MLC STUDY MANUAL
Short Description
FOR ACTURIAL STUDENTS...
Description
ACTEX MLC Study Manual
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Volume II • Spring 2013 Edition Johnny Li, Ph.D., FSA Andrew Ng, Ph.D., FSA No portion of this ACTEX Study Manual may be reproduced in any part or by any means without the written permission of the publisher.
MLC Study Manual
-17,, 81-~l 68-1 f0175 -x'12 J . r;;- ,L,, e ( v2Jr ·
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Volume II • Spring 2013 Edition Johnny Li, Ph.D., FSA Andrew Ng, Ph.D., FSA
ACTEX Publications Actuarial & Financial Risk Resource .Nlaterials
Since 1972
Copyright © 2013, by ACTEX Publications, Inc. ISBN: 978-1-56698-937-4 Printed in the United States of America.
No portion of this ACTEX Study Manual may be reproduced or transmitted in any part or by any means without the permission of the publisher.
Chapter 10: Multiple State Models
Chapter 10
Multiple State Models
1.
To state the assumptions underlying a multiple state model
2.
To model disability income model and permanent disability model as multiple state models
3.
To calculate transition and occupancy probabilities for multiple state models
4.
To calculate the premiums and reserves for multiple state models
In this chapter, we discuss multiple state models. We shall see how various non-traditional insurances can be analyzed under the framework of multiple state models.
In what follows we use the notation o(h) quite often. A function g is said to be o(h) if Jim g(h) = 0. A function is o(h) if it shrinks to 0 quicker than h. For example, g 1(h) = Ii and h->0 h g 2(h) = h 11 are o(h). However, g 3(h) =hand g4 (h) =sin hare not o(h) because the limits are I. It is obvious from the theorem of limits that if/ and g are o(h), then/+ g,f- g,fg, and cf(where
c is a constant) are also o(h). As a result, we write o(h) ± o(h) = o(h), o(h) x o(h) = o(h), c x o(h) = o(h).
Before we talk about the setting of a multiple state model, let us review the single decrement model and multiple decrement model in Chapters 1 - 7 and Chapters 8 - 9.
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Chapter 1 O: Multiple State Models
The Single Decrement Model In the ordinary single decrement model, there are two states, alive (0) and dead ( l ):
~~o_._A_li_ve~__.1--~~µ_x+_'~~~~~1~~l._D_e_a_d~---' Model l: The Single Decrement Model The single decrement model is characterized by the force of mortality:
d -Fx(t) l P(t T < h) Ji = dt = --lim t+hqx - ,qx = lim < x - t + =Jim hqx+t x+t SJt) Sx(t) h-;O h h-;O hPr(Tr > f) h-;O h ' and this means
"qx+t
=
µx+th + o(h) .
So you can see why the approximation" di q_w ~ µx+idt for small dt" holds. For this model, the probability for (x) to stay in state 0 (alive) for a period of length tis 1
Px =exp(-
J~µx+sds).
The probability for (x) to enter state l (dead) on or before time tis ,qx =
J~ ,pxµx+sds.
The Double Decrement Model In multiple decrements, what we need are the various forces of decrement. In the double decrement model, they are 2 modes of decrements, say, accidental death (1) and death due to other causes (2).
1. Accidental
0. Alive
2. Others
Model 2: The Double Decrement Model
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Chapter 10: Multiple State Models
In this model, we are interested in the joint distribution of ( 1) the time for the individual to leave state 0, (2) the state that the individual would enter after leaving state 0. The forces of decrement are the instantaneous rates of transition. In Chapter 8, we wrote Pr(t < T,. ~ t + dt, Jx = j I Tx > t) = dtq.~~; ~ Jl~~;dt for small dt ~ 0. Actually, the mathematically rigorous statement of the above is
/J) f x+t
= Jim Pr(t < Tx ~ t + h, Jx = h->0 h
J I T,. > t) = Jim ,, q_~~; /i->0
h
'
and this means (I)
(I)
h
hqx+t = Jl.r+I + 0
(h)
(2) (2) h (h) · ' hqx+t - Jlx+t +o
(Of course the two o(h)'s above are different!) The total force of decrement is µ_~:; = µ_~~1 +JI.~!;, and we have: Occupancy(= no change in state) probability: / p_~r) =exp(1
Transition probabilities: / q_~ > =
J~µ-~:l,ds)
J~ _, p~r) f.L.~~_,ds, / q_~2 > = J~ _, p_~r> Jl.~!l,ds
There are two characteristics for multiple decrement models. (a) Transition to any state is irreversible. (b) There is no need to specify the transition intensity after decrement has happened. Even if we change the model to a withdrawal model,
1. Withdrawal 0. Active 2. Dead
we would not specify what kind of decrements would the member follows after withdrawal. Of course the member would still die some time after withdrawal. But we simply do not model that because we are not interested in it.
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Chapter 1 O: Multiple State Models
The Disability Income Model A multiple state model models the transition rates between states. Transition to any state can be
reversible or irreversible. Consider the following disability income model:
0. Healthy
1. Disabled
2. Dead
Model 3: The Disability Income Model In this model, the insured can switch between state 0 and state 1. However, transition to state 2 is irreversible. We can change the above to a permanent disability model by disallowing transition from state 1 back to state 0.
In this chapter we want to analyze models with structures similar to Model 3. We shall first discuss a simple version, where state changes can only occurs at the end of each period. Then we shall discuss the more general case when state changes can only occur any time.
, 1 O. 1 Discrete-time Markov Chain
In this section we discuss discrete-time Markov chain, a model for which transitions can only occur at the end of each period. Such a model is useful in modeling health status, bonus malus system and credit rating in group health and non-life insurances. Let the state at time t be Y1. Let the state space, which is the set of all possible states, be E. For example, in the disability income model above, we have E = {0, 1, 2} and Yo
= 0.
As time progresses, Y switches between 0 and 1
and finally get struck in state 2. The following figure shows one of the possible paths of Y.
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Chapter 1 O: Multiple State Models
Y, 2 I
0
0
2 3 4 5 6 7 8 9
Transition Probability Matrix and the Markov Property For a discrete-time Markov chain, we assume that when the process is in state i at time t, the probability that it would be in state) at time t + I is pf. That is, Pr(Y1+i = j
I Y, =;) = pf.
It is
customary to place all probabilities into a transition probability matrix:
=
pI
00 P1 P:o P120 110
P1 Po
P102
p:I
P1
p;2
p:"
P121
p;2
P12,,
p;li
p;2
P1
2 Y2
Yi
P,
P1-i
~ ~ ~
Yo
1111
P2
Pi
0
On
P101
3 Y3
~ ~
t Yr
t- I Y1-i
t +I Y1+i
The sum of the probabilities on each row is I. Moreover, we assume for any t 2 0 and i,j EE.
The Markov Property Pr(Y1+i
=JI
Yr= i,
Yi-1 = i1-i, Y1-2 = i1-2, ... ) = Pr(Yi+i =JI Y, = i)
This above says that the conditional distribution of any future state Y1+i, given the past states {Yo, Yi, .. ., Yi-i} and the present state Y,, depends only on the present state Y; but not the past states.
Loosely speaking, "Markov" means given the present, you can forget about the past!
If P1 is constant with t, the Markov chain is said to be homogeneous. If P, is time-dependent, the Markov chain is said to be inhomogeneous.
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Chapter 1 O: Multiple State Models
In many actuarial applications, there would be one or more states that cannot be left once it is entered (examples include withdrawal, death, and bankruptcy). If you look at the following 3state transition probability matrix:
0
2
~ [~:~ ~.·~ ~:~] 2
0
1
0
then state 2 cannot be left once it is entered. Such a state is called an absorbing state. For an absorbing state, all elements in the row, except the one in the main diagonal t, are 0. The one in the main diagonal is 1.
Example 10.1 [Exam M 2005 Nov #4]
•
Kevin and Kira are modeling the future lifetime of (60). (i)
Kevin uses a double decrement model:
(ii)
Age
/(r)
d(I)
d(2)
x
x
x
60 61 62
1000 800 560
120 160
80 80
Kira uses a non-homogeneous Markov model: (a) The states are 0 (alive), 1 (death due to cause I), 2 (death due to cause 2). (b) P60 is the transition matrix from age 60 to 61; P61 is the transition matrix from age 61 to 62.
(iii) The two models produce equal probabilities of decrement. Calculate P6t·
lJOO (A) 0 0
l0.70 (D) 0 0
0.12 1.00
oz81
0
1.00
1.00
0~01
0
1.00
0.20
l080 (B) 0.~6 l060 (E) 0 0
0.12 0.16
00.08081
0
1.00
0.28 1.00
o.~21
0
1.00
l0.76 (C) 0 0
1.00
0~81
0
1.00
0.16
t The main diagonal of a square matrix is the diagonal which runs from the top left corner to the bottom right corner. i.:' Actex 2013
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Chapter 1 O: Multiple State Models
-
Solution
Obviously, both states l and 2 are absorbing. This means (B) must be wrong. For a life age (61 ), the probability that he/she would still be alive after one year is p~~) = 560/800 = 0.7.
The probability that he/she would die due to cause 1 within the year is q~:) = 160/800 = 0.2.
This means (D) is the correct answer. [ END]
The following is another actuarial application of homogeneous Markov chain. An insurance company classifies its insureds based on each insured's credit rating, as one of Preferred, Standard or Poor. Individual transition between classes is modeled by the following transition matrix: Preferred
l
Preferred Standard Poor
Poor
Standard
0.95
0.04
0.01
0.15
0.80
0.05
0.00
0.25
0.75
l
Very frequently we are not only interested in transition probabilities for a single period. For example, we may want to find the probability that an insured who is "Standard" at time 0 would end up in "Poor" after 2 years. This is called a 2-step transition probability.
One way to calculate such probabilities is to list out all possible paths: Path Standard ~ Preferred ~ Poor Standard ~ Standard ~ Poor Standard ~ Poor ~ Poor Sum
Probability 0.15 x 0.01=0.0015 0.8 x 0.05 = 0.04 0.05 x 0.75 = 0.0375 0.079
Similarly, we can calculate the probability that an insured who is "Standard" at time 0 would end up in "Preferred" after 2 years:
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Chapter 1 O: Multiple State Models
Path
Probability Standard~ Preferred~ Preferred 0.15 x 0.95 = 0.1425 Standard ~ Standard ~ Preferred 0.8 x 0.15 = 0.12 Standard ~ Poor ~ Preferred 0.05 x 0 = 0 Sum 0.2625 What is the probability for the insured to be in the state of "Standard" after 2 years? It is simply 1 - 0.079 - 0.2625 = 0.6585. Actually there is a very efficient way to calculate such probabilities if you know some elementary matrix algebra:
In case you have forgotten about matrix multiplication, read the following short example:
If A= [ I -1
3 O
],
7 -5
AB=[
B
=[~I -4
1 - 2], then
3
(1)(2)+(0)(-1)+(3)(-4) (-1)(2)+(7)(-l)+(-5)(-4)
l
l
(l)(l)+(0)(-2)+(3)(3) [-10 10 (-1)(1)+(7)(-2)+(-5)(3) - 11 -30.
In this chapter we would frequently calculate the product of two square matrices. Remember that for two square matrices A and B, the products AB and BA can be different!
Now let us look at 0.95 [
0.15
0.04 0.8
0.01][0.95 0.05 0.15
0.04 0.8
0.01] 0.05
0
0.25
0.75
0.25
0.75
0
=
[0.9085 0.2625
0.0725 0.6585
0.019] 0.079 .
0.0375
0.3875
0.575
What can you find from the matrix on the right? The 2-step transition probabilities are all in the second row of the matrix product P x P = P2 . If we want to find Pr(Y1+2 =JI Y, = 0 = 2 p;1, then we can simply look at the ij-th element of the matrix P
2
.
So, P2 is the 2-step transition
matrix. More generally, if we want to find Pr(f,+s =JI we only need to look up the ij-th element in
0) such that the state space Eis {O, 1, 2, .. ., n}. For i,j
E
E
and t ~ 0, let lj
µ_;+,
= lim h--tO
vx:,
=
"P. f1 00 OI II ds t Px - Jo s Px µx+s 1-s Px+s
because the integral only accounts for the possibility of not reverting back to state 0 at all. The possibility of there being one or more periods of sickness before time t, with healthy periods in between make up the difference of 1 p~I and the integral. For this model, while we can write down the KFE that governs 1Px: 00 d dt 1Px d OI { dt 1Px
00 (
OI
02 )
OI
IO
= -1Px Jl. = 0.05 and
we WIS. h to compute
(2) - 0 11 q 55 - . .
1(2) 02 µ 55 .7 - µ 55(2)+0.7 --
q 55 ,< l 2
l-0.7q55
Since both decrements are SUDO, 1(2))
0.05 = q;; 1) I - q 5~
1(1))
and 0. I l = q;; 2 l 1- q~
(
(
.
Upon subtraction, we get - 0 .06 -Putting q;; 1> = q;; 2 )
-
ql(I) -
55
q'(552 ) .
0.06 into the first equation, we get o.o5 = (q;; 2 > -0.06)(1- o.5q;; 2 i).
On solving the quadratic equation, we get q~; 2 i = 1.947 or 0.112994. In terms of the notation in multiple state model, q;5 =0.112994. (b ) The answer
.
IS
0.112994 0 1 = . 22699 . 1-0.7x0.112994
l I. This is a triple decrement model with constant force of decrements. The total (constant) force of decrement is 0.3 + 0.5 + 0.7 = 1.5.
0•5 (r) l - e -1.5 Px = qx = 1.5qx = = 0.2590 • 02
(2)
3
So the answer is (A). 12. (a) The total rate of exit from state 0 is v~ = µ_~ + µ~ = 0.06. By equation (10.5),
1
00
1P5o
=e
-0.061
2
·
00
Th us, 10P50 = e -0.6 · (b) The total rate of exit from state I is 0.05. Similarly,
11
1Px =e
-0.051
for any x::::: 0 and t::::: 0. By equation (10.6), 01 10 P5o
flO
= Jo
00 01
II
1P5o µ50+1 I0-1 P50+1
dt
= f:o e-0.061 x o.o2e-0.05(10-/)dt 05
= 0.02e- ·
(l - e-o. 1)
0.01 = 2e-05(1-e-01 ). The answer is thus given by I -e-0 6 - 2e-0·5(1- e-0 1 ) = 0.3358.
