Actuarial Society of India
EXAMINATIONS
June 2005
CT4 (103) – Models
Total Marks - 50
Indicative Solution
Q.1 (i)
a) Let U denote the process described by 2301 and V denote the process described by 2401. Then
5 6
( )
5 1
[ 6] [ 0] 0.0538
6!
P U P V e e
− −
= = = =
b) The arrivals of either train will be Poisson with parameter 6 per hour. In half an hour the number of arrivals will have a Poisson (3) distribution. Thus,
3 2
3 3
[ 3] 1 [ 0] [ 1] [ 2]
1 3 3 0.5768
2!
P U V P U V P U V P U V
e e e
− − −
+ ≥ = − + = − + = − + =
= − − − =
c) The waiting time T till the next is exponential with parameter 6. It is irrelevant whether one has just missed a train or not since the exponential distribution has the memoryless property. Thus, the required probability is:
6
1 4
0.2231 P T >4 =e− =
d) The distribution of the number of 2301’s in time t is Poisson (5t). Thus the probability of exactly one is e−5t5 .t However, we don’t know the time taken for the 2401 to arrive. The distribution of this is exponential with parameter 1. Thus, we condition on this time. As the time is continuous, we need to use the density function:
[ ]
0[ ]
5 0
exactly one 2301 before a 2401 arrives exactly one 2301 in time t
5 5 (integrated by parts)
36
t
t t
P P e dt
e dte dt
∞ −
∞ − −
=
= =
∫
∫
[10]
(ii)
As 1 in every 6 trains is on average a 2401, the answer is 1
6 . Alternatively, let the waiting time for a 2301 be denoted by T2 3 0 1 and that for a 2401 be denoted by T2 3 0 1. The probability of a 2401 arriving before a 2301 is then
2 4 0 1 2 3 0 1
[ ]
P T < T .
Conditioning on the 2401 waiting time, we thus have
2 4 0 1 2 3 0 1 2 3 0 1
0 5
0
[ ] [ ]
1 6
t
t t
P T T e P T t d t
e e d t
∞
−
∞ − −
< = >
=
=
∫
∫
[2]
(iii) The distribution of the number of Chittaranjan manufactured trains in time it t is Poisson with parameter 1 1 13
2 5 *3 t 6 t
+ =
. The waiting time will be thus be exponential with parameter 13
6 . Since the exponential distribution has the lack of
memory property, that a Chittaranjan manufactured train hasn’t come for over an hour is irrelevant. The expected waiting time is thus 13
6 hours or about 27.7 minutes.
[3]
Total [15]
Q.2
• A stochastic process X is stationary if the joint distributions of the n
1 2 m 1 2 m
t t t t t t
X , X ,...,X and X +k, X +k,...,X +k are identical for all
1, ,..., ,2 m 1, 2,..., m and all integers m.
t t t k t k t+ + k+t
• The process is weakly stationary of the expectations E[X ] are t constant with respect to t and the covariances Cov(X ,t Xt+k) depend only on the lag k.
• If t and t+u are in the set of permissible values, then the increment for time u will be Xt+u −Xt.
• For a discrete process the Markov property requires that:
1 2 m m
t t 1 t 2 t t t
P X =x| X =x, X =x ,...X = xm=P X =x| X =xm for all times t1< <t2 ...< <tm t and all states x1<x2 <...<xm <t.
• A discrete time martingale X satisfies two conditions: n 1. E[|X |] < n ∞ for all n
2. E[|X |n X , X ,...X ] = 0 1 m X for all m < n. m
Total [10]
Q.3 a) § At each step Aishwarya’s fund will change by a random amount Zn
n
n
where
1 with probability 1 2 1 with probability 1
2 Z
Z
= +
= −
So that Sn= Sn−1+ Zn will be a simple symmetric random walk. Initially
0 .
S =k The boundary conditions are such that
[ ]
[ ]
n 1
n 1
0 | 0 1 and
| 1
n n
P S S
P S K S K
−
−
= = =
= = =
So that we can consider it as a walk on (0, 1, 2…..K).
[3]
b) § Here,
[ ] [ ]
[ ]
n 0 1 1 1 0 1 1
1 1
| , ,..., n n n| , ,..., n
n n n
E S S S S E S Z S S S
S E Z S
− − −
− −
= +
= + =
if 0 < Sn−1 < K. In the case where Sn−1 = 0 or Sn−1 =K, the above property holds since thenSn =Sn−1.
[2]
c) § A stopping time for the process Sn is a positive valued integer random variable such that for all n, the indicator variable I{T n=} =1 if T = n and
{T n} 0
I = = otherwise is a function of the past and present values
0, 1,... n S S S only.
The optimal stopping theorem states that E[ST] = E[S0] if either 1. T is bounded, i.e. T ≤N for some constant N
2. S is bounded, i.e. |Sn|≤N for some constant N
[4]
d) § Let T be the stopping time until the process reaches 0 or K. Since Sn is a martingale and bounded, the optimal stopping theorem can be applied. This results in E[ST] = E[S0] = k.
