Faculty of Actuaries Institute of Actuaries

EXAMINATIONS

April 2001

Subject 104 — Survival Models

EXAMINERS’ REPORT

ã Faculty of Actuaries
ã Institute of Actuaries
 Subject 104 (Survival Models) — April 2001 — Examiners’ Report

Examiners’ Comments:

The last three examinations have seen very high numbers of candidates for Subject 104
together with disappointingly low pass rates and a large number of inadequately
prepared candidates obtaining very low marks.

Subject 104 requires a working knowledge of Subjects 101 and 102. The evidence from
examinations is that many candidates have but a superficial knowledge of key topics in
104, eg. the stochastic basis of life contingencies, the statistical basis of graduation,
proportional hazards models.

We now make some specific comments on the difficulties experienced by candidates in

each question in the April 2001 paper.

1 The main cause of lost marks was the failure to distinguish clearly between the
events whose probabilities were given in (a).

2 The majority of candidates either had very sketchy knowledge of Thiele’s

equation, or were unable to apply their knowledge of Thiele’s equation to the
special case described. Marks were lost in (iii) as a result of using the A67-70
rather the a(55) functions.

3 Marks were mainly lost in (i)(c) and (ii) as a result of incorrectly allowing for the
retrospective claims.

4 In (i) many candidates confused assessing if there was a linear relationship

o
between q x and q sx with the goodness of fit of q̂ x to q sx .

In (ii) marks were often lost because an expression for the weighting function
was just stated rather than determined.

In (iii) the impact of the linear relationships on the third differences of the rates
was rarely stated.

5 There were various common errors including the failure to value benefits payable
immediately on death and to allow for changing mortality. The most common
mistake was to use

a 70 − (1 + ½i)a 70

6 In (i) the most common mistake was the failure to say which calendar year was
the rate interval.

In (ii)(a) only a minority of candidates attempted to find the age definition at the
date of the census of Px(t) t=½, 1½, 2½. In (ii)(b) the assumption of a constant
force was rarely mentioned.

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 Subject 104 (Survival Models) — April 2001 — Examiners’ Report

7 The main causes of poor marks were:

(i) writing down rather than deriving the sampling distributions of

the test
statistic
(ii) not defining the test statistics
(iii) using ‘two tailed’ rather than ‘one tailed’ test rejection regions.

Much of the knowledge acquired in Subject 101 had apparently been forgotten.

8 Many candidates had no idea what was a random variable and what was an
expected value. Statements like

Var(a 40:20 ) = E[(a 40:20 )2 ] − (E[a 40:20 ])2

were common.

Some candidates failed to realise that a temporary immediate annuity is payable

in arrears. If the annuity is payable in advance several parts of the question
become much more straightforward.

9 (ii)(a) was only attempted by a minority of candidates. Some candidates

incorrectly assumed that an estimate of q70 was required, calculating E70=4.5 and
θ70=3 and then finding the actuarial estimate of q70.

In (ii)(b) many candidates failed to construct the likelihood from the data and
derive the maximum likelihood estimate. In many cases the actuarial estimate of
q70 was given instead of q̂ 70 .

10 Simple algebraic errors, eg. logging additive functions detracted from some good
attempts. There were few serious attempts at (iii) and (iv).

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 Subject 104 (Survival Models) — April 2001 — Examiners’ Report

1 (a) tpxII represents the probability of the event of being in state I at time/age
x + t given that the life was in state I at time/age x. Movements in the
interval (x, x + t) are not restricted.

tpxII represents the probability of the event of being in state I at time/age

x + t given that the life was in state I throughout the interval (x, x + t).

(b) In this model the probabilities are equal because there is no way of
returning to state I once a life has left this state.

∂( t V )
2 (i) = δtV + P − µx+t (tV − tV) = δ tV + P
∂t

i.e. t V ′ − δtV = P.

(ii) To solve, multiply by e−δt to get:

(e −δt
tV )′ = Pe −δt
,

e−δt tV = (−P/δ) e−δt + c

and 0V = 0 Þ c = P/δ

Hence tV = eδt (P/δ) (1 − e−δt)

= P(eδt − 1) / δ = Pst i , with 1 + i = eδ.

(Alternatively verify directly that Pst i , satisfies the equation and the
initial condition, 0V = 0.)

(iii) Ps20 0.04 = S = 20,000a65 = 20,000 × 9.790902

Þ P = 6,447.80

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 Subject 104 (Survival Models) — April 2001 — Examiners’ Report

3 (i) (a) P = Ax / ax :m

(b) Ax+t − 0

(c) ( Pax :m )
− Ax1:t / vt t px

(iii) Pax :m = Ax = Ax1:t + vt t px Ax +t

(
Þ Ax+t = Pax :m − A1x :t ) vt t px , as required.

