MLC STUDY MANUAL

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MLC

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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 mean

s

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

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, g1(h) = Ii and

h->0 h

g2(h) = h11 are o(h). However, g3(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.

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

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)

= Jim Pr(t < Tx

~

t

+

h, Jx =

J

I

T,. > t) = Jim ,,

q_~~;

f x+t h->0 h /i->0 h '

and this means

(I) (I) h (h)

h qx+t

=

Jl.r+I

+

0 ' hqx+t - Jlx+t +o (2) - (2) h (h) ·

(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(-

J~µ-~:l,ds)

Transition probabilities: /

q_~

1

>

=

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.

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.

Chapter 1 O: Multiple State Models 2 I

0

Y,

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:

00 P1 P101 P102 P1 On P:o

p:I

p;2

_p:"

p _I

=

P120 P121 p;2 P12,, 110 P1

p;li

p;2 P1 1111 Po Pi P2 P1-i P, ~ ~ ~ ~ ~ 0 2 3 t - I t t

+I

Yo Yi Y2 Y3 Y1-i Yr 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.

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 3-state transition probability matrix:

0

2

~ [~:~ ~.·~ ~:~]

2 0 0 1

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:

Age /(r) x d(I) x d(2) x 60 1000 120 80 61 800 160 80 62 560

(ii) 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) P 60 is the transition matrix from age 60 to 61; P 61 is the transition matrix from age 61 to

62.

(iii) The two models produce equal probabilities of decrement.

Calculate P6t·

lJOO

0.12

oz81

l080

0.12

0 081

l0.76

0.16

0~81

(A) 0 1.00 _{(B) 0.~6 0.16 0.08} (C) 0 1.00 0 0 1.00 0 1.00 0 0 1.00

l0.70

0.20

0~01

l060

0.28

o.~21

(D) 0 1.00 (E) 0 1.00 0 0 1.00 0 0 1.00

t The main diagonal of a square matrix is the diagonal which runs from the top left corner to the bottom right corner.

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 Standard Poor Preferred

l

0.95 0.04 0.01

l

Standard 0.15 0.80 0.05

Poor 0.00 0.25 0.75

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:

Chapter 1 O: Multiple State Models

Path Probability

Standard~ Preferred~ Preferred 0.15 x 0.95

=

0.1425 Standard ~ Standard ~ Preferred

Standard ~ Poor ~ Preferred Sum

0.8 x 0.15

=

0.12 0.05 x 0

=

0

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 O 3 ] ,

B

=[~I

-1 2], then

-1 7 -5

-4

3

AB=[

(1)(2)+(0)(-1)+(3)(-4) (l)(l)+(0)(-2)+(3)(3)

l

[-10 10

l

(-1)(2)+(7)(-l)+(-5)(-4) (-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 0.04 0.01][0.95 0.04 0.01] [0.9085 0.0725 0.019] 0.8 0.05 0.15 0.8 0.05

=

0.2625 0.6585 0.079 . 0.25 0.75 0 0.25 0.75 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 P2. So, P2 is the 2-step transition

matrix. More generally, if we want to find

Pr(f,+s

=JI f, =

i)

=.,

P.;,

we only need to look up the ij-th element in sPx

=

PxPx+I ... Px+s-1 ·

Chapter 10: Multiple State Models

Px+2 P.HS

x _{x+1 x+2 x+3 x+s-1 x+s x+s+ 1}

fr+s-1 fr+s fr+s+ I

This result can be thought of as a generalization of the relation

sPx

=

PxPx+IPx+2 ···P.as-l •

The above implies

?:I/

//////////////,l/'////////////,.-?,/'l'////h'/'///////,ij////////////,.-?,~'/'//////////~

~

FORMULA~

~

The Chapman-Kolmogorov (CK) Equation

~

~ ~

~ p

=

p p ~

~////////////////,l/////////////;~;//;/;;/;;~///,1///////////////,1///////////~

The order of multiplication on the right hand side of the CK equation matters!

Example 10.2 [Sample #151]

~

For a multiple state model with three states, Healthy (0), Disabled (1), and Dead (2):

(i)

Fork=

0, 1:

P.~~k

=

0.7,

P~!k

=

0.2,

P.~~k

= 0.1,

P.~:k

=

0.25

(ii) There are 100 lives at the start, all Healthy. Their future states are independent.

Calculate the variance of the number of the original 100 lives who die within the first two years.

(A) 11 (B) 14 (C) 17 (D) 20 (E) 23

Chapter 10: Multiple State Models

-

Solution

We are given the following information:

r

0.7 px

=

px+I =

O;J

0.2 ?

l

? 0.25 . ? ?

Since state 2 is an absorbing state, the last row must be [O 0 1]. By using the property that each

row of a transition probability matrix sums to 1, we get:

r

0.7 PX

=

PX+ I

=

0~1 0.2 0.65

0

0.11

o.~5

.

By the CK equation, 0.2 0.65 0 0.11[0.7

0.~5 O~l

where 0.22

=

0.7 x 0.1

+

0.2 x 0.25

+

0.1 x 1. 0.2 0.65 0

O.ll

r? ?

o.~5

=

~ ~

on

For a single life, the probability that he would die within the first two years is 0.22. If there are 100 independent lives, we let 2D0 as the number of deaths within the first two years. Then

2Do ~ B(l 00, 0.22)

(B(n, p) here means a binomial distribution with number of trials n and probability of success p)

and hence the variance is

Var(2D0)

=

100 x 0.22 x (1 - 0.22)

=

17.

So the correct answer is (C).

Calculating APV of Cash Flows

There are mainly two kinds of cash flows associated with discrete-time Markov chain:

( 1) Cash flow arising from being in a particular state.

(2) Cash flow arising from a transition from one state to another.

[ END]

It is easy to calculate APV of cash flows under a discrete-time Markov chain model. Study the following example:

Chapter 10: Multiple State Models

Example 10.3

~

C ons1 er a omogenous 1screte-t1me Mar ov c ain with 'd h d' . k h . P

=

[0.4 0.6]

.

The state space is

0.8 0.2

E

= { 0, 1}. The subject in now in state 0, and i = 5%.

(a) Suppose that there is a cash flow of 1 for being in state 0 now and the beginning of the next two years. Find the actuarial present value of the cash flow streams.

