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For More Information: email [email protected] or call 1(800) 282-2839 Chapter 1 – Actuarial Notation  1

CHAPTER 1 – ACTUARIAL NOTATION

There is a specific notation used for expressing probabilities, called “actuarial notation”. You must become familiar with this notation, as it is used extensively in the study material and on the exams.

The notation txp means “the probability of a person age x living t or more years”.

Example 1.1 – Interpret 57p

Answer – The probability of a person age 7 living five or more years.

Conversely, txq means “the probability of a person age x dying within the next t years”.

Example 1.2 – Interpret 57q

Answer – The probability of a person age 7 dying at or before age 12.

In the case where we are only looking for the probability over the course of one year we use the shorthand notation px or qx .

Example 1.3 – Interpret q0

Answer – The probability of a person age 0 dying at or before age 1.

Even though we are using a different notation we still follow the same rules of probability. One of the more common rules used is that if you have two (and only two) possible events, A and B, then PA() PB () 1. The probabilities of these two events are compliments of each other.

Example 1.4 – Given 510p  0.9 , find 510q

Answer – These two values are compliments of each other so we can say

510qp1 5 10  1 0.9  0.1

Because of the way we write the notation, it can be a little confusing to try to relate the different probabilities to each other.

Example 1.5 – Given 23p  0.9 and 45p  0.8 , find 63p

Answer – By rule, 63p  2345pp. This is true because in order for a person age 3 to live 6 or

more years (i.e. 63p ) he must first live 2 more years (i.e. 23p ) and then live another 4 more years

(i.e. 45p ). So in this case 63p 0.9 0.8 0.72 .

Example 1.6 – Given 46q  0.2 and 26p  0.9 , find 28p =>

Answer – Let’s solve this problem in steps:

p pp p I. Here 46 2628, so we first must find 46

2  Chapter 1 – Actuarial Notation

p 1 q p 1 0.2 0.8 II. By the property of complements, 46 46, so 46 . 0.8 0.9 p p  0.8889 III. Then we can say 28, so 28 .

Just because 20p pp 1 0 doesn’t mean that 20qqq 1 0. You must remember that 20q is the probability that a person age 0 dies within the first two years – the person could die in the first year or survive the first year and die in the second year. So we would write this as 20qqpq 0 0 1, where

q0 = probability of dying in the first year

p01q = probability of surviving the first year and dying in the second year

By extension we can also say 30q q 0 pq 0 1 2 pq 0  2.

Example 1.7 – Given 21p  0.6 and q1  0.1, find q2 Answer – We can solve this two ways: First way:

I. 21p pp 1 2

II. pq111  1 0.1  0.9

0.6 0.9 p2 p2  0.6667 III. 21p pp 1 2 => => q 1 0.6667  0.3333 IV. 2 Second way:

I. 21qqpq 1 1 2

II. 21qp1 2 1  1 0.6  0.4

III. pq111  1 0.1  0.9

IV. 0.4 0.1 0.9 q2 => q2  0.3333

The final type of notation is one that is used when we would like to “delay” the death or failure. For example, q means the probability of a person age 8 surviving for 3 years and then dying within the 3 48 next four years. The number to the left of the bar means that the person will survive at least that number of years before dying. We can calculate q using the equation qpq. Similarly, q is 3 48 3 48 38411 4 0 the probability of a person age 0 surviving for 4 years and dying within the next year ( qpq). 4 0404 Example 1.8 – Given p 0.9 and p 0.8 , find q . 20 22 2 20 Answer –

I. qpq 2 20 2 0 22

II. 22qp1 2 2  1 0.8  0.2 . III. So q 0.9 0.2 0.18 . 2 20

Chapter 1 – Actuarial Notation  3

Chapter 1 – Questions

Express the following using actuarial notation:

1) The probability of a person age 60 dying within 2 years.

2) The probability of a person age 10 living one or more year.

3) The probability of a newborn living at least 20 years.

4) The probability of a person age 2 dying within half a year.

5) The probability of a person age 5 living 3 years and then dying before age 10.

6) The probability of a newborn living 20 years and then dying within the next year.

Solve the following:

7) Given p0  0.8 , find q0

8) Given 22q  0.01, find 22p .

9) Given p1  0.8 and 32p  0.7 , find 41p

10) Given 73p  0.36 , p3  0.9 , and 24p  0.8 , find 46p

11) Given 24p  0.75 and q4  0.1, find p5 .

12) Given 31q  0.2 and q4  0.15 , find 41p .

13) Given 0.5p 2  0.7 and 1.5p 2.5  0.6 , find 22p .

14) Given p 0.8 and q 0.2 , find q 22 34 2 32

15) Given p 0.9 and p 0.7 , find q . 310 13  3 10

16) Given p 0.54 , p 0.5, and p 0.9 , find q 45 9  4  5 4

17) Given p 0.9 , p 0.7 , and p 0.6 , find q . 2  3  4  2 2

4  Chapter 1 – Actuarial Notation

Chapter 1 – Answers

Express the following using actuarial notation:

1) The probability of a person age 60 dying within 2 years.

Answer: 260q

2) The probability of a person age 10 living one or more year.

Answer: p10

3) The probability of a newborn living at least 20 years.

Answer: 20p 0

4) The probability of a person age 2 dying within half a year.

Answer: 0.5q 2

5) The probability of a person age 5 living 3 years and then dying before age 10.

Answer: q 3 25

6) The probability of a newborn living 20 years and then dying within the next year.

Answer: p 20 0

Solve the following:

7) Given p0  0.8 , find q0

Answer: qp001  1 0.8  0.2

8) Given 22q  0.01, find 22p .

Answer: 22pq1 22  1 0.01  0.99

9) Given p1  0.8 and 32p  0.7 , find 41p

Answer: 41ppp 132 0.8  0.7  0.56

10) Given 73p  0.36 , p3  0.9 , and 24p  0.8 , find 46p

Answer: 73p ppp 32446  => 0.36 0.9 0.8 46p => 46p  0.5

Chapter 1 – Actuarial Notation  5

11) Given 24p  0.75 and q4  0.1, find p5 .

Answer: 24p pp 4 5 => 0.75 0.9 p5 => p5  0.8333

12) Given 31q  0.2 and q4  0.15 , find 41p .

Answer: 41p  31pp 4 => 41p 0.8 0.85  0.68

13) Given 0.5p 2  0.7 and 1.5p 2.5  0.6 , find 22p .

Answer: 2ppp 2 0.5 2 1.5 2.5 0.7 0.6 0.42

14) Given p 0.8 and q 0.2 , find q . 32 34 2 32

Answer: qpq0.8 0.2 0.16 2 32 2 2 34

15) Given p 0.9 and p 0.7 , find q . 310 13  3 10

Answer: qpq0.9 0.3 0.27 3 10 3 10 13

16) Given p 0.54 , p 0.5, and p 0.9 , find q 45 9  4  5 4

Answer: We need to solve this in a few steps:

I. qpq 5 4549

II. 54p pp 445 => 54p 0.9 0.54  0.486 III. q 0.486 0.5 0.243 5 4 17) Given p 0.9 , p 0.7 , and p 0.6 , find q . 2  3  4  2 2

Answer: qpqppq0.9 0.7 0.4 0.252 2 2224 234

6  Chapter 1 – Actuarial Notation

Chapter 2 – Survival Models  7

CHAPTER 2 – SURVIVAL MODELS

A survival model is a model for a continuous, random variable T0 , which will “fail” at some age x. Some example of survival models include:

1) Duration of a battery 2) Duration of a baseball strike 3) Lifetime of a person

T0 = random variable for age of entity when it fails, assuming it starts at age 0. This would be the random variable used when modeling the survival of a person from birth.

Tx = random variable for the time of survival for an entity, assuming that it is alive at age x. For example, if a person aged 30 decides to purchase life insurance the, random variable would be T30 and an actuary would model the years survived past age 30.

The difference between T0 and Tx is as follows:

Tx is really the same variable as T0 -- it just starts x years later. We can say thatTxT0 x . For example, if a person aged 30 lives another 60 years, then T30  60 , but TT030 30 30 60 90 .

There are several functions of T0 that you need to be familiar with:

1) Cumulative Distribution Function (CDF) -- Ft00( ) Pr(T t ) . In actuarial notation this function

is represented by t q0 , which is the probability that a person aged 0 dies at or before time t.

2) Survival Distribution Function (SDF) -- St000( ) 1 F (t) Pr(T  t ) . In actuarial notation, this

function is represented by t p0 , which is the probability that a person aged 0 dies after time t. As

indicated by the above equation, St0 ()is the complement of Ft0 ().

