim

dx0+

F0(x+dx) F0(x)dx

= limdx0+

P[x


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However, we can find the conditional density, also known as the Force of

Mortalityvia

x= limdx0+

P[x x]dx

= limdx0+P[Tx

dx]

dx = limdx0+1

Sx(dx)

dx

= limdx0+

1 Sx(dx)dx

= limdx0+

1

S0(x+dx)S0(x)

dx =

1

S0(x) limdx0+S0(x)

S0(x+dx)

dx

= 1S0(x)

d

dxS0(x) =

f0(x)

S0(x)

(7)


Th F f M li


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In general, we can show

x+t= 1Sx(t)

d

dtSx(t) =

fx(t)

Sx(t) (8)

and integration of this relation leads to

Sx(t) =S0(x+t)

S0(x)

=e

x+t0 sds

ex

0 sds

=e

x+tx sds

=et

0 x+sds

(9)


E l 2 2


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Example 2.2


1 t120 1

6 for 0 t 120. Calculate x


E l 2 2


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Example 2.2


1 t120 1

6 for 0 t 120. Calculate xd

dxS0(x) =

1

6 1

x

120 5

6 1

120x= 1

1 x120 1

6

1

6

1 x120

56

1120

= 1

720

6x

(10)


G t L / M k h L


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Gompertz Law / Makehamss Law

One model of human mortality, postulated by Gompertz, is x=Bcx,

where (B, c) (0, 1) (1, ). This is based on the assumption thatmortality is age dependent, and that the growth rate for mortality is

proportional to its own value. Makeham proposed that there should alsobe an age independent component, and so Makehams Law is

x=A+Bcx (11)


G t L / M k h ss L


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Gompertz Law / Makehams s Law

Of course, when A= 0, this reduces back to Gompertz Law.


Gompertz Law / Makehamss Law
http://www.cdc.gov/nchs/data/nvsr/nvsr51/nvsr51_03.pdf


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Gompertz Law / Makehams s Law

Of course, when A= 0, this reduces back to Gompertz Law.By definition,

Sx(t) =ex+t

x sds =e

x+tx

(A+Bcs)ds

=eAt Bln ccx(ct1)(12)

Keep in mind that this is a multivariable function of (x, t) R2+

Some online resources:

CDC National Vital Statistics report, Dec. 2002 .


Comparison with US Govt data
http://www.cdc.gov/nchs/data/nvsr/nvsr51/nvsr51_03.pdfhttp://www.cdc.gov/nchs/data/nvsr/nvsr51/nvsr51_03.pdf


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Comparison with US Gov t data


Actuarial Notation


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Actuarial Notation

Actuaries make the notational conventions

tpx= P[Tx >t] =Sx(t)

tqx= P[Tx

t] =Fx(t)

u|tqx= P[u


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Actuarial Notation-Relationships

Consequently,

tpx+tqx= 1

u|tqx= upx u+tpxt+upx= tpx upx+t

x=1

xp0

d

dx(xp0)

(14)

Similarly,

x+t=1

tpx

d

dt

tpx

d

dt

tpx=x+t

tpx

x+t= fx(t)

Sx(t) fx(t) =x+t tpx

tpx=et

0 x+sds

(15)




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Actuarial Notation Relationships

Also, since Fx(t) =t

0 fx(s)ds, we have as a linear approximation

tqx=

t0

spx x+sds




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Actuarial Notation Relationships

Also, since Fx(t) =t

0 fx(s)ds, we have as a linear approximation

tqx=

t0

spx x+sds

qx= 1

0 spx

x+sds

=

10

es

0 x+vdv x+sds

1

0

x+sds

x+ 12

(16)



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Mean and Standard Deviation of Tx


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Mean and Standard Deviation ofTx

Actuaries make the notational definition ex := E[Tx], also known as the

complete expectation of life. Recall fx(t) = tpx x+t= d

dttpx, and

ex=

0

t fx(t)dt

=

0

t

tpx

x+tdt

=

0

t ddt

tpxdt

=

0

tpxdt


Mean and Standard Deviation of Tx


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Mean and Standard Deviation ofTx

Actuaries make the notational definition ex := E[Tx], also known as the

complete expectation of life. Recall fx(t) = tpx x+t= d

dttpx, and

ex=

0

t fx(t)dt

=

0

t

tpx

x+tdt

=

0

t ddt

tpxdt

=

0

tpxdt

E[T2x] =

0

t2 fx(t)dt=

02t tpxdt



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Example 2.6


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p.

Assume thatF0(x) = 1

1 x120 1

6 for 0 x 120. Calculate ex, V[Tx]fora.)x= 30 and b.)x= 80.


Example 2.6


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p

Assume thatF0(x) = 1

1 x120 1

6 for 0 x 120. Calculate ex, V[Tx]fora.)x= 30 and b.)x= 80.

Since S0(x) = 1 x120

16 , it follows that in keeping with the model where

survival is constrained to be les than 120,

tpx =S0(x+t)

S0(x) =

1 t120x

16

:x+t 1200 :x+t>120


Example 2.

6


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p

So,

ex=

120x0

1 t

120 x 1

6

dt=6

7 (120 x)

E[T2

x

] = 120x0

2t 1

t

120 x16

dt

=

6

7 6

13

2(120 x)2

(18)

and

(e30, e80) = (77.143, 34.286)

(V[T30], V[T80]) =

(21.396)2, (9.509)2 (19)


Exam MLC Spring 2007: Q8


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Kevin and Kira excel at the newest video game at the local arcade,Reversion. The arcade has only one station for it. Kevin is playing. Kira isnext in line. You are given:

(i) Kevin will play until his parents call him to come home.

(ii) Kira will leave when her parents call her. She will start playing assoon as Kevin leaves if he is called first.

(iii) Each child is subject to a constant force of being called: 0.7 perhour for Kevin; 0.6 per hour for Kira.

(iv) Calls are independent.

(v) If Kira gets to play, she will score points at a rate of 100,000 perhour.




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Calculate the expected number of points Kira will score before she leaves.

(A) 77,000

(B) 80,000

(C) 84,000

(D) 87,000

(E) 90,000




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Define

tpx= P [Kevin still there]

tpy = P [Kira still there](20)

and so

E [Kiras playing time] =

0(1 tpx) tpydt

=

0

1 e0.7t

e0.6tdt

=

0

e0.6t e1.3t dt

= 1

0.6 1

1.3= 0.89744 hrs.

(21)




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It follows that

E[Kiras winnings] = 100000

$

hrE

[Kiras playing time]= $89744.

(22)

Hence, we choose (E).


Numerical Considerations for Tx


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In general, computations for the mean and SD for Txwill requirenumerical integration. For example,

Table: 2.1: Gompertz Model Statistics: (B, c) = (0.0003, 1.07)

x ex SD[Tx] x+ ex0 71.938 18.074 71.938

10 62.223 17.579 72.22320 52.703 16.857 72.70330 43.492 15.841 73.49240 34.252 14.477 74.75250 26.691 12.746 76.691

60 19.550 10.693 79.55070 13.555 8.449 83.55580 8.848 6.224 88.84890 5.433 4.246 95.433100 3.152 2.682 103.152


Curtate Future Lifetime


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Define

Kx := Tx (23)and so

P [Kx=k] = P [k Tx


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E [Kx] :=ex =

k=0

k

P [Kx=k]

=

k=0

k (kpx k+1px)

=

k=1

kpxby telescoping series..