(b) Suppose that there is a cash flow of 1 if transition from state I to state 0 occurs at the end of the first, second and the third year. Find the actuarial present value of the cash flow streams.

- Solution

(a) Pr(.Y;

=

0

I

Y

₀

=

0)

=

pg

0 = 0.4,

Pr(Y₂ = 0

I

Y₀ = 0) = ₂

pg

0 = Pr(O ~ 0 ~ O)+ Pr(O ~ 1~0)= 0.4 x 0.4 + 0.6 x 0.8

=

0.64

Thus, the APV of the cash flow streams is

(b) Pr(

Yo=

1,

Y,

=

0

I

Yo=

0)

=

0,

1

+

0.4

+

0

·

6

~

=

1.96145. 1.05 1.05

Pr(Y1

=

1,

Y2

=

0

I

Yo=

0)

=

Pr(O ~ 1 ~ 0)

=

0.6 x 0.8

=

0.48

Pr(Y2 =I,

Y3

=

0

I

Yo=

0)

=

Pr(Y2

=

1

I

Yo=

O)P(Y3

=

0

I

Y2

=

1)

=

(1 - 0.64) x 0.8

=

0.288

Thus, the APV of the cash flow streams is

_o_

+

0.4~

+

0

·

28

~ =

o.684159. 1.05 1.05 1.05

[ END ]

Premiums and benefit reserves can likewise be calculated as in Chapter 5 and Chapter 6.

Example 10.4

~

You are given the following homogenous Markov chain model with three states, namely

Healthy (0), Disabled (1 ), and Dead (2). Transition between states occurs at the end of the year. The probabilities of moving among these various states are summarized by the transition matrix

below:

Chapter 10: Multiple State Models

0

2

0 r0.6 0.3 0.1

1

P

=I 0 0.6 0.4 2 0 0 I

An insurance company issues a 3-year insurance. A death benefit of$ I ,000 is payable at the end of the year of death. Level annual premium is payable for 3 years or until death, whichever occurs first, except that the premium is waived in the state of disablement. Interest rate is 6% per annum.

(a) Find the level benefit premium for an insured healthy at issue (t = 0).

(b) Using the premium calculated in (a), calculate the benefit reserve at t = 1 for

(i) an insured healthy at t = I, originally healthy at t = 0, and (ii) an insured disabled at t = 1, originally healthy at t = 0.

-

Solution

(a) Notice that once the insured enters state I, he/she cannot go back to state 0. Also, state 2 is an absorbing state. So it is a permanent disability model.

Pr(Y1 = 0 J

Yo=

0) = p~

0

= 0.6

Pr(Y2

= 0 J

Yo=

0) = Pr(O---+ 0---+ 0) = 0.6 x 0.6 = 0.36

APV, at time 0, of premiums =P(I + 0·6 + 0

·

3

~)

=I .886436P 1.06 1.06

Pr(Y1 = 2 J

Yo=

0) = p~ 2

= 0.1

Pr(Y2

= 2,

Y

₁

-:t=2

J

Yo=

0) = Pr(O---+ 0---+ 2) + Pr(O---+ 1---+ 2) = 0.6 x 0.1+0.3 x 0.4 = 0.18 Pr(Y3 =2,

Yi -:t=2, Y2-:t=2

I

Yo=O)

= Pr(O ---+ 0 ---+ 0 ---+ 2) + Pr(O ---+ 0 ---+ 1 ---+ 2) + Pr(O ---+ 1 ---+ I ---+ 2) =0.6 x 0.6 x 0.1+0.6x0.3 x 0.4

+

0.3 x 0.6 x 0.4 = 0.18

APV, at time 0, of benefits= 10oo(__QJ_+

O.I~

+

O.I~

)= 405.6704528 1.06 1.06 I .06

By the equivalence principle, we get P = 405.6704528 / I .886436 = 215.0459.

Chapter 10: Multiple State Models

(b) (i) If the insured is healthy at time 1, then Pr(Y2 = 2

I

Y1 = 0) =

pg

2 = 0.1

Pr(Y3 = 2,

Y2

*

21

Y1

= 0) = Pr(Y2 = 2,

Y,

:;t: 2

I

Yo=

0) = 0.18 (from (a))

1

v<

0 ) = 1

ooo

x

(__QJ__

+

0 · 1

~

)- P(1

+

0·6 ) = -82.231 1.06 1.06 1.06

Note again that negative reserve is possible.

(ii) If the insured is disabled at time 1, he/she can never be healthy again and there is no further premiums payable.

Pr(Y2 = 2

I

Y, = 1) = 0.4

Pr(Y3 = 2, Y2

*

21Y1=1) = Pr(l ~ 1 ~ 2) = 0.6 x 0.4 = 0.24

V(I) =} 000 X ( 0.4

+

0·24 ) = 590.9576.

I 1.06 1.062

[ END]

1 O. 2 Continuous-time Markov Chain

Now we come to the construction of a multiple state model for which state changes can occur any time. Consider a life aged (x) at time t

=

0. For each t, the life can be in one (and only one) of the many states. We let Y(u)

=

i to mean that the individual is in state i at age u. Then {Y(x

+

t): t 2:: O} would be a continuous-time stochastic process t. Let the state space of Ybe E. As time evolves, Y is constant when there is no change in state and it would jump when there is a change in state. For the disability income model, suppose that we start with a healthy life aged x at time

0. So, the process starts with Y(x) = 0. It then undergoes a series of jumps between 0 and 1 (one possibility being no jump at all). Ultimately Y(x

+

t) would jump to 2 and then get stuck in it:

Y(x

+

t)

~

•: •: •: •

1---: : : •: •: •: •: •

r · ---: _

---0 . - . - . - .

r. . .

-time t (age x

+

t)

A possible path of Y(x

+

t). t/s are (random) times of transition.

t A continuous-time stochastic process is a collection of random variables indexed by a continuous time variable. Another example is standard Brownian motion in Exam MFE.

Chapter 1 O: Multiple State Models

For a multiple state model, our first goal is to obtain the following probabilities:

- transition probability ,

p;;

=

Pr(Y(u

+

t)

=

j

I

Y(u)

=

i);

- occupancy probability ,

p;;

=

Pr(Y(u

+

s)

=

i for alls E

[O, t]

I

Y(u)

=

i).

