acst152 2015 lecture 3c life tables introduction
DESCRIPTION
ACST152 2015 Week 9 LeverageTRANSCRIPT
ACST152 Introduction to ACST152 Introduction to Actuarial StudiesActuarial StudiesLecture 3C 2015The Life Table
The Life TableThe Life TableA Life Table is used to work out
probabilities of survival and death for a given population
Let l0 be number of people born(assumed to be say 100,000)
lx = number of people alive at exact age x
The Australian Life TablesThe Australian Life TablesAustralian Life Tables (ALT) are
published by the Australian Government Actuary, based on census data of Australian population
Updated at regular intervals. Current version is 2005/2007 (ALT 2005/07)
Separately for males and females
Probability of survival : one Probability of survival : one yearyear
xageatalivenumberxageatalivenumber
llp
x
xx
11
Example: Use the Australian Life Table 2005-2007 for Males
Q. What is the probability that a male age 20 survives 1 year?
Q. What is the probability that a female aged 20 survives 1 year ?
Probability of survival Probability of survival exampleexampleMales l20 = 99008 and l21 =
98935 so p20 = 0.99926
Females l20 = 99272 and l21 = 99244
So p20 = 0.99972
Number of deathsNumber of deathsThe number of deaths between exact
age x and exact age x+1 is denoted dx.
◦dx = number alive at age x exact - number alive at age x+1 exact
dx = lx - lx+1
One year Probability of One year Probability of deathdeath.The probability that someone
aged exactly x will die before age x+1 is denoted qx
xxx
xx
x
xxx
x
xx pqor
llqor
lllqor
ldq
11 11
Example: Probability of Example: Probability of deathdeathExample: Use the Australian Life
Table 2005-2007 for Males
Q. What is the probability that a male age 20 exactly dies within 1 year?
What about a female aged 20?
q20 males = 1-p20 = 1 - 0.99926 = 0.00074
q20 females = 1 - 0.99972 = 0.00028
Probability of surviving t Probability of surviving t yearsyears
Example: What is probability that a person (M of F) aged 20 will survive to age 65?
Male 86931/99008 = 0.87802Female 91526 / 99272 = 0.92197
xageatalivenumbertxageatalivenumber
llp
x
txxt
Probability of dying in a Probability of dying in a specified age rangespecified age range
Q. What is the probability that a person aged x exactly will die between age x+t and x+s?
xageatalivenumbersxandtxagebetweendiewhonumber
x
sxtx
lll
Example:Example:What is the probability that a
person aged 20 will die between age 55 and 65? (Male & Female)
Male = 0.06581Female = 0.04069
Life ExpectancyLife ExpectancyLet Tx be a random variable which
represents the time until death for a person aged x exactly.
The expected value of Tx is called the life expectancy and is denoted as
E(Tx) =
oxeo
xe
Calculation of Life Calculation of Life ExpectancyExpectancyLet’s assume people die in the
middle of the year on average. For a person now aged x,◦ If he dies between x and x+1, Tx = ½◦ If he dies between x+1 and x+2, Tx = 1.5◦ If he dies between x+2 and x+3, Tx = 2.5◦ ...◦ If he dies between x+t and x+t+1, Tx =
(t+0.5)
Expected Value of TExpected Value of Txx
xw
tx txandtxbetweenDiesPtTE
0
)1()5.0()(
xw
t x
txtxx l
lltTE0
1)5.0()(
Life Expectancy ExampleLife Expectancy ExampleYou own a newborn mouse.Suppose that the life table for the
mouse is given below
If deaths occur at mid year on average, what is E(T)?