Curtate Future Lifetime


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E [Kx] :=ex =

k=0

k

P [Kx=k]

=

k=0

k (kpx k+1px)

=

k=1

kpxby telescoping series..

E

K2x

=

k=0

k2 P [Kx=k]

= 2 k=1

k kpx k=1

kpx

= 2

k=1

k kpx ex

(25)


Relationship between ex and ex


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Recall

ex=

0tpxdt=

j=0

j+1

jtpxdt (26)

By trapezoid rule for numerical integration, we obtain

j+1j

tpxdt

1

2(jpx+j

+1px), and so

ex

j=0

1

2(jpx+ j+1px)

=1

2+

j=1

jpx=1

2+ex

(27)

As with all numerical schemes, this approximation can be refined whennecessary.


Comparison of ex and ex


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Approximation matches well for small values ofx

Table: 2.2: Gompertz Model Statistics: (B, c) = (0.0003, 1.07)

x ex ex0 71.438 71.93810 61.723 62.223

20 52.203 52.70330 42.992 43.49240 34.252 34.75250 26.192 26.69160 19.052 19.550

70 13.058 13.5580 8.354 8.84890 4.944 5.433100 2.673 3.152


Notes


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An extension of Gompertz - Makeham Laws is the GM(r, s) formulax=h

1r(x) +e

h2s(x), where h1r(x), h2s(x) are polynomials of degree r

and s, respectively.Hazard rate in survival analysis and failure rate in reliability theory isthe same as what actuaries call force of mortality.



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Life Tables


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It follows that

lx+t=lx0 x+tx0 px0=lx0 xx0 px0 tpx=lx tpx


Life Tables


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It follows that

lx+t=lx0 x+tx0 px0=lx0 xx0 px0 tpx=lx tpx

tpx= lx+t

lx

(29)

We assume a binomial model where Lt is the number of survivors to age

x+t.


Life Tables


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So, if there are lx independent individuals aged x with probability tpx ofsurvival to age x+t, then we interpret lx+tas the expected number ofsurvivors to age x+tout of lx independent individuals aged x.Symbolically,

E[Lt| L0 =lx] =lx+t=lx tpx (30)


Life Tables


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So, if there are lx independent individuals aged x with probability tpx ofsurvival to age x+t, then we interpret lx+tas the expected number ofsurvivors to age x+tout of lx independent individuals aged x.Symbolically,

E[Lt| L0 =lx] =lx+t=lx tpx (30)Also, define the expected number of deaths from year x to year x+ 1 as

dx :=lx

lx+1 =lx1

lx+1

lx=lx (1 px) =lxqx (31)


Example 3.

1


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Table: 3.1: Extract from a life table

x lx dx30 10000.00 34.7831 9965.22 38.10

32 9927.12 41.7633 9885.35 45.8134 9839.55 50.2635 9789.29 55.1736 9734.12 60.56

37 9673.56 66.4938 9607.07 72.9939 9534.08 80.11


Example 3.

1


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Calculate:

a.) l40

b.) 10p30

c.) q35

d.) 5q30

e.) P [(30) dies between age 35 and 36]



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Example 3.

1


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Calculate:

a.) l40=l39 d39 = 9453.97b.) 10p30=

l40l30

= 0.94540


Example 3.

1


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Calculate:

a.) l40=l39 d39 = 9453.97b.) 10p30=

l40l30

= 0.94540

c.) q35 = d35l35 = 0.00564


Example 3.

1


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Calculate:

a.) l40=l39 d39 = 9453.97b.) 10p30=

l40l30

= 0.94540

c.) q35 = d35l35 = 0.00564

d.) 5q30 = l30l35

l30= 0.02107



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Fractional Age Assumptions


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So far, the life table approach has mirrored the survival distributionmethod we encountered in the previous lecture. However, in detailing thelife table, no information is presented on the cohort in between wholeyears. To account for this, we must make some fractional age

assumptions. The following are equivalent:


Fractional Age Assumptions


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So far, the life table approach has mirrored the survival distributionmethod we encountered in the previous lecture. However, in detailing thelife table, no information is presented on the cohort in between wholeyears. To account for this, we must make some fractional age

assumptions. The following are equivalent:UDD1For all (x, s) N [0, 1), we assume that sqx=s qxUDD2For all x N, we assume

Rx :=Tx Kx U(0, 1)Rx is independent ofKx.


Proof of Equivalence


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Proof:

UDD1

UDD2: Assume for all (x, s)

N

[0, 1), we assume that

sqx=s qx. Then

P [Rx s] =

k=0

P [Rx s, Kx=k]

=

k=0

P [k Tx k+s]

=

k=0

kpx sqx+k=

k=0

kpx s qx+k

=s

k=0

kpx qx+k=s

k=0

P [Kx=k] =s

(32)

and soRx U(0, 1).Albert Cohen (MSU) STT 455-6: Actuarial Models MSU 2011-12 39 / 283

Proof of Equivalence


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To show independence ofRx and Kx,

P [Rx s, Kx=k] = P [k Tx k+s]= kpx sqx+k=s kpx qx+k= P[Rx s] P[Kx =k]

(33)

UDD2UDD1: Assuming UDD2 is true, then for (x, s) N [0, 1)we have

sqx= P [Tx s]

=P

[Kx= 0,

Rx s]= P[Rx s] P[Kx= 0]=s qx

(34)

QED


Corollary


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Recall that sqx= l

xlx+slx . It follows now that

sqx=sqx=sdxlx

= lx lx+s

lx


Corollary


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Recall that sqx= l

xl

x+slx . It follows now that

sqx=sqx=sdxlx

= lx lx+s

lxlx+s=lx

s

dx

which is a linear decreasing function of s [0, 1)


Corollary


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Recall that sqx= lxlx

+s

lx . It follows now that

sqx=sqx=sdxlx

= lx lx+s

lxlx+s=lx

s

dx

which is a linear decreasing function of s [0, 1)qx=

d

ds[sqx] =fx(s) = spx x+s

(35)

But, since qx is constant in s, we have fx(s) is constant for s

[0, 1).

Read over Examples 3.2 3.5


Constant Force of Mortality for Fractional Age


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For all (x, s) N [0, 1), we assume that x+sdoes not depend on s, andwe denote x+s :=

x. It follows that

px=e1

0 x+sds =e

x

spx=es

0

xdu =e

xs = (px)s

spx+t=es

0

xdu

= (px)s

when t+s


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HW: 3.1, 3.2, 3.4, 3.7, 3.8, 3.9, 3.10


Contingent Events


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We have spent the previous two lectures on modeling human mortality.The need for such models in insurance pricing arises when designingcontracts that are event-contingent. Such events include reachingretirement before the end of the underlying life (x) .

However, one can also write contracts that are dependent on a life (x)being admitted to college (planning for school), and also on (x)

s externalportfolios maintaining a minimal value over a time-interval (insuringexternal investments.)


Contingent Events: General Case


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Consider a probability space (, F,P) and an event A F.If we are working with a force of interest s() and the time of event Was W, then we have under the stated probability measure PtheExpected

Present Value of a payoffK() contingent upon W

EPV = E[K()eW

0 s()ds] (37)

Actuarial Encounters of the Third Kind !!



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Recall...