Note that , p~ 2 , p~ because the events that correspond to, p: include those that leaves state i

and then re-enter state i again on or before time t. Only when state i cannot be re-entered again once it has been left (just like states 0 and I in the permanent disability income model) or when

state i is absorbing would 1 P.~

= ,

p~ hold.

We make the following assumptions for the multiple-state model, which is actually a

non-homogeneous continuous-time Markov chain (CTMC):

Assumptions of Continuous-time Markov Chain I. The Markov property:

Pr(Y(u

+

t)

=

j

I

Y(u))

=

Pr(Y(u

+

t)

=

j

I

Y(v) for v E

[O,

u]).

Loosely speaking, the above means that the value of ₁

p;;

does not depend on what has happened before the individual reaches age u. It only depends on the fact that the person is in state i at age

u.

2. The probability for 2 or more transitions within a time period of length h is o(h).

3. , P.~ is a differentiable function oft.

Assumption I may not be reasonable in a particular application. Suppose in the disability

income model that Y(u)

=

1, so that the person is sick at age u. Then Assumption 1 says that the probability of any future move after age u, either recovery or death, does not depend on information such as how long the life has been sick up to age u or how many periods of sickness the life has experienced before reaching age u.

Assumption 2 is a technical assumption needed for the derivation of differential equations as we

shall see later. It is usually not an unreasonable assumption. Assumption 3 is a technical assumption that guarantees the existence of the intensities to be defined shortly below.

Chapter 10: Multiple State Models

The Transition Intensity Matrix and Transition Probability Matrix

Just like in multiple decrements, in a CTMC the inputs are transition intensities. Suppose that there are

n

+ 1 possible states

(n

> 0) such that the state space

Eis

{O, 1, 2, .. ., n}. For i,j E

E

and t ~ 0, let

lj

µ_;+,

=

lim "P.<+t be the time-t instantaneous rate of transition from state i to state j for

i

-:f:.

j;

h--tO f1

11

vx:,

=

L

µ::,

be the time-t exit rate of state

i.

k=O,hi

The transition rates and exit rates are then put into a transition intensity (Q) matrix:

0

-vx+t µ_t+f 01 µx+I 02 µx+I On

IO

µX+I -v.<+1 I µx+I 12 µX+I In

Qx+t

=

_µx+t 20 _µX+I 21 -v 2 211

x+t µx+I

JlnO

,,,

n2 11

X+I µ_t+f µx+I -vx+i

The sum of elements in any row of Q.r+r is zero. A state that cannot be left once it is entered is called an absorbing (or cemetery) state and the row that corresponds to it contains only 0.

The transition probabilities 1

p_;

are put into a transition probability (P) matrix:

00 1Px 1Px 01 I po2 X I Pon x p'o II 12 p'" I x 1Px 1Px I X tpx

=

_1Px 20 _1Px 21 _1Px 22 _I p2n _X nO nl 1Px 1Px 1Px n2 I p"n X

The sum of elements in any row of 1Px is 1. For an absorbing state, all elements in the row, except the one in the main diagonal, are 0. The one in the main diagonal is 1.

Now let us formulate the three examples given in the introduction as CTMCs and see how their

Q

and

P

matrices look like.

Chapter 10: Multiple State Models

Single Decrement Model

There are only 2 states:

State 0

=

(x) is alive

State 1

=

(x) is dead (absorbing state)

We have µ.~~

1

= _Jt.<+1 and µ.~~

1

= 0. The transition intensity and probability matrices are

Double Decrement Model

There are 3 states:

State 0

=

(x) is alive

State 1

=

(x) dies from accident (absorbing state) State 2

=

(x) dies from other causes (absorbing state)

Transitions from state 0 to 1 and from state 0 to 2 are the only possible transitions. The

transition intensity and probability matrices are

[ (r) - - µX+I QX+I - 0 0 (I) µX+I

0

0

(2)1

µx+I

0 '

0 I q~I) 0

/q.~2)1

0

.

1

The primed rates, being unobservable rates, cannot be expressed as functions of ₁ P.~. So we need to define new symbols for them. The corresponding symbols are given in the following

table:

Standard

Multi

le State Force of decrement µx+t (i) µx+t Oi

Total force of decrement µ.x+t (r) µ.HI O• (

=

Vx+I 0 )

Survival probability 1 P.~r) _1Px 00

Transition probability tqx (1) 1Px Oi

Total transition probability I q.~r) _1Px O•

Independent survival probability 1Px 1 (i)

1 1Px

Absolute rate of decrement tqx 1(1) I q.:

Chapter 1 O: Multiple State Models

Example 10.5

~

For a three-state model with states numbered 0, 1, and 2, you are given:

(i) The only possible transitions are 0 to I and 0 to 2.

Cii)

J1~L

=

oJ,

t ~

o

(iii) JI.~:, = 0.4, t ~ 0 Calculate 2 P.~:

1

and 2 P.~:

1

•

Solution

The multiple state model is equivalent to a double decrement model with J1~~

1

= 0.3 and

( 2) 0 4 W k d (r) d (2) •

Jlx+i = . . e are as e to compute 2 qx+i an 2 qx+I .

q(rJ =

f

2 p<rlJ/'l dt=

f

2e-07_'0.7dt=l-e-14 _=0.7534 2 x+I

Jo

1 x+I x+l+t

Jo

(2) - f2 (r) (2) d - r2 -0.71 0 4d - 4 (1 -1.4) - 0 4305

2 qx+I -

Jo

I Px+1Jlx+l+1 t -

Jo

e . t -

7 -

e - . .