2,) = 2 J w, - x
0
define the maximum of T.ty by l = min( W1
-
we
0)2 - y 0
x, OJi - y). Then
f[1-(-l - + 1 Jr+ (w, - x)(w2 t2 ldt w, - x 0)2 -
E(T ) = xy 0
y
= /-_!_(_l_+ 2 w1 - x
If OJ1
-
x:::; OJi - y, then l
= OJ1 -
1 w2 - y
y)
3
J12 +_!_
/
•
3 (w1 - x)(w 2 - y)
x, and
J
3
1( 1 1 2 1 ( w, - x) E(T,,)=(w1 -x)-- - - + (w1 -x) + .J 2 w1 - x w2 - y 3 (w1 - x)(w 2 - y)
= w, -
x - 1 ( w,
2
- x)
2
6 0)2 - y
Furthermore, if w 1 - x = OJi - y, then we even have E(T ) = w, - x xy 3 ' and E(T-)
= w1 -
xy
2
x + w1 - x _
2
w, -
x = 2( w1 - x) .
3
3
You can find many questions based on these formulas in the chapter-end exercises.
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Chapter 11: Multiple Life Functions
What's next after De Moivre's law? Of course that would be independent lifetimes following constant force of mo1iality. In this case, you would need the following:
A Basic Fact about Independent Exponential Random Variables Let X1, X2, ... , X, 1 be independent, exponentially distributed random variable with rate A;. (That is, the density of X; is A;exp(-A;x).) Define the minimum as
X(I) = min(X1, X2, ... , ,Y,1). The survival function of X(1) is Pr(X(I) 2: x) =Pr( all X; 2: x) = Pr(X1 2: x) Pr(X2 2: x) ... Pr(X,1 2: x). Since Pr(Xi 2: x) = exp(-A;x), II
Pr(Xu> 2: x) =
f1 exp(-,.1,;x) = exp(-,.1,x) i=I
11
where A = LA; . So, X{I) is exponential with rate A. i=I
~
Example 11.6
For two independent lives (x) and (y), (x) is subject to a constant force of mortality of 0.02 and (y) is subject to a constant force of mortality of 0.03. Calculate Var(Try) and Cov(T"Y'T-;;).
-
Solution
Since T.r ~ Exp(rate 0.02) and Ty~ Exp(rate 0.03), their minimum
Tw follows Exp(0.05). As a
l l result, E(T.ry) = 0.0 = 20, Var(Try) = 0.0 52 = 400. 5
1
1
Cov(T.>., T-) = [E(T. )- E(T.Y )] x [E(T>,) - E(T"Y )] = ( -- - 20)(-- - 20) = 400. ,. xy " ·' · 0.02 0.03 [END]
It is rare that you are asked to compute the variance of T-;;,. If you are interested, read the following. Otherwise, jump to the next subsection on statuses involving the order of death.
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Chapter 11: Multiple Life Functions
r-------------------------------------------------------------------------------------------Variance of T-;y: Taking variance on T.xy + T; = T.x + r;,, we get
By making use of
Cov(T,Y, T-;y)
= Cov(T,, TY)+ [E(T, )- E(T,Y )] x [E(Ty) - E(T,Y )] ,
we get Var(Txy) + Var(T-;y) This means that even if
T.x
= Var(T,) + Var(Ty) -
and
2[E(Tx) - E(T,Y)] [E(Ty) - E(T,>')] .
r;, are independent, the symmetric relation does NOT hold for
variance: Var(T,Y) + Var(T-;y) =F Var(T,) + Var(TY) .
2. Statuses Involving the Order of Death
Here we further assume that P(T.x = !;.) = 0.
We define two statuses involving the order of deaths: I
(x y) fails at the time when x dies, if x dies before y. It does not fail if x dies after y. 2
(x y) fails at the time when x dies, if x dies after v. It does not fail if x dies before y.
I
2
Notice that (x y) and (x y) mean different things. While both fail only if x dies before y, the time of failure is
T.x
I
2
for (x y), and the time of failure is
r;, for (x y). So you can see that the number
indicates both the order and the time of death of which life status is being referred to.
For statuses involving the order of death we only talk about their CDF, which are called contingent probabilities (note that these CDF are strange because they may not be equal to I when t = oo).
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1
Chapter 11: Multiple Life Functions
We have 1
q.~y
= Pr(Tx < Ty and T., ~ t),
1
qx~,
= Pr(Tx < Ty and
Ty~ t).
Graphically, this means:
t
So, it is obvious that
, q·~>' = /qx~, + P(T., < t and Ty> t).
~
Example 11.7 [Exam 150 90 May #2) You are given: (i) qy = 0.25
(ii) q.;y = 0.12
(iii)qxy
(C)0.11
(E) 0.23
= 0.14
Calculate qx~,.
(A) 0.02
-
(B) 0.04
(D) 0.16
Solution
Statement (i) means (IV)+ (V) +(VI)= 0.25. Statement (ii) means (V) = 0.12. So, (IV)+ (VI)= 0.13 Statement (iii) means (IV) + (V)
= 0.14.
So (IV)= 0.02, (VI)= 0.11. We need (V) +(VI). So the answer is 0.12 + 0.11=0.23. [END ]
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Chapter 11: Multiple Life Functions
y;, > t) =Pr( Ty< t) -
Since Pr( Ty< f and
Pr( Tl< t and
y;,::::; t) = 1qx - q-;;,, we obtain 1
(Actually you can use this relation to do the previous example) If t = w, then since Pr( Ty < w and
y;, > w) ::::; Pr(Ty > w) = 0,
we have oo
q_~y =
oo
qx~ = Pr(Tx < J;,).
To calculate contingent probabilities in the general case, we may find the joint density of (Tx,
y;,)
and then integrate over the corresponding region. If lives are independent, we can simplify the calculation by conditioning on the life for which a number is on top of it. For example, :~////////////////////////////hY'/////////////hY'/////////////h'//////////////~:
~
~
;~
FORMULA%:
~
~ ~:
;;,;; Contingent Probabilities for Independent Lives
,.-,,
~
:~ ~ :~
q_~y = f: Pr(~, > T,
1
qxy = Jo Pr(T, , - ,Px;·)dt=0.75[(a>, -a,.)-(aX)' -axy.11) .n 1
)]
1
The final answer is obtained by summing over the 4 term: a~+ 0.5(ax -ax~)+ 0.75(ay -ay~)-0.25(axy -axy~).
[ END ]
~
Example 11.12
Consider Example 11. l 0 again. For the last survivor insurance, premiums are paid as follows: Level annual premiums 3P are payable as long as both insured are alive. When only one insured is alive, the premium is reduced to P. By using the equivalence principle, calculate P.
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Chapter 11: Multiple Life Functions
-
Solution
Method 1: (current payment technique) The condition "when only one of (x) and (y) is alive" occurs with probability
= kPxO- kPy)+ kpy(l- kPx) = kPx + kPy - kPxkPy.
kPxkqy + kqxkpy Thus the APV is 2
2
3P,l:Vk kP6o63 +PivkCP6o + kP63 - kP6okP63) k=O k=O 2
2
2
= 2PI vk k P6o:63 +PI vk k P6o +PI vk k P63 k=O
k=O
k=O
= P( 20 6o:63:3i + a6o3i + 0 63:3!) = 8.l 79829P Method 2: We can break the annuity down into two parts. The first part is a 3-year last survivor annuity that pays P. This has to be topped up by a 3-year joint life annuity that pays 2P so that the premium is 3P when both insured are alive. So, the actuarial present value of premiums is
P(2Ci 60 ,63 ,Ji +a 6063,Ji)= 8.179829P. By setting 8. l 79829P = 793.9163, we get P = 97.05781.
[ END ]
~
Example 11.13 [Exarn15089May#4]
A fully discrete last survivor insurance of 1 issued on two independent lives each age x. Net annual premiums are reduced by 25% after the first death. You are given: (ii) Axx = 0.55
(i) Ax= 0.40
(iii) ax
= 10.0
Calculate the initial net annual premium. (A) 0.019
-
(B) 0.020
(C) 0.022
(D) 0.024
(E) 0.025
Solution
The APV of benefits is A~
= 2Ax -Axx = 0.25
The APVofpremiums is P(a.u +0.75x2Cix J 1
From (i) and (iii) and Ax= 1 - dax , we get 0.4
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= P[ii_u +I.5(iix -ii.u)] = 1 - 1Od, and hence d = 0.06.
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Chapter 11: Multiple Life Functions
So, arx
= (1
- 0.55) I 0.06
The answer is P
= 0.25 I
= 7.5, and the APV of premiums
11.25
is 1 l .25P.
= 0.0222. [END]
4. Contingent Insurance 1
2
We can also define insurance for statuses (x y) and (x y). A continuous insurance written for 1
(x y)(first death contingent insurance) pays a unit payment immediately on the death of (x),
only if it dies before (y). If lives are independent, then we again condition on the life with a number on top of it:
Similarly, the APV of a contingent insurance payable at the end of the year of death is
='L...."" 11-l
A1 _
xy:nl
vk+1
1
k Pxyqx+k:y+k'
k=O
The probability qx~ky+k is usually calculated assuming UDO for each:
J
1
1
1 qx+k:y+k
= Jof 1Py+k1 Px+kµx+k+tdt = Jof (1- fqy+k )qx+kdt = q_')c
1
how terms can be arranged algebraically.
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Now let's see
Chapter 11: Multiple Life Functions
=
n
iv O
1 I
p )'I p x (A+ Bcx+i )dt x
= A Jof"v
1
Ip tydt +
.
=A( 1- ,2cx , c· + c>
f"v 1 Ip t)'B( ex + c>' )c 1 dt ex + c>' Jo . c
Jl"v
1 1
0
px>'dt+
1 1 ex f nv px>,[2A+B(cx +c>')c ]dt Cx +Cy Jo 1
(b) The whole trick of this part is to define a new age w from the relation II'
=
c From this, µty+t
ex+ c>' 2
= µx+t + µy+t = 2A + B(cx + c>')c = 2(A + Bcw+t) = 2µw+t = µww + 1• As a result, 1
the status (xy) is equivalent to the status (ww)! This also means tP>-y
aww:nj- = axy:nl- and A 1
....--~
-
= 1Pww· Thus,
= A_!_ -_ ~~
ww:nl
xy:nl
[END]
Similar results exist for the Gompertz mortality law.
5. M-thly payable multiple life benefits We have demonstrated in Example 11.2 that if Tt and Ty are independent and follow UDO, then
Tty is not UDO. As a result, we would NOT expect
/3( /11 ) '
(111) _ i A xy an d axy .. cmi -_ a ( m ) axy .. Axy - -:(;;;) -
I
which holds only if Tty is UDO. So, what are the corresponding expressions? It turns out that they are
(lll) - -i- A +-i (} l 2 2 Axy - -( 111 ) xy ·(Ill) +----+d(111) • I
I
a(lll) xy -- a(m)a xy - /J(m) -
I
/11
i
·(m)d(111)
I
(i
)I"' vk+l kPxyqx+kqy+k ' k=O
2 - + ~)~ + _!_ _ -d(111) vk+l kPxyqx+kqy+k • • ~ 111
I
k=O
The good news is that you need NOT remember the exact formulas. What you need to know is that the second term in the expression for A_~;>l is typically small because
L"'
vk+I
kPxyqx+kqy+k 1s
k=O
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the APV of an insurance that pays 1 at the end of the year of death of Cx) only if Cx) and (y) die in the same year. So, A;:::: _ ;:::: (1) By A_;
111
>
=
111
;
Ai::'> ;:::: _,_· Axy ;
a~ , By equivalence principle,
[ END]
Of course when the transition intensities are not constant, it can happen that the integrals can only be evaluated numerically.
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Chapter 11: Multiple Life Functions
A Model with No Common Shock Component Now we assume that µ~~i:y+i = 0 . 01 µx+t:y+I ~
State 0
State I
"
02 µ x+t:y+t
13 µx+I
•
23 µy+t
State 2
State 3
"
For this model, our goal is to
Step 1:
First express all the transition probabilities 1 p~,
U 7=
j) in terms of occupancy
probabilities;
Step 2:
Express various probabilities and benefits in terms of 1 p~, 's andµ 's, so that one can evaluate them analytically (when all forces of transitions are constant) or numerically.
Obviously, the occupancy probabilities in this model are given by 00 .. Pxy
11 1-S Px+s
01 02 = exp[ - Jors (µx+u:.v+u + µx+u:y+u )du
= exp( -
f
I 13 s flx+u
du ) '
1-s
J,
f
22 23 d ) p y+s =exp ( - ·'I µy+u u .
Then we can proceed to Step 1: To find 1 P.~;~ and 1 P.~;:, we use the approach as in the permanent disability model: 01 t Pxy
f1
00
01
11
= Jo s Pxy µx+s:y+.1· t-s Px+s
d
02 1Pxy -
S'
f1
J0
d
00 l 02 22 sPxyfx+s:y+s1-sPy+s S.
Also, 03 t Pxy
=I -
oo 1Pxy -
01 1Pxy -
02 1Pxy
'
Now we can proceed to Step 2: Annuities We have, as in the case of exponential common shock model, 00
tPxy
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1P-ry
=I -
03 1Pxy •
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tPx
00
= t Pxy +
01
1Pxy'
tPy
00
= 1Pxy +
02
1Pxy
from which annuities can be calculated. For example,
-
f"' I ood f"' I ( 00 Jo v I Pxy t' ax = Jo v 1Pxy +
axy =
01 )d
1Pxy
t'
-
axly =
f"' I 02d Jo v I Pxy t .
Insurances The density of the time to transit from state 0 to 2 is 1 p_~>~µ_~~i:y+i. As a result,
Axy02 -- Jof"' V
-1 -
Axy -
00 02
I 1
Pxy µx+t:y+tdt ·
Similarly, A2
xy
= Jof"' V
I
1
01 13
d
00 02
+
Pxy µx+t:y+t f ·
To find insurances on (x), we can use A x
= A-Ixy + A-2xy = Jof"° V
I (
1Pxyµx+1:y+1
01 13
1Pxyµ_.+1
)dt
•
Similar arguments lead to A,y
=
r
v' IP.~)~ (µ_~!1:y+1 + µ_~~t:y+t )dt'
~
Example 11.19 Express the following in terms of integrals for the model with no common shock: (a) I q_~y (b) I q;y
-
Solution
[ END]
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Chapter 11: Multiple Life Functions
If/titi
Exercise 11
Section I I. I Multiple Life Statuses
1. Joint and Last Survivor Statuses 1. For independent lives (x) and (y):
0.05 and
0.09
(i)
qx=
(ii)
Deaths are uniformly distributed over each year of age.
qy=
Calculate the following:
(a)
0.6qxy
(b)
0.6q--;),
(c)
f..lxy+0.6
(d)
JIxy+0.6
2. You are given: (i)
The future lifetimes of (40) and (40) are independent.