Here are some important properties of the CDF and SDF:

Property Interpretation The probability of a person dying at or before time F0 (0) 0 0 is 0. Meaning, we assume that a person survives past time 0. The two functions are complements of each other SF00(0)1(0)1  The probability of surviving past age t gets smaller limSt0 ( ) 0 t and smaller as the age gets higher. As t increases, the probability of a person dying at limFt00 ( ) 1 lim St ( )  1 tt  or before that age gets closer and closer to 1.

8  Chapter 2 – Survival Models

Given Sx() e.02x , calculate the following:

Example 2.1 – The probability of a person age 0 surviving past age 40.

Answer: This is Pt( 40) S (40)  e.02(40)  0.4493

Example 2.2 – The probability of a person age 0 dying before age 60.

Answer: This is Pt( 60)  F (60)  1 S (60)  1 e.02(60)  0.6988

Example 2.3 – The probability of a person age 0 dying between 20 and 30.

Answer: This is Pt(20 30)  S (20)  S (30)  e.02(20)  e .02(30)  0.1215

Example 2.4 – The probability of a person age 20 surviving past age 50.

Se(50) .02(50) Answer: This is Pt( 50 t 20)   0.5488 , since the person starts at age Se(20) .02(20) 20.

Another important function:

dd d 3) Probability Density Function (PDF) -- f ()tFtStSt () 1 () () => 00dt dt 0 dt 0

d f ()tSt () 00dt

As opposed to Ft0 ()and St0 (), which are interval measures, f0 ()t is a derivative and is therefore an instantaneous measure. We say that it is the instantaneous “rate of failure” at time t.

Ft0 ()is the integral of f0 ()t . Therefore,

t F ()tfxdx ( ) 00 0

So the probability Ft0 ()is represented by the area below the graph of f0  x between time 0 and t.

Chapter 2 – Survival Models  9

Although f0 x is not the probability measure, it can be used to approximate probabilities in very small intervals. Since St0 ()is the complement of Ft0 (), we can also say

 St() fxdx ( ) 00 t

Example 2.5 – fx( ) 0.02 e0.02x , for all x  0 . Find F(5) .

Answer:

55 5 I. Ffxdxedxe(5) ( ) 0.02 0.02x  0.02x  0 00 II. F(5) .9048  1  0.0952

Example 2.6 – fx( ) 0.03 e0.03x , for all x  0 . Find Ft().

Answer:

tt t I. Fx() fxdx () 0.03 e0.03x dx  e 0.03x  0 00 II. Ft() 1 e0.03t

1 Example 2.7 – fx() , for all 0100 x  . Find S(40). 100

Answer:

100 100 1 x 100 I. Sfxdxdx(40) ( ) 40 40 100 100 40 40 II. S(40) 1  0.6 100

f0 ()t is also called the unconditional density of failure at time t, since the only “condition” for the function is that the person exists at time t  0 . We contrast this with the conditional density function below:

f0 ()t 4) Hazard Rate Function (HRF) -- 0 ()t  . This function is the conditional density function, St0 ()

since it is conditional on survival to age t. In other words, it is conditional on St0 (). This becomes clearer when we consider the following law of probability: PAB(|) PB () PA ( B )

10  Chapter 2 – Survival Models

 Event A = survival to age t. Event B = dying the instant after age t.

 PA()() B f0 t=> probability of surviving to age t and then dying the instant after age t

 PB() S0 () t => probability of surviving to age t

 PAB(|) 0 () t=> probability of dying the instant after age t, given survival to age t

 PAB(|) PB () PA ( B )=> 00()tStft () 0 ()=>

f ()t  ()t  0 0 St() 0

The hazard rate is normally called the force of mortality and is written as t .

To summarize:

 f0 ()t => probability of dying the instant after age t.

 0 ()t => probability of dying the instant after age t, given survival to age t.

Example 2.8 – Find ()t if ft( ) 0.05 e0.05t , for t  0 .

Answer:

  I. We first need to find St() => St( ) 0.05 e0.05tt dt e 0.05 =>  t t St() 0  e0.05tt  e 0.05 ft( ) 0.05 e0.05t II. (t )  0.05 St() e0.05t

Notice that the force of mortality is a constant here and does not depend at all on t. This is true for all distributions that have the form et , where  is the force of mortality – these are all exponential distributions.

d Since f ()tSt (), by the chain rule for derivatives we can also say 00dt d  St0 () ft0 () dt d 00()tSt  ln (), which means that 0 ()t is the derivative of  lnSt0 ( ) . This St00() St () dt also leads us to the next important function:

t 5) Cumulative Hazard Function (CHF) -- ()txdxSt ( )  ln (). Raising both sides to the 00 0 0 power of e, we have the following useful result:

Chapter 2 – Survival Models  11

t 0 ()x dx 00()t St0 () e e

Example 2.8 – Calculate S(10) , given (x ) 0.006x .

10  0.006x 10  0.003x2 Answer – Se(10)00 e  e0.3 0.7408

Example 2.9 – (Exam 3, Fall 2003) For a loss distribution where x  2 , you are given:

z2 i) The hazard rate function: hx() , for x  2 2x ii) A value of the distribution function: F(5) 0.84

Calculate z.

Answer: This question combines the knowledge of three of the functions we have reviewed thus far:

I. Since Sx() 1 Fx (), we can say SF(5) 1 (5) 1 0.84 0.16

t 5 z2  hxdx()  dx   2x II. By definition, St() e0 ()t e2 so Se(5) 0.16 2 , where z is a constant.

zz22 lnx 5  ln 5 ln 2 III. Integrate => 0.16 ee222 (remember, we can take the expression with z out before integrating because it is a constant). z2 IV. ln 0.16 (ln 5  ln 2) => z2  4 => z  2 2

Moments

Moments of random variables are used to calculate expected values and variances:

  First Moment (Expected Value): E T t f t dt  S t dt . This is known as the 00 0  00 0

complete expectation of life at birth. We will refer to this as e0 . Since St00 t p, using

0  actuarial notation we can also write this as epdt . 00 t 0   Second Moment: E Ttftdt22 . 00  0 2  Variance: Var() T E T2 E T 00  0

12  Chapter 2 – Survival Models

Example 2.10 – Given the below definition of Sx(), find e0 :

Sx() 1 0.09 x, for 05 x 

Sx() e0.09x , for 520x

Sx() 0, elsewhere

Answer –

0 520 e p dx p p dx 00505xx 05

0 520520 e1 0.09 x dx  1 0.09 5 e0.09xx dx  1 0.09 x dx  0.55 e 0.09 dx 0   0505

520 0.09x 20 0 5 e exdxedxxx1 0.09  0.550.09x  0.045 2  0.55  3.875  0.55  5.2481  6.7615 0  0 0.09 05 5

Example 2.11 – Calculate the expected number of years survived for a person age 20 if x2 Sx() 1 and x  80 6400

Answer – This another way of asking for e20 , given survival to age 20. In other words, we are looking for:

80 80 x2 x3 0 1 dx x  e  6400 19200 20 20 20 36 => S(20) 0.9375 0.9375

Total years survived = 36 + 20 = 56

Chapter 2 – Survival Models  13

Median

Normally we think of the median as the “halfway” point for a series of numbers put in increasing order.

In the case of our random variable, T0 , there are infinitely many values, since a person can die at many different ages. So instead we must use the definition of a median to find the median of T0 . It is the age t 1 at which PrTt Pr Tt . Since Ft( ) Pr(T t ) and St() Pr(T t ), we can say oo2 00 00 1 that this is the age t at which St() Ft () 002

2 Example 2.12 – Find the median age for a survival model with f ()xxe 2 x for

1 Answer – We need to find the value of t such that Ft() 2

tt 1 22t I. f() x dx 2 xexx dx  e  0 2 00

2 2 II. 0.5 1 et => et  0.5 => t 2  0.6931 => t  0.8325

The median is just an example of a percentile – the 50th percentile. We can apply this same concept to any percentile. For example, if we wanted to find the 25th percentile in the above example then we would t 2 solve the equation 0.25  2xedx x 0

Examples of Survival Models

We will explore several distributions that can be used to model the survival of a random variable. These are called survival models.