Disability Income Model

Suppose the various rates of transition are labeled as follows:

01 Jl.t+I 0. Healthy 02 Jl.t+I 2. Dead 1. Disabled

State 2 is an absorbing state. The transition intensity and probability matrices are

01 - Jl.t+I - Jlx+t

l "' "'

01 Jlx+1 Jlx+I

"' l

l ""

1Px 1Px 1Px

"l

Q

10 10 12 12 p - 10 II 12

X+I = Jl.t+I - Jlx+I - Jl.t+I

Jl~+I

' I X - /~X 1Px 1Px ·

0 0 0 I

t /~ctex 2013 Johnny Li and Andrew Ng

I

SoA Exam MLC

Chapter 1 O: Multiple State Models

The permanent disability model can be obtained by setting µ_~~

1 =

0. The transition intensity and probability matrices for this model are

[ 01 02 01 - flx+t - µx+r flx+r Qt+/

=

0 - µ_~:/ 0 0 02 ] [ 00 µ·~;r _ rPx µ_t+/ ' I

PX -

0 0 0

The Chapman-Kolmogorov Equation

01 rPx II rPx

0

02] rPx 12 rPx · l

Similar to a discrete-time Markov chain, the CK equation holds in CTMC:

~//,/,.l~/////////////////'/////////////h/'/////////////h/'/////////////h/'////////~

/

%

~

FORMULA~

~

The Chapman-Kolmogorov Equation

~

%

~

%

t+spx

=

tpx sP.t+t

~

~/////./h'l////////////////////////////.//////////////.;'l.////////////////.////////A

Instead of giving you a proof of the CK equation (which is just an exercise in conditional probability), let us look at a particular case. In the double decrement model,

0

-[1

P.~rl

1Px - 0 0 The product 1Px s Px+t is

['r

t q_~I)

"'I "'

(I)

,qf

,~

.. ,

s qx+r l 0 0

which is obviously equal to 1+sPx.

(2)1

[

(r) rqx _ sPx+t 0 ' spx+t - 0 l 0 (IJ sq_t+/ l 0

(2)]

sqx+r

0

.

l sqx+t

"']

[

rPx sPx+t

"'

"'

(r) (I)

+

(I) I Px sqx+t tqx 0

=

0 l 0 0

"' "' + "']

,p,

,q,f

,q,

Chapter 10: Multiple State Models

10. 3 Kolmogorov's Forward Equations

This is the hardest part in this manual. We first study how the transition intensities govern the

local behavior of transition probabilities. Then we would see how P can be obtained from

Q.

To assist your understanding, we provide verbal interpretations for esoteric formulas along with

their mathematical derivation whenever possible. It is fine if you decide to skip the math. But please read the interpretations so that you can remember the formulas.

Interpretation of Transition Intensities

if

The definition of

µ';+1

says that for; -:F j,

µ_;+1

=

P.To

":w ,

or equivalently

This equation governs the local behavior of the CTMC: Suppose that at time t the life is in state

;, Then we can compute the approximate probability for the life to transit to state j (-:F

O

within an infinitesimally short interval.

What is the probability of not transiting to any other states but just stay in state ;? Since

II

L

"P;+

1

=

1 (recall that the row sum of P is always 1 ),

j=O

n n n

1iP.:+

1 =1-

L1iP;+1

=1-

_L[µ;+

1h+o(h)]=1-h

,Lµ_?+f

+o(h),

j=O,j# j=OJ;•i j=O,;#

(recall that o(h)

±

o(h)

=

o(h)) and hence

"P.:+

1

=

l -vx~

1

h

+

o(h) fort~ 0. ;~////////////////////////////h/"/////////////h/"/////////////h/"//////////////'/'.:

~

~

i;%

FORMULA~;

~

~

:~

Local Behavior

~:

~

1

;~

"P.?+

1

=

µ_?+

1h

+

o(h) for i -:F j (10.1)

~~

~ I'.

;%

ii - 1 i h (h) (10 2) /'.: '% /J p X+I - - V X+I

+

Q '

%'

~

1

:y/////////////h?'///////////ffh/"/////////////h/"/////////////h/"/////////////h'.

Chapter 10: Multiple State Models

Equations (10.1) and ( 10.2) enable us to establish a system of equations that determines 1

Pt.

The system of equations is known as Kolmogorov's forward equations (KFE). Define

~CPx)

as the matrix whose elements are

~(

1

pif).

Then

dt

dt

x

~/%'.-'/?'////;:'l/'l///////////////////..(~%///'l/////////,/Z/////.(/,////////'.~/..(/,///~

~

FORMULA~

~

Kolmogorov's Fonvard Equations

~

/::; Component form

~

~

~

~

d ij -

~

ik kj ij j

~

·, - ( p ) Li p

µ

-

p

v

with initial conditions ₀ P.~ = 1 and ₀ p_; = 0 for i

*

j r.

~

dt

I x - I x x+t I x x+t ~

~ k=O.hj ~

~

Matrix form

~

~

d

~

~

dt

C

Px)

=

1

Px

Q<+1 with initial condition

oPx =I

(the identity matrix)

~

~///h:///////////////,/~/b///////////,/:///////////////,/~//////,/,//////h:////////~

(Recall that an identity matrix is a square matrix with ones on the main diagonal and zeros

elsewhere.)

Again, the order of multiplication on the right hand side of the matrix form of KFE matters.

Though the textbooks for Exam MLC do not discuss the matrix form of KFE, we include them

as it is much easier to remember. The proof of KFE is given below and you may read it if you

are interested. However, the focus here is the various techniques on solving 1

Px

from KFE.

Proof

We derive the component form. To obtain

:t

(,p;), we differentiate 1

Pt

from first

ij lj

principlest. So, we first calculate the numerator in the difference quotient i+1iPx - iPx for

h

> 0.

h

i+h P.~ is the probability of transiting from Y(x)

=

i (at time 0) to Y(x

+

t

+

h) =j (at time t

+

h).

By the CK equation, 11 11 ij - " ik kj - " ik kj

+

ij ;J t+h Px - Li I Px h P.<+t - Li t Px h P.<+t 1Px h Px+r' k=O k=O.k* j By equations (10.1) and (10.2),

t The derivative of f(I) is the limit of the difference quotient [f(I + h)-f(t)] I hash tends to 0.

Chapter 1 O: Multiple State Models n I} I} -

"'"°'

ik [ kj h (h)]

u

[1 1 h (h)]

u

1+h Px - 1Px - L.J 1 Px µx+I + 0 + 1Px - Vx+1 + 0 - 1P x k=O.k* J ( n

J

- 1k kj ij ) j - h LI Px µ_r+/ - 1Px i x+I

+

o(h). k=O,k* j

S o, i+h

p; -

· h

,p;

·

=

L.J ,

~

Px ;k µ_;+f -k•

,p_:

;• v<+, 1·

+ - -

o(h) an t e component iorm 1s o tame d h .c- • b . db y ta mg k'

k=O,k*f h

limit. The two initial conditions follow obviously from definition.