(ii)
/40
= 79509, lso = 71881, ls1=70754.
(iii) Deaths are uniformly distributed over each year of age. Calculate the force of failure at duration 10.3 for the last survivor status of (40) and (40). 3. Explain why
min
q xy =m p xy qx+m:y+m ' while 11
min
qcannot be calculated from xy
m
pq--. xy n x+m:y+m
4. (2005 May #3) For independent lives (35) and (45): (i)
5p35
= 0.90
(ii)
(iii) q4o = 0.03
5p45
= 0.80
(iv) qso = 0.05 1
Calculate the probability that the last death of (35) and (45) occurs in the 6 h year.
(A)
0.0095
(B)
0.0105
(C)
0.0115
(D)
0.0125
(E)
0.0135
--·---·-·----
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5. (2004 Nov #28) For (80) and (84), whose future lifetimes are independent:
x Px
80 0.50
81 0.40
82 0.60
83 0.25
84 0.20
85 0.15
86 0.10
Calculate the change in the value 21 q 8084 if ps2 is decreased from 0.60 to 0.30. (A)
0.03
(B)
0.06
(C)
0.10
(D)
0.16
(E)
0.19
6. Consider the set up in Example 11.6. Calculate Var(T-). xy 7. (2003 Nov #39) You are given: (i)
Mortality follows De Moivre's law with co= 105.
(ii)
(45) and ( 65) have independent future lifetimes.
Calculate e4s:6s . (A)
33
(B)
34
(C)
35
(D)
36
(E)
37
8. (2001 Nov #33) You are given: (i)
(ii)
The survival function for males is S(x) = l -_..::_, 0 :-: :; x :-: :; 75. 75 . Female mortality follows De Moivre's law.
(iii) At age 60, the female force of mortality is 60% of the male force of mortality. For two independent lives, a male age 65 and a female age 60, calculate the expected time until the second death. (A) 4.33 (B)
5.63
(C)
7.23
(D)
11.88
(E)
13.17
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9. (2005 Nov #21) For a population whose mortality follows De Moivre's law, you are given: (i)
e 40,40 = 3 e 60:60
(ii)
e 20:20 = k e 60:60
0
Calculate k. (A)
3.0
(B)
3.5
(C)
4.0
(D)
4.5
(E)
5.0
10. (2006Nov#17) In a population, non-smokers have a force of mortality equal to one half that of smokers. For non-smokers, lx = 500( 110 - x), 0 ::; x ~ 110. Calculatee202s for a smoker (20) and a non-smoker (25) with independent future lifetimes. (A)
18.3
(B)
20.4
(C)
22.1
(D)
24.5
(E)
26.8
11. (1988 Exam 150 May #8) For two independent lifetimes (i)
(ii)
T.y and f;,,
you are given:
E(T.y) = E(f;,)
= 4.0 Cov(T,Y, T-;) = 0.09
Calculate E(T_yy).
(A) 2.0 (B)
2.8
(C)
3.7
(D)
4.0
(E)
4.3
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2. Statuses Involving the Order of Deaths 12. (2006 Nov #25) You are given:
(i)
The future lifetimes of (40) and (50) are independent.
(ii)
The survival function for (40) is based on a constant force of mortality, µ
(iii) The survival function for (50) follows De Moivre's law with
ffi
= 110 .
Calculate the probability that (50) dies within 10 years and dies before (40). (A)
10%
(B)
13%
(C)
16%
(D)
19%
(E)
25%
13. (1990 Exam 150 May #1 I) You are given:
(i)
Tr and Ty are independent.
(ii)
The survival function for (x) follows De Moivre's law with co= 95.
(iii) The survival function for (y) is based on a constant force of mortality µ. Determine the probability that (x) dies within n (< 95 - x) and predeceases (y). e-101
(A)
95-x ne-101
(B)
(C)
95-x 1-e-P"
µ(95-x) I - e-101
(D)
95-x -101
(E)
1-e -101 +e- -
95-x
14. (1990 Exam 150 Nov #30) You are given
(i)
(x) is subject to a uniform distribution of deaths over each year of age.
(ii)
(y) is subject to a constant force of mortality of 0.3.
( ii1")
qxyI
(iv)
Tr and Ty are independent.
=
0 •045 •
Calculate qx.
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= 0 .05
.
Chapter 11: Multiple Life Functions
(A)
0.052
(B)
0.065
(C)
0.104
(D)
0.214
(E)
0.266
15. (1985 Exam 150 Nov #18) If mortality follows De Moivre's law with co= 80, calculate "'q 607~. (Assume lifetimes are independent.)
(A)
0.55
(B)
0.60
(C)
0.65
(D)
0.70
(E)
0.75
16.(1991Exam150May#18)Youaregiven (i)
Mortality follows De Moivre's law with co= 100.
(ii)
Tso and Tss are independent.
(iii) G is the probability that (80) dies after (85) and before 5 years from now. (iv) His the probability that the first death occurs after 5 and before 10 years from now. Calculate G + H. (A)
41/144
(B)
7124
(C)
1/3
(D)
53/144
(E)
3/8
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Chapter 11: Multiple Life Functions
Section 11.2 Insurance and Annuities
1. Fully Continuous Insurances and Annuities 17. (2002 Nov #24) For a special fully continuous whole life insurance of 1 on the last-survivor of (x) and (y), you are given:
(i)
T_y and Ty are independent.
(ii)
µy+t
= µy+t = 0.07 't > 0
(iii) 8 = 0.05 (iv) Premiums are payable until the first death. Calculate the level annual benefit premium for this insurance .. (A)
0.04
(B)
0.07
(C)
0.08
(D)
0.10
(E)
0.14
18. For a special fully continuous last survivor insurance of 1 on (x) and (y), you are given:
(i) (ii) (iii)
T.t and Ty are independent. µy+t = 0.08, t > 0 µy+t
(iv) 8 (v)
=
{°.04 0.08
0 < t < 20 t;:::: 20
= 0.06 T-
z, = v '"
Calculate Var(Z1).
2. Fully Discrete Insurances and Annuities 19. Refer to the set up in the previous question. Let Z2 = vK.,+t. Calculate E(Z2).
------·----
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20. (2000 May #20) For a last-survivor insurance of 10,000 on independent lives (70) and (80), you are given: (i)
The benefit, payable at the end of the year of death, is paid only if the second death occurs during year 5.
(ii)
Mortality follows the following life table. x 70 71 72 73 74 75
(iii)
Ix 66161.54 63966.08 61646.62 59203.93 56640.51 53960.80
Ix
x 80 81 82 83 84 85
39143.64 36000.37 32845.41 29704.95 26607.34 23582.45
; = 0.03
Calculate the actuarial present value of this insurance. (A)
235
(B)
245
(C)
255
(D)
265
(E)
275
21. (2002Nov#17) For a temporary life annuity-immediate on independent lives (30) and ( 40): (i)
; = 0.06
(ii)
f 30 = 95013. 79, 150 = 89509 .00
(iii)
a30AO
= 14.2068' G40.so
= 12.4784
Calculate a 3040 i0i. (A)
6.64
(B)
7.17
(C)
7.88
(D)
8.74
(E)
9.86
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22. (1989 Exam 150 May #30) Z is the present-value random variable for a discrete whole life insurance of I issued to (x) and (y) which pays l at the first death and l at the second death. Given: (i)
ax= 9
ay= 13
(ii)
(iii)
i
= 0.04
Calculate E(Z).
(A)
0.08
(B)
0.28
(C)
0.69
(0)
1.08
(E)
1.15
3. Reversionary Annuities 23. (2005 May #36) You are pricing a special 3-year annuity-due on two independent lives, both age 80. The annuity pays 30,000 if both persons are alive and 20,000 if only one person is alive. You are given: (i)
pso = 0.91, 2Pso = 0.82, 3pso = 0.72
(ii)
i
= 0.05
Calculate the actuarial present value of this annuity.
(A)
78,300
(B)
80,400
(C)
82,500
(0)
84,700
(E)
86,800
24. (2006 Nov #3 7) You are pricing a special 3-year life annuity-due on two lives each age x, with independent future lifetimes. The annuity pays l 0,000 if both persons are alive and 2000 if exactly one person is alive. You are given: (i) (ii)
(iii)
= 0.04 qx+lx+1 = 0.01 i = 0.05 qxx
Calculate the actuarial present value of this annuity. (A)
27,800
(B)
27,900
(C)
28,000
(0)
28,100
(E)
28,200
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25. (2001 May #23) A continuous two-life annuity pays: l 00 while both (30) and (40) are alive; 70 while (30) is alive but (40) is dead; and 50 while (40) is alive but (30) is dead. The actuarial present value of this annuity is 1180. Continuous single life annuities paying l 00 per year are available for (30) and (40) with actuarial present values of 1200 and 1000, respectively. Calculate the actuarial present value of a two-life continuous annuity that pays l 00 while at least one of them is alive. (A)
1400
(B) . 1500
(C)
1600
(D)
1700
(E)
1800
4. Contingent Insurance 26. ( 1989 Exam 150 May #2) You are given (i)
(70) and (75) are independent lives.
(ii)
Mortality follows De Moivre's law with w = 100.
(iii)
a = 8.655. 75
Calculate A7 ~. 75 • (A)
0.2473
(B)
0.2885
(C)
0.3462
(D)
0.4167
(E)
0.6606
27. (1988 Exam 150 Nov #3) You are given (i)
Z is the present-value random variable for an insurance on the lives of (x) and (y), where Z=
(ii)
{
v7~ 0
(x) is subject to a constant force of mortality, 0.07.
(iii) (}1) is subject to a constant force of mortality, 0.09.
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(iv) (x) and (y) are independent lives.
(v)
o= 0.06.
Calculate E(Z).
(A)
0.191
(B)
0.318
(C)
0.409
(D)
0.600
(E)
0.727
28. ( 1986 Exam 150 Nov Afternoon #2) A life insurance on John and Paul pays: (i)
1 at the death of John if Paul is alive.
(ii)
2 at the death of Paul if John is alive.
(iii) 3 at the death of John if Paul is dead. (iv) 4 at the death of Paul if John is dead. John and Paul are both age x and they follow the same mortality law. Calculate the actuarial present value of the insurance provided.
(A)
7Ax -2A_u
(B)
7A x -4A xx
(C)
lOAX -2A_,._,
(D)
10Ax -4A,._,
(E)
lOAx -5A_n
29. (1989 Exam 150 Nov #22) Z is the present-value random variable for an insurance on the lives of Bill and John. This insurance provides the following benefits: (I) 500 at the moment of Bill's death if John is alive at that time; and (2) 1000 at the moment of John's death if Bill is dead at that time.
You are given: (i)
Bill's survival function follows De Moivre's law with co= 85.
(ii)
John's survival function follows De Moivre's law with co= 84.
(iii) Bill and John are both age 80. (iv) Bill and John are independent lives. (v)
i
= 0.
Calculate E(Z).
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(A)
600
(B)
650
(C)
700
(D)
750
(E)
800
30. Show that for two independent lifetimes
T.x and Ty under Gompertz's law,
= ex + d'.
where c"'
31. (1990 Exam 150 May #22) You are given: (i)
Mortality follows Gompertz's law for both (35) and (40).
(ii)
c
10
=4
= 0.6
(iii)
A35AO
(iv)
T3s and T40 are independent. -1
Calculate A 35 A 0 • (A)
0.2
(B)
0.24
(C)
0.28
(D)
0.30
(E)
0.32
32. (1986 Exam May afternoon #6) Mortality follows Gompertz's law with µx
= 0.5(1. lyt.
Find A4 ~ 50 I A4050 • (Assume (40) and (50) are independent.)
(A) 0.26 (B)
0.28
(C)
0.36
(D)
0.38
(E)
0.46
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33. (1987 Exam 150 May #22) A whole life insurance pays a death benefit of 1 upon the second death of (x) and (y). In addition, if (x) dies first, a payment of 0.5 is payable at the time of his death. Mortality follows Gompertz law. Calculate the net single premium for this insurance.
-
( "l c 2cw
A
(B)
2(:4, + A,)-A,,( 2 + ;;..
(C)
A..
"'
cw
l
where
cw = cx+y
l
where cw
= ex+ d'
where c11'
= ex+ d'
(1 + ~·. J
(D) 2cA, + A,J-A. ( 2 + ;;. (E)
= cx+y
where
l+-
(A)
A, + A, - A.( I -
;;..
l
34. (1989 Exam 150 Nov #11) You are given:
(i)
(x) and (y) are independent lives.
(ii)
(ww) is the equivalent joint equal age status for (.xy) assuming Makeham's law with
c=2. (iii) (z) is the equivalent single life status for (xy) assuming Gompertz's law with c = 2. Calculate z - w. (Remark: Assume the two lifetimes in (ww) are independent.) (A) -2 (B)
-1
(C)
0
(D)
1
(E)
2
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Chapter 11: Multiple Life Functions
5. M-th Payable Multiple Life Benefits
35. (1986 Exam 150 Nov Afternoon #16) Mortality for each life (x) and (y) follows De Moivre's law with co= 100. If (x) and (y) are independent, which of the following are true? ;
-
(i)
Axy---A xy'
(ii)
F:;:;; (t) = Fx (t)Fy (t)
(iii)
0
klsqxy
= kfJxy s qx+ky+k, where k is a positive integer and 0 :s; s :s; 1
(A) None (B)
(i) only
(C)
(ii) only
(D) (iii) only (E)
The correct answer is not given by (A), (B), (C) or (D)
36. A man and his wife are aged 30 and 27, respectively. The male survival distribution is modeled by Makeham's law with A= 0.0001, B = 0.0004 and c = 1.075. The female survival distribution is modeled by Gompertz's law with B = 0.0003 and c = 1.07. It is given that under an effective interest rate of 6% per year, a3,~:;30i = 13.1250 and a~/~~ = 13.92046. Assume lifetimes are independent. (a) Use the UDO approximation for the joint life status to calculate (b) Use the Woolhouse formula to approximate
a( 12 >;~~JOi
a02 l ;~~JOi
and a{~~.
and a;~?.
Compare the approximated values with the true values 12.70779 and 13.41516. 37. A man and his wife are aged 30 and 27, respectively. They are about to purchase an insurance policy that pays I 00,000 immediately on the first death. Premiums P are payable monthly in advance as long as both are alive and limited to 30 years. Mortality is as specified on the basis in the previous question. The interest rate is 6% per year effective. (a) Calculate the net (monthly) premium for the policy. (b) Calculate the gross (monthly) premium for the policy, assuming an initial expense of 250 and renewal expense of 3% of each premium (except the first).