Uniform Distribution

In the uniform distribution (also called De Moivre distribution), the PDF is constant over the entire distribution. This distribution is more appropriate for short ranges of time or ages. The general definition 1 for the PDF is ft() for atband ft() 0everywhere else. But in our case we always start 0 ba 0 at age 0 so a  0 and then

11 ft()  0 b 

t Since F ()tfxdx ( ) and therefore (remember,  is a constant), 00 0

14  Chapter 2 – Survival Models

t 1 t Ft() dx 0  0  

The CDF is therefore a linear function and the graph would look like this:

Here are some other important properties of the uniform distribution function:

tt   SDF => St() 1 F (t)  1  00  1 ft() 1  HRF =>  ()t 0  0   t St0 ()   t  tt2   First Moment => ET t f tdt  dt   00 0022 tt232  Second Moment => ET22 t f tdt  dt   00 0033

22222 2  w  2  Variance => Var() T00 E T E T 0  32 3412

Example 2.13 – Find the variance of a uniform model where e0  50

Answer –

 I. 50  =>  100 2 1002 II. Variance =  833.33 12

Chapter 2 – Survival Models  15

Exponential Distribution

t In the exponential distribution the PDF is defined as f0 ()te  , where  is a constant and must be greater than 0. Using integration by parts we can find the equation of the CDF, where tt F ()tfxdxedxe ( ) x 1 t . Here are the other functions that follow: 00 00

tt  SDF => St00() 1 Ft ()  1 (1  e )  e ft() et 0  HRF => 0 ()t t  St0 () e  1  First Moment => ET t et dt  0  0   2  Second Moment => ET t2  et dt  0  2 0 

2 1  Variance => Var() T E T2 E T  00 0  2

The hazard rate function (HRF) is of particular interest. You will notice that it is not dependent on t, which means that it is the same value for all age t. This means that the instantaneous probability of dying the instant after t, given survival to age t, is the same throughout the entire distribution. In other words, the exponential distribution is “memoryless” – the hazard rate at any time is not related to anything that happened up until that point. For example, a person age 20 and a person age 50 both have the same probability of living 80 years if the probability follows the exponential distribution. Contrast this with the uniform distribution, where the probability decreases as the person gets older.

Gompertz and Makeham Distributions

The last two distributions are fairly obscure and not tested nearly as much as the previous two distributions. The Gompertz distribution is defined by its force of mortality ()tBc t , where t  0 ,

t  Bcx dx B  (1ct ) B  0 , and c 1. The SDF is then defined as St() e0 eln c . This distribution is suggested as a model for human survival, since the force of mortality increases as t increases. In other words, the probability of instantaneous failure (i.e. death) increases as a person gets older. The Makeham distribution is similar to the Gompertz, except that its force of mortality is defined as ()tABc t . The reason for the extra constant is that Makeham believe that part of the hazard rate at any age is independent of the age itself. The SDF for this distribution is also fairly complicated and defined as t ()ABcdxx B  (1cAtt ) St() e0 eln c . The other functions (PDF, Moments) are even more complicated and not worth mentioning here.

16  Chapter 2 – Survival Models

Time-to-Failure Random Variable

Using the functions that we defined in the previous section, we can apply the same logic that we used for random variable T0 to random variable Tx . Recall that Tx represents the remaining lifetime for a person who has already survived to age x. We will define the functions below:

1) Survival Distribution Function (SDF) -- St()xx PrT ( t ). In actuarial notation, this is

represented by txp , which means the probability that a person age x will survive t more years.

Since TTx0 x , this is also the same as Pr() T00 x t T  x -- the probability that a person will survive x  t , given that they have survived past age x. Using the laws of probability we can

Pr()()() T00 x t T x Pr T 0  x t S 0 x  t say Pr() T00 x t T  x    tx p . So Pr() T000 x Pr () T x S () x

Sxt0 () txpSt() x Sx0 ()

Let’s apply this function to the uniform and exponential distributions

Example 2.14 – Find St()a time 10 for a uniform distribution, where 0 t  100 .

100 (t  10) St(10 ) 90  t Answer – St() p 100  10t 10 S(10)100 10 90 100

The above distribution “restarts” as another uniform distribution at time t 10 so that now it is a uniform distribution where 090t .

x2 Example 2.15 – Find p for a survival model with PDF fx() , where 03x . t 3 9

Answer –

I. First, let’s find the general formula for Sx()=> 3 3 x23xx 3 Sx() dx 1 x 92727x

2) Cumulative Density Function (CDF) -- Ft()xx PrT ( t ). In actuarial notation, this is

represented by txq , which means the probability that a person age x will die within the next t

Chapter 2 – Survival Models  17

Sxt000() Sx ()() Sxt years. Since Ftxx() 1 St (), we can say that Ftx () 1  . This Sx00() Sx () F ()()xt  Fx final fraction can also be written as 00. Sx0 ()

Example 2.16 – (Fall 2003, 5) The following information is given for a survival model:

 Mortality follows De Moivre’s Law 0

 e20  30

Calculate q20 .

Answer:

0  I. Let’s work backwards and solve for  => e  30  =>   60 20 2 21 20 1 II. qF(21) F (20) 20 60 60 60

ddSxt0 () 3) Probability Density Function (PDF) -- ftxx() Ft () 1 . We treat dt dt S0 () x

everything except for the Sxt0 () as constant, since it is the only part of the expression that d Sxt0 () dt f0 ()xt involves t, so this becomes ftx ()(the negatives cancel out). Sx00() Sx () III.

4) Hazard Rate Function (HRF) – By definition, the hazard rate x ()t is the same as 0 ()x  t , since both of them represent the instantaneous rate of failure that is conditional on survival to age x  t .

0 0

The variable eETx  x represents the expected future lifetime for a person age x (in contrast to e0 ,

0  which represents the expected future lifetime for a person age 0). Just as epdt , we can also say 00 t 0 0  Sxt() 0  Sxt() epdt . Since p  0 , we can say edt 0 . In this type of scenario Sx() is xtx tx x  0 0 Sx0 () 0 Sx0 () a constant so this can be simplified as

18  Chapter 2 – Survival Models

0 1  eSxtdt() x  0 Sx0 ()0

Since the random variable Tx is just T0 shifted over x units the variance is still given by 2 Var() T E T2 E T -- the variance is not affected by a shift by a constant. 00  0

Example 2.17 – Find e30 and Var() T30 , if x is uniform on 0,80 .

Answer – As mentioned earlier, the uniform distribution “restarts” at time x so we can use a shortcut and say that this is really the same as a uniform distribution on 0,50, starting at time 0.

0 50 I. e 25 30 2 502 II. Var( T ) 208.33 30 12

Example 2.18 – A survival model follows an exponential distribution with fx( ) 0.04 e0.04x . 0

Find e40 and Var() T40 .

0 0.04x I. Let’s use the above formula to solve for e40 => Sx() e => 0 11 eStdtedt(40 ) 0.04(40 t) 40 0.04(40) Se(40) 00 0 1 e0.04(40) e e0.04(40) edt 0.04(ttt ) edtedt  0.04( )  0.04( ) 40 0.04(40) 0.04(40) ee000 0 00

This is the same formula for e0 , which means that ee040 . In other words, survival models that follow an exponential distribution have the same future expectation no matter what the starting point is. This is consistent with what we said earlier about the “memoryless” characteristic of the exponential distribution. In an exponential 0 1 distribution, e  , so x  0 1 e 25 40 0.04

II. The variance of T40 is the same as the variance of T0 , so 1 Var( T ) 625 40 0.042

Chapter 2 – Survival Models  19

Occasionally we will only be interested in the discrete or integral part of Tx . The random variable Kx is defined as the curtate (i.e. discrete) duration at failure random variable. It is the greatest integer that is * * less than Tx . For example, if Tx  5.6 , then K x  5 . K x is defined as KKxx 1. So if failure occurs * in the first time interval (e.g. in the first year of life) then K x  0 and K x 1. In other words to find Kx * we always round down and to find K x we always round up. ex represents the curtate expectation of life at age x. Since we are only dealing with discrete (integral) value here, we use a different notation for ex  * * => epx   tx. By definition, the expectation for K x is => eexx 1 . t0

An interesting result is the relationship between ex and ex1 : epppp......  ppp ...... xxxxx0123 xxx 23 eppppxxxxx10 1 2 3...... ppp xxx 12 13  1  ......

Multiply ex1 by px =>

pexx112131234 p x p x  p x  p x ......   p x  p x p x  ......

