Stochastic Modelling 2000-2004

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Faculty of Actuaries Institute of Actuaries

EXAMINATIONS

10 April 2000 (pm)

Subject 103 — Stochastic Modelling

Time allowed: Three hours

INSTRUCTIONS TO THE CANDIDATE

1. Write your surname in full, the initials of your other names and your

Candidate’s Number on the front of the answer booklet.

2. Mark allocations are shown in brackets.

3. Attempt all 11 questions, beginning your answer to each question on a

separate sheet.

Graph paper is not required for this paper.

AT THE END OF THE EXAMINATION

Hand in BOTH your answer booklet and this question paper.

In addition to this paper you should have available

Actuarial Tables and an electronic calculator.

Faculty of Actuaries103—A2000 Institute of Actuaries


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103—2

1 Let X n = Y 1 + Y 2 + ... + Y n be a simple random walk with step distribution

P[Y j = 1] = p = 1 − P[Y j = −1].

Derive an expression, for each λ > 0, for the two values of γ such thatM n = e

n X n− +λ γ is a martingale with respect to the natural filtration of X n. [4]

2 Let X and Y be jointly normal random variables with means µ X , µY , variances

σ σ X Y 2 2, and correlation coefficient ρ. Derive an expression for the values of

the coefficients α and β, given that the conditional expectation E[ X Y ] is of theform α + βY .

[5]

3 The total number of claims received by an insurance company is described byan inhomogeneous Poisson process with rate λ(t).

Write down the Kolmogorov forward equations for this process and show that,

as in the homogeneous case, the solution is of the form

P 0 j (s, t) =( ( , ))

!

( , )m s t e

j

j m s t−

where m(s, t) = z st λ(x )dx . [5]

4 According to a model used in econometrics, three economic time series, X , Y and Z , are related to one another by the equations

X n = X n−1 + θ X Z n−1 + e1,n

Y n = Y n−1 + θY Z n−1 + e2,n

Z n = θZ Z n−1 + e3,n

where each of θ X , θY and θZ lies in the interval (−1, +1) and the randomvariables {ei,n : n ≥ 1, i = 1, 2, 3} may be assumed to be uncorrelated and to havemean zero.

State, giving your reasons:

(a) whether X , Y and Z are I (0) or I (1)

(b) whether each of X , Y and Z individually satisfies the Markov property

(c) whether the vector-valued process {( X n , Y n , Z n) : n ≥ 1} satisfies theMarkov property

(d) whether X and Y are cointegrated [5]


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103—3 PLEASE TURN OVER

5 Let Bt be a standard Brownian motion, and let Ft = σ( Bs , 0 ≤ s ≤ t) be its naturalfiltration.

(i) Derive the conditional expectations E[ ] Bt s2F and E[ ] Bt s

4F , where s ≤ t.

You may assume that the fourth moment of a random variable with

distribution N (0, σ2) is 3σ4. [4]

(ii) Hence construct a martingale out of Bt4. [3]

[Total 7]

6 (i) Apply the inverse transform method to generate an observation fromthe density

f 1(x ) =1

1 2( )+ x (x > 0)

using a pseudo-random number u in the range 0 < u < 1. Explain how

this can be extended to generate an observation from the symmetrised

form of the same density

f 2(x ) =1

2 12

( )+ x (x ∈ R) [3]

(ii) The Cauchy distribution has density function

f (x θ) = θπ θ( )x 2 2+ (x ∈ R)

where θ is a positive parameter.

Show that

f (x θ) ≤ Cf 2(x ) for all x ∈ R

as long as C ≥ 2π

(θ + θ−1). Hence devise a method based on Acceptance-

Rejection sampling for generating observations from the Cauchy

distribution. [4]

[Total 7]


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103—4

7 (i) (a) Calculate the autocovariance function {γ k : k ≥ 0} andautocorrelation function {ρk : k ≥ 0} of a first-order Moving

Average process

X t = µ + et + β1 et−1 ,

where {et : t ≥ 0} is a sequence of uncorrelated, zero-mean random

variables with common variance σe2.

(b) State the conditions on the values of the parameters such that

the process is invertible. [5]

(ii) A sequence of observations x 1 , x 2 , ..., x n has sample variance $γ 0 = 14.5,sample lag-1 autocovariance $γ 1 = 5.0. Show that there is more than onefirst-order moving average process which can be fitted to these data,

but verify that only one of the fitted processes is invertible. [4]

[Total 9]

8 The members of a health insurance scheme are classified as contributors orbeneficiaries; a member who is a contributor in one period becomes a

beneficiary in the next period if he or she becomes seriously ill, and this

happens with probability 0.1. The probability of a serious illness continuing

into the next period is 0.2. The rules of the scheme specify that any member

who is a beneficiary for three successive periods must become a contributor for

the next period; if the illness still persists the member may thereafter revert

to being a beneficiary.

(i) (a) Construct a discrete time Markov chain to model this health

scheme, introducing if necessary various classes of beneficiaries

and contributors (a five state model is suggested).

(b) Draw the transition graph.

(c) Write down the transition matrix of the chain. [6]

(ii) Explain whether the above Markov chain is irreducible, periodic or

both. [2]

(iii) (a) Calculate the stationary probability distribution of the chain.

(b) Determine the proportion of beneficiaries among the

membership in the stationary régime. [4]

(iv) Let b be the average gross payout per beneficiary and c the average

gross payout per contributor per period; this means that the nett

payments are b − f and c − f respectively, where f is the membership feeper period (assumed to be uniform over members and over time).

(a) Explain how b, c and f should be related if the scheme is to be

viable.

(b) Calculate the average profit per period per member in the

stationary régime if b = 600, c = 150 and f = 300. [3][Total 15]


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9 A stationary second-order autoregressive process X , which may be assumed tobe in equilibrium at time 0, is defined by

X t = µ + α1( X t−1 − µ) + α2 ( X t−2 − µ) + et ,

where {et : t ≥ 1} is a sequence of independent, zero-mean Normal random

variables, each with variance σe2.