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Section 11.3 Dependent Life Models (Sample Question 122A- C) For Questions 38 to 40, use the following information: You want to impress your supervisor by calculating the expected present value of a last-survivor whole life insurance of I on (x) and (y) using multiple-state methodology. You defined states as State State State State
0 = both are alive 1 = only (x) is alive 2 =only (y) is alive 3 = neither is alive
You assume: (i) Death benefits are payable at the moment of death. (ii) The future lifetimes of (x) and (y) are independent. 01 02 13 23 - 0 06 >0 ( iii) µx+t:y+r - µx+t:y+r - µx+t:y+I - µx+t:y+t · ' ( 03
• ) ( IV
µx+1:y+1 --
(v)
o= 0.05
0' t> - 0
Your supervisor points out that the particular lives in question do not have independent future lifetimes. While your model correctly projects the survival function of (x) and (y), a common shock model should be used for their joint future lifetime. Based on her input, you realize you should be using µ~!i:y+i = 0.02, t ~ 0. 38. To ensure that you get off a good start, your supervisor suggests that you calculate the expected present value of a whole life insurance of 1 payable at the first death of (x) and (y). You make the necessary changes to your model to incorporate the common shock. Calculate the expected present value for the first-to-die benefit.
(A) 0.55 (B)
0.61
(C)
0.67
(D) 0.73 (E)
0.79
39. Having checked your work and ensured it is correct, she now asks you to calculate the probability that both have died by the end of year 3. Calculate that probability.
(A) 0.03 (B)
0.04
(C)
0.05
(D)
0.06
(E)
0.07
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SoA Exam MLC
Chapter 11: Multiple Life Functions
40. You are now ready to calculate the expected present value of the last-to-die insurance, payable at the moment of the second death. Calculate the expected present value for the lastto-die benefit. (A)
0.39
(8)
0.40
(C)
0.41
(D)
0.42
(E)
0.43
41. (2000 May #34) For a last-survivor whole life insurance of 1000 on (x) and (y): (i)
The death benefit is payable at the moment of the second death.
(ii)
The independent random variables shock model.
(iii)
T.r *has an exponential distribution with a failure rate of 0.03, t 2 !;.* has an exponential distribution with a failure rate of 0.05, t 2
(iv)
T.r *, T;.*
and Z are the components of a common 0. 0.
(v)
Z, the common shock random variable, has an exponential distribution with a failure rate of 0.02, t 2 0.
(vi)
o= 0.06
Calculate the actuarial present value of this insurance.
(A)
0.216
(8)
0.271
(C)
0.326
(D)
0.368
(E)
0.423
42. (2004 Nov #20) The mortality of (x) and (y) follows a common shock model with components T.r *, T;.* and Z.
T.r *, T;.*
(i)
and Z are independent and have exponential distributions with respective forces µ1, µ2 and .-1.
(ii)
The probability that (x) survives 1 year is 0.96.
(iii) The probability that (y) survives 1 year is 0.97. (iv) .-1=0.0l Calculate the probability that both (x) and (y) survive 5 years.
(A) 0.65 (8)
0.67
(C)
0.70
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Chapter 11: Multiple Life Functions
(D)
0.72
(E)
0.74
43. (2007 May #20) For professional athletes Derek and A-Rod: (i)
Professional athletes are subject to a constant total force of mortality µ = 0.001.
(ii)
Professional athletes are subject to a constant force of mortality due to crashes of the team airplane ;1 1> = 0.0002.
(iii) Mortality of athletes on the same team follows a common shock model, where all team members die if the team plane crashes. (iv) Future lifetimes of athletes on different teams are independent. (v)
Derek and A-Rod are on the same team now, but after one year will play for different teams.
Calculate the probability that both Derek and A-Rod survive two years. (A)
0.9958
(B)
0.9960
(C)
0.9962
(D)
0.9964
(E)
0.9966
44. (Sample Question #194) For a multiple state model of an insurance on (x) and (y): (i)
The death benefit of 10,000 is payable at the moment of the second death.
(ii)
You use the states: State 0 = both alive State I =only (x) is alive State 2 = only (y) is alive State 3 = neither alive
(iii)
,u~!iy+i = µ_~~i:y+i = 0.06, t ~ 0
(1'1'1')
Jl. T;>+m) = Tt;>+,,,, Letting A
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Chapter 11: Multiple Life Functions
we get Pr(B I A)= Pr(Txy+m < n) = nqxy+m· However, letting A =Cf; 2 m}, B ={Tg; < /11 + n} would not lead to a similar result. The reason is that A is a disjoint union of three events {Tx 2 /11 and Ty 2 111}, { Tx 2 /11 and Ty< m} and {Tx < m and f;. 2 m}, while the given condition for the conditional death probability qx+my+m is {Tx 2111 and Ty 2 m}. 11
4. We use symmetric relation because we have the probabilities for four life statuses (35), (40), (45) and (50), from which we can calculate the deferred probabilities for (35) and (40): 51q35:45 = 51q35 + 51q45 - 51q35:45 = sP3s q 40 + sP 45 q so - sP35:45 q 40:50 = 0.9 x 0.03 + 0.8 x 0.05 -0.9 x 0.8 x (I - p 4050 ) = 0.067 -0.72(1-0.97 x 0.95) = 0.01048 So, the answer is (B). 5. We use symmetric relation to shift the last survivor status to joint life status, for which the deferred probability can be more easily handled: 21q80:84 = 21q80 + 21q84 - 21q80:84 =
2P80
(I -
P82)
+ 21q84 -
2P80
2P84 q 82:86
= 0.2-0.2p 82 + 21 q 84 -0.2 x 0.03(1- P 82 P 86 ) = 0.194- 0.2 p 82 + 21 q 84 + 0.0006 p 82 = 0.194+ 2lq84 -0.1994p82 If ps2 is decreased from 0.6 to 0.3, 21 q80,84 would increase by 0.1994 So the correct answer is (B).
x
0.3 = 0.05982.
6. By Var(Tr;) + Var(Tg;) = Var(T") + Yar(T,,)-2[E(T")- E(Try)][E(Ty)-E(T,y)], Yar(Tg;) = Yar(T") + Var(Ty )- Var(Try) - 2[E(Tr) - E(T,,,)][E(T,,) - E(Try)] =50
7.
2
+(1~ 0 r-20 2 -2(50-20)(1~ 0 -20)= 21 ~ 00
E(T4s)= 105-45 =30, E(T6s)= 105-65 =20 2 2 The maximum of T4565 is min(105 -45, 105 - 65) = min(60, 40) = 40. To find E(T4565 ), we use the short-cut formula:
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Chapter 11: Multiple Life Functions
E(T45 .65 ) .
40 1 40 2 = - - - - = 15.56. 2 6 60
By symmetric relation, E(T4565 ) = 30 + 20 - 15 .56 = 34.44, which is closest to (B). 8. The limiting age for males is 75. We find the limiting age for females. So let it be 1 force of mortality at age x for a life following De Moivre's law is - - . As a result,
OJ.
The
OJ-X
--=0.6x
w-60
and
OJ=
1
75-60
,
85. The expected time until the second death is, by symmetric relation, E(T;~') + E(T6~) - E(r;;:fo) .
65 85 60 1 . . = 5 , E(T601 ) = = 125 . , an d smce t h e maximum o fT"' 65 :60 2 2 is min(75 - 65, 85 - 60) = min(l 0, 25) = I 0,
. 1y, E(T"') Ob v1ous = 65
75
E(T"'J) = 65 60 ·
.!_Q _ _!_I 02 = 4_!_. 2
6 25
3
The correct answer is given by 5 + 12.5 - 4_!_ = 13.167, which is (E). 3 9. Suppose that the limiting age is we have E(T-) = xy
OJ.
For the short-cut formula for E(T-) when OJ1 -x = OJi - y, xy
2(0J-X) . So statement (i) means 3
2
2
J(OJ-40) = 3x J(OJ-60), which means
By putting
70. Statement (ii) means 2 2 -(OJ-20) = k x-(OJ-60). 3 3 70, we get k = 5.
OJ=
OJ=
I 0. For non-smokers, the lifetime distribution follows De Moivre's law with a limiting age of f
NS
110. Hence, 1 p 25 = 1-
f
=I - - . 110- 25 85 Since the force of mortality of smokers is S -[
I
P20 -
F;· = 2µ.:s,
-(1-
NS]2 1P20 -
2
f
110 - 20
)
-(1--{) 2
-
90
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Chapter 11: Multiple Life Functions
E(TSNS) = f85(1--t )2(1-!_ 20.25 Jo 90 85
=
r
\it
J:'(t- :5 8::0)(1-:} +
85 = f (1-0.033987t + 0.0003849t 2 -
Jo
= 85 - 0.0169935(85
2
)
t }t 3
688500
+ 0.0001283(85 3 )
854 -
2754000
= 22.06 and hence the answer is (C). I I. Recall that Cov(T,Y, T-;;) = Cov(Tr, TY)+ [E(TJ- E(Try )][E(Ty) - E(Try )] . When T.~ and Ty are independent, the covariance term on the RHS is 0. Putting all the information in (ii) and (iii) into this equation, we have 0.09 = [4 - E(Txy)]2 and hence E(T.xy) = 3.7. So the answer is (C). 12. We need to compute 10 which has a density of
q! 040 . Since the
number is attached to 50, we condition on (50),
I f 50 (t)=- for0'
+ /l )dt
= (JLy + /l) fo"" e-01e-(fly+AJ1 (1-e-fl,t)dt
= (Jly + /l)(
Jly
1 + /l + o
flx(µy
flx
J
1 + µ>' + ll + o
+ /l)
46. (a) The reversionary annuity only pays 1 unit if the continuous-time Markov chain is in state I. The transition probability is 01 f1 00 01 lid 1 Pxy = Jo s Pxy µx+s:y+s 1-s Pxy S =
I:
e-0.18s O. le-0.12(1-s)ds
= _QJ_ e-o 121 (I _ e-0.061) 0.06 5 ( e -0.121 =3
-e -0.181)
As a result, the reversionary annuity has an APV of aylx
= fo°'e-0051 xl(e-0121 -e-Ol81)dt =
(b) Since
1 Px
=
2(-
1 1 - - - - ) = 2.5575 3 0.17 0.23
5 5e--0.121 _ 2e--0.181 oo +poi =e-0.181 +-(e-0.121 -e-o.181)=-----1Pxy I -'Y 3 3
the APV of the 10-year continuous life annuity on (x) is 10 e-0.051 X-(5e-O.l21 1 Q _ = -2e-O.l81)df x:lOI o 3
1
= _!_(5(1- e-u) - 2(1- e-23)) = 5.404961 0.17 0.23 3 (c) Obviously,
axylOI-
io 1-e-23 = f e-0·051 xe- 0181 dt= =3.911918. Jo 0.23 1
I
02 = Jof
Pxy
=
0~µ02., . p22ds x+s.J+.1 1-s xy
s Px)
J: e-o 18s 0.0 8e-o.12(1-s)ds
= 0.81P.~:
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As a result ,
5 --0.121 _ e-0.181 4 _ p = poo + po2 =e-0.181 +-(e-0.121 -e-o.181)xO 8 =-e _ _ __ I y I xy I xy 3 . 3
a-
y:I~
= f 10 e-0.051 x_!_(4e-0121 -e-018/)dt Jo 3 23 = _!_( 4(1-e-l.7) - (l-e- )] = 5.106353 3 0.17 0.23
By symmetric relation, Zi:;:loi = ax:loi + ay:loj - a,y:loj = 6.599396 . (d) The APV of the contingent insurance is -I f"' I 00 02 d f"' -0.051 -0.181 0 08d 0.08 8 Axy =Jo v iPxyµx+1:y+1 t =Joe e . t = 0.23 = 23.
(e) Method l: Similar to (d), -2
Axr
f"'
01 13 Pxyµx+1:y+1dt -0.121 -0.181) = f"' e -o 051 5( e - e 0. l 2dt Jo 3
= Jo
I
V
1
_l_)
= 5 x 0.12 (-1- = 0.306905 3 0.17 0.23
A, = /i;y + /i~
= o.65473
Method 2: Similar to (b),
a x
= f"' e-0051 X_!_(5e-0121 Jo 3
-2e-018t)df
2 5 = _!_(-- - -- ) = 6.90537 3 0.17 0.23 Ax = l-0.05ax = 0.65473 (f) Similar to Method l in (e), 51
-2 Axy:SJ =
=
flO I 01 13
Pxyfl.
0-(-0.082618M))~ o.o 5 .Jo.068034M
Pr(z > 0.082618M )
~ o.o 5
.J0.068034M 0.082618M
z l. 645
.J0.068034M M z 26.97. Hence, the answer is (B). 24. Let 0L be the prospective loss random variable for a single contract. We have
Jr)
000 + d V K'"+I o L -100 ' 000 v K,"+I - Jra.. Kw+ll -(100 '
1rd '
E(L0 ) =(100,000+ 1r)A60 - n, d .d and
.
1r )' (2 A
Var(0 L)= ( 100,000+ d
60
2
-A60 ).
Let 0L* be the aggregate prospective loss random variable for 10,000 policies. We have E(oL*) = IO,OOOE( 0L) and Var(0 L*) = J0,000Var(0 L). We require
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Chapter 12: Interest Rate Risk
Pr(
0
Pr(z >
L* > 0) = 0.01 O-E(.L*) )=0.01 ~Var( 0 L*)
-E(.L*)
2.326
~Var( 0 L*) -10,000( (100,000+; )A60
-
;
)
2.326
2
10,000(toO,OOO+; )' ( A60 -A:,) -100((100,000+
tr
0.06/1.06
)o.36913-
tr
)
0.06/1.06 = 2.326
( 100,000+ 0.06tr/1.06 )'-'0.11141-0.36913' tr" 3379.
Hence, the answer is (C). 25. The uncertainty associated with future lifetimes of policyholders is diversifiable, because some policyholders live longer than expected while some live shorter than expected. The uncertainty associated with interest rates and that associated with death probabilities are non-diversifiable, because any error in estimating interest rates or death probabilities would affect all policies in a portfolio. Hence, the answer is (A).
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~12-3~ -.
Chapter 12: Interest Rate Risk
,+
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Chapter 13: Profit Testing
Chapter 13
Profit Testing
1.
To calculate emerging surplus, profit vector and profit signature
2.