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The equivalent interest rate i :=e 1 per yearThe discount factor v := 11+i =e

per year

The nominal interest rate i(p) =p (1 +i)1p 1 compounded p

times per yearThe effective rate of discount d := 1 v=i v= 1 e per yearThe nominal rate of discount d(p) :=p

1 v1p

compounded p

times per year


Whole Life Insurance: Continuous Case

Consider now the random variable


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Consider now the random variable

Z=vTx

=eTx

(38)which represents the present value of a dollar upon death of (x). We areinterested in statistical measures of this quantity:

E[Z] =Ax := E[eTx

] =

0 et

tpxx+tdt

E[Z2] = 2Ax := E[e2Tx]

=

0e2ttpxx+tdt

Var(Z) = 2Ax

Ax

2

P[Z z] = P

Tx ln (z)

(39)


Whole Life Insurance: Yearly Case


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Assuming payments are made at the end of the death year, our randomvariable is now Z=vKx+1 =eKx and so

E[Z] =Ax := E[vKx+1] =

k=0

vk+1P[Kx=k]

=

k=0

vk+1

k|qx

E[Z2] =

k=0

v2k+2k|qx

Var(Z) =2

Ax (Ax)2

P[Z z] = P

Kx ln (z)

(40)


Whole Life Insurance: 1m thly Case


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Instead of only paying at the end of the last whole year lived, an insurance

contract might specify payment upon the end of the last period lived. Inthis case, if we split a year into m periods, and define

K(m)x =

1

mmTx (41)

For example, ifKx= 19.78, then

K(m)x =

19 m= 119 12 = 19.5 m= 219 3

4= 19.75 m= 4

19 912 = 19.75 m= 12


Whole Life Insurance: 1m thly Case


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It follows that we needr 0, 1m

, 2m

, ..., m1m

, 1, m+1m

,...

P K(m)x =r= P r Tx


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Recursion Method

One of the computational tools we share directly with quantitative finance


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One of the computational tools we share directly with quantitative financeis the method of backwards-pricing. In option pricing, we assume the

contract has a finite term. Here, we assume a finite lifetime maximum of < . It follows that

A1 = E

vK1+1

= E

v1

= v (45)


Recursion Method

One of the computational tools we share directly with quantitative finance


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One of the computational tools we share directly with quantitative financeis the method of backwards-pricing. In option pricing, we assume the

contract has a finite term. Here, we assume a finite lifetime maximum of < . It follows that

A1 = E

vK1+1

= E

v1

= v (45)

At age 2, we have P[K2 = 0] =q2 and soA2 = E

vK2+1

=q2

v+p2

E v(1+K1)+1

=q2 v+p2 v E

vK1+1

=q2 v+p2 v2(46)


Recursion Method


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In general, we have the recursion equation for a life (x) that satisfies

Ax=vqx+vpxAx+1

A1 =v(47)

in the whole life case, and

A(m)x =v

1m 1

mqx+v

1m 1

mpxA

(m)

x+ 1m

A(m)

1m

=v 1m

(48)

in the 1m

thly case.


Recursion Method


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Recall that for Makehams law we have, respectively

1m

px=e A

m Bc

x

ln (c)

c

1m1

1px=eA Bc

x

ln (c)(c1)

(49)

and for the power law of survival, S0(x) =

1 xa for some a, >01m

px=

x 1

m

x

a

1px=

x 1 x

a (50)


Recursion Method


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For any positive integer m, it follows that for Makehams law:

A(m)x =v

1m

1 e

Am Bc

x

ln (c)

c

1m1

+v 1m e

Am Bcxln (c)

c 1m1

A

(m)

x+ 1m

A(m)

1m

=v 1m

(51)



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Term Insurance: Continuous Case


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Consider now the case where payment is made in the continuous case, anddeath benefit is payable to the policyholder only if Tx n. Then, we areinterested in the random variable

Z =eTx1{Txn} (53)

and so

A1x:n = E[Z] =

n0

ettpxx+tdt

2

A1x:n = E

Z

2= n

0 e2t

tpxx+tdt

(54)


Term Insurance: 1m thly Case

Co side a ai the case he e the death be efit is a able at the e d of


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Consider again the case where the death benefit is payable at the end of

the

1

m

thly

period in the death year to the policyholder only ifK

(m)x +

1m n. Then, we are interested in the random variable

Z =e(K(m)x +

1m

)1K

(m)x +

1mn

(55)

and so

A(m)1x:n = E[Z] =mn1

k=0

v k+1

m km| 1

mqx

2A(m)1x:n = E

Z2

=mn1

k=0

v2k+2

m km| 1

mqx

(56)


Pure Endowment


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Pure endowment benefits depend on the survival policyholder (x) until atleast age x+n. In such a contract, a fixed benefit of 1 is paid at time n.This is expressed via

Z =en

1{Txn}

A 1x:n = E[Z] =vn

npx

=ennpx

(57)


Endowment Insurance

E d i i bi i f i d


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Endowment insurance is a combination of term insurance and pure

endowment. In such a policy, the amount is paid upon death if it occurswith a fixed term n. However, if (x) survives beyond n years, the suminsured is payable at the end of the nth year. The corresponding presentvalue random variable is

Z=e min{Tx,n}

E[Z] =Ax:n

=

n0

ettpxx+tdt+en

npx

=A1x:n +A

1x:n

(58)

This can be generalized once again to the 1m

thlycase and for E[Z2]


Deferred Insurance Benefits

Suppose policyholder on a life (x) receives benefit 1 ifu Tx


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( )Then

Z=eTx1{uTx


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By definition, we have

Ax=A1x:n +n|Ax

=A1x:n +vnnpxAx+n(60)


Relationships


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By definition, we have

Ax=A1x:n +n|Ax

=A1x:n +vnnpxAx+n(60)

What about relationship between Ax and Ax?


Employing UDD: Ax vs Ax


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If expected values are computed via information derived from life tables,

then certainly Axmust be approximated using techniques from previouslecture.




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If expected values are computed via information derived from life tables,

then certainly Axmust be approximated using techniques from previouslecture.

Recall that by the definition of spxand the UDD, we have

spxx+s=fx(s)

= d

dsP[Tx s]

= d

ds(sqx) =

d

ds(s qx)

=qx

(61)



It follows that using the UDD approximation leads to


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Ax=

0ettpxx+tdt=

k=0

k+1k

ettpxx+tdt

=

k=0

kpxvk+1

10

eesspx+kx+k+sds

k=0

kpxvk+1qx+k

1

0eesds

=

k=0

vk+1P[Kx=k] 1

0eesds=Ax

1

0eesds

=Ax i



It follows that using the UDD approximation leads to


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Ax=

0ettpxx+tdt=

k=0

k+1k

ettpxx+tdt

=

k=0

kpxvk+1

10

eesspx+kx+k+sds

k=0

kpxvk+1qx+k

1

0eesds

=

k=0

vk+1P[Kx=k] 1

0eesds=Ax

1

0eesds

=Ax i

A(m)x i

i(m)Ax=

i

m (1 +i) 1m 1 Ax

(62)

Albert Cohen (MSU) STT 455 6: Actuarial Models MSU 2011 12 65 / 283

Claims Accelaration Approach : A(m)x vs Ax


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Consider now a policy that pays the holder at the end of the 1mthly

periodof death. In this case, the benefit is paid at one of the times r where

r

Kx+ 1

m, Kx+

2

m, ..., Kx+

m

m

(63)

and so under the UDD,

E [Tpayment| Kx=k] =m

j=1

1

m

k+ j

m

= k+

m+ 1

2m (64)


Claims Accelaration Approach : A(m)x vs Ax

Once again, it follows using the UDD aproximation


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g g p

A(m)x = E[v

K(m)x +

1m ] =

k=0

v k+1

m km| 1

mqx

k=0

vE[Tpayment|Kx=k]P [Kx=k]

=

k=0

vk+m+1

2m k|qx= (1 +i)m1

2m

k=0

vk+1k|qx

= (1 +i)m1

2m

Ax

(1 +i)

12

Ax

(65)

as m . Note that using the UDD approximation, both AxAx

and A(m)x

Axareindependentofx.