A solution using commutation functions is also possible.

4 (i) Plot qˆ x against qxs and look for an approximate straight line fit.

(ii) At each age there will be a different sample size/exposed to risk, Ex . This
will usually be largest at ages where many term assurances are sold e.g.
25 to 50 and smaller at other ages.

The estimation procedure should pay more attention to ages where there
are lots of data. These ages should have a greater influence on the choice
of a and b than other ages.

Other relevant comments also received marks.

So weights wx α Ex

−1
Suitable choice is wx = éë Var ( qˆ x ) ùû

Ex
=
qx (1 − qx )

Ex
; as qx ; 10 −2
qx

Ex E2
These weights can be estimated by = x
qˆ x θx

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 Subject 104 (Survival Models) — April 2001 — Examiners’ Report

(iii) The graduated rates qox are a linear function of the rates in the standard
table qxs . The standard table rates will already be smooth.

Smoothness is based on the size of the third differences of the graduated

rates ∆3 qo , which because the relationship is linear will be equal to
x

b∆3 qxs . ∆3 qxs will already be acceptably small because the standard table
rates will already be smooth.

5 Expected Present Value of Benefits

D65 ì ′
D70 ü
5,000 ía5 4% + ′ ý = 19,910.340
a70
D45 î ′
D65 þ

where ′ = a(55)

assuming weekly payments are a good approximation to continuous payment.

Marks for evaluations

1 − (1.04)−5
a5 4% = = 4.54028
log e (1.04)

′ ; a70 + ½ = 8.463
a70

D65 2,144.1713
= = 0.37689
D45 5,689.1776

′
D70 43,852
= = 0.71196
′
D65 61,593

( M 45 − M 55 )
10,000 A45:10 ; 10,000 (1 + ½i ) = 371.4054
D45

D55 ( M 55 − M 65 )
25,000 A55:10 ; 25,000 (1 + ½i ) = 1,732.3690
D45 D45

Expected Present Value of Premiums. Annual Premium P.

Pa&&45:20 = 13.488P

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 Subject 104 (Survival Models) — April 2001 — Examiners’ Report

Equation of Value

13.488P = 19,910.340 + 371.4054 + 1,732.3690 = 22,014.114

P = £1,632.126

6 (i) x = CY of death − CY of birth

= age, x, on birthday in CY of death
= age next, x, on 1 January in CY of death
= age next, x, on 1 January before date of death

So Calendar Year Rate Interval starting, for lives classified x, on

1 January on which the life is aged x next birthday.

Age range at start of calendar year x − 1 to x.

(ii) (a) Px(t) census at t of those x next on previous 1 January would

correspond to the classification of deaths

but ages in the censuses used are ages on 1 July

So (x − 1, x) on 1 January
is (x − ½, x + ½) on 1 July = date of census

So required x in Px(½), Px(1½), Px(2½) is x nearest birthday at date

of census

(b) Need Birthdays uniform over calendar year

to get average age at start of rate interval, x − ½

Need force constant over (x − ½, x + ½)

So µˆ x + f will be x + 0, f = 0

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 Subject 104 (Survival Models) — April 2001 — Examiners’ Report

7 (i) H0: The observed rates qox are consistent with coming from a population in
which the premium rate basis qxp are the true rates.

(ii) Ex initial exposed to risk at age x in the investigation

Chi-squared

(a) The test statistic is

( )
2
Ex qox − E x qxp
åx E x qxp

where Ex is the initial exposed to risk at age x.

(b) If H0 is true, then

Ex qox − Ex qxp
~ N (0,1)
E x qxp approx.

for each age x using the Central Limit Theorem and assuming that
1 − qxp @ 1.

Then if the observations at each of the n ages are independent

( Ex qox − Ex qxp )2
åx Ex qxp
~
approx.
χn2

(c) Reject H0 if the Observed Value of the test statistic > χ2n (1 − α) at
α% level of significance. This is a one-tailed test.

Grouping of Signs

(a) The test statistic is G, the observed number of positive groups of

signs among n1 positive and n2 negative deviations, where the
deviation at age x = E qo − E q p .
x x x x

(b) If H0 is true, then the deviations will be allocated randomly to

groups.