To calculate profitability measures of a cash flow stream
In Chapter 7 we discussed the calculation of profit in a particular policy year for a group of independent and identical policies. In this chapter we expand our discussion on the calculation of profit to a variety of profit measures .
•
13. 1 Profit Vector and Profit Signature
In Chapter 7, the profit over a period was defined as •
the accumulated value (using the period effective interest rate) of the sum of the reserve at the beginning of the period and the cash flows that occur at the beginning of the period
less •
the sum of the value of the reserve at the end of the period and the cash flows that occur at the end of the period.
Expected profit is the profit calculated using the values that were anticipated for the next time period prior to the start of that time period, while the actual profit is the profit calculated using the values that were observed during the time period.
Suppose that we look at a fully discrete policy, and that there are N policies in force at the beginning of the (h + 1)th policy year. The insurance company holds a reserve of h yg for each insured.
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Chapter 13: Profit Testing
Total reserve = N,,V" at time h
- Death benefit Nqx+hbh+I - Settlement expense Nqx+11Ei.+1 - Total reserve needed= (Npx+h) h+I V" time
h
h +I
+ Contract premium NG11 - fraction of premium paid for expense c1,NG,, - per pol icy expense Ne,, The expected profit that emerges in the (h +!)th policy year is then Pr,,+ 1= N[,, V" + G1,(l - CJ,) - e11](1 + i) - N[(bh+I + Ei,+1)qx+h + Px+h h+I V"] . Now we want to make a few observations concerning the calculation above. (I) The premium G,, is calculated under some specific assumptions on mortality, expenses and interest. The assumptions are collectively called the premium basis. (2) The reserves 11 V" and h+I V" are calculated under another set of assumptions on mortality, expenses, and interest, which is usually more conservative than the premium basis. This basis is called the reserve basis. (3) In the calculation of profit, we can use another set of assumptions on mortality, expenses and interest. This basis is called the profit testing basis. (4) It is customary to exclude acquisition costs in the first year in the calculation of the profit in the first policy year. The acguisition costs are incorporated as the negative of Pr0, so that it would stand out clearly. Since we would handle the effect of N later, we now let N = I for every policy year. This means that Pr,,+ 1 is the profit for a single policy in force at time h. So, the profit can be written as Pr,,+ 1 = [,,Ji"+ G11(l - CJ,) - e,,](I + I) - [(b11+1 + Eh+1)qx+h + Px+h h+I V"] = G11(1 - c,,)- e,,](l + i)- (bh+I + Eh+1)qx+h surplus emerging at time h + I
+
(1 + i)hV" -px+hh+I V"
change in reserve in year h + 1
for h = 0, 1, 2, .... For the first policy year, c,,G,, and e,, may not be included if they are treated as acquisition costs. For Pr0 , it is set equal to the negative of the sum of all acquisition costs.
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Chapter 13: Profit Testing
;~//////////////l"///////////////l"///////////////l"///////////////?'///////.b'////1:
:% :% ~ ~ :% ~ :% ~ ;% ~ ;% :% ~ :% ~ :% ~
. ·.. ·. . . ·....· .·. F 0 R M U. L A
%' ~ ~
Profit Vector and Profit Signature
% ~
The column vector
~ ~
~ ~:
Pr= (Pro, Pr1, Pr2, Pr3, ... )'
~ ~:
is called the profit vector for the contract. The column vector IT= (Ilo, I11, I12, Il3, ... )'=(Pro, Pr1, PxPr2, 2PxPr3, ... )' is called the profit signature for the contract.
%: ~ %; % %: ~
;~///////////////l"///////////////l"///////////////l"///////////////l"/////////////h: (Recall that A' is the transpose of matrix A.) The profit signature can be thought of as the expected (undiscounted) emerging profit at issue of the policy. Exam MLC may use this term to test your understanding. 0) suggests that it is a cash flow at time h + I, so that its
Notice that while the notation Pr,,. 1 (h
~
appropriate discount factor should be
11+ 1Ex
factor turns out to be v"+I
hPx·
= v"+J
h+IPx.
this is not the case! The correct discount
Look carefully at the structure of Pr,,+1:
[,, V" + G,,(1 - ch) - e1,](1 + i) - (bi.+1 + E11+1)qx+h + Px+hh+I V" time
h+I
h Conditioning on the survivorship at timeh
We have conditioned on survivorship at time h when we define Prh+J, and the probabilities
q,+h
and Px+il have already appeared in the second term when we calculate the expected loss at the end of the period. If we incorrectly multiply
Prh+I
by i.+1P.r =
1,Px Px+h
when we calculate its
expected value at time 0, we would have double counted the probability in (h, h +I).
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Chapter 13: Profit Testing
~
Example 13.1
We revisit Example 7.7 again. Suppose that the profit testing basis is the same as the premium basis, except that the settlement expense is 4. Calculate the profit vector and profit signature for the contract. -
Solution
Recall that the premium for this policy is G = I 02.42811, and the reserves are 0V =
0, 1V = 3.133132, 2V = 6.8058589, 3V = 0.
We now calculate the profit vector: Pro= negative of sum of initial expenses= -I 0 - 0.25G = -35.6070275 Pr1=Gx1.06-(0.08x1004+0.921V)=25.37!3152 Pr2 = ( 1V + 0.95G- 2) x 1.06 - (0.09 x I 004 + 0.912V) = 7.79289509 Pr3 = (2V + 0.95G- 2) x l.06 - (0.1 x 1004) = 7.8393172 So the profit vector is
Pr= (-35.6070275, 25.3713152, 7.7928951, 7.8393172)'. Noting that q40 = 0.08, q41 =0.09,
= 0.10, the profit signature is
q 42
IT= (-35.6070275, 25.3713152, 7.7928951 x 0.92, 7.8393172 x 0.92 x 0.91)'
= (-35.6070275, 25.3713152, 7.1694635, 6.5630764)'. [ END ]
Extension to Multiple State Models
The calculation of profit can be extended to multiple state models with cash flows payable at the end of every period. Here we assume that state changes are irreversible. Examples of such multiple state models include the multiple decrement model in Chapter 8, the permanent disability model in Chapter I 0, and the dependent life model in Chapter 11. The double decrement model with withdrawal and permanent disability model are particularly important.
We can calculate profit that emerges in the (h + I )th policy year, conditional on the state at time h being i:
.<
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Chapter 13: Profit Testing
n
Pren=[ vu> +Gu>(J-cu>)-ecn](I+i)"Cbw> h+I h h Ir h L..J h+I +EM+ h+I
ft+\
vu>)pijx+li
)""O
In the expression above, >](l + i) [ h vu>+ ou>(1-cu>)-ec h h " 1
is the accumulated value of the sum of the reserve at the beginning of the period and the cash flows that occur at the beginning of the period, and
" "Cbwi L...i h+I + Ewi /r+I +
h+I
vui) p x+h ij
f""O
is the sum of the expected value of the reserve at the end of the period and the cash flows that occur at the end of the period. After calculating the state-dependent profit vector, the profit signature can be obtained as follows:
" 'V ITO = Procoi, IT , -- L...i
1-1
Pxo1 p r,(1) ·
J=O
IT= (ITo, IT,, IT1, ... )'.
" 01 Notice also that IT 1 = 'V = 0p _,.00 PrcI 0 > = Prcoi because the insured must be m state 0 .L...JO p x Pru> t I j=O
during the first year.
Traditional Insurance Policies with Withdrawal
To be concrete, let us illustrate the above using a model with two modes of decrements: (I) withdrawal, and (2) death. In case of death, death benefit (together with settlement expenses) would be paid. In case of withdrawal, a withdrawal benefit (cash value of the policy) would be paid. In case of survival, a reserve is needed. So, (O) Prh+I
-
-
[
h
v(O)
+ G" (I -
ch
)-
eh
](I +1') - [ h+I vc•J Px+h (') +
h+I
C" (I) (b h+I + E h+l )qx+h (l) l h > 0 rqx-t-h + ' -
and 0> - 0 c2 > -- 0 ' h > Pr,,+, - ' Pr1r+1 - I
because the policy would be terminated immediately after withdrawal or death. For h = 0, we only need Pri 0 >=-acquisition costs. The profit signature can be computed from
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(cB-6)
________....,'--'"',.,,....._____________________________________________________________________ Chapter 13: Profit Testing
IT 1i --
(c) h-1P.r
p
r,,(O) , h > - 2·
Permanent Disability Model I
Instead of giving you equations, let us study an example.
~
Example 13.2
Consider the 3-year insurance with a 1000 death benefit payable at the end of the year of death in Example 10.4, and treat the set up there as the reserve basis. The reserves are found to be: oJIOl = 0, 1 J)Ol = -82.231, 2JIOl = -120.706, 3J)Ol = 0 1 Jill=
590.9576, 2Jl'l = 377.358, 3Jill= 0
k0 21 =0fork=
1,2,3.
Since the company does not hold negative reserves, the reserves under state 0 are all treated as zeros. Suppose that the yearly level premium of the policy is 260, and that under the profit testing basis, the following transition probability matrices are used:
0.25
0
P0 = 1 2
[°:'
0
2
0 0.75 0
'°']
0
0.~5 , P, = P, = 1 2
[':'
2 0.2 0.75
l
0 l o.~5 .
0
The interest rate is 8%. There is a settlement expense of 5, an initial expense of 50% of the first premium, and renewal expenses of 10% of the second premium and 5% of the third premium (the renewal expenses are payable even if premium is waived). Calculate the profit signature for the policy.
-
Solution
Pr,C'l= 260 x 1.08- [Ox 0.7 + 590.9576 x 0.25 + 1005 x 0.05] = 82.8106
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Chapter 13: Profit Testing
Pri'>= (260-26) x 1.08 - [Ox 0.7 + 377.358 x 0.2 + 1005 x 0.1] = 76.7484 Prj 0 l = (260 - 13) x 1.08 - I 005 x 0.1 = 166.26 Pri'> = (590.9576 - 26) x 1.08 - (377.358 x 0.75 + 1005 x 0.25) = 75.8857 Prj'>= (377.358-13) x 1.08- 1005 x 0.25 = 142.2566 (The profits under state 2 are all Os.) So, TI0 = -130, IIi = 82.8106, II2 = 0.7 x 76.7484 + 0.25 x 75.8857 = 72.6953 2
Pr(Y2 = 0 I Yo= 0) = 0.7 = 0.49, Pr(Y2 =I I Yo= 0) = 0.7 x 0.2 + 0.25 x 0.75 = 0.3275
Il3
= 0.49 x
166.26 + 0.3275 x 142.2566 = 128.0564.
The profit signature is TI= (-130, 82.8106, 72.6953, 128.0564)'. [ END]
Extension to Policies with Continuous Benefit In the previous example we have assumed that benefits are payable at the end of the year. What if benefits are payable immediately on death? In this case, a crude approximation is to assume that deaths occur in the middle of the year, and adjust the expected death benefits by (I + i)
112
•
This means we use
Example 13.3
,
Consider a I-year term insurance on (58). $100,000 is payable at the moment of death, while the quarterly premium of200 is throughout the term. The policyholder is subject to Makeham's law of mortality with parameters A = 0.0002, B = 3 x I 0-6, and c = 1.12. Under the reserve basis, -
expenses are ignored,
-
the mortality rates of a 60-year old is used,
-
the interest rate is 6%.
This results in the following quarter-end reserves:
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Chapter 13: Profit Testing
025 V=2.982944,
osV=4.035478,
01sV=3.071047
Under the profit testing basis, -
initial expenses are 300 plus 50% of the first quarterly premium; renewal expenses are 10% of each subsequent premium; settlement expense is 0.
-
the mortality rates of a 58-year old is used.
-
the annual effective interest rate is 8%.
Assuming deaths occur at the middle of each quarter, calculate the profit signature.
-
Solution
First we find all survival probabilities required in the calculation. By
Bex , ,Px =exp(-At---(c -1)), 1nc we get
x
58.25 0.999390199
58 0.999405833
0.25 x
58.5 0.999374117
58.75 0.999357572
We then also calculate the following:
t 1Ps8
I
I
0.25 o.999405833
I
I
0.75 0.998171264
o.5 o.998796394
Obviously, Pro= -300 - 0.5 x 200 = -400. Now we can calculate the profit emerging at each quarter end. 118 Pro.2s = 200 x 1.08 114 - 02sP58 x 025V- 02sqs8 x 100,000 x 1.08
= 140.913083
(Note: if the death benefit is payable at the end of each quarter, then Pr025 = 200 x 1.08 114 - 025p 58 x 025 V - 0 25 q 58 x 100,000 = 141.4874377) 114 118 Pr0 .5 = Co. 25 V + 200 x 0.9) x 1.08 - 0 25p 58 25 x o.5 V - o 25q58 25 x 100,000 x 1.08 = 120.935086 118 114 Pro 75 = (os V + 200 x 0.9) x 1.08 - 02sP58 s x o.75 V - o.isqs8s x 100,000 x 1.08 = 121.348214 118 Pr1=(o 15 V+200 x 0.9) x 1.08 114 - 02sq5875 x 100,000 x 1.08 = 121.763679 The profit vector is (-400, 140.913083, 120.935086 x 0.999405833, 121.348214 x 0.998796394, 121.763679 x 0.998171264)' = (-400, 140.913083, 120.863230, 121.202159, 121.541005)'. [END ]
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Chapter 13: Profit Testing
-
13. 2 Profit Measures
After we have calculated the profit signature, we can compute a variety of profit measures under the framework of discounted cash flow analysis and capital budgeting. All these measures are generalizations of profit measures for deterministic cash flows. So let's have a brief review of these profit measures in the simplest deterministic setting.
Consider an investment that leads to cash flows of Co, C1, C2, ... , Cn at times 0, 1, 2, ... , n.
1. Net present value (NPV) Given a risk discount rater (This is not the effective rate of discount d. We call it a risk discount rate, just because it is used to discount (contingent) cash flows.), the NPV of the cash flow stream is n
NPV(r) =
C
L ', k=o(l+r)
If the NPV is positive, then the investment is deemed profitable.
2. Internal rate of return (IRR) The internal rate of return is the zero of the equation NPV(r)=
" c L ' , =0. k=o(l+r)
If {Ck} changes sign only once, then the IRR is unique. Otherwise, there may be multiple IRRs, or no !RR (this is the case when all cash flows carry the same sign).
Very frequently an investor or a company would specify a minimum discount rate (or
hurdle rate) which is also known as the "required rate of return." If the IRR exceeds the hurdle rate, then the investment is deemed profitable. This rule works well if C,'s are initially negative, and then become positive.
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Chapter 13: Profit Testing
3. Discounted payback period (DPP) Given the hurdle rate r, the discounted payback period (also known as the break-even period) is the smallest value of m such that
c
I
Ill
k
k
~O.