Variable Insurance Benefits

Upon death, we have considered policies that pay the holder a fixed


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amount. What varied was the method and time of payment. If, however,

the actual payoff amount depended on the time Txof death for (x), thenwe term such a contract a Variable Insurance Contract.

Specifically, if the payoff amount dependent on Tx is h(Tx), then

Z=h(Tx)eTx

E[Z] =

0

h(t)ettpxx+tdt

IAx :=

0

tettpxx+tdt

IA

1x:n :=

n0

tettpxx+tdt

(66)


Example 4.

8

Consider an nyear term insurance issued to (x) under which the deathbenefit is paid at the end of the year of death The death benefit if death


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benefit is paid at the end of the year of death. The death benefit if death

occurs between ages x+k and x+k+ 1 is valued at (1 +j)k

. Hence,using the definition i := 1+i1+j 1,

Z=vKx+1(1 +j)Kx

E[Z] =

n1k=0

vk

+1(1 +j)k

k|qx

= 1

1 +j

n1k=0

vk+1(1 +j)k+1k|qx

= 1

1 +j

n1k=0

k|qx1+i1+j

k+1 = 11 +j A1x:n

(67)


Homework Questions


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HW: 4.1, 4.2, 4.3, 4.7, 4.9, 4.11, 4.12, 4.14, 4.15, 4.16, 4.17, 4.18


Life Annuities

A Life Annuity refers to a series of payments to or from an individual asl th t i till li F fi d t i d t ll


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long as that person is still alive. For a fixed rate iand term n , we recall

the deterministic pricing theory:

an i = 1 +v+...+vn1 =

1 vnd

an i =v+...+vn

= an i 1 +vn

=1

vn

i

an i =

n0

vtdt=1 vn

a(m)n i =

1

m 1 +v 1m +...+vn

1m= 1 v

n

d(m)

a(m)n i =

1

m

v 1m +...+vn

1m +vn

=

1 vni(m)

(68)


Whole Life Annuity Due

Consider the case where 1 is paid out at the beginning of every periodtil d th O t d i bl i


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until death. Our present random variable is now

Y := aKx+1 =1 vKx+1

d (69)

and so

ax= E[Y] = E

1 vKx+1d

=

1 Axd

(70)

V[Y] =V 1 vKx+1

d = 1

d2V[1] +

1

d2V[vKx+1]

= 0 +2Ax A2x

d2

(71)

Alb t C h (MSU) STT 455 6 A t i l M d ls MSU 2011 12 72 / 283

Whole Life Annuity Due

The present value random variable can also be represented as


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Y =

k=0

vk1{Tx>k} (72)

As P[Tx >k] = tpx, we have the alternate expression for ax

ax = E[Y] = E

k=0

vk1{Tx>k}

=

k=0E[vk1{Tx>k}]

=

k=0

vkkpx=

k=0

k|qxak+1

(73)

Alb t C h (MSU) STT 455 6 A t i l M d l MSU 2011 12 73 / 283

Term Annuity Due

Define the present value random variable

aK 1 : Kx {0 1 2 n 1}


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Y = aKx+1 :Kx {0, 1, 2, ..., n 1}an :Kx {n, n+ 1, n+ 2,...}

Another expression is

Y = amin{Kx+1,n} =1 vmin{Kx+1,n}

d (74)

and so

ax:n = E[Y] =1 E vmin{Kx+1,n}

d

=1

Ax:n

d

=n1t=0

vttpx=n1k=0

k|qxak+1 +npxan

(75)


Whole Life and Term Immediate Annuity


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Define Y

=

k=1vk

1{Tx>k}. Then we have an annuity immediate thatbegins payment one unit of time from now. It follows that

ax= ax 1V[Y] =V[Y]

(76)

Also, if we define Y =amin{Kx,n}, then

ax:n =n

t=1vttpx= ax:n 1 +vnnpx (77)


Whole Life Continuous Annuity


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Define

Y = aTx =1 vTx

=

0et1{Tx>t}dt

ax= E[Y] =1 Ax

=

0ettpxdt

(78)

Note that if= 0, then ax= ex

Alb C h (MSU) STT 455 6 A i l M d l MSU 2011 12 76 / 283

Term Continuous Annuity


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Define Y = amin{Tx,n} .Then

Y =1 vmin{Tx,n}

ax:n = E [Y] =1 Ax:n

=

n0

ettpxdt

(79)

Alb C h (MSU) STT 455 6 A i l M d l MSU 2011 12 77 / 283

Deferred Annuity

Consider now the case of an annuity for (x) that will pay 1 at the end ofeach year, beginning at age x+uand will continue until death agex + T We define |a to be the Expected Present Value of this policy It


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x+Tx. We define u|axto be the Expected Present Value of this policy. It

should be apparent that

u|ax= ax ax:u

=

t=u

vttpx

=vttpx

t=0

vttpx

=vttpxax+u

(80)

holds in the discrete case, and similarly in the continuous case,

u|ax = ax ax:u (81)Albert Cohen (MSU) STT 455-6: Actuarial Models MSU 2011-12 78 / 283

Term Deferred Annuity

For the cases of an annuity for (x) that will pay 1 at the end of each year,


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beginning at age x+uand will continue until death age x+Txup to aterm of length n, or annuity-due payable 1

m

thly. Then

u|ax:n =vu

upxax+u:n

u|a(m)x =v

uupxa

(m)x

+u

(82)

respectively.These combine with the previous slide to reveal the useful formulae:

ax:n = ax vnnpxax+na(m)x:n = a

(m)x vnnpxa(m)x+n (83)


Guaranteed Annuities

There are instances where an age (x) wishes to buy a policy wherepayments are guaranteed to continue upon death to a beneficiary. In this


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payments are guaranteed to continue upon death to a beneficiary. In this

case, define the present random variable as Y = an +Y1, where

Y1 =

0 :Kx {0, 1, 2, ..., n 1}aKx+1 an :Kx {n, n+ 1, n+ 2,...}

and so

E[Y1] = E

aKx+1 an

1{Kxn}

= n|ax=v

nnpxax+n


Guaranteed Annuities

There are instances where an age (x) wishes to buy a policy wherepayments are guaranteed to continue upon death to a beneficiary. In this


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payments are guaranteed to continue upon death to a beneficiary. In this

case, define the present random variable as Y = an +Y1, where

Y1 =

0 :Kx {0, 1, 2, ..., n 1}aKx+1 an :Kx {n, n+ 1, n+ 2,...}

and so

E[Y1] = E

aKx+1 an

1{Kxn}

= n|ax=v

nnpxax+n

E[Y] := ax:n = an +v

n

npxax+n

and E[Y(m)] := a(m)

x:n = a

(m)n +v

nnpxa

(m)x+n

(84)


Example 5.