(i) (a) Obtain an equation for γ 1 in terms of γ 0 and γ 2 by substituting for X t in the equation γ 1 = Cov( X t , X t−1).

(b) Derive similar equations for γ 2 and γ 0.

(c) State the autocorrelation function ρk of X for k = 0, 1, 2. [5]

(ii) Suppose that the equations derived in (i) for ρ1 and ρ2 are used as thebasis of an estimation procedure: estimates $α

1

and $α2

are defined to be

the solutions of those equations when ρ is replaced by a suitably-defined sample autocorrelation function r.

Solve these equations. [3]

[Total 8]

10 (i) (a) Define standard Brownian motion Bt , t ≥ 0 and give its transitionprobability density.

(b) Write down the transition probability density of general

Brownian motion W t = σ Bt + µt. [4]

Let S t defined represent a share price at time t.

(ii) Solve the stochastic differential equation

dS t = µS tdt + σS tdBt . [5]

(iii) Calculate, given the parameters µ = 25% p.a., σ = 20% on an annualbasis, the probability that the share price will exceed 45 in four months’

time given that its current price is 38. [4]

(iv) Calculate the probability that the share price will exceed 45 at any

stage during the next four months given that its current value is 38.

[You may use the formula

P[max( ) ]0≤ ≤

+ >s t

s B s yλ = Gt y

te G

y t

t

yλ λλ−F H G

I K J +

− −F H G

I K J

2,

where y ≥ 0 and G denotes the normalised Gaussian probabilitydistribution function.] [4]

[Total 17]


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103—6

11 Patients arriving at the Accident and Emergency department (state A) wait foran average of one hour before being classified by a junior doctor as requiring

in-patient treatment (I), out-patient treatment (O) or further investigation (F).

Only one new arrival in ten is classified as an in-patient, five in ten as out-

patients.

If needed, further investigation takes an average of 3 hours, after which 50% of

cases are discharged (D), 25% are sent to receive out-patient treatment and

25% admitted as in-patients.

Out-patient treatment takes an average of 2 hours to complete, in-patient

treatment an average of 60 hours. Both result in discharge.

It is suggested that a time-homogeneous Markov process with states A, F, I, O

and D could be used to model the progress of patients through the system,

with the ultimate aim of reducing the average time spent in the hospital.

(i) Write down the matrix of transition rates, {σij : i, j = A, F , I , O, D}, of

such a model. [2]

(ii) Calculate the proportion of patients who eventually receive in-patient

treatment. [1]

(iii) Derive expressions for the probability that a patient arriving at time

t = 0 is:

(a) yet to be classified by the junior doctor at time t, and

(b) undergoing further investigation at time t [4]

(iv) Let mi denote the expectation of the time until discharge for a patient

currently in state i.

(a) Explain in words why mi satisfies the following equation:

mi =1

λ

σ

λi

ij

i

j

j i D

m+∉∑{ , }

where λi = ij j

σ∑ .

(b) Hence calculate the expectation of the total time until discharge

for a newly-arrived patient. [4]

(v) State the distribution of the time spent in each state visited according

to this model. [1]

The average times listed above may be assumed to be the sample mean waiting

times derived from tracking a large sample of patients through the system.

(vi) Describe briefly what additional feature of the data might be used tocheck that this simple model matches the situation being modelled. [2]


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103—7

(vii) The hospital management committee believes that replacing the junior

doctor with a more senior doctor will save resources by reducing the

proportion of cases sent for further investigation. Alternatively, the

same resources could go towards reducing out-patient treatment time.

(a) Outline briefly the calculations that would need to be performed

to compare the options.

(b) Discuss whether the current model is suitable as a basis for

making decisions of this nature. [4]

[Total 18]


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EXAMINATIONS

April 2000


EXAMINERS’ REPORT



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Subject 103 (Stochastic Modelling) — April 2000 — Examiners’ Report

Page 2

1 E[ M n+1F

n]= E F[ ]( )e n X

nn

1 1

= e en X Y n

n n

( ) ( )[ ]1 1E F = 1( 1) [ ]n n X Y nn

e e e

E F

= e M enY

n

E[ ]1

= e M pe p en

( ( ) ).1

Hence the condition for martingale:

pe + (1 p)e = e .

Multiply by e and solve quadratic equation in unknown e:

e =e e p p

p

2 4 1

2

( ).

2 Assume E[ X Y ] = + Y and determine , using:

(i) orthogonality condition E{( X E[ X Y ])Y } = 0.

(ii) E{E[ X Y ]} = E[ X ].

(i) gives E[ XY ] E[Y ] E[Y 2] = 0;

Since the correlation coefficient is

=E E E[ ] [ ] [ ]

, XY X Y

X Y

we have E[ XY ] = X Y + X Y and (i) yields

Y Y Y

( )2 2 = XY + X Y .

(ii) gives + Y = X .

Solve the two simultaneous equations to get

=

X

Y

, = X

X

Y

Y


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Page 3

3 0 1 2 3 ( ) ( ) ( )

.t t t

A(t) =

F

H

GG

GG

I

K

J J

J J

( ) ( )

( ) ( )

( )

.

t t

t t

t

Forward equations:

t P (s, t) = P (s, t) A(t), t s.

tP

00(s, t) = (t) P

00(s, t) and P

00(s, s) = 1 imply that P

00(s, t) = exp( z

s

t (u)du) =

e

m(s,t)

.

For j > 0, we have

t P

0 j(s, t) = (t) P

0, j1(s, t) (t) P

0 j(s, t) with initial

condition P 0 j(s, s) = 0.

Verify that the form of P 0 j(s, t) given in the question satisfies this equation:

LHS = ( jm(s, t) j1 m(s, t) j)e

j tm s t

m s t ( , )

!( , ) ,

RHS = (t)m s t e

jt

m s t e

j

j m s t j m s t( , )

( )!( )

( , )

!

( , ) ( , )

1

1

The observation that

tm s t( , ) = (t) is sufficient to finish the verification.