Add px =>

px peppppxxxxxx1  234   ......  e x

Therefore, we can say

eppex  xxx1

Example 2.19 – If ex  7 and px  0.9 for all x, find ex2

Answer – We can extend the rule above and say epx  xxxx22 p pe 2 =>

22 7 0.9 0.9  0.9 ex2 => ex2  6.53

20  Chapter 2 – Survival Models

Chapter 2 – Formula Summary

Important Functions

Ft00( ) Pr(T t ) =>

Cumulative Distribution Function (CDF) F00()()xt Fx txq  Sx0 ()

St000( ) 1 F (t) Pr(T  t ) =>

Survival Distribution Function (SDF) Sxt0 () txpSt() x Sx0 () d f ()tSt () => 00dt Probability Density Function (PDF) t  F ()tfxdx ( ) , St() fxdx ( ) 00 00 0 t

f0 ()t Hazard Rate Function (HRF) 0 ()t  St0 () t ()txdxSt ( )  ln () => 00 0 0 Cumulative Hazard Function (CHF) t 0 ()x dx 00()t St0 () e e  E T t f t dt  S t dt , 00 0  00 First Moment (Expected Value) 0  epdt 00 t 0  E Ttftdt22 Second Moment 00  0 2 Var() T E T2 E T Variance 00  0

Recursive Relationship eppex  xxx1

Uniform Distribution

t 1 t Ft() dx Cumulative Distribution Function (CDF) 0  0     t Survival Distribution Function (SDF) St() 0  11 Probability Density Function (PDF) ft()  0 b 

Chapter 2 – Survival Models  21

1 Hazard Rate Function (HRF)  ()t  0   t  First Moment (Expected Value) ET 0 2  2 Second Moment ET2  0 3 2 Variance Var() T  0 12

Exponential Distribution

t Cumulative Distribution Function (CDF) F0 ()te 1

t Survival Distribution Function (SDF) St0 () e

t Probability Density Function (PDF) f0 ()te 

Hazard Rate Function (HRF) 0 ()t   1 First Moment (Expected Value) ET 0  2 Second Moment ET 0  2 1 Variance Var() T  0  2

22  Chapter 2 – Survival Models

Chapter 2 – Warmup Questions

1. For a new life, ()xx4 , 0 50. 50  x  Find e .

2. If ()xx2 , 0 100, find PX20  50 . 100  x

3. If X is DeMoivre on [0, ) and VX( ) 468.75 , what is ?

 4. X is exponential and e30  40. Find  .

(30 x )2 5. If fx() , 0 x 30, find p . 9, 000 t 5

6. If fx()20  x , 0 x 20, find  (10) . 200

7. The lifetime of a machine component is modeled by  Sx() 1x , 0  x  and  0.    If (40) 2 (20) , find  .

8. For a population of 20 year olds, 2 are nonsmokers and 1 are smokers. The remaining lifetime 3 3 of a nonsmoker is uniform over [0, 80) and for a smoker it is uniform over [0,50) . For a person selected at random from those who survive to age 40, what is the probability that he will survive 5 more years?

2 9. Given Sx( ) 1x , 0  x 100 , find the probability that a person aged 20 will die between  100 the ages of 50 and 60.

10. If fx()20  x , 0 x 20, find e . 200 5

11. If ()xx2x , 0 20, find VX ( ) . 400  x2

12. The lifetime, X, of a machine is modeled as follows: A. For 030,x it follows the DeMoivre model on [0,100). B. For x  30, it follows the constant force model with   0.02. Find E().X

Chapter 2 – Survival Models  23

Solutions to Chapter 2 – Warmup Questions

1. First, find the cumulative hazard function  (x ) :

xx x ()xudu () 4  4ln(50  u ) 0050  u 0 4 4 ln(50 x )  ln(50)  ln 1 x  50 4 ln 1 x 4 Sx() e()x e  50  1 x  50 50  50 50 xx4550 eSxdx0 () 1 dx 1 10 00 50 5 50 0

2. The cumulative hazard function is

x x 2 (xduu )2  2ln 100   ln 1 x 0 100 u 0  100 2 Sx() e()x 1 x  100 22 PX[20 50]  S (20)  S (50)  120  1 50  100  100 (0.8)22 (0.5) 0.39

2 3. For X uniform on [0, ) , VX  . 12

2  468.75,2 5,625. 12   75.

  4. For an exponential distribution, een  0 due to the memoryless property of the exponential.  1 eEX0 () for the exponential distribution. 

1 40 0.025. 

x xx(30tt )23 (30 ) (30  x ) 3 5. Fx() ftdt () dt  1 , 0 x 30 009,000 27,000 27,000 0 (30 x )3 Sx 27,000 3 St(5)(305) t t 3 t p5  1 S(5)(30 5)3  25

24  Chapter 2 – Survival Models

x xx (20tx )22 (20 ) 6. Fx() ft () dt20  t dt  1 00200 400 400 0 (20 x )2 Sx() 400 20  x fx()   ()x 200 2 Sx()(20 x )2 20 x 400

(10)  1 5

Sx()   1 7. ()xSxSx , () 1xx , ()1 1 Sx()       1  1 x  ()x      x 1 x    (40) 2 (20) 2 40 20 20 2  80    60

8. For nonsmokers, pn 1.20  3 Their remaining lifetime at age 40 is linear over [0, 60). 20 20 80 4 For smokers, ps 1.20  3 Their remaining lifetime at age 40 is linear over [0, 30). 20 20 50 5 From a population of l at age 20, there are 2 l nonsmokers and 1l smokers. At age 40, 20 3 20 3 20 there are 32ll 0.5 nonsmokers and 31ll 0.2 smokers. Among 40 year olds, 5 43  20 20 53  20 20 7 are nonsmokers and 2 are smokers. 7 For nonsmokers, pn 1.511  540 60 12 For smokers, ps 1 55  540 30 6 For the total population,

p 51125 540 712   76   75 84  25 28

Chapter 2 – Survival Models  25

9. The probability that a 20 year old will die between the ages of 50 and 60 is

30pp 20 40 20. 2 1 t  20 St(20)   p 100 t 20 S(20) 2 1 20  100 2 1 50  p 100 0.25 25 30 20 2 0.64 64 1 20  100 2 1 60  p 100 16 40 20 0.64 64 pp9 30 20 40 20 64

x x (20tx )22 (20 ) 10. Fx()20  t dt 1 0 200 400 400 0 (20 x )2 Sx() 400 (20t 5)2 St(5) 2 pt400 1,015t t 5 S(5)2  15 (20 5) 400 15  15 15 tt2315 e5  p dt115 dt 00t 5  15 3 15 0

2x 11. ()x  , 400  x2 x ()xdu2u 0 400  u2 x 2 ln(400 u2 )  ln 400  x 0 400 2 ln 400x 2 Sx() e()x e 400 400  x 400

26  Chapter 2 – Survival Models

fx() Sx ()  x 200 20 20 23 EX() xfxdx () x dx 20 40 00200 600 3 20 20 34 EX()22 x f xdxx dx 20 200 00200 800 2 VX( ) 20040  22.22  3 

12. For X DeMoivre on [0,100) , Sx() 1x . S(30) 0.7. 100 If X is constant force with   0.02, Sx() e0.02t .

For the lifetime of the machine

 x 1, 030x Sx()  100 0.02x 30 0.7ex , 30 

30 E()XSxdx 1x dxe 0.7 0.02x 30 dx 00 100  30 In the second integral let ux 30 30 2  EX() xx  0.7 e0.02u du 200 0 0 25.5 0.7(50)  60.5

Chapter 2 – Survival Models  27

Chapter 2 – Questions

12 x 1) (Fall 2004, 8) Given Sx() 1 , for 0 x  100 , calculate the probability that a life 100 age 36 will die between ages 51 and 64.

A. Less than 0.15 B. At least 0.15, but less than 0.20 C. At least 0.20, but less than 0.25 D. At least 0.25, but less than 0.30 E. At least 0.30

2) (Fall 2005, 10) You are given the survival function Sx()as described below:

x  Sx() 1 for 040x 40  Sx()is zero elsewhere.

Calculate e25 , the complete expectation of life at age 25.

A. Less than 7.7 B. At least 7.7, but less than 8.2 C. At least 8.2, but less than 8.7 D. At least 8.7, but less than 9.2 E. At least 9.2

3) (Fall 2005, 11) Individuals with Flapping Gum Disease are known to have a constant force of mortality  . Historically, 10% will die within 20 years. A new, more serious form of the disease has surfaced with a constant force of mortality equal to 2 . Calculate the probability of death in the next 20 years for an individual with this new strain.

A. 17% B. 18% C. 19% D. 20% E. 21%

28  Chapter 2 – Survival Models

xt 58 4) (Fall 2005, 12) For a life age x, the force of mortality is   , for x  58 and t  0 . xt 200 Calculate q , the probability that (60) will die between ages 70 and 71. 10 60

A. Less than 0.015 B. At least 0.015, but less than 0.030 C. At least 0.030, but less than 0.045 D. At least 0.045, but less than 0.060 E. At least 0.060

5) (Spring 2006, 11) Eastern Digital uses a single machine to manufacture digital widgets. The machine was purchased 10 years ago and will be used continuously until it fails. The failure rate of the machine, ()x , is defined as:

x2 ()x  for x  4000 , where x is the number of years since purchase. 4000

Calculate the probability that the machine will fail between years 12 and 14, given that the machine has not failed during the first 10 years.

A. Less than 1.5% B. At least 1.5%, but less than 3.5% C. At least 3.5%, but less than 5.5% D. At least 5.5%, but less than 7.5% E. At least 7.5%

x2 6) (Spring 2007, 6) You are given the survival function Sx() 1 , for x 100 . Calculate 10000 0

e30 .