4


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A pension plan member is entitled to a benefit of 1000 per month, inadvance, for life from age 65, with no guarantee. She can opt to take alower benefit with a 10year guarantee. The revised benefit is calculatedto have equal EPV at age 65 to the original benefit. Calculate the revisedbenefit using the Standard Ultimate Survival Model, with interest at 5%per year.


Example 5.

4

Let Bdenote the revised monthly benefit. Then the two options are

12000 per year, paid per month with Present Value Y1


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12Bper year, paid per month with Present Value Y2

Hence E[Y1 Y2] = 0 implies

12000a(12)65 = 12Ba

(12)

65:10

= 12B

a(12)10 +v

1010p65a

(12)75


Example 5.

4




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12000a(12)65 = 12Ba

(12)

65:10

= 12B

a(12)10 +v

1010p65a

(12)75

B= 1000 a(12)65

a(12)10 +v

1010p65a

(12)75

= 1000 13.087013.3791

= 978.17


Example 5.

4




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12000a(12)65 = 12Ba

(12)

65:10

= 12B

a(12)10 +v

1010p65a

(12)75

B= 1000 a

(12)65

a(12)10 +v

1010p65a

(12)75

= 1000 13.087013.3791

= 978.17

V[Y1 Y2] = 0?

(85)



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Linearly Increasing Annuities

If then the annuity is payable continuously, with payments increasing by 1th


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at each year end and the rate of payment in the tth

year constant andequal to t fort {1, 2, ..m,.., n},then h(t) =m1{mt


113/359

Evaluating Annuities Using Recursion

By recursion, we observe


114/359

ax= 1 +vpx+v22px+v33px+....

= 1 +vpx

1 +vpx+1+v2

2px+1+v3

3px+1+....

= 1 +vpxax+1

a

(m)

x =

1

m +v

1m

1m pxa

(m)

x+ 1m

(89)


Evaluating Annuities Using Recursion

By recursion, we observe


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ax= 1 +vpx+v22px+v33px+....

= 1 +vpx

1 +vpx+1+v2

2px+1+v3

3px+1+....

= 1 +vpxax+1

a

(m)

x =

1

m +v

1m

1m pxa

(m)

x+ 1m

(89)

Consider the case where there is a maximum age in the model, and so

a1 = 1

a(m) 1m

= 1m

(90)



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Evaluating Annuities Using UDDIt follows that

(m) 1 A(m)x


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ax = d(m)=

1 ii(m)

Ax

d(m)

= i(m) iAx

i(m)d(m)

= i(m) i(1 dax)

i(m)d(m)

= id

i(m)d(m)ax i i

(m)

i(m)d(m)

:=(m)ax (m)ax=

id

2ax i

2

(93)


Evaluating Annuities Using UDD


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For term annuities, we have

a(m)x:n = a

(m)x vnnpxa(m)x+n

=(m)ax (m) vnnpx ((m)ax+n (m))

=(m) ax vnnpxa(m)x+n (m) (1 vnnpx)=(m) ax:n (m) (1 vnnpx) ax:n m 1

2m (1 vnnpx)

(94)


Woolhouses Formula


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Consider a function g : R+ R such that limtg(t) = 0, then

0 g(t)dt=h

k=0

g(kh) h

2 g(0) +

h2

12 g

(0) h4

720 g

(0) +... (95)


Woolhouses Formula

Define

g(t) =vttpx


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g(t) = tpxet vttpxx+tg(0) = x

(96)

and so for h= 1,

ax

k=0

g(k) 12

+ 1

12g(0)

=

k=0

vkkpx

1

2

1

12

(+x)

= ax12 1

12(+x)

(97)


Woolhouses Formula

1


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Correspondingly, for h= m ,

ax 1m

k=0

g

k

m

1

2m+

1

12m2g(0)

=

k=0

v km km

px 12m

112m2

(+x)

= a(m)x 1

2m 1

12m2(+x)

(98)


Woolhouses Formula


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Equating the previous two approximations for ax, we obtain

a(m)

x a

xm

1

2m m2

1

12m2 (+

x) (99)


Woolhouses Formula


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For term annuities, we obtain the approximation

a(m)x:n ax:n

m

1

2m (1vn

npx)m2

1

12m2 (+xvn

npx(+x+n)) (100)


Woolhouses Formula

Letting m , we get


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ax ax12 1

12(+x)

ax:n ax:n 12

(1 vnnpx) 112

(+x vnnpx(+x+n))(101)

For ax with = 0, the approximation above reduces further to

ex (ex+ 1) 12 1

12x (102)

NB: For life tables, we can compute these quantities using theapproximation x 12[ln (px) + ln (px+1)]


Select and Ultimate Survival Models

Notation:

Aggregate Survival Models: Models for a large population, where

d d l h


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tpxdepends only on the current age x.

Select (and Ultimate) Survival Models: Models for a select groupof individuals that depend on the current age x and

Future survival probabilities for an individual in the group depend onthe individuals current age andon the age at which the individual

joined the groupd>0 such that if an individual joined the group more than d yearsago, future survival probabilities depend only on current age. So, afterdyears, the person is considered to be back in the aggregatepopulation.

Ultimately, a select survival model includes another event upon whichprobabilities are conditional on.


Select and Ultimate Survival Models

N i


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Notation:

d is the select period

The mortality applicable to lives after the select period is over isknown as the ultimate mortality.

A select group should have a different mortality rate, as they have beenoffered (selected for) life insurance. A question, of course, is the effect onmortality by maintaining proper health insurance.


Example 3.

8

Consider men who need to undergo surgery because they are suffering

f ti l di Th i li t d d


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from a particular disease. The surgery is complicated and

P[survive one year after surgery] = 0.5

(d, l60, l61, l70) = (1, 89777, 89015, 77946)(103)

Calculate P[A],P[B],P[C], where

A= {(60),about to have surgery, will be alive at age 70}B= {(60),had surgery at age 59, will be alive at age 70}C =

{(60),had surgery at age 58, will be alive at age 70

}


Example 3.

8

l


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P[A] = P[(60),about to have surgery, alive at age 61] l70l61

= 0.5 7794689015

= 0.4378

P[B] = l70l60

= 0.8682

P[C] = l70l60

= 0.8682

(104)


Select Survival Models

S[x ]+s (t)P

[( + s) selected at ( ) s r i es to( + s + t)]


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S[x]+s(t) = [(x+s) selected at (x), survives to(x+s+t)]tq[x]+s= P[(x+s) selected at (x)dies before(x+s+t)]

[x]+s= force of mortality at (x+s) for select at (x)

= limh0+

1 S[x]+s(h)h

tp[x]+s= 1 tq[x]+s=S[x]+s(t)

=et

0 [x]+s+udu

(105)

For tx+s>x+t x x0. Then

yxtp[x]+t= ly

l[x]+t

stp[x]+t=l[x]+s

l[x]+t

(107)


Example 3.9

Proof.

yxtp[x]+t= yxdp[x]+d dtp[x]+t

= yxd px+d dt p[x ]+t


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= y x dpx+d d tp[x]+t=

lylx+d

lx+dl[x]+t

= lyl[x]+t

stp[x]+t= dtp[x]+t

dsp[x]+s

= lx+dl[x]+t

l[x]+slx+d

=l[x]+s

l[x]+t

(108)


Example 3.11

A select survival model has a select period of three years. Its ultimatemortality is equivalent to the US Life Tables, 2002 Females of which anextract is shown below. Information given is that for all x 65,