D

Actually, KFE simply says that "net rate change

=

rate in - rate out." As t increases to t

+

dt,

why would ,

p;

change?

(I) On one hand, it is possible for Yto transit from any other states (except}) to state}. This is

ti

approximately equal to

L ,

p~k µ_~

1

dt if we ignore multiple transitions. k=O,k* j

(2) On the other hand, it is possible for Y to leave state j. The total exit rate is ,

p_;

vx{,dt .

You need to know how to solve KFE analytically in simple cases and numerically in the general

case. First we focus on solving KFE analytically.

Example 10.6

~

Consider the single decrement model. Use KFE to derive the expression for 1Px·

- Solution

For the single decrement model,

Tx

is the time spent in state 0. That is, it is the smallest t such that Y(x

+

t)

=

1. To find the distribution and survival function of

Tx,

we use KFE, which says

:J~' '~}['~' '~l~W µ~HJ

with initial conditions oPx

=

1 and oqx

=

0. The matrix form is equivalent to 4 component equations, the first two being:

(00)

:t

1 Px =-1PxFr+1 (01) :(1 qx = 1Pxµ.<+I (i .. e. fx(t) = tPxµx+t)

Chapter 10: Multiple State Models

(Equations (10) and (11) are

__!

0 = 0 and

__!

1 = 0, which trivially hold.)

dt dt

The solution for (00) with oPx = l is , Px =exp(-

J~

µx+sds). (See C 1-13 if you have forgotten how this result is derived.) By integrating both sides of (0 l) (first replacing t bys)

d

dssqx

=

sPxµx+s

from time 0 to time t, we get

Since oqx = 0, we finally obtain , q x

=

J~

s Pxµ.wds .

[ END]

Example 10.7

~

Consider a multiple decrement model with n decrements. Use KFE to analyze the model and derive the expressions for , p;r) and ,

q;n .

-

Solution

Similar to a double decrement model, the

Q

matrix is

( r) (I) (n)

- µx+I µx+I µx+I

0 0 0 0 0 0 KFE says (r) (I) (11) I P.~r) (I) (11) _-µ(T) Ji;~, (n) 1Px I qx ,qx ,qx ,qx x+I µx+I d 0 0 0 0 0 0 0

=

dt 0 0 0 0 0 0 0

and the initial conditions are ₀ p_;r)

=

1 and ₀ q_;i)

=

0.

The component form for the (00)-th element is

-.!,

p<rl

=

-,p~r)

µ;r+l,, which gives

dt x .. .

Chapter 10: Multiple State Models

(r}_ ( f' (r)d)

I Px - exp -

Jo

JI.HS s .

The component form for the (Oj)-th element (j > 0) is

.i,

qu> = p<r> 11<n, which gives

dt x 1 x rx+1

(}) =

f'

(r) (J) ds

t qx

Jo

s Px flx+.1·

by integrating both sides and using ₀ q_~il = 0.

[ END]

Frequently, KFE is not that easy to solve. However, for a state that cannot be re-visited once it has been left (for example, state 0 in the previous two examples), we have the relation ₁ P.~ =

1 P.~ and it turns out that 1 p~ can be easily obtained. Both examples hint to

1

P.~

=exp(-

J:

vx~sds)

, and it turns out that this equation holds in general.

"///////~////////?///////////////,///////~///////,/////////////////////~/////////, :~//////////////////////////////I///.////.////////.//////////////.///////////////.//~

'.~

FORMULA~;

~

~

:%

Occupancy Probabilities ~:

~

~

'%

₁₁

P.~+1 =l-vx~

₁

h+o(h) (10.3)

;r:;

~

~

~

,r;. ₁

p;

-

=exp -

( I i )

f

vr+sds (10.4)

~:

,,,,

;/,:

.

Jo .

1::

~

-~

;~/////////////h'l"/////////////h'l"/////////////h'l'/////////////h'l"////////////~;

While equations (10.2) and (10.3) look alike, their solutions are different. These equations only govern local behavior in infinitesimal sense.

Read at least the first paragraph of the fo !lowing proof for (l 0 .3):

Proof We know that ₁₁ P.~+i ~ ,,p~+t. But what makes up the difference between them? It is the

probability of those events for which Y makes two or more transitions between ages

x

+

t and x

+

t

+

h such that it is back into state i at age x

+

t

+

h. From assumption 2., this probability is o(h). As a result,

Chapter 10: Multiple State Models

ii ii (h)

h Px+t - 1iP.<+1

=

0 •

Equation (10.3) then follows from equation (10.2).

Then we establish a differential equation for ₁

p;

and solve for the solution. Notice that

Ii ii ii ii

[l

I h (I )] t+h Px

=

1Px h Px+t

=

1Px -V_<+t

+

0 1 and hence Ii if (h) t+h Px - 1 Px Ir i 0 h

=

-1Px v.<+t

+h ·

Passing to limit, we get

and hence 1

P.~

=

oP.~

exp(-

J~ vx~_,ds).

However, obviously

oP.~

=

1 and the result is proven.

D

The Permanent Disability Model

Armed with KFE and the analytic solution to occupancy probability, we can finally tackle the permanent disability model analytically.

The P matrix is 0. Healthy

02

µ_t+/

01 µx+I 2. Dead I. Disabled 12 µx+I

Model 4: The Permanent Disability Model

t:~ Actex 2013

r

00 _ 1Px 1Px - 0 0 01 1Px II ,Px

0

021

1Px 12 1Px · 1

Chapter 10: Multiple State Models

Since neither state 0 nor state 1 can be re-entered once it has been left,

1

p_~

0 = 1

p_~

=exp(-

J:

(µ_~!s

+

µ~!Jds)

and 1

p;

1 = 1

P~

=exp(-

J:µ!!sds).