A. Less than 40 B. At least 40, but less than 45 C. At least 45, but less than 50 D. At least 50, but less than 55 E. At least 55

Chapter 2 – Survival Models  29

1 7) (Fall 2008, 12) You are given the force of mortality ()x  . Calculate the 3(100 x ) probability that a life aged 70 will die between ages 75 and 80.

A. Less than 0.04 B. At least 0.04, but less than 0.06 C. At least 0.06, but less than 0.08 D. At least 0.08, but less than 0.10 E. At least 0.10

8) (Fall 2008, 13) You are given the curtate expectation of life at age x and x + 1:

 ex  5.000

 ex1  4.530

Calculate qx .

A. Less than 0.091 B. At least 0.091, but less than 0.093 C. At least 0.093, but less than 0.095 D. At least 0.095, but less than 0.097 E. At least 0.097

9) (Spring 2008, 13) The future lifetime of (x) is described by the following survival function:

   x  Sx()      40   1.1

Find e30 , the expected future lifetime for (30)

A. Less than 5.00 B. At least 5.00, but less than 5.25 C. At least 5.25, but less than 5.50 D. At least 5.50, but less than 5.75 E. At least 5.75

30  Chapter 2 – Survival Models

10) (Spring 2008, 15) You are given the following survival function:

7 Sx() e5x

Calculate ()x , the force of mortality.

A. 5x7 B. 35x6 7 C. 35x65e x D. 5ln(35)x76x

7 35xe65 x E. 7 1 e5x

11) (Spring 2009, 1) An actuary is modeling the failure of a product over the next 20 years. You are given that the product’s lifetime follows a constant force model with ()x  0.05. Calculate the expected lifetime of a new product over the next 20 years.

A. Less than 10.0 B. At least 10.0, but less than 11.0 C. At least 11.0, but less than 12.0 D. At least 12.0, but less than 13.0 E. At least 13.0

12) (Fall 2009, 2) You are given the following information about the discrete random variable K()x , the time interval of failure for a life aged x:

 Kx() k, if kTx k1, where Tx is the future lifetime for a life age x. k  kxpr

 3 qx  0.657  r is a fixed but unknown constant between 0 and 1.

Calculate ex , the curtate expectation of life.

A. Less than 2.50 B. At least 2.50, but less than 2.75 C. At least 2.75, but less than 3.00 D. At least 3.00, but less than 3.25 E. At least 3.25

Chapter 2 – Survival Models  31

13) (Fall 2007, 31) You are given:

 x p0  0.75  Sx( u ) 0.70  q  0.04 u vx

Calculate Sx() u v.

A. Less than 0.665 B. At least 0.665, but less than 0.675 C. At least 0.675, but less than 0.685 D. At least 0.685, but less than 0.695 E. At least 0.695

14) (Spring 2010, 5) A piece of electronic equipment is built with three cooling fans. The equipment will only function if at least two of the three fans are working.

The lifetimes of the fans are subject to hazard rates that are independent, constant, and identical. The hazard rate is equal to 0.098.

Calculate the probability that the electronic equipment will work for at least five years.

A. Less than 0.50 B. At least 0.50, but less than 0.55 C. At least 0.55, but less than 0.60 D. At least 0.60, but less than 0.65 E. At least 0.65

15) (Fall 2005, M, 13) The actuarial department for the SharpPoint Corporation models the lifetime of pencil sharpeners from purchase using a generalized DeMoivre model with

 x Sx() 1 , for   0 and 0  x   .  A senior actuary examining mortality tables for pencil sharpeners has determined that the original value of  must change. You are given:

i. The new complete expectation of life at purchase is half what it was previously. ii. The new force of mortality for pencil sharpeners is 2.25 times the previous force of mortality for all durations. iii.  remains the same,

32  Chapter 2 – Survival Models

Calculate the original value of  .

A. 1 B. 2 C. 3 D. 4 E. 5

16) (Spring 2005, 30) Acme Products will offer a warranty on their products for x years, where x is the largest integer for which there is no more than a 1% probability of product failure.

Acme introduces a new product with a hazard function for a failure at time t of 0.002t .

Calculate the length of the warranty that Acme will offer on this new product.

A. Less than 3 years B. 3 years C. 4 years D. 5 years E. 6 or more years

17) (Spring 2005, 29) For a life aged (x), the m-year deferred probability of death is:

m 1275 for m  0,1,....,50

Calculate the probability that (x) lives at least 23 full years, but less than 30 full years.

A. Less than 0.11 B. At least 0.11, but less than 0.13 C. At least 0.13, but less than 0.15 D. At least 0.15, but less than 0.17 E. At least 0.17

Chapter 2 – Survival Models  33

18) (Spring 2006, 10) The force of mortality is given as

2 ()x  for 0 x  110 110  x

Calculate the expected future lifetime for a life aged 30.

A. Less than 20 B. At least 20, but less than 30 C. At least 30, but less than 40 D. At least 40, but less than 50 E. At least 50

19) (Fall 2009, 4) You are given the following information for two lives, (x) and (y), both aged 4:  Both lives’ survival functions are independent and identically distributed.  Mortality is uniformly distributed on the interval 4,14

Calculate the probability that the second death occurs in the interval 10,11 .

A. Less than 8.0% B. At least 8.0%, but less than 9.5% C. At least 9.5%, but less than 11.0% D. At least 11.0%, but less than 12.5% E. At least 12.5%

20) (Spring 2010, 1) You are given the curtate expectation of life at age x and x + 1:

eex xx1 q

Calculate px .

A. Less than 0.20 B. At least 0.25, but less than 0.50 C. At least 0.50 but less than 0.75 D. At least 0.75 E. No real solution exists

34  Chapter 2 – Survival Models

21) (Spring 2010, 2) You are given the following survival function:

1 x 2 Sx() 1 , 090x 90

Calculate 10 25q 30 .

A. Less than 0.15 B. At least 0.20, but less than 0.25 C. At least 0.25, but less than 0.30 D. At least 0.30, but less than 0.35 E. At least 0.35

22) (Spring 2010, 3) Your company offers extended warranty coverage for MP3 players. The policy provides for the replacement of MP3 players that fail within three years of purchase of the MP3 player. The lifetime of a particular model of player is subject to a constant hazard rate and the expected lifetime is five years.

On January 1, coverage is purchase for 1000 MP3 players.

Calculate the expected number of MP3 players replaced in the first year.

A. Less than 50 B. At least 50, but less than 150 C. At least 150, but less than 250 D. At least 250, but less than 350 E. At least 350

23) (Fall 2010, 1) You are given the following density function:

x3 fx() , for 010x c

Calculate the force of mortality at x = 5.

A. Less than 0.035 B. At least 0.035, but less than 0.040 C. At least 0.040, but less than 0.045 D. At least 0.045, but less than 0.050 E. At least 0.050

Chapter 2 – Survival Models  35

24) (Spring 2011, 1) You are given:

 The force of mortality,   0.10

Calculate the variance of Tx(), the future lifetime of x.

A. Less than 70 B. At least 70, but less than 90 C. At least 90, but less than 110 D. At least 110, but less than 130 E. At least 130

2 x 3 25) (Spring 2011, 2) Given that Sx() 1 for 010 x  . 10 Calculate the probability that a life aged (1.5) survives at least six time units.

A. Less than 0.441 B. At least 0.441, but less than 0.442 C. At least 0.442, but less than 0.443 D. At least 0.443, but less than 0.444 E. At least 0.444

26) (Spring 2011, 3) You are given the following information:

 Deaths are uniformly distributed between integer ages.

 q50  0.10

 q51  0.20

0 ____ Calculate e50:1.25 , the temporary 1.25-year complete expectation of life of a person aged 50.

A. Less than 1.09 B. At least 1.09, but less than 1.12 C. At least 1.12, but less than 1.15 D. At least 1.15, but less than 1.18 E. At least 1.18

36  Chapter 2 – Survival Models

27) (Fall 2011, 1) You are given the following:

 For a group of lives ABC, p30  0.8  For a second group of lives XYZ which is subject to an additional risk, the force of

mortality is x ()tk , where x ()t is the force of mortality for group ABC and k is a positive constant.  The probability of someone age (30) dying within 1 year is 50% higher for group XYZ than it is for group ABC.

Calculate k.