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p[x], p[x1]+1, p[x2]+2

= (0.999, 0.998, 0.997). (109)

Table: 3.5: Extract from US LIfe Tables, 2002 Females

x lx70 8055671 7902672 77410

73 7566674 7380275 71800


Example 3.11

Calculate the probability that a woman currently aged 70 will survive to


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Calculate the probability that a woman currently aged 70 will survive toage 75 given that

1 she was select at age 67:

2 she was select at age 68





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Example 3.11

5p[701]+1 =l[69]+1+5

l[69]+1

= l75

l[69]+1l


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l[ ] l[ ]

= l75

l[69]+3(1p[69]+1)(1p[69]+2)

= l75

l72 (1p[69]+1)

(1p[69]+2)

=71800

77410 0.997 0.998 = 0.9229

5p[70]= l75l73

(1p[70]) (1p[70]+1) (1p[70]+2)

=7180075666

0.997 0.998 0.999 = 0.9432

(111)


Example 3.12

Given a table of values for q[x], q[x1]+1, qxand the knowledge that themodel incorporates a 2

year selct period, compute


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4p[70]

3q[60]+1


Example 3.12




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4p[70]

3q[60]+1

4p

[70]=p

[70]p

[70]+1p

[70]+2p

[70]+3=p

[70]p

[70]+1p

72p

73=

1 q[70] 1 q[70]+1 (1 q72) (1 q73)


Example 3.12



p


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4p[70]

3q[60]+1

4p[70]

=p[70]

p[70]+1

p[70]+2

p[70]+3

=p[70]

p[70]+1

p72p73

=

1 q[70] 1 q[70]+1 (1 q72) (1 q73)

3q[60]+1=q[60]+1+p[60]+1q62+p[60]+1p62q63

=q[60]+1+ 1 q[60]+1 q62+

1 q[60]+1 (1 q62) q63(112)


Example 3.13

A select survival model has a two-year select period and is specified asfollows. The ultimate part of the model follows Makehams law, where(A, B, c) = (0.00022, 2.7

106, 1.124):


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x= 0.00022 + (2.7 106) (1.124)x (113)The select part of the model is such that for 0 s 2,

[x]+s= 0.92sx+s (114)

and so for 0 t 2,

tp[x] =et

0 [x]+sds = exp

0.92t

1 0.9t

ln(0.9) + c

t

0.9t

ln

0.9c

(115)


Example 3.13

It follows that given an initial cohort at age x0, that is given lx0 , we cancompute the entries of a select life table via


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g g gcompute the entries of a select life table via

lx =px1lx1

l[x]+1=

lx+2

p[x]+1

l[x] = lx+2

2p[x]

(116)


Homework Questions


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HW: 3.1, 3.2, 3.4, 3.7, 3.8, 3.9, 3.10, 5.1, 5.3, 5.5, 5.6, 5.11, 5.14


What is a Premium?


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When entering into a contract, the financial obligations of all parties mustbe specified. In an insurance contract, the insurance company agrees topay the policyholder benefits in return for premium payments. The

premiums secure the benefits as well as pay the company for expensesattached to the administation of the policy


Premium Types

A Net Premiumdoes not explicitly allow for companys expenses, while aOffi G P i d Th b Si l P i


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OfficeorGross Premiumdoes. There may be aSingle Premiumor or aseries of payments that could even match with the policyholders salaryfreequency.

It is important to note that premiums are paid as soon as the contract issigned, otherwise the policyholder would attain coverage before paying forit with the first premium. This could be seen as an arbitrage opportunity -non-zero probability of gain with no money up front.


Premium Types

Premiums cease upon death of the policyholder. The premium paying


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term is the maximum length of time that premiums are required.Certainly, premium term can be fixed so that upon retirement, say, nomore payments are required.

Also, the benefits can be secured in the future (deferred) by a singlepremium payment up front. For example, pay now to secure annuitypayments upon retirement until death.


Assumptions

Recall the life model used in Example 3.13 : The select survival model hasa two-year select period and is specified as follows. The ultimate part ofthe model follows Makehams law, where

(A, B, c) = (0.00022, 2.7 106, 1.124):


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x= 0.00022 + (2.7 106) (1.124)x (117)The select part of the model is such that for 0 s 2,

[x]+s= 0.92sx+s (118)

and so for 0 t 2,

tp[x] =et

0 [x]+sds = exp

0.92t

1 0.9tln(0.9)

+ct 0.9t

ln

0.9c

(119)Albert Cohen (MSU) STT 455-6: Actuarial Models MSU 2011-12 114 / 283

Assumptions

Furthermore, we can extend the recursion principle when using a select life


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gmodel to obtain

ax= 1 +vpxax+1

a[x]+1= 1 +vp[x]+1ax+2a[x] = 1 +vp[x]a[x]

(120)


Basic Model

In general, an insurance company can expect to have a totalbenefit paid


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out, along with expense loading and other related costs. We represent thistotal benefit as Z. Similarly, to fund Z, the company can expect thepolicyholder to make a single payment, or stream of payments, that has

present value P Y. Here, Prepresents the level premium P and Yrepresents the present value associated to a unit payment or paymentstream.


Future Loss Random Variable

For life contingent contracts, there is an outflow and inflow of money

during the term of the agreement. The premium income is certain, butsince the benefits are life contingent the term and total income may not


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since the benefits are life contingent, the term and total income may notbe certain up front. To account for this, we define the Net Future LossLn0 (which includes expenses) and the Gross Future Loss L

g0 (which does

not includes expenses) as

Ln0 =PV[benefit outgo] PV[net premium income]Lg0 =PV[benefit outgo] +PV [expenses]

PV[gross premium income](121)


Example 6.2

An insurer issues a whole life insurance to [60], with sum insured Spayable immediately upon death Premiums are payable annually in


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payable immediately upon death. Premiums are payable annually inadvance, ceasing at 80 or on earlier death. The net annual premium is P.What is the net future loss random variable Ln0 for this contract in terms oflifetime random variables for [60]?


Example 6.2

An insurer issues a whole life insurance to [60], with sum insured Spayable immediately upon death. Premiums are payable annually in


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payable immediately upon death. Premiums are payable annually inadvance, ceasing at 80 or on earlier death. The net annual premium is P.What is the net future loss random variable Ln0 for this contract in terms oflifetime random variables for [60]?

Ln0 =SvT[60] Pa

min{K[60]+1,20} (122)


Equivalence Principle

Absent a risk-neutral type pricing measure, insurers price theseevent-contingent contracts by setting the average value of the loss to bezero. Symbolically, this is simply (for net premiums) find P such that


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E [Ln0] = 0 (123)

Note that this value Pdoes not necessarily set Var[Ln0] = 0

Returning to our general set-up, we see that the equivalence pricingprinciple can be summarized as

P= E[Z]

E

[Y]

(124)



As an introductory example, consider >0 and a contract where (underno selection)


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Z =vTx

Y = aTx

tpx=et

(125)



As an introductory example, consider >0 and a contract where (underno selection)


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Z =vTx

Y = aTx

tpx=et

(125)

Hence, we have a unit whole-life insurance payable immediately upon deathof (x), where mortality is modeled to be exponential with parameter .



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We obtain

Px= E

vTx

E

= Axa

= Ax

1 A


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E

aTx

ax 1 Ax

=

0 etetdt

1 0 etetdt=

+

1 +

=

(126)

HW: repeat the above calculation ifS0(x) = x

for a finite lifetime

model with maximal age .