•=o(l+r)
Loosely speaking, DPP is the time until the investment starts to make a profit. Ifno such m exists, then the investment never pays back.
4. Profit margin Profit margin is the NPV of net cash flows as a percentage of the NPV of the revenues. Suppose that the revenue cash flows are Ro, R1, R1, ... , Rn. Then the profit margin is NPV(r)
"
R
k=O
(I+ r)'
I
'
Miscellaneous Information: Calculator Trick
To find NPV when n is large, you can use BA II plus. Suppose that you have the following cash flow stream: Co= -450, C1 = 25, C2 = 320, Ci= 160 and r = I 0%. The following series of keystrokes calculates the NPV. First clear the cash flow worksheet by pressing IQ'.] ~ndl ICLR WORKj. Step I: Press IQ'.], the display should read CFO= Step 2: Press 450 ~ IEnte~ l±J, the display should read CO I
=
Step 3: Press 25 ~ l±J l±J (the 2nd arrow means we ignore FO I) the display should read C02 = Step 4: Press 320 IEnterl l±J l±J, the display should read C03
=
Step 5: Press 160 IEnterl, Step 6: Press~- The display should read I= 0. Press I 0 ~ l±J lcP1j. The NPV is -42.59955.
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Chapter 13: Profit Testing
To find the IRR when n 0, the discounted payback period is 2 years. (d) The premium is 260. The APV of premiums is 07 260(1 + · + 0.4;) = 530.7438. I.I I.I So the profit margin is I 01.5719 I 530.7438 = 19.1 %. (e) 101.5719/ 130= 78.I%. 2. (a) The profit signature is (--400, I40.913083, 120.863230, 121.202159, 121.541005)'. TheNPV is - 400 +
140.913083 120.863230 121.202159 121.541005 + + +---1.1025 I.lo.> 1.1°'' I.I
76.166.
(b) The NPV of revenues is 200(1 + 0.999405833 + 0.998796394 l.I 0·" 1.1° 5
+0.998 I 7I264) = 771.499756. 1.1°-"
So, the profit margin is 76.166 I 77I .499756 = 9.8725%. (c) We let r be the IRR. The yield equation is -400+ 140.913083 + I20.863230 + I21.202159 + 121.541005 = 0. 015 0 (l+r) 0·" (I+r) ' (l+r) l+r However, in this form we cannot use a financial calculator to solve for r. So, we get i be the interest rate per quarter. That is, I + r =(I + ;)4. Then -400+ I40.913083 + 120.863230 + I21.202I59 + 121.541005 = 0. l+i (l+i) 2 (l+i) 3 (l+i) 4 A financial calculator would give i = I 0.244649% as the solution. The annual effective 4 yield is r =(I+ i) - I= 47.716%.
1-
1
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Chapter 13: Profit Testing
3. From the Illustrative Life Table, we find q40 = 0.00278. This means p 40 = 0.99722. 0 99722 ). So, the NPV of the premiums is 100(1 + · l+r The expected profits given are said to be calculated assuming that the policy is in force at the beginning of the year. So the profit vector is (-275, 80, 238)'. The profit signature is (-275, 80, 238 x 0.99722)' = (-275, 80, 237.33836)'. . -275 + ~ + 237.33836 S O, t he Npv IS l+r (l+r) 2 Since the profit margin is -8.66%,
•
-275 +~ + 237.33836 - -0.0866x 100(1 + 0.99722) l+r (l+r) 2 l+r 237.33836 + 88.635925 266.34 = 0 2 (l+r) l+r By using a financial calculator, r is found to be 12.494%. [Alternatively, by using quadratic formula,-'-= 0.888939626 or-1.26 (rejected). l+r This would again giver= 0.12494. However, if the policy has a term greater than 2 years, you cannot solve the equation directly but have to rely on a financial calculator.] 4. Since expected profits given are said to be calculated assuming that the policy is in force at age 50 when the policy is issued, the profit signature is IT= (-275, -76, 122, 250, 218)'. From the Illustrative Life Table, the death probabilities are x
qx This gives 1
t pso
0.99408
2 0.98769801
3 0.98081375
So, the APV of the premiums, under a risk discount rate of r, is 100(1 + 0.99408+0.98769801x1.2 + 0.98081375 x l.2J. l+r (l+r)' (l+r) 3 . . 76 122 250 The NPV of the policy 1s -275---+ + + 2 l+r (l+r) (l+r) 3 Since the profit margin is 11.37%,
r(•,
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Chapter 13: Profit Testing
-275-__l!!___+ 122 + 250 + 218 3 4 l+r (l+r)' (l+r) (l+r) =O.ll 37 x!OO(l+ 0.99408 + 0.98769801xl.2 + 0.9808!375xl.2) 3 l+r (l+r)' (l+r) or -286.37
87.302690 + !08.523848 + 236.617777 + 218 =0. 3 4 1+r (l+r)' (l+r) (1 +r)
By using a financial calculator, r is found to be 14.9996%. 5. Statement (vi) in the question specify the expenses of 100, including the one incurred at time o+, are not acquisition expenses. Hence they are incorporated in the calculation of Pr 1, Pr2 , etc. Pro= -!000 Pr 1 = (14500- 100) x 1.06 - (0.014 x 1000000 + 0.986 x 700) = 573.8 Pr 2 = ( 14500 + 700 - l 00) x l .06 - (0.0 l 5 x I 000000 + 0.985 x 700) = 316.5 Pr3 = (l 4500 + 700 - l 00) x 1.06 - (0.0 l 6 x l 000000 + 0.984 x 0) = 6 The expected present value of profits at issue, under a risk discount rate of l 0%, is -1000+
573 8 316 5 · + ; x0.986+--;x0.986x0.985 =-216. 1.1 I.I I.I
6. The meaning of "immediately after the 7th year end" is t = 7. We need to calculate the sum of the APVs of Pr 8 , Pr9 and Pr 10 • From the Illustrative Life Table,
q61 = 0.02544, q68 = 0.02779. This gives
P61= 0.97456, 2P61=0.97456 x 0.97221=0.947477. Assuming that lapses happen only immediately before premium is due (that is, at the beginning of the year), the APV is vPr + 0.97456 x 0.95v2Pr + 0.947477 x 0.95 2v3Pr 8
=
140 1.1
10
9
+
13
I.I
~
x0.925832+
13
I.I
~ x0.8550980=314.09.
Thus the correct answer is closest to (C). 7. We treat all transitions as irreversible or else the question cannot be done. The expected profit is (6000 + 700- IO) x 1.06 12 - (0.9965 x 6200 + 0.0015 x (15000 + 50000) + 0.002 x 200000)
-'-
= 6690 x 1.06 12 - 6675.8 = 46. 764. So the answer is (B).
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8. The expected profits given in (v) are conditional on the survival to age 40, so these are profit signatures. By comparing (iv) and (v), we get P, note that under i = 5%, a(l2) = 1.000197, Al2) = 0.466508 and hence, assuming UDO over each year
of age, the APV is 49.433048S (1.000197 x 13.213 - 0.466508) = 630.22662S.
Step 4:
Now we equate (2) and (3) to obtain the smallest contribution rate: 7340.9695cS = 630.22662S. On solving, we get c = 8.59%. [ END]
The following is a more realistic example which makes use of a salary scale. We also illustrate how dates can be dealt with.
~
Example 15.6
An employer establishes a DC plan. The contribution rate is set using the following assumptions: -
Contributions are payable monthly in arrears at a fixed percentage of the salary rate at that time.
-
Contributions earn investment returns of I 0% effective per year.
-
The salary scale is given by Sx = 1.045·' and salaries are assumed to increase continuously.
-
Upon retirement, the employee will use the proceeds to buy a I 5-year certain 20-year life annuity due, payable annually. The annuity is priced at an assumed interest rate of 3% effective per year.
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Chapter 15: Pension Mathematics
A new entrant has a date of birth of I April 1985. The valuation date is I July 2010. Calculate the smallest contribution rate required to meet a target replacement ratio of 70% if the entrant retires on 31 December 2044. Assume a Makeham survival law µx =A + Be', with A = 0.0004,
B=4x 10-6 ,c= 1.13.
-
Solution
Step I:
Since the contribution is proportional to salary rate (but not salary), we project the future salary rates. Let the annual salary rate of the entrant at the valuation date be S. Since the new entrant is aged 2010.5 - 1985.25 = 25.25 now, the (annual) salary rate after k months is S s""'+•m-os s 25.25-0.S
= S(l .045)"".
The entrant plans to retire on 2045, which is (2045 - 2010.5) x 12 = 414 months from now. Sok= I, 2, ... , 414.
Step 2:
Let the contribution rate be c. Then the accumulated value of all contributions is k
414
414
S "' ( _!._ 045J' 34 'L CL S(!. 045)12 (l.1)_ -• =~(1.1)
112
12_
,;,
'°'
I.I S !)"' 0.95 415112 - 0.951112 =~(I 12 . 0.95 1112 -1 = 432.4425cS
12
12
Note that the rate obtained in Step I is an annual rate, and hence we have to divide the rate by 12 to obtain the monthly salary.
Step 3:
The entrant plans to retire on 2045 when he is 59.75 years old. S(l .045) 34 •
The final-year salary is S s 58 ·75 s 25.25-0.5
So, the target pension benefit per year is 0.7S(l.045) 34 = 3.126453S, and the APV of the pension is 3.126453S(ii15I
+,, ii 1
-
59.75.S 1
).
To calculate this annuity, we note that under
a Makeham survival law, , p 5975 =exp(- At
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5975
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Chapter 15: Pension Mathematics
15 0.770129
t , p59_75
16 0.740014
17 0.707435
18 0.672378
19 0.634884
Thus,
..
..
ai5j+ 151 a 5975 ~ =
1-1.or"
1 ' 1 +I--,, p 5915 = 12.296073+2.140495=14.43657 0.03I1.03 ,. 15 1.03
and the APV of the pension is 45. l 353S. Step 4:
We equate (2) and (3) to get 432.4425cS = 45.1353S. Hence, the smallest coniribution rate is c = 10.44%. [ END)
Now suppose that the investment rate of return in the accumulation period is only 8%. Then the accumulated value of all contribution would become k
414 S(l 045)12 cI · •·I 12
S 12- = ~ o.08)" o.08)-414-•
(I
414 045)"12 I -·12 k•I J.08 415112 1112 = cS (l.0 )"' (1.045/1.08) -(J.045/1.08) 8 12 (1.045I1.08) 1112 -1 5
= 292.8653cS The corresponding smallest contribution rate would be
45.1353 292.8653
15.41%, which is about
50% higher than the original contribution rate! If the original contribution rate is used, then the yearly payment of the life annuity that the retiree can purchase is 292.8653cS /14.43657 = 20.28635cS = 2.1179S
. l . . l . f w h1c 1 imp 11es a rep acement rallo o
2. l l 79S S(l .045) 34
47.4% only. An apparently small difference
between the assumptions used to set the contribution rate can make a huge difference to the actual replacement ratio.
So you can see the risk of a DC plan borne by the plan members. In the DB case, such risk is taken by the employer, whose contributions are adjusted when the shortfall becomes apparent.
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Chapter 15: Pension Mathematics
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15. 4 DB Plan and Service Table
For a defined benefit plan pension plan, the retiree receives a pension benefit that is a function of his or her age of entry, time of retirement, and salary received over part or the entire career. In Section 15.2 we discussed final salary plan and career average salary plan. Now let us go into these and a few more plans in some further detail.
Suppose that a person enters a company at age x exact and retires at age x + t exact. His annual salary over age (x, x + 1), (x + 1, x + 2), ... , (x + t - I, x + t) are denoted by AS.,, ASx+ 1,
... ,
ASx+1-l·
Final salarv plan For a final salary plan, the annual retirement income is a fraction (100d%) of the final salary: Annual retirement income= d(ASx+i-1)
Final m-year average salary plan For an m-year average salary plan, the annual retirement income is a fraction of the average of the final m years of salary. Typically, m = 3, 4 or 5. Annual retirement income= d x ASrn-1 + ASx+1-2 + · · · + ASw-m
m
Career average salary plan For a career average salary plan, the annual retirement income is . . Annua l retirement income
= d x AS x+I- I + AS x+I- 2 + ... + AS
.T
t
There are some other variants as well. For example, some plans have annual retirement income expressed as a multiple of the number of years of service t.
If the retirement income benefit is payable monthly for the life of the retiree, then the total liability of the company, evaluated at the time when the employee retires, would be (Annual retirement income) x 'a;~>,
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Chapter 15: Pension Mathematics
where the superscript r in the annuity symbol emphasizes that the mortality for which the annuity is computed is the selected mortality for a retiree, which may or may not be the same as the general population.
For regulatory purposes, most companies need an estimate of the APV of the retirement benefit at a time point when their employees are still in service. Suppose that at the valuation date, an employee is at age x + h exact: ASx
x
ASx+I
x+I
x+2 ... x+h-lx+h
x+h+t
As in the figure above, both the time of retirement, as well as the salary after time h, are unknown. As a result, to calculate the annual retirement income and the APV of retirement benefits at time h, we need to ( 1) project future salary, (2) build a model for the time ofretirement and compute the probability of retiring at age x + h + I for various values oft,
(3) compute the APV based on (1) and (2).
Let us explore each step in some detail below.
Step I: Projeclingfuture salary This is exactly what we have done before. If we know the past salary at the valuation date (that is, the salary earned over age (x + h - 1, x + h)), we can project future salaries by using Projected salary over (x + h + k, x + h + k + 1) =
ASx+h-I
s x+h+k
.
S_f+h-1
For a final salary plan, if the employee retires at age x + h + t, the projected annual retirement income would be d x AS
x+h-1
s x+h+1-1
'
S x+h-1
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Chapter 15: Pension Mathematics
For a final 3-year average salary plan, the projected annual retirement income is
cf x ASx+h-1
sx+1r+1-1
+ s x+1i+1-2 + s x+h+1-J 3s x+h-1
for t :2::: 3.
For a career average salary plan, the project annual retirement income is d (As s,+i.+H + s,.,,.,_, + ... + s.,.,, AS AS AS ) x+h-1 + .-r+h-1 + x+h-2 + ' '' + x ' t s .r+li-1
In the expression above, the first part in the parentheses is the sum of the future projected salaries, while the second part is the sum of past salaries. (Note: Sometimes you may be given the current salary AS.,.,, and you have to change the formulas slightly. For example, for the final 3-year average salary plan, the projected annual retirement income would become AS
S x+h+l-1 x+h
+ S x+h+t-2 + S x+h+1-J ) ' 3sx+h
~
Example 15.7
Consider an employee entering a company at age 30 now, with a salary of 40,000 over the coming year. A defined benefit pension plan guarantees an annual retirement benefit of 2% of final 5-year average salary at normal retirement age 60 for each year of service. The salary function is as given in the hypothetical salary scale given in Table 15.1 Calculate the projected annual retirement income ifthe employee retires at age 65 exact.