Thus,
0 0
0
[ ] 0. [ 0 | ] . [ | ]
(1 [ 0 | ])
T T T
T
E S P S S k K P S K S k
K P S S k
= = = + = =
= − = =
Since the probability of ruin is just P S[ T =0 |S0= k], and we know that [E ST]=k, we can thus obtain ( T 0 | 0 ) K k
P S S k
K
= = = −
[3]
e) § If Aishwarya is greedy then K is very large. If we get K → ∞ the, ( T 0 | 0 ) K k 1
P S S k
K
= = = − →
In other words, she is certain to be ruined if she does not quit. [2]
f) § Vivek needs to be super rich so that he will not go broke. The stopping
time therefore only depends on Aishwarya’s position. [1]
Total [15]
Q.4
a) 1 2
1
2
(0,1) and (0,2). Thus,
( 1) 1 (1) 0.159
( 1) 1 1 0.240
2
B N B N
P B P B
≥ = −Φ =
≥ = −Φ =
: :
[3]
b)
( ) [ ] [ ] [ ]
[ ]
( )
2
2
2 2 2
2
4
(0,2) Thus
P Z>z , , 2 ,
2
Now P Z>z 2 1
2
2 1
( ) 2 1 z > 0.
2 2 2
z
B N
P Z z B z P Z z B z P Z z B z P B z
z
z z
f z e
z π
−
= > > + > < = > >
= >
= −Φ
∂
= ∂ −Φ = Φ = :
[4]
c) E X( T)=E X( 0) if T is the stopping time and Xt is a martingale. Bt is a
martingale.
Thus, E B( Tx)=E B( 0)=0 for all x.
Tx is the first time the process Xt hits x. Thus XTx =x. Thus
( ) 0 ( ) 0
( )
Tx x
Tx x
Tx x
x
B T x
B x T
E B E x T
E T x µ
µ
µ µ
+ =
= −
= ⇒ − =
⇒ =
[3]
Total[10]
*******************
Actuarial Society of India
EXAMINATIONS
June 2005
CT4 (104) – Models
Total Marks - 50
Indicative Solution
Q.1 (i) x+t x
[
x]
0
lim1 P T t+h|T >t h
µ h
= → × ≤
[1]
(ii)
[ ]
[ ] [ ]
{ }
x x
x x
0
f (t) d F (t) dt
d P T t
dt
lim1 P T t+h P T t
h
x
h→
=
= ≤
= × ≤ − ≤
[1]
Q.2 a) (i)
x x ( ) F ( ) .
Thus
1 .
t x x x
x t t x
x x
x t
t x x
d d d
f t t q t q q
dt dt dt
p
q q
p t q
µ µ
+
+
= = = =
=
= =
−
by definition of uniform distribution of deaths. [1]
(ii) .
Thus
1 1
1 1 ( )
1 1
t x s x t s x s
t x t s x s
s x
t x s x t x
t s x s t s x s
s x s x
x x x
x x
p p p
p p
p
p p p
q p
p p
sq tq t s q
sq sq
− +
− +
− + − +
=
=
= − = − = −
− − + −
= =
− −
[2]
b) p56 =0.990581 p57 =0.989503 and p58 =0.988314
(i)
( ) ( )
2 56.75 0.25 56.75 57 0.75 58 56
57 58
0.75 56
udd: p p p p
= p 1 1 0.75*
1 q
0.990581
(0.989503) (1 0.75 0.011686) 1 0.75 0.009419
0.9785044
q q
= × ×
× − × −
−
= × × − ×
− ×
= [2]
(ii) Constant force of mortality:
2 56.75 0.25 56.75 57 0.75 58
0.25 0.009464 0.75 0.011755
log log (1 )
Thus
0.989503 0.9785
e px e qx
p p p p
e e
µ
− × − ×
= − = − −
= × ×
= × ×
=
[2]
Total [7]
Q.3 Poisson distribution of claims
Mean = nq and std deviation = nq
Thus 5% confidence interval will be equivalent to 5% of the likely death claim or
2
2
1.96 nq 0.05 nq or nq 1.96
0.05
1.96 1
or n *
0.05 0.002 768320
× = ×
=
=
=
[4]
Q.4
The likelihood for each life is
a) 1 2 3 4 5 6 7
p 40 0.3p40.4 0.6q40.3 0.9q40 q40 0.5q40.5 0.4p 40 The total likelihood is the product
( )( )( )( )( )( )( )
( ) [ ]
( )( )( )
( )( )( )
40( )
40 0.3 40.4 0.6 40.3 0.9 40 40 0.5 40.5 0.4 40
40 40 40
40 40 40 40
40 40 40
4
40 40 40 40
40 40 40
1 1 1
0.3 0.6 0.5
1 1 0.9 1 0.4
1 0.4 1 0.3 1 0.5
1 1 0.7 1 0.4 0.27 1
1 0.4 1 0.3 1 0.5
q q q q q q q
q q q
q q q q
q q q
q q q q q
q q q
− − −
= − × − − × − × × × − × −
− − − × −
= =
− − −
( )
( )( )
404 40
40 40
1 0.7 0.27
1 0.3 1 0.5
q q
q q
− ×
− −
[2]
[2]
b) The likelihood for each life is proportional to, assuming constant force of mortality '40