4 (a) Z is stationary, i.e. I (0), as it is a first-order autoregression;

X is not stationary but X is just a linear combination of Z and e1, so is

stationary; this implies that X is I (1). The same goes for Y .

(b) Z satisfies the Markov property on its own; X and Y do not, since theydepend on values of Z .

(c) ( X , Y , Z ) is Markov; indeed, it is a vector autoregression.

(d) X and Y are not cointegrated. Although both are I (1), any linearcombination W =aX + bY satisfies W

n = W

n1 + W Z n1 + e3,n which does notdefine a stationary process.


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Page 4

5 (i) E[ ] Bt s

2F = E[( Bt Bs + Bs)2Fs]

= E[( Bt Bs)2 + 2( Bt Bs) Bs + Bs s

2F ]

= E[( Bt Bs)2

Fs] + 2 BsE[ Bt BsFs] + Bs

2,

by the property of conditional expectations which allows one to “take outwhat is known”. Moreover, by independence of the increments, the above is

E[( Bt Bs)2] + B

s

2 = t s + Bs

2 .

Similarly,

E[ ] Bt s

4F = E[( Bt Bs + Bs)4Fs]

= E[( Bt Bs)4

+ 4( Bt Bs)3

+ 6( Bt Bs)2

Bs

2

+ 4( Bt Bs) s B

Bs

3

+ Bs s4

F ]

= E[( Bt Bs)4] + 6 2 B

s E[( Bt Bs)

2] + Bs

4 ,

where we used the independence of increments property as well as the fact

that moments of odd order of N (0, 2) vanish. Finally

E[ ] Bt s

4F = Bs

4 + 6(t s) Bs

2 + 3(t s)2.

(ii) From above

E[ ] B tBt t s

4 26 F = Bs

4 + 6(t s) Bs

2 + 3(t s)2 6t(t s + Bs

2 )

= Bs

4 + 6 2sBs

+ 3(t s)2 6t(t s) = Bs

4 6 2sBs

+ 3(s2 t2)

B tB tt t

4 2 26 3 is a martingale.

6 (i) u = F 1(x) =

x

x1 is solved by x = F u

1

1 ( ) =u

u1 .

For the symmetrised version, the simplest thing is to multiply x by a

variable y which takes 1 depending on whether another pseudo-randomuniform number v is in the range (0, 0.5) or (0.5, 1).

(ii) By symmetry we only need consider x > 0, so we find maxx>0 2 1 2

2 2

( )

( ).

x

x

Differentiating the logarithm of this fraction and setting equal to 0, we get2

1 x =

22 2

x

x , with solution x = 2. Substituting this value in, we obtain

the required value of C .


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Page 5

Let g(x) = f x

Cf x

( )

( )

2

, which we observe is less than or equal to 1 everywhere.

The method of Acceptance-Rejection sampling goes as follows: use (i) togenerate a variable y from density f

2. We accept y as a valid observation

from f (x) with probability g( y), otherwise reject it. (Generate a uniformvariable u, and reject if u > g( y).) If we reject it, go back and generateanother y from f

2 , and continue to do the same until eventual acceptance.

7 (i) (a) 0 = Var(et + 1 et1) = ( )1 1

2 2 e and

1 = Cov(et + 1 et1 , et1+ 1et2)

= 1 e

2, with k = 0 for k > 1.

This gives 0 = 1,

1 =

1 / (1 +

1

2 ), k = 0 otherwise.

(b) Invertibility requires that 1 < 1, so that the sum X t 1 X t1 +

1

2

2 X

t ... converges. and e are irrelevant.

(ii) We need to solve ( )11

2 2 e = 14.5,

1

e

2 = 5.0. Eliminating e

2 , we

have 11

2 = 2.91 , or

1 = ½(2.9 2 9 42. ) = 2.5 or 0.4.

1 = 2.5 corresponds to

e

2 = 2, whereas 1 = 0.4 corresponds to

e

2 =

12.5.

For invertibility, solve 1 + 1z = 0. In the first case, z = 0.4 (no

good); in the second, z = 2.5 (OK).

8 (i) (a) States:

C : healthy contributor

C : contributor but ill B

1, B

2, B

3: beneficiary, with index giving duration of illness

(b) Transition

graph:

C

B1

B2

B3

C

0.1 0.8 0.8 0.8

0.8

0.2 0.2

0.9

0.2

0.2


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Page 6

(c) Transition matrix (states ordered C , C , B

1, B

2, B

3):

P =

0 9 0 01 0 0

0 8 0 0 2 0 0

0 8 0 0 0 2 0

0 8 0 0 0 0 2

0 8 0 2 0 0 0

. .

. .

. .

. .

. .

F

H

GG

GGG

I

K

J J

J J J

(ii) The chain is irreducible by inspection: every state is accessible from everyother state. State C is clearly aperiodic because of the one-step loop from C to C ; because of irreducibility, every other state must be aperiodic too.

(iii) (a) = P reads

c = 0.9c + 0.8(c + 1 + 2 + 3)

c = 0.23

1

= 0.1c + 0.2c

2

= 0.21

3

= 0.22 .

Discard first equation and choose c as working variable:

3 =1

02.c = 5c

2

=1

02.3 = 5

3 = 25c

= c(1248, 1, 125, 25, 5)

1 =1

02. 2 = 52 = 125c

c =1

01.1

02

01

.

.c = 101 2c = 1248c .

Find c by normalisation: c(1248 + 1 + 125 + 25 + 5) = 1,

c =1

1404.

(b) Proportion of beneficiaries: 125 25 51404

= 11.04%.


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Page 7

(iv) (a) Average profit per period per member in stationary régime is

Z = ( f c) 1248 1

1404

F H G

I K J

+ ( f b)125 25 5

1404

F H G

I K J

= f c 12491404

155

1404 b .

For Z > 0 you need f > c1249

1404

155

1404 b .

(b) With the given data

Z = 300 150 1249 600 155

1404

= 100.32.