Also, since 00 01 02 1 I Px

+

1Px

+

1Px

= '

11 12 } 1Px

+

1Px

= '

the only unknown is ₁ p~1 •

:'./,"////////h/'///,%/'////////////h'.%/'////////////h'.%/'////////////h'.%/'/////////////~

'%

%'

;%

F O R M U L A % :

~

~

~ ~

;%

The Permanent Disability Model ~:

~

~

~

1

p_~

0 =exp(-

J:

(µ_~!s

+

µ_~:Jds}

1

P_!

1 =exp(-

J:

µ_!!,ds) (10.5)

~i

~ ~ ~

%

;%

01

=

f1

00 01 II ds (10.6) ~: 1

%

1 Px

Jo

s Px µx+s 1-s Px+s

%'

~

~

:~/////.-0////////,/,/////////////fi'.ij'//////////////,%/'//////////////,/,//////h'l'////~;

Example 10.8

~

Prove equation (10.6) and explain its meaning intuitively.

- Solution

(a) To prove equation (10.6), we need to show that it satisfies the KFE and the corresponding

initial condition.

Putting t = 0 into the integral on the RHS of (10.6), we get 0 (because the upper and lower

limits are both 0). So, the initial condition is satisfied.

The KFE for the model is

l

00 d 1Px

-

0

dt 0 01 1Px II 1Px 0

Extracting the component form for the (01 )-th element, we have

d 01 00 01 01 12

dt I Px

=

1Px µ_Ht - 1Px µ_Ht •

'(:, Actex 2013

I

Johnny Li and Andrew Ng

I

SoA Exam MLC

Chapter 10: Multiple State Models

To verify that equation (10.6) is the solution of(*), we differentiate (10.6) with respect to

t.

By Leibniz's rule for differentiation under the integral sign, which says that

d fb(x) fb(x)

a

- f(x,y)dy= f(x, b(x))b'(x)-f(x, a(x))a'(x)

+

-f(x,y)dy,

dx a(x) a(x)

ax

~

f' 00 OJ 11 d

=

00 OJ 11

+

f'

~(

00 01 11 )d dt

Jo

s Px µ x+s 1-s Px+s

s

1Px µ_t+/ 1-1 Px+1

Jo

at

sPx µ_t+s 1-S Px+s

s

=

00 01

+

f' 00 01 (

~

11 )ds. 1Px µ_t+/

Jo

s Px µx+s

at

1-S Px+s By equation (10.5),

a

11

a (

ft-s 12 ) (

f'-s

12 )

a (

f/-s

12 ) 11 12

at

1-S Px+s

=

at

exp -

Jo

µx+s+u du

=

exp -

Jo

µ_t+I/ du

at -

Jo

µ_t+s+u du

=

1-sP x (-µ_t+i> '

so that

d f' 00 II OJ d _ 00 01

+

f' 00 01 ( II 12 )d

dt

Jo

s Px 1-s Px+sµx+s S - 1Px µ_t+t

Jo

s Px µ_t+s -1-.1-Px+sµx+t S

=

00 OJ _ f' 00 01 11 ds X 12

1Px µ_t+t

Jo

s Px µx+s 1-s Px+s µx+t'

which is the same as (*).

(b) To transfer from state 0 to state I on or before time

t,

there must be a time point (says) for the transition to occur. In [O, s), Yhas to stay in state 0. In (s, t], Yhas to stay in state 1.

0 As a result, x 0 00 sPx ~ µx+s 01 d S 1-s Px+s II x+s x+s +ds s s

+

ds age time 01

=

f' 0o Ii _ 01 _ds

=

f'

oo

01 _ 1_1 ds. 1 Px

Jo

s Px 1-s P.t+.1 µx+s

Jo

s Px µ_t+s 1-.1 P.>+s [ END ]

While we have a closed form solution for , p_~1 , only in the case when we have constant transition intensities would it be possible to compute , p_~1 easily. If the transition intensities are time-varying, you may need to use numerical integration method to calculate , p_~1 •

Chapter 10: Multiple State Models

Example 10.9

~

Calculate IO p~g, Io p~~ and Io p~; for the permanent disability model if

(a) µ_~I = 0.028, µ~2 = 0.023, and µ_!2 = 0.032;

(b) µ_~I= 4 x 10-4

+

3.5 x 10-6exp(0.15x), µ_~2 = µ_!2 = 8 x 10-5exp(O.lx). For (b), use Simpson's rule with 4 subintervals.

- Solution

(a) In this case all transition intensities are constant. The

Q

matrix is -r-0.051 Qx+t - 0 0 0.028 0.0231 - 0.032 0.032 .

0

0

. 00 ( rIO ) -0 SI

By

equation (10.5), Io p₆₀ =exp - Jo 0.05 ldt =

e ·

= 0.60050.

By

equation (10.6), I --0.0I91 OI =

f1.

OOµOI . .

p"

ds =

f

1e-0051.10.028e-0032(1-1)ds=0.028e-0032t

-e

'

1P60 j_{0 .,P6o} 60+s1-.1 60+s Jo 0.019 1 --O.I9 oI = 0 028e-0·32 -

e

0 18517 IoP6o · _{O.Ol 9 = ·} ·

By

equation (10.5),

s

p~g

=exp(-

L'c

4x10-4

+

3.5x10-6 e0.I5x60 e0.I5t )dt -

r

(8x10-s eo Ix60 eo It )dt)

[ 4 10 _4 3.5xl0-6e9(o.I5s l) 8xl0-5 e6(o.Is l)] = exp - x s e - - e - . 0.15 0.1

As a by product, putting s = 10, we get Io p~g = 0.29616. Also,

II =exp _

( !

IO-f . (8X1 o-5 e0.Ix(60+s)e0.It )dt = exp(-8X1 o-4 ) e6+0.Is (eO.I(IO-s) -1)].