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It follows that

Px = d Ax

= d

k=0e

(k+1) (kpx k+1px) (k 1)


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Px d1 Ax d1 k=0e(k+1) (kpx k+1px)

=d(1 e)

k=0e

(k+1)ek

1

(1

e)

k=0

etek

=d(1 e) e 1

1e(+)

1 (1 e) e 11e(+)

= (1 e) e

(128)



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Example 6.5

By the EPP, the fact that d= 1 11.05 , and tables 6.1 and 3.7, we have

100000 A[45]:20 =P a[45]:20


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P= 100000 A[45]:20

a[45]:20=

100000

1 da[45]:20

a[45]:20

= 100000

1 d

a[45] l65l[45] v20a[65]

a[45] l65l[45] v20a[65]

= 100000

0.383766

12.94092

= 2965.52

(129)


New Business Strain

Starting up an insurance company requires start-up capital like most othercompanies. Agents are charged with drumming up new business in theform of finding and issuing new life insurance contracts This helps to


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form of finding and issuing new life insurance contracts. This helps todiversify risk in the case of a large loss on one contract (more on thislater.)

However, new contracts can incur larger losses up front in the first fewyears even without a benefit payout. This is due to initial commisionpayments to agents as well as contract preparation costs. Periodicmaintenance costs can also factor into the premium calculation.


An Example..

Consider offering an n

year endowment policy to an age (x) in the

aggregate population where the benefit B is paid at the end of the year ofdeath oron maturity. There are periodic renewal expensesofrper policy.


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y p p p p y


An Example..



aggregate population where the benefit B is paid at the end of the year ofdeath oron maturity. There are periodic renewal expensesofrper policy.


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y p p p p y

Then the premiumP is calculated via the EPP as

Pax:n =B Ax:n +rax:n P=B Ax:n

ax:n+r

(130)

and we see that periodic expenses are simply passed on to the consumer!


Another Example..



aggregate population where the benefit B is paid at the end of the year ofdeath oron maturity. There are periodic renewal expensesofr per policy


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andan inital preparation expense ofzper contract.


Another Example..



aggregate population where the benefit B is paid at the end of the year ofdeath oron maturity. There are periodic renewal expensesofr per policy


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andan inital preparation expense ofzper contract.

Then the premiumP is calculated via

P=B Ax:nax:n

+r+ z

ax:n(131)

and so the initial preparation expense is amortized over the lifetime of thecontract.


Example 6.6

An insurer issues a 25year annual premium endowment insurance with


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sum insured 100000 to a select life aged [30]. The insurer incurs initialexpenses of 2000 plus 50% of the first premium and renewable expenses of2.5% of each subsequent premium. The death benefit is payableimmediately upon death. What is the annual premium P?



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Some more worked examples

Consider the following net loss random variables:

Ln0 =vTx Pamin{Tx,t}


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Ln0 =vmin{Tx,n} Pamin{Tx,t}


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Ln0 =vmin{Tx,n} Pamin{Tx,t}Ln0 = v

Kx+1 Pa(133)


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L0 =v Pamin{Kx+1,t}

What are the fair premiums under the EPP?






Kx+1 Pa i {K 1 }(133)


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L0 =v Pamin{Kx+1,t}


P=Ax

ax:t






Kx+1 Pa i {K +1 }(133)


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L0 v Pamin{Kx+1,t}


P=Ax

ax:t

P=Ax:nax:t





Ln0 =vmin{Tx,n} Pamin{Tx,t}Ln0 =v

Kx+1 Pa i {K +1 t}(133)


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0 amin{Kx+1,t}


P=Ax

ax:t

P=Ax:nax:t

P= Ax

ax:t

(134)


Refunded Deferred Annual Whole-Life Annuity Due

Consider the case where a nyear deferred annual whole-life annuity due


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of 1 on a life (x) where if the death occurs during the deferral period, thesingle benefit premium is refunded without interest at the end of the

year of death. What is this single benefit premium P?



The net loss random variable is

Ln

0=PvKx+1

1{

Kx+1

n}

+vn1{

Kx+1>

n}

aKx+1n

P (135)

and by the EPP we have


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0 =PA1x:n + n|ax P (136)



The net loss random variable is

Ln

0=PvKx+1

1{

Kx+1

n}

+vn1{

Kx+1

>n}

aKx+1n

P (135)

and by the EPP we have


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0 =PA1x:n + n|ax P (136)

This implies that

P= n|ax

1 A1x:n

=

Ax:n

A1x:n

1 A1x:n ax+n(137)


Profit

Consider a 1year term insurance contract issued to a select life [x], withsum insured S= 1000, interest rate i= 0.05, and mortality

q[x] = P[T[x] 1] = 0.01It follows that L0, the future loss random variable calculated at the time of


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issuance, is

L0 = 1000v1

1{T[x]1} P


Profit

Consider a 1year term insurance contract issued to a select life [x], withsum insured S= 1000, interest rate i= 0.05, and mortality

q[x] = P[T[x] 1] = 0.01It follows that L0, the future loss random variable calculated at the time of


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issuance, is

L0 = 1000v1

1{T[x]1} P P= E[1000v11{T[x]1}] = 1000v P[T[x] 1]

=(1000)(0.01)

1.05 = 9.52

(138)



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Profit

In general, we have the event that the insurer turns a profit on this groupof policies is{Profit} = D[x] N q[x], and so as N ,

P[Profit] =

P[D[x] N q[x]]

= P[D[x] E[D[x]]]


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= P

D[x] E[D[x]]

VarD[x] 0

(0) =12

by the CLT.

(140)

HW:Compute

E

Profit | D[x] N q[x]N P (141)


Profit

Consider now a whole-life contract issued to [x] with sum insured S andannual premium P. Then

P[Profit] =

P[L0


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d

= P vK[x]+1 < P

P+d S= P

K[x]+ 1>

1

ln

P

P+d S

= P K[x]+ 1> 1

ln

P

P+d

S= 1

ln ( PP+dS)

p[x]

(142)


Conditional Sums

Define L0(k) =PV[Loss | K[x] =k]. For a contract with a term n, itfollows that

L0 = E[PV[Loss]] = E

n1k=0

PV[Loss | K[x] =k] 1{K[x]=k}


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k=0

+ E PV[Loss | K[x] n] 1{K[x]n

}=

n1k=0

PV[Loss | K[x] =k] P[K[x] =k]

+L0(n) P[K[x] n]

=n1

k=0

L0(k)

k|q[x]+L0(n)np[x]

(143)

Q: Can we use this for the case n= ?Albert Cohen (MSU) STT 455-6: Actuarial Models MSU 2011-12 138 / 283

Example 6.9

A life insurer is about to issue a 25

year endowment insurance with a

basic sum insured S= 250000 to a select life aged exactly [30]. Premiumsare payable annually throughout the term of the policy. Initial expenses are1200 plus 40% of the first premium and renewal expenses are 1% of the


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1200 plus 40% of the first premium and renewal expenses are 1% of thesecond and subsequent premiums. The insurer allows for a compound

reversionary bonus of 2.5% of the basic sum insured, vesting on eachpolicy anniversary (including the last.) The death benefit is payable at the

end of the year of death. Assume the Standard Select Survival Model withinterest rate 5% per year.