-
Solution
The salary over age (30, 31) is known to be 40,000. The projected 5-year average salary is thus s 59 + s 58 + s 57 + Ss6 + S 55 5s30
X
4 0,000
= 5.86 + 5.53 + 5.21+4.91+4.62 X 40,000 = 209, 040 . 5x I
The projected annual retirement benefit is 2% x 209,040
= 4,180.8
for each year of service.
Since the employee would have worked for 30 years when he retires at age 60, the projected annual retirement income is 4, 180.8 x 30 = 125,424. [ END]
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Chapter 1 S: Pension Mathematics
Step 2: Modeling the time of retirement Many employees would quit before the normal retirement age. Reason for dropping out include withdrawal, die in service, early retirement, disability and so on. On the contrary, some employees would work after reaching the normal retirement age. A multiple state model is useful in modeling the various decrements. Consider a model with 5 states: State 0: Active employee State I: Withdrawn (w) State 2: Permanently Disabled (i) State 3: Retired (r) State 4: Died (d) We assume that transitions are irreversible. The following shows the multiple state model: State I: Withdrawn
State 2: Disabled
State 0: Active
State 4: Died
State 3: Retired
Typically, we would use a multiple decrement model combined with three single decrement models to simplify the above multiple state model.
-.- .
.• .
1-. - . - . - . - . - . - . - . - . - . - . - . : ·.........-: . - . - . - . - . - . - . - . - . - . - . - . - . - . - . - . - . - . - . - . - . - . - .
~
'·
~~~~~~~~~--
'.
State l: Withdrawn
State 4: Died
"
State 2: Disabled
State 4: Died
State 3: Retired
State 4: Died
State 0: Active
.-·-·
"·,_ ""
- . -.-.-.-·-. -·- ·-·-·-·- ·- ·- ·-. - . - .-·-··' ' -.-·-. - ·-· - ·-. -·-. -· -·- ·- ·-·-. - .-.
...
State 4: Died
" -.
-
.
.
.-
'·
multiple decrement model t1 Actex 20 13
..-
_•. ..-
.-
3 single decrement models
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Chapter 15: Pension Mathematics
The three single decrement models specifies the probability of death I force of mortality for a person who is in state I, state 2 and state 3. To model retirement benefits, we need not model all the forces of transitions.
Normally, active employees can transit to state 3 before and after the normal retirement age a. Transitions that happen before a are treated as early retirement, while transitions that happen after a are treated as normal retirement. The pension we have been discussing is thus called normal retirement benefit. This should be distinguished from early retirement benefit, which is provided for employees who have worked for some minimum number of years but retire earlier than the normal retirement age. A typical rule would allow early retirement at age 55 (the early retirement age) or after 10 years of service, whichever is later. DB plans often use a simple formula for reducing the benefit, which as a reduction of 100c% for each year that retirement precedes the normal retirement age. For example, in Example 15.7 (where the normal retirement age is 60), if the rule above is applied, then if a employee retires at age 59, the project annual retirement income is
s,, + s,, + s,, + Sss + Ss<
X
40,000 X 0.02 X 29 X (I- c).
5s 30
If a employee retires at age 58, the project annual retirement income is
s,, + s,. + s,, + S 54 + S 53
X
40,000 X 0.02 X 28 X (I- 2c),
5s 30
and so on. If an employee chooses to quit before the early retirement age, a withdrawal is resulted. While there may still be withdrawal benefit, the amount can be significantly smaller than early retirement benefits if withdrawal happens within the first few years of service. The concept of earning a withdrawal benefit is called vesting.
Withdrawal benefit
x
vesting
Early retirement benefit
early retirement age
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Normal retirement benefit
normal retirement age a
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Chapter 15: Pension Mathematics
For pension mathematics, the multiple decrement model used is usually called a service table. In Exam MLC Tables, there is an Illustrative Service Table, a part of which is reproduced below: d(d)
l(c) x
x 30 31 32 33 34 35 36 37
100000 79910 65454 55524 49761 45730 42927 40893
100 80 72 61 60 64 64 65
58 59 60 61 62 63
25149 24505 23856 19991 18106 15130
297 316 313 298 284 271
69 70
2040 987
49 17
x
d(w)
d(i)
d(>)
x
x
x
19900 14376 9858 5702 3971 2693 1927 1431
0 0 0 0 0 46 43 45
0 0 0 0 0 0 0 0
148 199 0 120 213 0 0 0 3552 ·-·-·-·-·-·-·-·-·-·-·-·-·-·-·-· 0 0 1587 0 0 2692 0 0 1350
·-·-·-·-·-·-·-·-·-·-·-·-·-·-·-·
0 0
1004 970
0 0
Table 15.2 Illustrative Service Table
In the table, withdrawal constitutes the majority of decrements for young employees. Retirement starts at age 60. After age 60, no one would exit by mode (i) and mode (w). Everyone is forced to retire at age 70. Step 3: Calculating the APV ofprojected retirement benefits
After modeling the time of retirement, we can calculate the APV of projected retirement benefits by using the method illustrated in Chapter 9. Let R(x, h, t) be the projected annual retirement income for an employee who enters service at age x, is currently age x + h, and plans to retire t years from now. Then the APV at time t would be R(x, h, t) 'ii~">:
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Chapter 15: Pension Mathematics
Annual retirement income R(x, h, t) ASx
ASx+I
ASx+h-1 '---y--J
0 x
time age
x+ I
h+t x+h+t
h x+2 ... x+h-1 x+h
valuation date
retirement date
If we have forces of decrement for (i), (w), (d) and (r), then the APV at the valuation date h would be
f
o:i
a-.T-h v
r
(r)
f
(r)
r .. (12)
P.t+hµx+Jr+IR(x, h, t) ax+li+ldt.
Example 15.8
,
An employee who is now 40 exact entered service at age 30 and may retire at any time between ages 60 and 70. His current annual salary is I 0,000. Salary is assumed to be revised continuously and s40+1 =e0021 for any t. The valuation force of interest is 5%. You are given: (i)
The projected annual retirement income is 2% of the total number of service times the salary just before retirement.
(ii)
The retirement benefit is in the form of a continuous annuity payable for life.
(iii) 'a,= l6e-0°"' forallx~60 (iv)
(v)
(0 µ40H
-{0.03 0.13
(') -{ 0
µ40H -
O.J
0 20
=I 0,000,
1j;>
= 0 dJ;~, = 800, fork= 0, I, 2, .. ., 9
(iii) i = 0.06 (iv)
'a, 0+,
(v)
The retirement benefit is paid as a continuously whole life annuity.
=JO fork= 1,2, .. ., 10
Calculate the expected present value at age 60 of the retirement annuity benefit. 15. A pension plan provides a retirement benefit of 20 per month for each year of service. The earliest retirement age is 65, and all retirements occur before age 70. An employee age 50 entered service at age 30. Assuming retirements occur midyear and the annuity is paid continuously, which of the following is the expected present value of the retirement annuity? (A)
m I (') ' (20 + k +- ) ,_ 20 "L.Jv k+l/1 kPsoqsO+k 0 so+k+112
(B)
240l:v'+ 111
,~,
2
19
~· 20
(C)
240
l
,p:;>q; +,(20+k+-) 'a 0
50
+,. 112
2 l 0 +, (20 + k +-) 'aso+k+ 111 2
:I v'+ p:;>q; k-15
112
,
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SoA Exam MLC
Chapter 15: Pension Mathematics
20
(D)
240Lv' ,pj;>q;,+,(IO+k) 'aso+• k=t5 20
(E)
240l:v' ,pj;>q;,+,(20+k) 'aso+• k=JS
16. Suppose ( 40) entered service at age 30, has a current annual salary of 50,000 and total past salary of 400,000. Express the expected present value at age 40 for the retirement annuity benefits given by the following descriptions. Write the expected present value as a sum assuming that all retirements occur at midyear. (a) I 000 for each year of service including the fractional part, (b) 60% of the final 3-year average salary, (c) 2% of the final 3-year average for each year of service including the fraction part, or (d) 2% of career salary. Assume that the normal retirement age is 60 and the retirement annuity is payable monthly for life. Also assume that you are given a salary function sx for integer-valued x. Use linearinterpolation to calculate sy for non-integer-valued y. 17. Consider a newly hired employee age 30, earning 80,000 in the first year of service. Regular salary increases are assumed to be 3%; in addition, employers are assumed to receive merit increases of 5% at each of their first three employment anniversaries. The pension benefit formula is I% of the final five year average salary per year of service. (a) What is the projected final five-year average salary, ifthe employee retires at age 65? (b) What is the projected pension benefit at age 65? (c) Find the employee's replacement ratio. ( d) What would a I% career average benefit be as a percentage of this final five-year average benefit?
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Chapter 15: Pension Mathematics
Solutions to Exercise 1S I. The salary received in year of age 34.5 to 35.5 is exactly 100,000. The salary between ages 58 and 60 is estimated to be s +s59 34 65 35 7 100 000 · + · = 705 263.2. 100 000 58 ' ' 9.975 ' s34.5
2. The annual rate of salary at age 60 is approximated by the salary between ages 59.5 and 60.5. The answer is 36 225 I 00,000 s,, 5 =I 00,000 · = 363,157 .9. s, 4.5 9.975 3. Birthday: September I 1958 (=year 1958.75) Today: January 1 2010 (=year 2010), and the member is 51.25 years old. The annual salary rate of 60,000 given in the question corresponds to the earnings in age (50.75, 51.75), computed from year 2009 July 1to2010 June 30. The estimated expected.earnings is 6 0,000 S51.2 5 S 5015
"'
60,000 S 51 X 0.75 + S 52 X 0.25 s 50 x0.25+s 51 x0.75
= 60 000 3.63 x 0.75 + 3.86 x 0.25 ' 3.4lx0.25+3.63x0.75 3 6875 = 60 000 • - 61888.11 ' 3.575 4.
2 94 40,000 s62 + s" + s 64 = 40000 x 1. 3s 39 3(1 .74)
168122.61
5. The monthly salary over year of age ( 49~,
6
50~) is 6000. The expected salary in the final 6
year of work is 7 70 · 6000xl2~=72000x s 4 ,~ .1-x3.2l+~x3.41
6
6
164,185.587.
6
The expected target pension benefit per year is 164, 185.587 x 0.65
= 106,721.
6. Let the salary rate at age 25 be S. The salary rate in the coming months are projected to be k
S(l .05)", where k = 1, 2, ... , 420 (= 12 x 35). The projected fund at retirement is
' x + h, for h being a positive constant? This is nothing but a question in conditional probability:
Pr(X> x + h IX> x) = Pr(X> x + h andX> x) I Pr(X> x) =
Pr(X> x + h) I Pr(X> x)
This is the same as Pr(X> h). Putting in another way, given Xis greater than x, the excess of
X over x, which is X - x, has the same distribution as the original X: Pr(X- x > h IX> x) = Pr(X> h). It seems that X has forgotten that it has already exceeded x.
-
2. 4 Joint Distribution
The set up of two jointly distributed random variables appears in Chapter 11 when we study the distribution of two lifetime random variables. For other chapters, you nearly do not need anything for joint distributions.
Two independent random variables For X and Y being independent, the joint density and cdf of X and Y are simply given by the product of the marginal densities and cdfs:
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SoA Exam MLC
Appendix 2: Review of Probability
Fx,r(x, y) = Pr(X :". x and Y ;'. y) = Fx(x)F r(y),
a'
i'u(x,y)= -Fvy(x,y)=fx(x)fr(y).
.
Bxi)y ·'·
Also, we have E(XY)
= E(X)E(Y)
and Var(aX + bY)
= a 2 Var(X) + b2 Var(Y).
Two dependent random variables The chance that you need to use the formula for dependent random variables in Exam MLC is very low. What you need to know are just two formulas to compute variance and covariance. If X and Y are dependent, then Var(aX + bY)
= a 2 Var(X) + 2abCov(X,
+ b2 Var(Y),
Y)
where Cov(X, Y)
= E[(X -
/lx)(Y - /Ir)]°' E(XY) - E(X)E(Y).
(In the above, fix= E(X) and /ly = E(Y).)
•
2. 5 Conditional and Double Expectation
The conditional expectation of a random variable X, given that Y
= y,
is computed from the
conditional probability mass function or conditional density of X given Y =y: Ixpx r(x I y)
Xis discrete
1
E(XIY=y)=
(
:,
Lxfx r (x Iy)dx
Xis continuous
1
(The conditional probability mass function and conditional density function can be computed from
Pxir (x I y) = P(X = x I Y = y) =
p xi· (x, y)
~~~
Pr(Y)
if X and Y are both discrete and
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SoA Exam MLC
Appendix 2: Review of Probability
-
f XJY ( X IY ) -
fXY(x,y) f,(y)
if X and Y are both continuous. For Exam MLC, the conditional distribution is usually given.) The conditional expectation of a function of a random variable X given Y = y is similarly defined as
Xis discrete Xis continuous Rules for ordinary expectations hold for conditional expectations. For example,
E(aX+bZ+cl Y=y)=aE(XI Y=y)+bE(ZI Y=y)+c, Var(aX + b I Y = y) Var(XI Y = y)
Notice that E(X I Y
=
= a 2Var(XI Y =y),
= E(X2 I Y = y) -
[E(XI Y = y)]2.
y) is a function of y. It is not a random variable. The random variable
version of conditional expectation is obtained by replacing y with Y. For example, if E(X I Y = y) = 2y, then E(X I Y) = 2Y.
The double expectation formula and variance decomposition formulas say that E[g(X)]
= E[E(g(X)IY)],
Var(X) = E[Var(XI Y)] + Var[E(XI Y)].
~
Example 2.1
Suppose that fx 1,(xl y)=_!_e-xly forx,y> 0, and Y-Exp(l). Find E(X), Var(X) and Cov(X, Y).
y
-
.
Solution
We know that conditional on Y = y, Xfollows Exp(! / y). This means E(XI Y=y)=y
and
Var(XI Y=y)=/.
and
Var(XI Y) = Y2 .
The random variable versions are E(XI Y) = Y We also note that since Y - Exp(!), E(Y)
= 1 and Var(Y) = 1.