µ
1 2 3 4 5 6 7
'40
e
−µe
−0.3 'µ40e
−0.4 'µ40µ '
40e
−0.6 'µ40µ '
40e
−0.8 'µ40µ '
40e
−0.4 'µ40µ '
40e
−0.4 'µ40Thus the total likelihood is the product
( )
40
40 40
40
3.9 ' 4
40
4 3.9 ' 3.9 ' 3
40 40
40
40
L e ( ' )
Differentiating
L 3.9 ' e e 4 '
'
Equating to zero 3.9 ' 4 0
' 1.0256 or
µ
µ µ
µ
µ µ
µ µ µ
−
− −
∝
∂ = − +
∂
− + =
=
[3]
[3]
Total [10]
Q.5
(i) Partial likelihood is
k T
j T
j 1 i
i R(tj)
exp( x )
L( ) exp( x )
β β
= β
∈
=
∏ ∑
where k is the number of deaths assumed to occur at distinct times t is the tj th lifetime
R(t ) denotes the set of lives at risk at time j t . j [1]
(ii) The partial likelihood is
( ) ( )
3
4 2
e e 1 1 e 1
7+7e 6+6e 6 3e 4 2e 2 2e 1 e
e
504 1+e 2 e
β β β
β β β β β β
β β
β
× × × × ×
+ + + +
= + [5]
(iii)
( ) ( )
3
4 2
L e
504 1 e 2 e
log L =3 4 log(1+e ) 2log(2 e ) const Differetiating with respect to , we get
d log L 4e 2e
d 3 1 e 2 e
β β
β β
β β
β β
β
β
β β
= + +
− − + +
= − −
+ +
setting this equal to zero and rearranging (y = eβ)
2
4 2
1 2 3
or 3y 6
Thus y 1.257 using only +ve value = log y = 0.2287
y y
y y
y β
+ =
+ +
+ =
=
[5]
(iv) We need to use Breslow approximation to get the new partial likelihood which is
2
e e 1 1 e 1
8 8e 7 7e 7 3e 5 2e 3 2e (2 e )
β β β
β × β × β × β × β × β
+ + + + + + [3]
Total [14]
Q.6 a)
(i) Policy year rate interval. [1]
(ii) Assume birthdays uniformly distributed over the policy year.
Lives will be thus on the average 1
x+2 at the start of the rate interval (for qx estimate)
Thus x + f = x + 1, age at the middle of the rate interval (for µx estimate)
[2]
(iii) The assumption of uniform birthdays over the policy year now no longer holds.
Lives are now 1 3
1 4 4
x+ − = +x at the start of the rate interval (qx type) and 1 14 x+
at the middle of the rate interval (µx type). [2]
b)
(i) P = number of policyholders in force at 1.1.2000+t aged x nearest birthday on x,t
previous policy anniversary.
Company I
Let P' = number of policyholders in force at 1.1.2000+t aged x nearest birthday. x,t
( )
e 1 x 0 x,t
E = P' dt 1 P'(x,0) P'(x,1)
=2 +
∫
assuming P' varies linearly over the calendar year x,t Since deaths are recorded as age nearest birthdays at death.
( )
P'(x,t) 1 P( 1, ) P( , )
2 x t x t
= − +
assuming policy anniversaries are uniformly distributed over the calendar year and birthdays are uniformly distributed over the policy year.
Thus
( ) ( ) ( ) ( )
x
E 1 1,0 , 0 1,1 ,1
4
e = P x− +P x +P x− +P x
Company II Similarly
( )
1
x 0
E '( , ) dt 1 '( ,0) '( 1,1)
2
e =
∫
P x t = P x +P x+where P x t'( , ) is the number of policyholder in force at 1.1.2000+t aged x nearest birthday on 1.1.2000.
Since deaths are recorded as age nearest birthday on 1.1.2000
( )
'( , ) 1 '( 1, ) '( , )
P x t = 2 P x− t +P x t with the same assumptions as the company I.
Thus x
( ) ( ) ( ) ( )
E 1 1,0 ,0 ,1 1,1
4
e = P x− +P x +P x +P x+
[3]
[3]
Total [6]
(ii)
For company I, the above observed rates apply to age x - 1 x x for q and x for
2 µ . No
assumption is required.
For company II, the observed rates apply to age x x 1 x for q and x+ for
2 µ .
Assumption: Birthdays are uniformly distributed over the calendar year. [2]
Total [13]
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