A. Less than 0.05 B. At least 0.05, but less than 0.10 C. At least 0.10, but less than 0.15 D. At least 0.15, but less than 0.20 E. At least 0.20

28) (Fall 2011, 2) You are given the following information:

x4 Fx() , for 09x 81

0 Calculate ex:4 , the expectation of life over the next four years for a life age four.

A. Less than 1.7 B. At least 1.7, but less than 2.0 C. At least 2.0, but less than 2.3 D. At least 2.3, but less than 2.6 E. At least 2.6

Chapter 2 – Survival Models  37

Chapter 2 – Answers

1) This is another way of asking, “What is the probability of dying between the ages of 51 and 64, given survival to age 36?”. In actuarial notation we can write this as q => the probability 1513 36 of a person aged 36 surviving for 15 years and then dying within the next 13 years. The steps to solve this are as follows: SS(51) (64) I. q  1513 36 S(36) S(36) (1 0.36)12  0.64 12  0.8 ; S(51) (1 0.51)12  0.49 12  0.7 ; S(64) (1 0.64)12  0.36 12  0.6 SS(51) (64) 0.7 0.6 0.1 0.125=> The answer is A. S(36) 0.8 0.8

2) The question is asking for the expected future lifetime for a person aged 25. However, we know from the above definition of Sx()that the person must die by age 40, since Sx()(i.e. probability) 40  Sxdx() is zero after 40. So we must solve the equation ET 25 : 25 S(25) 40 40 xx2 I. 1dx x 25 40 80 25 4022 25 40 25 20 17.1875  2.8125 80 80 40 Sxdx()  2.8125 2.8125 II. ET25 7.5 => The answer is A. 25 S(25)25 0.375 1 40

3) The key here is to notice that the individuals have a constant force of mortality. The one distribution with a constant force of mortality is the exponential distribution. The formula for the probability of survival in the exponential distribution is St() et . Once we have established this, we solve the question in the following steps: I. The question states that 10% will die within 20 years. This is another way of saying that the probability of surviving to age 20 is 0.9. So we can say Se(20)20 0.9 II. The probability of survival under the new force of mortality => 2 Se'(20)20(2 ) e 20  (0.9) 2 0.81 III. The probability of death in the next 20 years =>

20 qFx '(20) 1 S '(20) 1 0.81 0.19 => The answer is C.

38  Chapter 2 – Survival Models

4) The key here is to notice that you need to substitute the 60 for x in the equation for force of mortality, since that is the initial age of the life. Then you will use the force of mortality to find q , which is equal to SS(70) (71) : 10 60 60tt 58  2 I.   xt 200 200 tt x2 1 x2  ()xdx dx 2x 200 200 2 II. St() e00 e e  SS(70) 0.7047, (71) 0.66199 => SS(70) (71) 0.04271 => The answer is C.

q 5) In actuarial notation the question is asking for the following value: 212 10p 0 t x2  dx x3  4000  I. Sx() e0 e12000 1000  12000 10pS 0 (10) e  0.9200 1728 2744  12000 12000 II. 212qS(12) S (14) e e 0.8659  0.7956  0.0703 q 0.0703 III. 2120.0764 => The answer is E. 10p 0 0.9200 100 Sxdx() 0  6) You are looking for e  30 . 30 S(30) 100 3 100 100 x2 x I. S( x ) dx 1 dx x  66.6667  29.1 37.5667 10000 30 30 30000 30 302 II. S(30) 1  0.91 10000 100 Sxdx() 0  37.5667 III. e 30 41.2821=> The answer is B. 30 S(30) 0.91

Chapter 2 – Survival Models  39

SS(75) (80) 7) The question is looking for the value of q , which can be written as . It is 55 70 S(70) the probability of dying between the ages of 75 and 80, given survival to age 70. t 1 dx  3(100x ) I. St() e0

tt 111 dx dx 1 3(100xx ) 3 100 ln(100t ) ln100 St() e00 e e3

13 1 1 100tt  100 13 ln(100t ) ln100 ln ln  3 100 100 100  t St() e3 e e . 100 13 13 13 30  25  20 II. SSS(70) 0.6694, (75)   0.6299, (80)   0.5848 100  100  100 SS(75) (80) 0.6299 0.5848 III. 0.0674 => The answer is C. S(70) 0.6694

8) As mentioned in the reading, since epxx 23 p x p x..... and

epxx112131 p x   p x  ....., we can say eppex  xxx1 : I. 54.535.53pp  p=> p  0.9042 x xxx

II. qpxx1  1 0.9042  0.0958 => The answer is D.

1.1 40  x 9) If we substitute in for the variables we get Sx() . Since Sx()must be greater than 40 or equal to zero (it is a probability measure), the maximum x, or age, is 40. So we must solve 40 Sxdx() 0  e  30 : 30 S(30) 1.1 10 I. S(30) 0.2176 40 1.1 40 40 40 40xx 1 (40 )2.1 1 10 2.1 II. Sxdx( ) dx 1.0364  1.1 1.1 30 30 40 40 2.130 40 2.1 40 Sxdx() 0  1.0364 III. e 30 4.763 => The answer is A. 30 S(30) 0.2176

40  Chapter 2 – Survival Models

10) t   ()x dx t t 7  I. Sx() e5x e0 => 5()x7   xdx => 5()x7    xdx 0 0 II. We can take the derivative of both sides in order to get rid of the integral => 35x6   (x ) => The answer is B.

x   ()x dx x 11) To answer this question, we must use the formulas Sx() e0 and ExSxdx  () . 0

xx  (xdx ) 0.05 dx I. Sx() e00 e e0.05x

x 20 20 0.05xx 0.05 II. ExSxdxedxe( )  20 20(0.3678 1) 12.6424 => The 00 0 answer is D.

 12) Recall that the equation for ex , is eppppxtxxxx 23...... In this case we must see t1 if we can find a pattern in the probabilities:

I. 33pqxx1  1 0.657  0.343 3 2 II. 3 prx 0.343 => r  0.7 . So px  0.7 and 2 px 0.7 0.49 234 III. So ex 0.7 0.7  0.7  0.7  ...... This is geometric series where the first term is 0.7 and the ratio between terms in also 0.7. The formula for the sum of a geometric a series is S  , where a is the first term and r is the ratio between successive terms. 1 r 0.7 0.7 In this case, Se  2.3333 => The answer is A. x 10.70.3

13) It’s always easier to substitute numbers for variables to make these types of problems easier to solve. Let x  5, u 10 , and v  20 .

I. The value we are looking for is SS(5 10 20) (35) , or 35p 0 . We are given p  0.75 , SSp(5 10) (15)   0.70 , and qpq 0.04 . 50 15 0 10 20 5 10 5 20 15

150p  10550pp => 0.70 10p 5  0.75 => 10p 5  0.9333 qpq => 0.04 0.9333 q => q  0.0429 10 20 5 10 5 20 15 20 15 20 15

20pq 151 20 15  1 0.0429  0.9571

II. 35p 0 20pp 15 15 0 => 35p 0  0.9571 0.70 0.6699 => The answer is B.

Chapter 2 – Survival Models  41

14) This is an extension of a joint-life/last-survivor problem, but it requires a bit of innovative thinking to solve the problem: I. The equipment will work if all three fans are working or if two of three fans are working. 32 This is equivalent to 50ppq 3  50 50. The reason why we multiply the last expression by three is because there are three ways to arrange two fans working and one not working. ( .098)(5) 3 (.098)(5)(3) II. 50pe => 50pe0.2299

2 ( .098)(2) ( 0.098)(5) 350pqe 50 3 1 e  0.4362 32 III. 50ppq3  50 50 0.2299 0.4362 0.6661 => The answer is E.

15) I. Let’s calculate the new force of mortality from the first force of mortality =>  1 d  x   Sx() 1 dx     1   Sx() x xx  x 1 11  

The new survival model has the same form, except that  changes. Let’s call it  ' =>

 '   2   x

We know that 21 2.25 , so,

 ' 2.25  =>  '  2.25 x x

II. Now let’s calculate the new expected value from the old expected value =>

1   xx  Ex()1  1 dx   1 1  0 0

  Ex() Ex() 1  1 2  ' 1

  By the first bullet point =>  => ' 12 1 ' 12 1

42  Chapter 2 – Survival Models

III. We have two equations with two unknowns and we need to solve for  =>

2.25 1 2  1 => 2.25  1 2 2 => 0.25  1 =>   4 => The answer is D.

16) You know the hazard rate the probability that you need to solve for, so you need to use the x   ()tdt 0 equation xtpSxe() to solve for x:

x  0.002tdt  2 2 I. Sx() 1 0.010.99   e0 => 0.99  e0.001x => 10.05  x => x  3.17 => The answer is B.

17) The problem is asking for the value of 23 7 qx , which is the probability of living 23 years and then dying within the next seven years.

I. The person could live 23 years and then die the next year, live 24 years and then die the next year, etc., so 23 24 25 ... 29 qqqq.... q   0.1427 => The 23 7xxxx 23 24 25 29 x 1275 answer is C.