So,
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SoA Exam MLC
Mock Test 1
!
d 00 =-.,px 0 I 00 +' 0 08 1PxOI -,px In component form, d dt , starting at t = 6. 1 01 0 -,p dt • =o.1,p~• -co.1+0.0011),p:. From statement (iii) 6 p: 0 = 0.629270 and 6 p: 1 = 1- 0.629270- 0.029503 = 0.341227 . So, 0 6.1 p: "'0.629270 + (-0.1x0.629270 + 0.08 x 0.341227) x 0.1=0.625707116
I 0.
61
p~ "'0.341227 + (0.1x0.629270- 0.106 x 0.341227) x 0.1 = 0.343902693
6.2
p~ "'0.343902693 + (0.1x0.625707116- 0.107 x 0.343902693) x 0.1 = 0.346480
1
1
[Chapter 9] Answer: (B) There is a I 0% surrender charge. So, the answer is simply
0.9'°W. =
0.9(1-~J = 0.9(1-~) = 0.192857. P,. 70 10
11.
[Chapter 14] Answer: (E) For convenience, we refer December 31, 2011 to have
I -
1 and December 31, 2012 tot. We
AV,= (A VH + P, - EC, - Col,)(! + ;;) = (16700 + 1000- I 00- 170)(1.05) = 18301.5. The cash value of the pol icy on December 31, 2012 is given by CV,= max(AV 1 -SC" 0) = max(l8301.5- 1000, 0) = 17301.5. 12.
[Chapter 12] Answer: (B) Let 0L be the prospective loss random variable for a single contract. We have O
L = 100000vK'"+1 -n:ii- =(100000+ n:)vK,"+ 1- n: K60 +11 d d'
E( 0 L) = ( 100000 + ;
)A
60 -
;
and
n:
2
2
2
Var( 0 L)= ( IOOOOO+d ) (A 60 -A60 ).
Let 0L* be the aggregate prospective loss random variable for 10000 policies. We have E(0L*) = IOOOE(0L) and Var(oL*) = 1000Var(0L).
We require
0) = 0.01
Pl Z>
l
O-E(,L*)J=0.01
~Var( 0 L*)
-E(,L*) - 2.326 ~Var( 0 L*) -1,ooo( ( 100,000 + ;
)A,
0 -
;
)
2.326 1,000( I 00,000 + dJr)' ( 2 A,0
-
A,20 )
. -1,000((100,000 +
/r )o.369 I 3 /r ) 0.06/1.06 0.06/1.06 -.==="================ 2.326
1,000(100,000+
13.
Jr
0.06/1.06
)' (0.17741-0.36913 2 )
[Chapter 7] Answer: (C) Let G be the gross premium. The APV of expenses is (O. lG + 20+3.5x100)ii 40 The equivalence principle gives
•
(O. IG + 20 + 3.5 x I OO)ii 40 +I OOOOOA40 = Gii, 0 . Dividing both sides of the equation above by ii 40 , we get (O. lG + 20+3.5x100) + 1OOOOOP40 = G 0.9G = 100 x 7.85 + (20 + 350) G = 1283.33 So the correct answer is (C). 14.
[Chapter 2] Answer: (A) The calculation of the required probability involves two steps. First, we need to know the composition of the population at age 70. -
Suppose that there are /30 persons in the entire population initially. At time 0 (i.e., at age 30), there are 0.5/30 nonsmokers and 0.5/30 smokers.
-
· of m · d"1v1"d ua ts w ho can survive · to age 70 1s · e-o ·tox= 0.002 for accidental death.
(iii) 1f2> = 0.01 for non-accidental deaths (iv)
o= 0.06
Calculate the single benefit premium for this insurance. (A) 9,630 (B)
9,660
(C)
9,690
(D) 9,720 (E)
9,750
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SoA Exam MLC
Mock Test 2
For Questions 20 and 21, consider a fully discrete life insurance of 1,000 issued to (50). (i)
Mortality follows the Illustrative Life Table.
(ii)
i
= 0.06
(iii) The following expenses are allocated to this policy at the beginning of each year:
first year renewal years
20.
Percentage of remium 25% 8%
per 100
per policy
2 0
10
2
Find the gross premium for the policy. (A)
18.8
(B)
20.9
(C)
22.5
(D) 24. l (E)
21.
25.2
Find the expense reserve at the end of 10 years. (A)
-27
(B)
-24
(C)
-20
(D) -17 (E)
-15
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Johnny Li and Andrew Ng
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SoA Exam MLC
Mock Test 2
22.
Let ,v:,. be the benefit reserve at the end of year t for a fully discrete whole life insurance of l on (x). You are given:
Vio = 0.6
(i)
10
(ii)
sV40 = 0.45
Find the benefit reserve
15
Vio.
(A) 0.48
(B)
0.56
(C)
0.64
(D) 0.72 (E)
23.
0.78
For a Markov model for an insured population: (i)
Annual transition probabilities between health states of individuals are as follows: State 0: Healthy State 1: Sick State 2: Terminated 0
T"
1 0.4
2 (ii)
0
2 0.1 0.5 0
"'] 0.1 I
The mean annual healthcare cost each year for each health state is: Healthy: 250
Sick: 3000
Terminated: 0
The healthcare cost, if any, is paid at the middle of each year. (iii) Transitions occur at the end of the year. (iv) 8=0.10 Calculate the expected healthcare cost over the current and the next two years for an individual who is now sick. (A) 5,000 (B)
5,010
(C)
5,020
(D) 5,030 (E)
5,040
'.[') Actex 2013
j Johnny Li and Andrew Ng
j SoA Exam MLC
(T2-B)
Mock Test 2
'--,
24.
Find
20iAx
given
(i)
Ax=
0.35
(ii)
Ax+20 =
(iii)
,,-.
0.42
A,2oj = 0.5
(A) 0.08 (B)
0.11
(C)
0.13
(D) 0.15 (E)
25.
0.18
An employer establishes a defined contribution plan to Albert who joins the company at age 45. On withdrawal from the plan before retirement age, 60, the proceeds of the invested contributions are paid to Albert or his survivors. The contribution rate is set using the following assumptions: (i)
The employee will use the proceeds at retirement to purchase an annuity immediate, payable monthly.
(ii)
The following salary scale is used to project the annual salary of Albert:
s,
x 45 46 47 48 49
x
2.51 2.67 2.84 3.00 3.19
50 51 52 53 54
Sx 3.37 3.58 3.76 4.04 4.19
x
Sx
55 56 57 58 59
4.32 4.51 4.78 4.92 5.00
(iii) Contributions are payable annually in arrears at 15%. (iv) Contributions are assumed to earn investment returns of 10% per year. (v)
21
ii~~
= 10.65
Calculate the replacement ratio r if Albert relies solely on this pension plan for retirement purposes and does not have any savings prior to joining the company.
(A) 0.24 '
AJ...- . xy.1~
If the force of interest is set equal to 0 (i.e. v =I) and the terms of the policies are set equal to infinity (i.e. n = oo), A:Y~ = E[ vr, l(T, < TY, T, < oo )] = E[I(Tx + 1oooq;1i + 36(0.4)q;2i or 1.1 G = 36 p'.'l + 2000q'.'l + 72(0.4)q;2i Upon subtraction, we get l.l(G-1r)= 1000q'.'l+36(0.4)q'.'J 1.1(9.36) = 1ooq;2 i + I4.4q;2 i On solving, we get q;2 i = 0.09. 24.
[Chapter 14] Answer: (A) According to Statement (viii), it is assumed in calculating the cost of insurance that I,= 01-x for 0 Sx < m(i.e., De Moivre's law). Hence, •
q,
=
1
w-50-t
.
This is a specified amount (Type A) policy. As there is no corridor requirement, we can write the cost of insurance for year t as •
Col 1 = q,v,/FA-AV,)=
1
1
x-x(lOOOOO-AV,) w-50-t 1.05
The account value at the end of year 1 is given by AV 1=(AV 0 +P1-EC 1-Col 1)(1 +ii°). Hence, we have 22387.76 = ( 20000 + 0.95 x3000-50
1 1 x--x(l00000-22387.76))(1.05) w-50-1 1.05
which gives w = I 0 I. When the actuary calculates the cost of insurance for year 2, he increases w by 10. So mortality follows Ix= 111 - x for 0 S x < 111. The cost of insurance for year 2 is given by Col, =q;v,,(FA-AV2 ) 1 1 - - -_-- -_- x-1.- x (100000-A V2 ). 1 11 50 2 05
The account value at the end of year 2 is given by
., =JCfrf>"1or O ~ t < I . Th en , p 20 l(w)
Also,,q 20
l
0
t
=0
f(W)
Oje-""' ',,Id) dt -)e-""' ' r20+1 J 20 r20+1 112
1(w)
=(I - !fu_) x (l --J0.75) +(I -q;b"'>j(-J0.75 - 0.75) 3 = 0.25-0.160684q;b"')
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SoA Exam MLC
Mock Test 4
Since withdrawa I is discrete,
2
1(w) (w)
q 20
=
2->2 0->0->2->2 0->0->0->2
" 1700)" Pr(Z >
) = Pr(Z > 2.28) = 1- cD(2.28) .
.J439750.08 21.
[Chapter 9] Answer: (C) The benefit reserve at the end of 5 years, which is the expected prospective loss at age 55, IS
1000 5 V50:151 -
= 1000(1- ii,,,iQi .•
J
a so:i5i
=I ooo(I-
12.2758-0.48686 x 9.8969 ) 13.2668 - (753396418950901) x 9.896911.06 15 7 57395 ) = 238.3336 9.790894
= 1000(1- .4
Let S be the reduced pure endowment. When the policy is paid up after 5 years, the reserve is used to support 1oooA;,iDi + s
'°E"
= (305.14- 0.48686 x 439.8) + 0.48686S = 91.01897 +0.48686S
Setting the above to 238.3336 and solving, we get S = 302.6. 22.
[Chapter 2] Answer: (D) We are given:
x
qx+2
x+2
0.30445 0.32834 0.35360 0.38020
86 87 88 89
qrxl+r
q1x1
84 85 86 87
By using the information from statements (i) and (ii), we obtain
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SoA Exam MLC
Mock Test 4
x 84 85 86 87
q[xl
0.152225 0.16417
qrx1+1
qx+2
x+2
0.152225 0.16417 0.1768
0.30445 0.32834 0.35360 0.38020
86 87 88 89
The trick to solve this question is to make use of the fact that of 1[861 and1r 871 would evolve to /g9. Firstly, /s9=lrs11P1s11P1s11+1 =irs11(1-0.16417)(1-0.1768). Similarly, /39 '." i1s6J P(86J Pf86J+I P1ss1 = I 000 x (I - 0.152225)( I - 0.16417) x (I - 0.35360) As a result, I 000 x (I - 0. I 52225)(1 - 0.16417) x (I - 0.35360) = lrs11 (I - 0.16417)(1 - 0.1768)
. I I 000(1 - 0.152225)(1 - 0.35360) 0 n so Ivmg, we get 1871= I - 0.1768 23.
665. 6 97.
[Chapter 7] Answer: (A) The level expense premium is P' = 390.20 - 323.12 = 67.08. The expense reserve at time 3 is 0. By using the recursion relation for expense reserves, (2V"+P,-O.IG-3)(1 +i)=qx+2xO+px+2x1V' 211" = -P, + 0.1G+3 = -67.08 + 39.02 + 3 = -25.06.
24.
[Chapters 13 and 14] Answer: (A)
At time 0 Initial expenses: 200 Pro: -200 [Note: According to the Supplementary Notes of the textbook Actuarial Mathematics for Life Contingent Risks, initial expenses (but not other items) are incorporated in Pro.] At time I A Vo: 0 1950 AV 1: P1: 3000 0 (since initial expenses are incorporated in Pro) £ 1: EDB1: 1.2 x 0.00592 x (80000 + 1950 + 100) = 582.8832 [Note: This is a specified amount plus the account value (Type B) policy. In the absence of any corridor factor requirement, the death benefit is $80000 plus the time-I account value of$1950. The value 0.00592 is obtained from the Illustrative Life Table.] (I - 1.2 x 0.00592) x 0.1x(1950-800+100) = 124.112. ESB,: [Note: The cash value CV 1 is the account value A V 1 = 1950 less the surrender charge SC 1 = 800.]
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Mock Test 4
(1 - 1.2 x 0.00592)(1 - 0.1 )(1950) = 1742.5325 (0 + 3000 - 0)(1.1 )- 582.8832 - 124.112 - 1742.5325
=
850.4723
The answer is thus (A). Of course you can continue to calculate Pr2 and Pr3 :
At time 2 AV1: AV2: P2: E2: EDB2: ESB2: EAV2: Pr2:
1950 4100 3000 50 + 0.01 x 3000 = 80 1.2 x 0.00642 x (80000 + 4100 + 100) = 648.6768 (1-1.2 x 0.00642) x 0.2 x (4100- 800 + 100) = 674.7613 (I - 1.2 x 0.00642)(1 - 0.2)(4100) = 3254. 7309 (1950 + 3000- 80)(1.1)-648.6768- 674.7613 - 3254.7309
= 778.83 l
At time 3 4100 AV2: 6400 AV3: 3000 P3: 50 + 0.01 x 3000 = 80 £3: 1.2 x 0.00697 x (80000 + 6400 + l 00) = 723.486 EOB3: (1 - 1.2 x 0.00697) x 1 x (6400- 800 + 100) = 5652.3252 ESB3: (1 - 1.2 x 0.00697)(1 - 1)(6400) = 0 EAVi: [Note: The surrender rate fort= 3 is 100%, which means all policyholders surrender at the end of the term. Hence, EAV 3 must be zero.] ( 4100 + 3000 - 80)(1. l) - 723.486 - 5652.3252 - 0 = 1346. l 888 Pr3 : Therefore, the profit vector is given by (-200, 850, 779, 1346)'.
25.
[Chapter 15] Answer: (C)
26000 x
s., + s" + s" JS,,
9 26000 x l 0. J = 83828.9. 3xl.13
I
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SoA Exam MLC
,________M_o-ck_T_e•_t_S__
~-----------------------------~ ~~~ .....
Mock Test 5 I.
You are given: 0
(i)
e7141= 3.6
(ii)
q70 = 0.0054 2
(iii) For integer-valued x, 1 Px = 1 - t q.,, 0 = p;~d) p;~w) = 0.997138907 x 0.95 = 0.947281962, ,(J)
= [p''0 1]q:;' lq::.' = (
l
("' I
P60
0.95
-
-
[p(x) ]l/"')
View more...
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