Example 6.9

In this case, we have the reversionary bonus exponentially grow the sum

insured:

L0 =B0(K[x]) +E0 P0(K[x])


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B0(k) = 250000 1.025kvk+1 fork {0, 1, 2, ..., 24}

B0(25) = 250000 1.02525

v25

E0 = 1200 + 0.39P

P0(k) = 0.99Pamin {k+1,25}

(144)


Example 6.9

For 1 +j= 1+i1.025 , we have j= 0.02439 and so

E[L0] = E[B0(K[x])] +E0 E[P0(K[x])]

=24

k=0

B0(k)k|q[30]+B0(25)25p[30]+E0 0.99Pa[30]:25


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=24

k=0

250000 (1.025)t(1.05)t+1 k|

q[30]+B0(25)25p[30]

+E0 0.99Pa[30]:25=

250000

1.025 A1

[30]25j+B0(25) 25p[30]+ 1200 + 0.39P 14.5838P

(145)

Under the EPP, we have P= 97644.44.


Example 6.9

For 1 +j= 1+i1.025 , we have j= 0.02439 and so

E[L0] = E[B0(K[x])] +E0 E[P0(K[x])]

=24

k=0

B0(k)k|q[30]+B0(25)25p[30]+E0 0.99Pa[30]:25


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=24

k=0

250000 (1.025)t(1.05)t+1 k|

q[30]+B0(25)25p[30]

+E0 0.99Pa[30]:25=

250000

1.025 A1

[30]25j+B0(25) 25p[30]+ 1200 + 0.39P 14.5838P

(145)

Under the EPP, we have P= 97644.44.

HW:Replicate Table 6.3 in the text by using a spreadsheet program.Compare this example with Example 6.10.


Portfolio Percentile Premium Principle

Assume once again that a company is about to issue insurance to N

independent lives [x], each with loss L0,i for i {1, 2,.., N} . In this case

L0 =N

L0 i


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L0 =i=1

L0,i

E[L0] = E N

i=1

L0,i

=N

i=1

E [L0,i] =N E [L0,1]

Var[L0] =N Var[L0,1]

(146)



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Example 6.11

An insurer issues whole life insurance policies to select lives aged [30]. Thesum insured S= 100000 is paid at the end of the month of death andlevel monthly premiums are payable throughout the term of the policy.Initial expenses, incurred at the issue of the policy, are 15% of the total ofthe first years premiums Renewal expenses are 4% of every premium


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the first year s premiums. Renewal expenses are 4% of every premium,including those in the first year. Assume the SSSM with interest at 5% per

year.Calculate the monthly premium Pusing the EPP and

Calculate the monthly premium Pusing the PPPPsuch that= 0.95 and N= 10000.


Example 6.11

For the EPP calculation, we have

E[PV(Premiums)] = 12Pa(12)[30] = 227.065P

E[PV(Benefits)] = 100000A(12)[30] = 7866.18


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[30]

E[PV(Expenses)] = (0.15)(12P) + (0.04)(12Pa(12)[30])

= 10.8826P


Example 6.11

For the EPP calculation, we have

E[PV(Premiums)] = 12Pa(12)[30] = 227.065P

E[PV(Benefits)] = 100000A(12)[30] = 7866.18


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[ ]

E[PV(Expenses)] = (0.15)(12P) + (0.04)(12Pa(12)[30])

= 10.8826P

E[PV(Premiums)] = E[PV(Benefits)] + E[PV(Expenses)]

P= 36.39

(149)


Example 6.11

For all i

{1, 2,.., N

}, we have the i.i.d. PV(Loss) random variables

L0,i= 100000vK

(12)[30]

+ 112 + (0.15)(12P)

(12)


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(0.96)

12Pa(12)

K(12)[30]

+ 112

E[L0,i] = 100000A(12)[30] + (0.15)(12P) (0.96)(12Pa

(12)[30])

= 7866.18 216.18P

(150)


Example 6.11

To find the variance, we rewrite

L0,i=

100000 +

(0.96)(12P)

d(12)

v

K(12)[30]

(0 96)(12P)


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+ (0.15)(12P) (0.96)(12P)d(12)

Var[L0,i] =

100000 +

(0.96)(12P)

d(12)

2

2A(12)[30]

A

(12)[30]

2= (100000 + 236.59P)2 (0.0053515)

(151)


Example 6.11

Collecting our results, we now have

0.95 == P[L0


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Var[L0]

= N E[L0,1]Var[L0,1]

=

10000 216.18P 7866.18

(100000 + 236.59P) 0.0053515

(152)


Example 6.11

It follows that

1.645 = 1(0.95)

10000 216.18P 7866.18(100000 + 236.59P) 0.0053515

P 36 99

(153)


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P= 36.99

For general N, we have216.18P 7866.18

(100000 + 236.59P) 0.0053515 =1.645

N(154)

and as N

, we have P

36.39, recovering the EPP premium as

expected.


Independent Exponential RVs

Imagine that a fully continuous whole life insurance is offered to N

individuals aged [x] with T(i)[x]

exp() for all i

{1,.., N

}. For each

insured, the i.i.d. loss random variables are

L0,i =SvT

(i)[x] Pa

T(i) =

S+P

e

T(i)[x] P

(155)


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,T

(i)[x]

( )

and so for L0 =N

i=1L0,i, the PPPP seeks to determine P such that

P

Ni=1

S+P

e

T(i)[x] P


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P [Yi >y] = P eT

(i)[x] >y= 1 y

.


Independent Exponential RVs

In this case, we can rewrite this as

P 1N

Ni=1 e

T(i)[x]

y] = P eT

(i)[x] >y= 1 y

.

HW Using convolution techniques, find the above probability

P

1

N

N

i=1

Yi < P

P+S

= . (158)

Are there any ergodic theory results that we can use?


Capped Maximal Loss

Another principle is to find Psuch that for any (0, 1), the probabilitythat any contract suffers a loss of


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Capped Maximal Loss

Another principle is to find Psuch that for any (0, 1), the probabilitythat any contract suffers a loss of


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=P

mini=1..N

T

(i)

[x]

> 1

lnP+

P+

=

e[1

ln ( P+P+ )]N

=

P+

P+S

N



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Comments on PPPP

Notice that the PPPP only guarantees that the probability of a loss is1 .

It says nothing about the size of what that loss could be if it

arises.

This is a big problem if the loss is extremely large and bankrupts theinsurer. It may seem very unlikely, but recent economic events have


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shown otherwise.

Further improvements to this model can be seen in the ERM forStrategic Management (Status Report) by Gary Venter, posted onthe SOA.org website

Also, there is a close link, perhaps to be explored in a project, with

VAR in the financial world. Click here for an informative article in theNY Times TM for an article on VAR and the recent financial crisis.


Homework Questions

HW: 6.1, 6.2, 6.5, 6.7, 6.8, 6.12, 6.14, 6.15
http://www.soa.org/library/monographs/other-monographs/2008/april/mono-2008-m-as08-1-venter.pdfhttp://www.soa.org/library/monographs/other-monographs/2008/april/mono-2008-m-as08-1-venter.pdfhttp://www.soa.org/library/monographs/other-monographs/2008/april/mono-2008-m-as08-1-venter.pdfhttp://www.soa.org/library/monographs/other-monographs/2008/april/mono-2008-m-as08-1-venter.pdfhttp://www.nytimes.com/2009/01/04/magazine/04risk-t.html?pagewant

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