18) This looks like De Moivre’s Theorem, but unfortunately we cannot use that shortcut here because of the 2 in the numerator. I. Find the general equation for Sx() =>

tt 2  ()x dx dx 110x St() e00 e

2 110t 2 ln 110 110  t St() e 110 110 1102 110 110 110  t 2 2 S( t ) dt dt 110 t dt 12100 220 t t dt 0    30 30110 30 30 II. e30 2 2  S(30)110 30 80 6400  110 110 2 t3 0 12100tt 110 3 30 e30 27.93 => The answer is B. 6400

Chapter 2 – Survival Models  43

19) The key here is to figure out that the lives follow a De Moivre distribution with  14 4 10 : I. The probability of one person surviving the first six years and then dying in the interval 61 10,11 is 0.15 . 10 4 6 The probability of the a person dying in the first six years is 10.4 . 10 In order for the second death to occur in the interval 10,11 , the first death must occur in the first six years. This could happen two ways – Person A could die first or Person B could die first. We write this as 2 (0.15)(0.4)  0.12 . We multiply by two because either person could die first. II. The other possibility is that both people die in the interval 10,11 , which means that they both survive six years and then die within the next year. This is written as 2 61 0.0225 10 4 III. The total probability is 0.12 0.0225 0.1425 => The answer is E.

20) We need to use the recursive relationship eppex  xxx1 to solve the problem

I. eppex xxx1 =>

qppqx xxx

11pxxxpp p x 2 1pppxxx p x 2 ppxx310 => Solve using the quadratic formula => 3(3)4(1)(1)2  p 0.3820 => The answer is B. x 2(1)

44  Chapter 2 – Survival Models

21)

I. 10 25qpq 30 10 30 25 40 => S(40) p  10 30 S(30) 1 40 2 1 90 10p 30 1 0.9129 30 2 1 90 S(65) qp11  25 40 25 40 S(40) 1 65 2 1 90 25q 40 11  0.2929 => 40 2 1 90

10 25qpq 30 10 30 25 40 (0.9129)(0.2929)  0.2674 => The answer is C.

0 1 22) In an exponential distribution ex  =>  1 5  =>   0.2 

I. The amount of MP3 players replaced in the first year is equal to 1000q0 => 1 0.2dx 0 1000qp00 1000 1 1000 1 e   0.2 1000qe0  1000 1 181.27 => The answer is C.

Chapter 2 – Survival Models  45

f ()x 23) We need to solve for c and use the relationship ()x  to solve for the force of mortality. Sx() 10 10 x3 I. Ffxdxdx(10) ( ) 1 => 00c 10 x4  1 4c 0 2500 1 => c  2500 c x3 fx() 2500 x4 II. Sx() 1 Fx ()  1 => 10000 x3 ()x  2500 x4 1 10000

53 (5)2500 0.0533 => The answer is E. 54 1 10000

1 24) The variance of an exponential distribution with a constant force of mortality is =>  2 11 100 => The answer is C.  220.1

S(7.5) 25) p  61.5 S(1.5) 2 7.5 3 1 2 10 0.25 3 61.5p 220.4423 => The answer is C. 1.5 330.85 1 10

46  Chapter 2 – Survival Models

26) 0 11 1 I. e50  pdt11 qdt tqdt (the last step is true because it is a tt50 50  50 00 0 uniform distribution) => 0 1 2 1 etdttt50 10.1  0.05  0 0 0 e50  0.95 0 0.25 0.25 0.25 ____ II. e51:0.25  pdt11 qdt tqdt tt51 51   51 00 0 0.25 0 0.25 ____ 2 etdtt51:0.25 10.2  10.2   0 0 0 ____ e51:0.25  0.24375 00 0 ______III. eepe50:1.25 5050  51:0.25 0 ____ e50:1.25 0.95 0.9 0.24375  1.1694 => The answer is D.

27)

I. p30  0.8 => 0.8  e   0.2231

II. For group ABC, q30  0.2

For group XYZ, q30 1.5(0.2) 0.3 => 0.7  e(0.2231k ) 0.3567 0.2231 k k  0.1336 => The answer is C.

Chapter 2 – Survival Models  47

28) 88 S() x dx 1 F () x dx 0  44 I. ex:4  SF(4) 1 (4) 8 8 x2  x3  1 dx x  0    4 81  2434 ex:4  42 0.8025 1 81 0 2.1564 ex:4 2.6871 => The answer is E. 0.8025

48  Chapter 2 – Survival Models

Chapter 9 – Practice Exams  295

CHAPTER 9 – PRACTICE EXAMS

Practice Exam #1

1) (Spring 2012, 1) You are given the following information: x2  fx() , for 0 x c X c

Calculate (2.3) .

A. Less than 1.05 B. At least 1.05, but less than 1.15 C. At least 1.15, but less than 1.25 D. At least 1.25, but less than 1.35 E. At least 1.35

2) (Spring 2013, 1) You are given the following information:

1, x  0   Sx() 0.8, x 1  x1 rx,  2,3,....

 221q  0.288  Sx(1)() Sx for all x

Calculate S(2)

A. Less than 0.56 B. At least 0.56, but less than 0.62 C. At least 0.62, but less than 0.68 D. At least 0.68, but less than 0.74 E. At least 0.74

296  Chapter 9 – Practice Exams

3) (Spring 2012, 2) You are given the following information:

 pk  0.8 , for k  7

 pk  0.7 , for 78k

 e9  1.5

Calculate e6 .

A. Less than 2.40 B. At least 2.40, but less than 2.50 C. At least 2.50, but less than 2.60 D. At least 2.60, but less than 2.70 E. At least 2.70

4) (Spring 2012, 3) You are given the following information:

 Mortality follows the Illustrative Life Table.  Assume the Uniform Distribution of Deaths between integer ages.

Calculate the median future lifetime for a life aged 24.5 years.

A. Less than 52.50 B. At least 52.50, but less than 52.75 C. At least 52.75, but less than 53.00 D. At least 53.00, but less than 53.25 E. At least 53.25

5) (Spring 2012, 4) You are given the following information:

 The future lifetimes of (30) and (50) are independent and identically distributed.  The probability that the first death occurs more than 20 years from now is 0.30

 10p 60  0.40

Calculate 30p 30 .

A. Less than 0.605 B. At least 0.605, but less than 0.655 C. At least 0.655, but less than 0.705 D. At least 0.705, but less than 0.755 E. At least 0.755

Chapter 9 – Practice Exams  297

6) (Spring 2013, 4) You are given the following information for two independent lives aged 45 and 65:

 Mortality for the life age 45 follows the Illustrative Life Table.  Mortality for the life age 65 follows: fy() 0.2 e0.2 y for y  0

Calculate 10 p _____ 45:65

A. Less than 0.925 B. At least 0.925, but less than 0.935 C. At least 0.935, but less than 0.945 D. At least 0.945, but less than 0.955 E. At least 0.955

7) (Spring 2012, 5) You are given the following information:

 Individual X is 50 years old and has a family history of cancer.  Individual Y is 60 years old and has no family history of cancer.  Mortality for individuals with no family history of cancer follows De Moivre’s law with  120 ,  An individual with a family history of cancer is subject to a force of mortality equal to three times the force of mortality of an individual with no family history of cancer.  The future lifetime of X and Y are independent.

Calculate the probability that both individuals X and Y live more than three years and at least one dies within the following five years.

A. Less than 0.232 B. At least 0.232, but less than 0.242 C. At least 0.242, but less than 0.252 D. At least 0.252, but less than 0.262 E. At least 0.262

298  Chapter 9 – Practice Exams

8) (Spring 2012, 6) You are given the following information for a double decrement model:

1   (1) ()t  , t  0 x 5 1   (2)()t  , t  0 x 20  T is the time-until-decrement random variable for (x)  J is the cause-of-decrement random variable for (x)

Calculate ETJ1 , the expected time to failure from decrement 1.

A. Less than 3.5 B. At least 3.5, but less than 6.0 C. At least 6.0, but less than 8.5 D. At least 8.5, but less than 11.0 E. At least 11.0

9) (Spring 2012, 7) You are given the following double decrement table:

x (1) (2) () () (1) (2) qx qx px  x dx dx 61 0.015 0.935 100,000 62 0.016 0.944 63 0.030 1,589 64 0.020 0.960 1,681

(2) Calculate 22q 61

A. Less than 0.040 B. At least 0.040, but less than 0.042 C. At least 0.042, but less than 0.044 D. At least 0.044, but less than 0.046 E. At least 0.046

Chapter 9 – Practice Exams  299

10) (Spring 2012, 8) You are given the following information:

 Workers transition through the labor force according to the following matrix:

0.90 0.07 0.03  Q  0.05 0.80 0.15 0.15 0.15 0.70  State 0: Employed full-time  State 1: Employed part-time  State 2: Unemployed  Individual Y is employed full-time at time t = 0.  Individual Z is employed part-time at time t = 0.  Individuals Y and Z transition through the labor force independently.