9 (i) (a) 1 = Cov( X t , X t1) = Cov(1 X t1 + 2 X t2 + et , X t1) = 10 + 21+ 0, since

et is independent of X

t1.

(b) Similarly 2 =

11 +

20 and

0 =

11 +

22 + Cov( X t , et). A further

application of the same technique gives Cov( X t, e

t) =

e

2.

Thus 1 =

1

21

0 and

2 =

2

1

2

21

F

H GI

K J

0.

(c) k is found by the relation k = k / 0.

(ii) We have 1 = r

1( )1

2 and

2

1

2

21

= r2, which are solved by

1 =

r r

r

1 2

1

2

1

1

( ),

2 =

r r

r

2 1

2

1

21

.

10 (i) (a) Bt defined by following properties:

Independent increments:

Bt B

s independent of B

a , 0 a

s

whenever s t.

Stationary Gaussian increments: Bt Bs ~ N (0, t s).

Continuous sample paths: t Bt continuous.


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Page 8

Transition density to go from x at time s to y at time t:

gts( y x) =1

2

22

( ).( ) / ( )

t s

e y x t s

(b) {W s = x, W t = y} = { Bs + s = x, Bt + t = y} = { Bs =x s

,

Bt =

y t

}. Hence transition density of W is

1

gts y x t s F

H G I

K J

( ).

(ii) By Itô’s lemma

d(log S t) =

1 1

2

1

2

2

S dS

S dS

t

t

t

t

F

H G

I

K J ( )

= µdt + dBt 2

2dt.

Hence

log S t = log S 0 +

F

H GI

K J 2

2 t + Bt ,

and finally

S t = S et B

t

0

2

2

F

H GI

K J

.

(iii) P[S t > bS 0 = a] = P

B t b

at

F

H GI

K J

L

NMM

O

QPP

2

2log

= P B b

at

t

F

H GI

K J F

H G

I

K J

L

NMM

O

QPP

1

2

2

log

= 12

2

F

H GI

K J F

H

GGGG

I

K

J J J J

G

b

at

t

log

.


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Page 9

Here a = 38, b = 45, = 0.25, = 0.2, t = 13

year.

So, above quantity is

1 G(0.800) = 1 0.7881 = 0.2119.

(iv) P Pmax max log0

00

2

2

1

F

H GI

K J F

H G

I

K J

L

NMM

O

QPPs t

ss t

sS b S a B

s b

a

= G

t b

a

t

b

aG

t b

a

t

F

H GI

K J

F

H

GGGG

I

K

J J J J

R

ST

UVW

F

H GI

K J

F

H

GGGG

I

K

J J J J

2

2

2

2

2 2 2log

exp log

log

.

The first term is 0.2119 by (iii).

The second term is the product ofb

a

F H G I K J

2 2

2

= 6.9893 with G(2.128)

= 1 G(2.128) = 1 0.9833.

So the result is finally 0.2119 x 0.0167 = 0.3286.

11 (i) The generator matrix of the process would be

F

H

GGGGG

I

K

J J J J J

1 0 4 01 0 5 0

0

0 0 0

0 0 0

0 0 0 0 0

1

3

1

12

1

12

1

6

1

60

1

60

1

2

1

2

. . .

.

(ii) The probability of ever visiting state I is1

10

4

10

1

4 =

1

5.

(iii) (a) d

dt p t

AA( ) = p AA(t), which has solution p AA(t) = e

t.

(b) Similarly,d

dt p t

AF ( ) =

1

3 p AF (t) + 0.4 p AA(t), so that

d

dt{et/3 p AF (t)} = 0.4e

t/3 p AA(t) = 0.4e2t/3,

giving p AF (t) = et/3 0.6(1 e2t/3).


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Page 10

(iv) (a) The equation arises as follows: when the process arrives in state i the

subsequent holding time has mean i

1, after which the process

jumps to a different state, choosing state j with probability pij = ij / i(independent of the length of the holding time). The total time toreach state D is therefore the time until the first jump plus the time

from arriving in the new state until hitting D (unless the new state isD).

(b) We have m I = 60, mO = 2, m F = 3 +1

4 60 +

1

4 2 = 18.5,

m A = 1 + 0.1 60 + 0.5 2 + 0.4 18.5 = 15.4 hours.

(v) The time-homogeneous Markov model has exponential holding times, so thedistribution is completely determined by the expectation.

(vi) A simple check on whether the Markov model fits the data is therefore toverify that the distributions of holding times are at least roughlyexponential, and a simple way of doing that is to compare sample standarddeviations with sample means. More detailed comparisons might bepossible, depending on the size of the data set.

(vii) (a) Calculations required in the first case would include working out theexpected duration of stay if the change were implemented, whichinvolves solving the equations in (iv) again. For the second situation,

just replace mO in the original calculation. New parameter values

will need to be guessed. Whichever model comes out better should becompared with the initial situation, to determine whether theimprovement was worth the additional resources.

(b) Model suitability: on the one hand the required decision is couched interms of expectations, which lend themselves well to Markov processtreatment. On the other, the fundamental problem in the system isqueue length, which can never be successfully modelled by a processwhich tracks only a single individual at a time. (A network of queuing processes would be a much better model.)


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EXAMINATIONS

11 S ept ember 2000 (pm)


Ti me al l owed: T hr ee hours

I N S T R U C T I O N S T O T H E C A N D I D A T E

1 . W ri t e you r su rn a m e in f u l l , t h e in i t ia ls of you r ot h er n am es a n d you r

Cand i dat e’s N um ber on th e fr ont of the answer bookl et.

2 . M a rk a l loca t ion s a re sh own in b ra cket s.

3. Attempt all 11 questions, beginn i ng your answer t o each questi on on a

separate sheet.

G r a p h p a p er i s n o t r eq u i r ed f o r t h i s p a p er .

A T T H E E N D O F T H E E X AM I N A T I O N

H and in BOT H your answer bookl et and th i s questi on paper .