Age x lx dx0 100 201 80 402 40 303 10 104 0
E(T) = 0.5 * 20/100
+ 1.5 * 40/100+ 2.5 * 30/100+ 3.5 * 10/100
= 1.8 years
Shortcut formula for Life Shortcut formula for Life expectancyexpectancy
...5.25.15.0)( 32211
x
xx
x
xx
x
xxx l
lll
lllllTE
...)(*5.2)(*5.1)(*5.01)( 32211 xxxxxxx
x lllllll
TE
...*5.01)( 321 xxxxx
x lllll
TE
...15.0)( 321 xxxx
x llll
TE
Mouse example : shortcutMouse example : shortcutQ. Repeat mouse example using
shortcut formula
E(T0) = 0.5 + [80+40+10] / 100= 1.8 years
Tute exerciseTute exerciseOn a spreadsheet, use the above
formula to calculate male and female life expectancy at age 20, and confirm that this matches the ALT expectation of life at that age. (Note ALT tables are on iLearn in spreadsheet format)
Life expectancy in ALTLife expectancy in ALTThe life expectancies at older ages in
the Government Actuary’s tables are slightly higher than those we would calculate using our formula.
Our formula assumes everyone is dead at age 110.
It appears that the Australian Govt Actuary has allowed for a small probability of survival past age 110.
Iterative formula for E(TIterative formula for E(Txx))Q. Suppose our mouse has a life expectancy
of 1.8 years when new born.A year later, will its life expectancy be 0.8
years ?
A. No, it will be higher than 0.8E(T1) = 0.5 + [40 + 10]/80
= 1.125 Tute exercise: Explain why.
Iterative formulaIterative formulaTute exercise: Show mathematically
and by general reasoning that
◦ ex= ½ + px (1/2 + e x+1)
Hence if Joe (male) has a life expectancy of 40.71 at age 40, and p40 = 0.99855 and p41 = 0.99846 find the life expectancy at age 41 and 42.
Answers to Example40.71 = ½ + 0.99855(1/2+e41)
e41= 39.77
39.77 = ½ + 0.99846(1/2+e42) e42 = 38.83
[Note with repetition of iterative formula, rounding off can lead to an accumulation of errors.]
Multiple LivesMultiple LivesSometime we want to find
probabilities involving multiple lives:◦What is the probability that both Joe and
Mary are still alive in t years?◦If there are three people A B and C alive
now, what is the probability that exactly one person is alive in t years ?
◦If there are two people alive now (A and B), what is the probability that A dies first?
Probability RulesProbability RulesIf A and B are independent eventsP(A and B) = P(A) * P(B)
If A and B are independent eventsP(A or B) = P(A) + P(B) – P(A and B)
• If A and B are independent events• P(Exactly one of A and B) • = P(A) * [1-P(B)] + P(B) * [1-P(A)] • = P(A) + P(B) – 2 P(A) P(B)
Example: Multiple LivesExample: Multiple LivesJoe is age 21 and Bill is age 23
when they graduate Uni.What is the probability that both
are alive for their 40th reunion ?Use ALT 05/07 rates (male)
AnswerAnswerThe probability that Joe is alive = P(J)P(J) = l61/l21 = 90269/98935 The probability that Bill is alive = p(B)P(B) = l63/l23 = 88756/98784
Assuming independence,P(Both alive) = P(J) * P(B) = 0.81978
Example: Multiple LivesExample: Multiple LivesA superannuation fund is required
to pay pensions to ex-employees and their spouses.◦If both husband and wife are alive at
year end, then pay $10,000◦If only one is alive, pay $7,000
Joe is age 65; his wife Mary is 60.Q. What is the expected value of
the payment due at the end of the year?
AnswerAnswerThe probability that Joe is alive = P(J)P(J) = l66/l65 males = 0.98800The probability that Mary is alive = P(M)P(M) = l61/l60 females = 0.99565Assuming independence,P(Both alive) = P(J) * P(M) = 0.98370P(J alive and M dead) = P(J) * [1-P(M)] = 0.00430P(M alive and J dead) = P(M) * [1-P(J)] = 0.01195Expected value of payments = 10000*0.9837 + 7000 *(0.00430+0.01195)= 9951
Tute ExerciseTute ExerciseIf there are three people A, B,
and CAnd P(A), P(B) and P(C) are the
probabilities that each one is alive in one year,
What is the probability that exactly one is alive at the end of the year?
Binomial DistributionBinomial Distribution If you have a group of people who all have
the same probabilities of death, then you can use the binomial distribution.
Example: There are 10 males all age 60 now. (a) What is the probability that all are still
alive at age 70? (b) What is the probability that just one
person has died before reaching age 70?
AnswersAnswers