Then n1 positive deviations can be divided into g groups in

æ n1 − 1 ö
ç ÷ ways
è g −1 ø

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Subject 104 (Survival Models) — April 2001 — Examiners’ Report

These g groups of positive signs can be allocated amongst the n2

negative deviations in

æ n2 + 1 ö
ç ÷ ways
è g ø

The unrestricted number of ways of arranging n1 + n2 positive and

negative deviations is

æ n1 + n2 ö
ç ÷
è n1 ø

So the sampling distribution of the test statistic, G, is

æ n2 + 1 ö æ n1 − 1 ö
ç ÷ç ÷
è g ø è g −1 ø
P[G = g] = g = 1, 2, 3, …n2 − 1
æ n1 + n2 ö
ç ÷
è n1 ø

Alternatively, if n1 + n2 > 20 (approx.) then

æ n (n + 1) (n1n2 )2 ö
G ~ Nç 1 2 , 3 ÷
approx.
è (n1 + n2 ) (n1 + n2 ) ø

using the Central Limit Theorem and E[G] and Var[G].

(c) Reject H0 if the Observed Value of the test statistic, G ≤ k* where

k* is smallest integer such that

P[G ≤ k*] ≥ 0.05

(or as similar statement based on the N(0, 1) distribution). This is

a one-tailed test.

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 Subject 104 (Survival Models) — April 2001 — Examiners’ Report

8 (i) W = amin( K
40 ,20)
K 40 ≥ 0

or W = aK K40 = 0, 1, 2, …20
40

= a20 K40 > 20

(ii) (a) E[W] = a40:20 A67/70 ultimate 5%

l 60v20
= a&&40:20 + −1
l 40

30,039.787 (1.05)−20
= 12.760 + −1
33,542.311

= 12.0975

5% tables give: a19 = 12.0853 a20 = 12.4622

So W < 12.0975 if life dies in (40, 60)

l 60 30,039.787
=1− =1−
l 40 33,542.311

= 1 − 0.89558 = 0.10442

(b) The probability function of the discrete random variable W is very

skew so P[W < E[W]] = the very small “left hand tail area”.

fW(w)

E[W]

W
a1 a2 a18 a19 a20

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 Subject 104 (Survival Models) — April 2001 — Examiners’ Report

ìïaK +1 − 1 K 40 = 0,1, 2, ... 20

(iii) Now W = í 40
îa21 − 1
ï K 40 > 20

ì1 − v K40 +1
ï − 1 K 40 = 0,1, 2, ...19, 20
ï d
= í
21
ï1 − v − 1 K 40 > 20
ï
î d

ì1 1 K 40 +1
ï −1 − v K 40 = 0,1, 2, ...19, 20
ïd d
= í
ï1 − 1 − 1 21
v K 40 > 20
ï
îd d

1
So Var(W) = Var(Y) where
d2

ìïv K 40 +1 K 40 = 0,1, 2, ...,19, 20

Y = í 21
îv
ï K 40 > 20

20 k =∞
Now E[Y] = å
k =0
vk +1 kq40 + åv
k =21
21
kq40

= A40:21

Where A is determined at 5% p.a.

20 k =∞
E[Y2] = å (v
k =0
k +1 2
) kq40 + å (v
k =21
21 2
) kq40

′
= A40:21

Where A ′ is determined at (1.05)2 − 1 = 10.25% p.a.

2
æ 1.05 ö 0.05
So Var(W) = ç ÷
è 0.05 ø
{ A′ 40:21
2
− A40:21 } where d = 1.05

′ (
= 441 A40:21 2
− A40:21 )
The result can be derived in a similar way using

amin( K = amin( K −1
40 ,20) 40 +1, 21)

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 Subject 104 (Survival Models) — April 2001 — Examiners’ Report

9 (i) (a) b−aqx+a = 1 − b−a px+a

Now b px = a px b−a px+a

b px
So b−aqx+a = 1 −
a px

(b) If deaths are uniformly distributed over (x, x + 1) then

t qx = t . qx

So t px = 1 − t . qx

1 − bqx (b − a )qx
Then b−aqx+a =1− =
1 − aqx 1 − aqx

(ii) (a) The likelihood for each life is

Life i 1 2 3 4 5 6
Likelihood p70 p
0.6 70.3 q
0.5 70.5 0.4 p70 0.9 q70 q70

And the total likelihood is the product

(1 − q70) (1 − 0.6 q70.3) 0.5 q70.5 (1 − 0.4q70) 0.9q70q70

Using (i)(b) we can write

æ 0.6q70 ö æ 0.5q70 ö
(1 − q70 ) ç1 − ÷ç ÷ (1 − 0.4q70 ) 0.9q70q70
è 1 − 0.3q70 ø è 1 − 0.5q70 ø

3
(1 − q70 )(1 − 0.9q70 ) (1 − 0.4q70 ) 0.45q70
=
(1 − 0.3q70 )(1 − 0.5q70 )

(b) The likelihood for each life is proportional to (assuming constant

force µ70 ) .