I n a d d i t i on t o t h i s pa p er you sh ou l d h a ve a va i l a b le Act u a r ia l

Tables and an electronic calculator.

Fa culty of Actua ries

103—S2000

Inst itut e of Actua ries


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103—2

1 There a re N individuals in a population, some of whom have a certain infectiontha t spreads a s follows. Conta cts betw een t wo members of this popula tion occur

in a ccordance with a P oisson process ha ving ra te λ . When a conta ct occurs, it is

equally likely to involve any of the P = N 2

e j

pairs of individuals in the population.

If a conta ct involves a n infected a nd a non-infected individua l, then w ith

probability p the noninfected individua l becomes infected. Once infected, a n

individual remains infected throughout. Let X (t ) denote th e num ber of infected

members of the populat ion a t t ime t .

(i) S t a te w het her X (t ), t > 0 is a cont inuous-time Ma rkov jump process. If so,

write down its state space and transition rates; i f not, explain how it can

be expressed in t erms of a different process w hich i s Ma r kov. [2]

(ii) Derive a n expression for the expected time until a l l members a re infected,

st a r t in g fr om a sin g le in fect ed in dividua l [2]

[Total 4]

2 During a long motorwa y journey a child a muses himself by noting down , a t theend of ea ch minute, the lane in which the ca r is tra vell ing. The motorwa y ha s

three lanes and the journey lasts N minutes.

(i) Describe how to f it a three-sta te t ime-homogeneous Ma rkov chain model

t o t h e d a t a , w r i t in g d ow n f orm u l ae f or t h e e st i m a t e s of t h e t r a n s it i on

pr oba bilit ies. [2]

(ii) Describe one test wh ich could be a pplied to determine w hether the process

possesses the Ma rkov property . [2]

[Total 4]

3 A sta nda rd Ornstein-U hlenbeck process ma y be defined a s a sta tiona ry zero-mean Gaussian process {U

t : t ∈ R} w ith a utocova riance function given by

Cov(U t , U

s ) =

τθ

θ2

2e t s − − (s , t ∈ R).

(i) S h ow t h a t t h e pr oces s {U n

: n = 1, 2, ...}, obtained by observing U only at

in t eger t im es, is a fir st -order a ut or eg ression . [2]

(i i) De r iv e e xp r es s ion s f or t h e p ar a m e t e rs α a n d σ2 of the autoregression interms of θ a n d τ2. [2]

[Total 4]


20/189


4 (i ) S t a t e I t ô ’s l em m a a s i t a p pl ie s t o a s t och a s t i c p r oce ss {X t : t ≥ 0} a nd a

function f (X t ) which is not explicitly dependent on t . [2]

(i i) App ly I t o ’s L e m ma w i t h f (x ) = x 4 to calcula te t he stocha stic dif ferentia l

d B t ( ),4 w h e r e B

t is st a n da r d B row n ia n m ot ion . [3]

(i ii ) H e n ce e xp r es s t h e I t ô i n t eg r al z 03t s s B d B in terms of B t a n d of a n or di n ar y

integral involving B s . [2]

[Total 7]

5 Consider a homogeneous Ma rkov chain with sta te space S = {1, 2, 3} a ndt r a n s it i on m a t r i x

P =

¼ ½ ¼

½ ½

¾ ¼

0

0

F

H

G

G

I

K

J

J

.

(i) C a lcula t e t h e 3-st ep t ra n sit ion m a t r ix. [2]

(ii) Ca lculat e, for each of the follow ing initia l conditions, the probabili ty tha t

the chain w ill be in sta te 3 when i t is observed a t t ime n = 3 g iv e n t h a t :

(a ) t h e ch a i n is in s t a t e 1 a t t im e z er o

(b ) t h e ch a i n is i n s t a t e 1 a t t im e z er o a n d in st a t e 2 a t t i m e 1

(c) t h e pr ob ab i li t ie s of b ei n g i n s t a t e s 1, 2, a n d 3 a t t i m e z er o a r e

given by 1431 931, a n d 831 respect ively [4]

(iii) How w ould your a nsw ers to (a ), (b) a nd (c) cha nge if the time of the

observat ion w ere n = 300 instea d of n = 3? [2]

[Total 8]


21/189

103—4

6 The da ily closing price of a sha re is observed every tr a ding day for a yea r,yielding a sequence of values {s 1 , . . . , s n }. A model is required for t he purposes of

predicting fut ure va ria bility of th e sha re price. The model suggest ed is a

Brownian one.

(i) Explain briefly, on purely theoretical grounds, wh ich of the two models

I : S t

= µ + αt + σB t

I I : log (S t ) = µ + αt + σB

t

y ou w ould expect t o provide a bet t er fit . [1]

(i i) Re fe r t o F i g u re s 1 a n d 2 b el ow .

(a ) E x p la i n b r ie fl y w h e t h e r y ou r ch os en m od el a p pe ar s t o p r ov id e a

g ood f it t o t h e d a t a .

(b) S t a te on e of the t ests you could ca rry out on the da ta to a scerta in

w h et h er t h e m odel fit s a deq ua t ely . [3]

(i ii ) (a ) De s cr i be h ow a Lé v y p r oce ss m ode l d if fe r s f r om a B r ow n i an m od el .

(b) Outline the dif ficulties would you encounter in pra ctice i f you were

fit t in g a L évy pr ocess m odel t o t he da t a provided. [3]

[Total 7]

Figure 1: the share price S n

Figure 2: log-transformed share price log S n


22/189


7 A client wishes to model th e beha viour of a st ocha stic process {X t : t ≥ 0} wh ich

represents t he avera ge annua l return for a part icular cla ss of asset. After a

number of observat ions th e client ha s determined th a t Corr(X t , X

t −1) = 0.7 a nd

Corr(X t , X

t −2) = 0.5. He thinks tha t one of the tw o models

I : X t = µ + 0.7(X t −1 − µ) + 0.5(X t −2 − µ) + e t

I I : X t = µ + e t + 0 .7e t −1 + 0 .5e t −2

w ill be best, but ca nnot decide w hich. He ha s simulat ed both processes from t ime

t = 1 t o t i m e t = 200, but has not obtained the results he expected, so is seeking

your advice.