Calculate the probability that either Individual Y or Individual Z, but not both, will be unemployed at t = 2.

A. Less than 0.254 B. At least 0.254, but less than 0.256 C. At least 0.256, but less than 0.258 D. At least 0.258, but less than 0.260 E. At least 0.260

Chapter 10 – Answers to Practice Exams  319

CHAPTER 10 – ANSWERS TO PRACTICE EXAMS

Practice Exam #1 (Answers) f ()x 1) In order to solve this problem we need to solve the equation ()x  . However, first we need Sx() to solve for Sx() . c x2 I. F()cdx 1 0 3c c x3 1 9c 0 c3 1 9c x3 c2  9 => c  3 => Fx() 27 x3 II. Sx() 1 Fx ()  1 => 27 2.33 S(2.3) 1  0.5494 27 2.32 f (2.3) 0.5878 9 f (2.3) (2.3) 1.0699 => The answer is B. S(2.3) 2)

I. 221qpq 2 1 2 3

221qp 2 11 2 p 3 SSSSSSS(3) (5) (3) (3) (5)  (3) (5) 221q 1   SSSS(1) (3) (1) (3)  S (1) SSSSSSS(3) (5) (3) (3) (5)  (3) (5) 221q 1   SSSS(1) (3) (1) (3)  S (1) rr24 q 0.288 221 0.8 rr240.2304 rr420.2304  0 => This is a quadratic formula with x  r 2 => 1 1 4(0.2304) xr2 0.36 2 r  0.6 II. S(2) 0.61 0.6 => The answer is B.

320  Chapter 10 – Answers to Practice Exams

3) We need to use the recursive formula eppexxxx 1 to solve this problem.

I. eppppe662636369    =>

\e6 0.8 (0.8)(0.8)  (0.8)(0.8)(0.7)  (0.8)(0.8)(0.7)(1.5)

e6 0.8 0.64  0.448  0.672

e6  2.56

4) We are looking for the value of x such that x p24.5  0.5 .

 x24.5 I. x p24.5 0.5 =>  24.5 Because the mortality follows the UDD, the amount of survivals at age 24.5 is just equal to the average of the amount of survivals at ages 24 and 25 =>

 24.5 0.5 9576288 9565017 9570652.5  II. 0.5  x24.5 9570652.5

 x24.5  4785326.25 => We need to find the age at which there are exactly 4785326.25 survivors. We see from the Illustrative Life Table that this occurs somewhere in between ages 77 and 78. We need to use the UDD assumption to find the exact point in time when there are this many survivors =>  x 24.5 77 77x 24.5 77 78 4828182 4785326.25 x 24.5 77  77 0.1439 4828182 4530360 x 24.5 77.1439 x  52.64 => The answer is B.

5) This is a last survivor question, where they are giving is the value of 20p 30:50 .

I. 20p 30:50  0.3

20pp 30 20 50 0.3

40p 30  0.3

II. 40p 30 30pp 30 10 60

0.330p 30 0.4

30p 30  0.75 => The answer is D.

Chapter 10 – Answers to Practice Exams  321

6) This is a last survivor question where 10p _____11 10qqq _____ 10 45 10 65 : 45:65 45:65

10d 45 45 55 9164051 8640861 I. 10q 45   0.0571 45 45 9164051 FF(75) (65) II. q  => 10 65 S(65) y y Fy() 0.2 e0.2yy dy e 0.2 1 e  0.2 y  0 0 Sy() 1 Fy ()  e0.2 y =>

0.2(75) 0.2(65) FF(75) (65) 11ee  e0.2(65) e 0.2(75) 1e0.2(10) 0.8647 Se(65) 0.2(65) e 0.2(65)

III. 10pqq_____ 1 10 45  10 65  1 0.0571 0.8647  0.9506 => The answer is D. 45:65

7) This is a joint-life problem which is looking for the value of35q 50:60 .

I. 3 5qpq 50:60 3 50:60 5 53:63 =>   t To find p we can use the shortcut for De Moivre’s Theorem where St() => 360  60 3 57  120 60 60 => p 0.95 360 60 60 3 13 70  x The force of mortality for Y is ()x  3 => St()  => 70 x 70  x 70 33 70 3 67 350p 0.8769 70 70 3 57 67 350:60350360ppp 0.8330 60 70

II. 5qpp 53:631 5 53  5 63 => 3 52 62 553:63q 1  0.2771 57 67

III. 3 5qpq 50:60 3 50:60 5 53:63 (0.8330)(0.2771)  0.2308 => The answer is A.

322  Chapter 10 – Answers to Practice Exams

8) The question is asking for the “expected value” for decrement 1, which means you will need to solve  the equation E Ttftdt(,1) . 0  0 () (1) I. f (,1)tp tx x () t =>

11 1  ft(,1) e520 0.2 e0.25t 5  II. E Ttftdt(,1) => 0  0  ET 0.2 te0.25t => We can use a shortcut here if we make an adjustment to the 0  0 coefficient in the integeral =>  ET 0.8 0.25 te0.25t 0  0  1 ET0.8 0.25 te0.25t 0.8 3.2 => The answer is A. 0   0 0.25

(2) 9) It is probably easiest to fill in the missing pieces in the table and then calculate 22q 61 : I. Age 61 => () () (1) (2) 1p61qqq 61 61 61 (2) 0.065 0.015 q61 (2) q61  0.05 => (1) ( ) (1) dq61 61 61 (1) d61 100000 0.015 1500 (2) ( ) (2) dq61 61 61 (2) d61 100000 0.05 5000 II. Age 62 => () () (1) (2) 62 61dd 61  61

() 62 100000 1500 5000 93500 =>

() () (1) (2) 1p62qqq 62 62 62 (2) 0.056 0.016 q61 (2) q61  0.04 => (1) ( ) (1) dq62 62 62

Chapter 10 – Answers to Practice Exams  323

(1) d62 93500 0.016 1496 (2) ( ) (2) dq62 62 62 (2) d62 93500 0.04 3740 III. Age 63 => () () (1) (2) 63 62dd 62  62

() 63 93500 1496 3740 88364 =>

(1) (1) d63 1589 q63 () 0.018 =>  63 88364 (2) ( ) (2) dq63 63 63 88364  0.03  2647.92 =>

() () (1) (2) pq6311 63  qq 63  62

() p63 1 0.018  0.03  0.952 IV. Age 64 => () () (1) (2) 1p64qqq 64 64 64 (1) 0.04q64 0.02 (1) q64  0.02 => () () () 64 63p 63 88364  0.952  84027.08 => (2) ( ) (2) dq64 64 64 84027.08  0.02  1680.54

x (1) (2) () () (1) (2) qx qx px  x dx dx 61 0.015 0.05 0.935 100,000 1,500 5,000 62 0.016 0.04 0.944 93,500 1,496 3,740 63 0.018 0.030 0.952 88,264 1,589 2,647.92 64 0.02 0.020 0.960 84,027.08 1,681 1,680.54

() (2) (2) (2) ( ) (2)  63 2dd 63 2 63 V. 22qpq 61 2 612 63 () () ()  => 61 63  61 2647.92 1680.54 q(2) 0.043 => The answer is C. 22 61 100000

324  Chapter 10 – Answers to Practice Exams

10) I. To find the transition probabilities between time 0 and time 2, we need to multiply the transition matrix by itself =>

0.90 0.07 0.03 0.90 0.07 0.03 0.818 0.1235 0.0585  0.05 0.80 0.15 0.05 0.80 0.15 0.1075 0.666 0.2265 0.15 0.15 0.70 0.15 0.15 0.70 0.2475 0.2355 0.517

The probability of Individual Y transitioning from employed full-time to unemployed is in Row 1 Column 3 => 0.0585 The probability of Individual Z transitioning from employed part-time to unemployed is in Row 2 Column 3 => 0.2265 II. They both cannot be unemployed so we must find the probability of Individual Y being unemployed and Individual Z not unemployed or Individual Y being employed and Individual Z unemployed => (0.0585)(1 0.2265) (0.2265)(1  0.0585) 0.2585 => The answer is D.

11) We need to figure out which combinations will give us an increase to 103 after two transitions and the corresponding probabilities: I. Transition to State 1 and remain in State 1 => 100 1.04 1.04 108.16 => (0.10)(0.05) 0.005 Transition to State 1 and then to State 2 => 100 1.04 1.02 106.08 => (0.10)(0.65) 0.065 Transition to State 2 and then to State 1 => 100 1.02 1.04 106.08 => (0.80)(0.30) 0.24 Transition to State 2 and remain in State 2 => 100 1.02 1.02 104.04 => (0.80)(0.50) 0.40 II. Total Probability = 0.005 0.065 0.24 0.40 0.71 => The answer is C.