Life i 1 2 3 4 5 6
−µ70 −0.6µ70 −0.4 µ70 −0.4 µ70 −0.7 µ70 −0.8 µ70
Likelihood e e e µ70 e e µ70 e µ70

And the total likelihood is the product

L ∝ e −3.9µ70 (µ70 )3

∂L
Then = − 3.9e −3.9µ70 (µ70 )3 + e −3.9µ70 3µ70
2
∂µ70

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 Subject 104 (Survival Models) — April 2001 — Examiners’ Report

Then −3.9µˆ 70 + 3 = 0

3
µ̂70 = = 0.7692
3.9

The same result can be obtained using the log likelihood.

Then ML estimator of q70 ; q̂70 = 1 − e−0.7692 = 0.5366

10 (i) Write times in rank order, label groups and label first absences from
work.

Determine cumulative probabilities and contributions to partial likelihood

from first absences.
2+ F eβ 7 + 8eβ
eβ
4 F eβ 7 + 7eβ
7 + 7e β
6+ M 1 7 + 6eβ
eβ
7 F eβ 6 + 6eβ
6 + 6e β
8+ F eβ 6 + 5eβ
10+ F eβ 6 + 4eβ
1
11 M 1 6 + 3eβ
6 + 3eβ
12+ F eβ 5 + 3eβ
13+ M 1 5 + 2eβ
1
15 M 1 4 + 2eβ
4 + 2eβ
16+ M 1 3 + 2eβ
eβ
17 F eβ 2 + 2eβ
2 + 2eβ
19+ M 1 2 + eβ
1
20 M 1 1 + eβ
1 + eβ
21+ F eβ eβ

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 Subject 104 (Survival Models) — April 2001 — Examiners’ Report

Partial Likelihood, L

eβ eβ 1 1 eβ 1
. . . . .
7(1 + e ) 6(1 + e ) 3(2 + e ) 2(2 + e ) 2(1 + e ) 1 + eβ
β β β β β

e 3β
=
84(1 + eβ )4 6(2 + eβ )2

So Loge L = l (β) = 3β − 4 loge (1 + eβ) − 2 loge (2 + eβ) + c

∂l 4 eβ 2eβ
(ii) Then =3− −
∂β 1 + eβ 2 + eβ

Let x = eβ then x̂ is given by

4 xˆ 2xˆ
3− − =0
1 + xˆ 2 + xˆ

3(1 + xˆ )(2 + xˆ ) − 4 xˆ (2 + xˆ ) − 2xˆ (1 + xˆ ) = 0

6 + 9xˆ + 3xˆ 2 − 8xˆ − 4 xˆ 2 − 2xˆ − 2xˆ 2 = 0

3xˆ 2 + xˆ − 6 = 0

−1 ± 1 − 4 ( −6.3)
x̂ =
6

−1 ± 73
=
6

73 − 1
Estimate must be +ve, so = 1.25733
6

So β̂ = loge 1.25733 = 0.2290

∂2l æ −4eβ . eβ + (1 + eβ ) . 4eβ ö æ −2eβ . eβ + (2 + eβ ) . 2eβ ö

(iii) Now = − ç ÷−ç ÷
∂β2 è (1 + eβ )2 ø è (2 + eβ )2 ø

æ 4 eβ 4 eβ ö
= −ç β 2
+ ÷
è (1 + e ) (2 + eβ )2 ø

ˆ
At eβ = 1.25733, this has value − (0.98700 + 0.47401) = −1.46101

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 Subject 104 (Survival Models) — April 2001 — Examiners’ Report

1
So asymptotic standard error = +
1.46101

= 0.8273

(iv) 95% Confidence Interval for β̂ is approximately

0.2290 ± 2 × 0.8273

i.e. (−1.43, 1.88)

This interval includes β = 0, so the model

λ(tx) = λ0(t) exp{0 × x} = λ0(t)

is compatible with the observed data i.e. same model applies to males and
females. There is no significant difference between rates.

OR

Using a one sided alternative hypothesis and testing

H0: β = 0 against H1: β > 0 we have:

Under the Null hypothesis

0.2290 − 0
= 0.28
0.8273

should be N(0, 1).

This value is not in the critical region (1.64, ∞) and so there is no reason to
reject the null hypothesis. Conclusions as above.

OR

A likelihood test can be used

−2( l(0) − l(βˆ )) = − 2( −4.9698 + 4.9315) = 0.0765

which is χ12 if H0: β=0 is true. This value is not in the critical region so
there is no reason to reject H0.

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