(i ) (a ) O u t lin e a s ui t a b le m e t h od of s im u la t in g a s econ d -or d er

a utoregression, a ssuming you have a ccess to a reliable strea m

{u k : k ≥ 0} of pseudo-ra ndom numbers uniformly dist ributed over

th e ra nge [0, 1].

(b ) E x pla in w h y m ig h t it b e d es ir a b l e t o e ns u re t h a t t h e s t r ea m {u k }ca n be re-used if n ecessa r y . [4]

(ii) S t a t e w hy n eit her of t he sugg est ed m odels is suit a ble. [1]

(i ii ) (a ) D e r iv e t h e l a g -1 a n d la g -2 a u t ocor r el a t i on s , ρ1 a n d ρ2, of a second-order a utoregressive process

X t = µ + α1 (X t −1 − µ) + α2 (X t −2 − µ) + e t .

(b ) F i nd v a l u es of t h e pa r a m et e r s α1 a n d α2 wh ich w ould provide asuitable AR(2) model for {X

t : t = 0, 1, 2, … }. [5]

[Total 10]


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103—6

8 Consider a time-homogeneous Ma rkov jump process {X (t ) : t ≥ 0} w i t h t w o s t a t e sdenoted by 0, 1, and transition rates σ0,1 = λ , σ1,0 = µ.

(i ) S t a t e Kolm og or ov ’s f or w a r d e q u at i on f or t h e p r ob ab i li t y P 0,0(t ) t h a t X is in

s t a t e 0 a t t i m e t , given t h a t it st a r t s in st a t e 0. [1]

(ii) S how t ha t P 0,0(t ) = µ

λ µ

λ

λ µ

λ µ

+

+

+

− +e t ( ) . [3]

(iii) L et O t denote th e tota l am ount of t ime spent in sta te 0 up until t ime t ,

w hich ma y be expressed a s O t = z 0

t s I d s , w h e r e I s =

1 if 0

0 if 0

s

s

X

X

=

≠

. D erive,

using t he result in pa rt (ii), a n expression for E[O t X (0) = 0], the expected

occupat ion t ime in st a te 0 by time t for th e tw o-sta te cont inuous-time

Ma rkov ch a in st a rt ing in st a t e 0. [2]

(iv) Write down t he expected occupat ion time in sta te 1 by time t for t he tw o-

st a t e con tin uous-t im e Ma r kov ch a in st a r tin g in st a t e 0. [1]

(v ) A h e a l t h i n su r a n c e s ch e m e l a b el s m e m be r s a s “h e a l t h y” (state 0) or

“u n h e a l t h y” (s ta te 1) a t a ny t ime. When in stat e 0, members pay

contributions at rate α; wh en in sta te 1 they receive benefit a t r a te β.Expenses amount t o a consta nt γ per member per unit time.

(a ) E x p la i n h ow t h e a b o v e m od el ca n b e u s ed t o ca l cu l a t e α in t ermsof β a n d γ .

(b ) S t a t e t h e a s s u m pt i on s w h i ch y ou m a ke i n a p p ly i n g t h e m od el .

(c) D i scu ss w h e t h er t h ey a r e li kely t o b e s a t i sf ied in pr a ct i ce. [4]

[Tot a l 11]

9 Su ppose th a t th e evolution of th e price of a n asset follow s the lognorma l model

log(S t ) = Y

t = y + µt + αB

t w h e re B

t denotes the standard Brownian motion and µ

is a nega tive drift . The asset w ill be liquidat ed at the stopping t ime

T α = inf{t : Y t = a } when its value reduces to e a , where a is some number less than

y . Consider now the exponential V t = exp(u Y

t − c (u )t ).

(i ) D e riv e t h e con d it i on on c (u ) under w hich {V t : t ≥ 0} is a m a r t in g a le. [3]

(ii) S t a t e t h e opt ion a l s t oppin g t h eor em a n d ex pla i n h ow it is u sed . [3]

(iii) Derive the moment genera ting function f (y , v ) = [ u vT

e −

E Y 0 = y ] of t heba nkruptcy t ime for positive v by a pplying t he optiona l stopping t heorem

t o t h e m a r t i n ga l e V t . [5]

[Tot a l 11]


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10 Consider a surviva l model wit h tw o sta tes “alive” (A ) a n d “dead” (D ), w ith t ime-

dependent tra nsition r a te from A t o D e q u a l t o µ(t ) = µt . The time para meter, t ,represents the age of the individual under consideration.

(i ) C a l cu la t e t h e t r a n s it i on pr ob a b ili t y P AA

(s , t ), defined by

P AA

(s , t ) = P(X (t ) = AX (s ) = A ). [2]

(i i) S h o w , b y m a kin g u s e of t h e f or m u la

E(X ) = z ∞0 P[X ≥ x ] d x

for a positive random variable X , that the expected future lifespan of an

individual a ged s is

E[R s ] =

1 ( )1

( )

G s

g s

− µ

µ µ ,

w h e r e G is the sta nda rd G a ussia n proba bility distribution function a nd g

is it s den sit y . [4]

(iii) I t is desired to ca libra te the above model so tha t t he expected future

lifespa n of an individual a ged 70 is 6 years. Derive a n a pproxima tion t o

th e corresponding va lue of µ, using the double inequality

1 1 1 13x x

G x

g x x − ≤

−≤

( )

( ). [5]

(iv) A company w ishes to test the va lidity of the above model. They assumetha t the t rue force of morta lity from a ge 70 onw a rds is of the form

µ(t ) = a + bt and intend to test whether a = 0. The testing method will beto simula te one sa mple of size 1000 when a = 0 a n d a n o t h e r w h en a ≠ 0,then t o see which most resembles th e dat a w hich t he company ha s

collected.