IO-s P6o+s ₀

Chapter 10: Multiple State Models

B _{yusmgt etwoexpress1ons,wecompute} · h · 00 OI II d

sp60 µ 60+.rio-sP6o+s ats=0,2.5,5, 7.5an 10:

00 OI II

Integrand

s _"P6o µ60+.r IO-s P6o+s

0 1 0.028760794 0.574322903 0.016518 2.5 0.836362337 0.041664711 0.629457925 0.021935 5 0.655364538 0.060439801 0.708083524 0.028047 7.5 0.469158484 0.087757395 0.823608884 0.03391

10 0.296160642 0.127504259 1 0.037762

Simpson's rule with h = 2.5 then gives

IO

p~~ ~

2 ·5 [0.016518

+

4(0.021935

+

0.03391)

+

2(0.028047)

+

0.037762) 3 . = 0.27813. By Io p~g

+

IoP~~

+

10P~~ = 1 , Io P~~ ~ 0.42571. [END]

The Disability Income Model

Finally, we analyze the disability income model. This model is very different from the multiple decrement and permanent disability models because it is possible to transfer from state 1 back to state 0. As a result,

OI >

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:

{ d 00 00 ( OI 02 ) OI IO dt 1Px

=

-1Px Jl.<+t

+

µx+t

+

1Px Jl.r+t ₀₀ l OI O d OI 00 OI OI ( IO I2 ) , o P x

= ,

o Px = ,

dt

1Px

=

1Px µx+t - 1Px µx+t

+

Jlx+t and { d IO IO ( OI 02 ) II IO dt 1Px

=

-1Px µx+f

+

µx+t

+

1Px Jl.r+t IO O 11 l ' 0 Px

= '

0 Px

= '

d 11 IO OI II ( IO I2 ) di 1Px

=

1Px Jl.t+t - 1Px µx+t

+

µx+t

Chapter 10: Multiple State Models

it is not easy to write down the solutions for the transition probabilities in closed form. However, we can solve them numerically using Euler's method. Let's look at the first two coupled system of equations. We can use the recursion formulas

00 00 [ 00 ( 01 02 ) 01 10 ]h (k+l)h Px ~ k!Px + -khp x µx+kh + µx+kh + khPx µ_<+kh '

01 ~ 01 [ 00 01 01 ( 10 12 )]h (k+l)h Px ~ khPx

+

khPx µ_<+kh - khPx µx+kh

+

µx+kh '

The initial values (for k

=

0) are 0 p_~

0

=

I, 0 p~'

=

0. We can then calculate [ "p_~

0_{, "p_~'}

],

[ _{21i P x ,} 00 _{211 Px} 01 ] _{, .. ·}

Typically h has to be very small for Euler's method to give accurate answers. For example, if t

=

I 0 years, then one may take h

=

0.0 I and the time grid would have I 000 steps.

Example 10.10

~

Consider a disability income model for which the transition rates are constant at all ages with

; }1

=

5%,

µ

02 = 3%,

µ

10 =I% and

µ

12 = 7.5%. 5 --0.06/

+

4 --0.105/

(a) Show that /

p~g

= e e

9

(b) By using Euler's method with h

=

0.1, approximate the values of ₀₃ p~g and ₀₃ p~~.

- Solution

(a) The expression given satisfies the initial condition ₀ p~0 =I. Putting this expression into the first equation in the KFE

{

~

00 - - 00 ( 01 02 )

+

01 10 dt 1Px - 1Px µx+t

+

µx+t 1Px µx+t d 01 00 01 01 ( 10 12 ) , dt 1Px

=

1Px µx+t - 1Px µx+t

+

µx+t we get -O.Je--0.061

-0.42

e--0.1051 9 Se -o 061

+

_4e --0 1051 9 0.08

+

0.0 I ,p_~'

- - - =

and hence _I p _X 0'

=

_!_Q(e-0061 -e-0·'051) . This expression satisfies the initial condition 9

0 p~'

=

0, and now we check if it satisfies the second equation in the KFE. The RHS is

Chapter 10: Multiple State Models

5

e-oo6r

+

4e-o.1os1

- - - x 0 . 0 5

9

1 O( e-0.061 _ e-o.1os1) e-0.061 7 e-o.1osr x 0.085 = +

-9 15 60

5

e-o.061

+

4

e-o.1os1

which is equal to the LHS. So,

1

p~~

=

is indeed the solution ofKFE.

9

(b) We discretize the KFE to get the following:

{

Ck+1)1i

P~

0

:::;

k1iP.~

0

+

[-0.08k1iP~

0

+

O.Olk1iP~

1 ]h Ck+1)1i

P~

1 :::;

k1iP.~

1 + [0.05

khP~

0 - 0.085

khP~

1 ]h With h

=

0.1 and the initial conditions, we get

{ o.i

p~

0 >::: 1+[-0.08(1)+0.01(0)] x 0.1 = 0.992 0 I P.~1 :::; 0 + [0.05(1) - 0.085(0)] x 0.1 = 0.005 { 0.2 P.~ 0 >::: 0.992 + [-0.08(0.992) + 0.01(0.005)] x 0.1 = 0.984069 0.2 P.~ 1 >::: 0.005 + [0.05(0.992)-0.085(0.005)] x 0.1=0.0099175 { 03 P.~ 0 :::; 0.984069 + [-0.08(0.984069) + 0.01(0.0099175)] x 0.1 = 0.97620637 03 p~1 >::; 0.0099175 + [0.05(0.984069)-0.085(0.0099175)] x 0.1 = 0.01475355

The true values are given by 0.976308 and 0.014633. The percentage errors are about 1 % in both cases.

[ END]

10. 4 Calculating Actuarial Present Value of Cash Flows

In a multiple state model, cash flows can arise from many different situations. For example, in a

permanent disability insurance plan, some or all of the following may be provided:

- an annuity while the insured is disabled (state 1)

- a lump sum at the moment of becoming disable (transition from state 0 to state 1)

- a lump sum at the moment of death (transition into state 2)

In general, we have the following cash flows in a multiple state model:

Benefits cash flows:

( 1) Cash flow arising from a specific transition (e.g. transition from state 1 to state 2)

Chapter 1 O: Multiple State Models

(2) Cash flow arising from leaving a specific state (e.g. every time when

Y

leaves state 0) (3) Cash flows that arises when entering a specific state (e.g. every time when Y enters states I)

Premium cash flows:

( 4) Cash flow that arises when

Y

is in a specific state (e.g. when

Y

is state 0 in the first year or at the beginning of every year)

In what follows, assume that the life age (x) is in state i at time 0.

(I) Suppose that

b

1 is payable at time t if a transition from state

k

to state

l

happens at time t

fort

::::; m.