Explain how to simulate a value from the proposed distribution, for

arbitrary values of a a n d b . [5]

[Tot a l 16]


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103—8

11 The movement s of a consum er price index ar e t o be subjected to time series

a na lysis w ith t he a im of foreca sting future beha viour. The index is ca lculat ed

monthly.

(i) Expla in wh ether you would expect to f it a model wh ich included (a ) a

t ren d t erm , (b) a sea son a l effect . [3]

The values {x t : 1 ≤ t ≤ n } are th e residuals wh ich rema in once a ny trend or

seasona l va ria tions ha ve been r emoved. An ARIMA(1, 1, 1) model is to be fitt ed

t o t h e {x t }.

(i i) (a ) As su m i n g t h e AR I M A(1, 1 , 1) m od el i s cor r ect , w r i t e d ow n a n

equation for X n + 1

in t erms of th e w hite n oise process

{e t : 1 ≤ t ≤ n + 1} a nd th e observa tions {x

t : 1 ≤ t ≤ n }.

(b) S t a t e t h e pa ra m et er s of t h e m odel. [2]

(iii) The B ox-J enkins procedure defines th e k -step-ahead forecast for X to be

( )x k n = E(X n + k x n , x n −1 , . .., x 1).

(a ) Derive the 1-step-a head and 2-step-a head foreca sts for X for the

ARIMA(1, 1, 1) model, assuming that the values of the parameters

and the value of e 0

are known exa ctly.

(b ) E v a l u a t e t h e pr ed ict i on v a r i a n ce Va r (X n + 1

− ˆ (1))n x , a g a i n a s s u m in gt ha t t he va lues of t he pa ra met er s a re kn ow n. [5]

(iv) The most elementa ry form of the technique know n as exponential

smoothing produces at time n a 1-step-ahead forecast x n

* defined by

x n

* = x n

+ ξ( ),*x x n n − −1

for some ξ ∈ (0, 1) w hich m a y be chosen by th e user.

Show tha t, for part icular va lues of the a utoregressive and moving a verage

par a meters, t he B ox-J enkins foreca sts a bove coincide w ith th e foreca sts

pr oduced by expon en t ia l sm oot h in g. [2]

(v ) (a ) S t a t e w h e t h er t h e AR I M A(1, 1, 1) m od el i s I (0), I (1) or neith er.

(b ) D i s cu s s w h e t h e r t h e r e i s a d if fe re n ce b et w e en a n I (0) model a nd a n

I (1) model in terms of the conditional distribution of X n + k

given {x t :

1 ≤ t ≤ n } for lar ge va lues of k . [3]

(v i) I t i s s u g ge st e d t h a t a s a l a r i es in d ex m i gh t b e coi n t eg r a t e d w i t h t h e

consu mer price index.

(a ) E x pla i n w h a t is m ea n t by t h e s ug g es t ion .

(b) C om men t on w het her it is a rea son a ble suggest ion . [3]

[Tot a l 18]


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EXAMINATIONS

S eptember 2000


EX AM INERS ’ RE P O RT

Faculty of Actuaries

Institute of Actuaries


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Subject 103 (Stochastic Modelling) — September 2000 — Examiners’ Report

Page 2

1 (i) This is clea rly a Markovian birth process. The sta te spa ce is {0, 1, . . ., N }.

G i v en t h a t w e h a v e m infected and N − m hea lthy individuals , thenumber of “dangerous” pair contacts is m (N − m ). Thus, the rate

σm ,m +1 = λ p ( )

.

m N m

P

−

(ii) The expected total infection time is the sum of the reciproca l rates

1

1

1.

( )

N

m

P

p m N m

−

=λ −

(Since 1

( )m N m − =

1 1 1,

N m N m

+ −

th is time ma y a lso be expressed a s

1

1

1 1.

N

m

N

p m

−

=

−λ

)

2 (i) L et N i j

denote th e number of minutes when t he ca r w a s in lane i a t t h e

s ta r t of th e mi n ute a n d i n la n e j a t t he end. The estima te of the

tra nsition probability p i j

is N i j

/N i + , where N i + = Σ j N i j .

(i i) The problem here is the alterna tive hypothesis. I t w ould be possible to

test w hether t he distribution of X n + 1 conditional on X n = i w a s r e a l ly

independent of X n −1. Or one might t est, using a st a nda rd goodness-of-fit

test, whether the distribution of the number of consecutive minutes spent

in lane i rea lly w a s geometrica lly distributed with para meter determinedby i .

3 (i) S ince {U n

} is Ga ussia n a nd sta tiona ry, i t is determined uniquely by its

mean a nd a utocova riance functions. For k > 0 w e h a v e γ k

= Cov(U n

, U n −k )

=2

2

τθ

e −θk , so tha t t he ACF is ρk

= e −θk a n d t h e v a r i a n ce γ 0 =2

2

τθ

.

(i i) Compa re this wit h the corresponding va lues for an AR(1): γ 0 =2

21

σ

− α a n d

ρk

= αk for k > 0.

The tw o are seen t o mat ch a s long a s α = e −θ a n d σ2 = (1 − e −2θ)2

.2

τθ


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Subject 103 (Stochastic Modelling) — September 2000 — Examiners’ Report

Page 3

4 (i) I f X satisfies d X t = Y

t d B

t + Z

t d t , then f (X

t ) sa tisfies

d [f (X t )] = f ′(X

t ) Y

t d B

t + { f ′(X

t ) Z

t + ½ f ′′(X

t ) 2 }t Y d t .

(ii) (x 4)′ = 4x 3 , (x 4)′′ = 12x 2.

4( )t d B = 34 t B d B t +

12

. 212 t B d t = 34 t B d B t +

26 .t B d t

(iii) 40 ( )t

s d B = 3 20 04 6 .t t

s s s B d B B d s +

Therefore 30t

s s B d B = 4 231

04 2 .t

t s B B d s −

5 (i) P =

1 2 11

2 0 2 ,43 1 0

P 2

=

8 3 51

8 6 2165 6 5

, P 3

=

29 21 141

26 18 20 .6432 15 17

(ii) (a ) 313P = 14

64=

7

32= 0.21875.