The probability to get

b

1 in (t, t

+

dt)

is the probability that there is a transition from

state

k

to state

l

in (t, t

+

dt),

which is , p_~k µ_;~

1

dt. So the APV is

f"'b I ik kl d

Jo /

V 1 Px µ_<+1 t ·

(2) Suppose that

b

1 is payable at time

t

if

Y

leaves state

k

at time t

fort::::; m.

By making use of

(I), the APV is

(3) Suppose that

b

1 is payable at time t if

Y

enters state lat time t

fort::::; m.

By making use of (I),

the APV is

For

b

1 =I,

Actuarial Mathematics for Life Contingent Risks

has the following notation:

/jil_

=

f

111

v'"

ik kl

dt

x:m[

Jo

L.. 1 P x

µ_t+/

·

k"'I

( 4) Suppose there is an annuity of I per year payable continuously

form

years while the life is state}. The APV of this annuity is, assuming a constant interest rate, is

aif-

=

f"'

v'

ij

dt.

x:ml

Jo

I Px

Similarly, if the annuity is payable at the beginning of each year, we have

m-1

.. ;; L

I ij

a -

x:m[

=

v

/ Px'

1=0

Cha pt er 1 O: Multiple State Models

Example

10.11

~

Consider the permanent disability model with transition intensities as specified as in Example

10.9 (a). Assume that the effective interest rate is 5% per year. For a life age (x) that is healthy,

calculate the actuarial present value of

(a) an insurance that pays 1000 at the moment when the insured becomes disabled;

(b) an annuity due that pays 10 at the beginning of each month if the insured is healthy, for at

most 10 years;

( c) an annuity due that pays 100 at the beginning of each month if the insured is disabled, for at

most 3 years starting from the time when the first annuity payment is made.

-

Solution

(a) The density of the time to transit to state

1

from state

0

is 1 p_~

0 _{µ_~L.}

Since

1

p_~0

=

e-0°511

by equation (10.5) (or Chapter 8),

1

p~0 µ_~!

1

=0.028e-a0511• The actuarial present value of the insurance is

1000

f"'-

1-0.028e-00511dt

=

28

=

280.5888.

Jo

1.051

In

1.05

+

0.051 (b) The actuarial present value of the annuity is

120 119 ₁ _0.0511 119(e-0.051]&

1

-(e

1

-:~']"

10:L

e

12 ₌

1o:L - -

_{= 10 = 762.3716.} 1=0 1.05 1112 1=0 1.05 (e-0.051

lii

(c) Method I: 1 -1.05

We first calculate the APV of an insurance that pays a dollar benefit at the end of the month

of becoming disabled. The probability of getting the payment at time k (where k

=

1/12, 2/12, ... ) is I 00 01 -0.051(k--) 28 k-1112Px 1112Px+k-1112=e 12 x19e 0.032 0.019 12 (I - e 12 ) . So the APV is

Chapter 10: Multiple State Models

oo l _0_051<~_!_> _{28 _ 0.032} 0.019

A<12)01 _ " e 12 12 x-e 12 (l-e 12 )

x -f:::l.05 1112 19

28 _0032 _0019 l oo l _00511

= - e 12 (I - e 12 ) " e 12

19 1.051/12

f;t

1.051/12

28 _o.032 0019 l I

=19e 12 (l-e 12 )l.051112 !-(e-0.051/1.05)1112

= 0.27964505

Suppose that the first annuity payment starts at time t. The APV, at time t, of the life annuity due, is

120oa<

12

_>~

1

_=100~

1 _{e-0032 k 112 =100 l-(e--0·}032_{/l.0 5)}3 _{=3207.716824.} x+t:31 ~ I.05k112 l-(e-0032 I _l.05)1112

Since the above is independent of

t,

we can treat the above as a constant benefit function. The APV, at time 0, of the benefits, is

3207.716824A<12 >01 =3207.716824 _x x 0.27964505 = 897.0221.

Method 2:

The maximum number of annuity payment is 36. Let us fix a time point t (which is a multiple of 1112). For l 00 to be payable at

t,

the insured has to be in state l, but not in state 1 for more than 3 years. As a result, the probability of a payment at tis, fort= 0, 1/12, .. ., 35/12, is

Pr(Y(x+t)= l

I

Y(x)=O)=

iP.~1

=

~:e-00321(1-e-00191).

Fort?:.3,itis

Pr(Y(x

+

t) = l n Y(x

+

t- 3) = 0

I

Y(x) = 0) = 1_3 p~

0

3 P.~~

1

_

3

•

Since all transition rates are independent of age, the above can be reduced to 00 POI - --0.051(1-3) 01

1-3 Px 3 x -

e

3 Px · So, the APV of the annuity is

Cha pt er 1 O: Multiple State Models 35 I 28 -0 o32k -0.051k "' I -0.051k

IOOL

k/12 - ( e 12 - e 12 )+100 3p~1

L

k/12 e 12 eo.051x3 k=O 1.05 19 k=361.05 36 36 36

(e~"'T

(e-'"T

(el_,;;'

r

I- - - I

-2800 1.05 1.05 _{+ 2800 e-0.096 (I - e-0.051) e0.051x3} =

19 I I 19 I

(e~"T

(e-"'7

e-'"T

I- - - I- - - I -1.05 1.05 1.05 ' = 897.0221 [ END]

Example 10.12

~

Consider the disability income model in Example I 0.10 where the transition rates are constant at

all ages with µ01

=

5%, µ02

=

3%, µIO= I% and µ12

=

7.5%. Assume that the force of interest is 4%.

(a) Calculate the actuarial present value of a disability benefit of 200 per annum paid

continuously to a life now aged 40 exact and disabled, during this period of disability,

payable to a maximum age of 60.

(b) The company issues a disability income benefit policy, which pays an inflation-adjusted

disability benefit of 2000e0·051 immediately every time the policyholder becomes disabled at

time

t.

Calculate the actuarial present value of the benefit for a healthy life aged 40.

(c) The policyholder of the policy in (b) pays a level premium of rate P continuously while he is

healthy, up to age 50. Find

P

using the equivalence principle.

(d) A company also issues a special disability income policy with a benefit of 360 per annum

payable continuously while the insured is disabled, provided that the he has been disabled

continuously for at least one year. Benefit payments under this policy cease at age 60.

Calculate the actuarial present value of the benefit for a healthy life aged 40.