(b) 223P = 2

16=

1

8= 0.125.

(c) 3 3 313 23 3314 9 831 31 31

P P P + + = 14 14 9 20 8 1731 64× + × + ×× = 51231 64× = 8 .31

(iii) B y t im e n = 300 the effects of the sta rt ing point ha ve worn off. The

a nswer is therefore indistinguishable from the sta tiona ry proba bility π3

i n

all three cases.

It is ea sily observed t ha t the distribution in (c) is s ta tionary, so tha t

π3 =

8

31.

6 (i) The daily cha nge in va lue of a sha re is genera lly on a sca le consistent

with the va lue of the sha re: this tends to indica te th a t model II is

preferable.

(i i) (a ) M od el I I d oes a p pea r t o f it b et t e r t h a n m od el I ; t h e S dataset does

indeed exhibit la rge varia tions w hen it is a t a high level, a nd

smaller ones when low.

However, the f i t does not a ppear a ll tha t good, as B rown ian

increments a re normally distributed, so are seldom a s large a ssome of the jumps which a ppear in t his dat a set.


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Subject 103 (Stochastic Modelling) — September 2000 — Examiners ’ Report

Page 4

(b ) I f t h e B r o w n i a n m od el w e r e a c cu r a t e, t h e d a y -t o -d a y i n cr em e n t s

s n − s

n −1 should be independently normally distributed with

constant mean and variance:

a test of normality (Anderson-Darling, Kolmogorov-Smirnov, χ2

goodness-of-fit) w ould do fine; a t est of independence (ba sed onsa mple ACF , or the D urbin-Wa tson st a tist ic) w ould a lso be a good

suggestion.

(iii) (a ) A L évy process is th e sum of three independent component s: a

deterministic part of the form µ + αt , a Brownian part of the formσB

t a nd a pure jump part w hich ma y be thought of as a compound

P oisson process.

(b) One problem would be in estima ting the distribution of the jump

sizes, pa rt icula rly w ith only 250 observa tions. Even if a fa mily

w ere a ssumed for t he distribut ion (e.g. double exponent ial), t herewould be the a dditiona l diff iculty tha t sma ll jumps would not be

detecta ble a ga inst th e background of the G a ussia n noise.

7 (i) (a ) F ir st t h e u k need to be tra nsformed so tha t their distribution is

someth ing suita ble for t he w hite noise sequence of a t ime series,

since at the very least the mean of the sequence needs to be zero.2(0, )e N σ is th e sta nda rd choice: one meth od of achieving this is to

define, for each integer t ,

e 2t

= 2 2 12 log s in (2 )e t t u u +σ − π

e 2t + 1

= 2 2 12 log cos (2 ),e t t u u +σ − π

but there a re others, such a s th e polar method, inverse tra nsform

method or acceptance-rejection sampling.

The va lues of t he e t can now be fed into the formula to give the

values of the X t , whichever model is in use.

(b) The a bility to re-use a pseudo-ra ndom number sequence is

importa nt wh en comparing t he a bility of different m echa nisms to

control a process w hich is a ffected by ra ndomness: in order t o

ensure fair compa rison of the m echa nisms, t he must be subjected

to t he sa me degree of “random ” input.

(ii) The models do not possess th e correct correla tion stru cture.


30/189


Page 5

(iii) (a ) ρ1

= Corr(X t ,X

t −1) = α1Corr(X t −1 , X t −1) + α2Corr (X t −2 , X t −1) = α1 + α2 ρ1 .

Hence ρ1

= α1

/(1 − α2)

ρ2

= Corr(X t ,X

t −2) = α1Corr(X t −1 , X t −2) + α2Corr(X t −2 , X t −2) = α1 ρ1 + α2 .

(b ) We h a v e 0.7 = ρ1

= α1

/(1 − α2)

a nd 0.5 = ρ2

= α2

+ 21α /(1 − α2) = α2 + 0.7α1. Two equa tions in tw o

unknown s. Solution: α1

= 35

,51

α2

= 1

.51

(2 ma rks for t he observa tion t ha t α1

= 0 .7 a n d α2

= 0 is very close

to giving t he right a nswer, a s i t gives ρ2

= 0.49.)

8 (i) 0,0 ( )P t ′ = µP 0,1(t ) − λ P 0,0(t ), or a more genera l form such a s 0,0 ( )P t ′ =ΣP

0,k(t )σ

k,0

(ii) S in ce P 0,1

(t ) = 1 − P 0,0

(t ),

w e h a v e 0, 0 ( )P t ′ = µ(1 − P 0,0(t )) − λ P 0,0(t ). Any solution method w ill do,

e.g. d

d t [e (λ + µ)t P

0,0(t )] = µe (λ + µ)t , solved by P

0,0(t ) =

µλ + µ

+ Ce −(λ + µ)t , w i t h C

being determined by th e fa ct tha t P 0,0

(0) = 1.

(iii) E0

O t

= E0 0

t I

s d s = 0

t E

0 I

s d s = 0

t P

0,0(s )d s

=2( )

t µ λ

+λ + µ λ + µ

(1 − e −(λ + µ)t )

(iv) Since the process must be in sta te 0 or sta te 1 a t a ll t imes, the solution is

just t − E0

0t =

2( )

t λ λ

−λ + µ λ + µ

(1 − e −(λ + µ)t ).

(v ) (a ) As su m i n g a m e m be r w h o i s i n it i a l l y h e a l t h y , e xp ect e d ou t g oi n gs

(including expenses) by time t a nd expected income by time t , a r e

respectively

γ t + β ( )2

(1 )( )

t t e

− λ+µ λ λ − − λ + µ λ + µ

a n d ( )2

(1 ) .

( )

t t e

− λ+µ µ λ α + −

λ + µ λ + µ


31/189


Page 6

In the long run, then, as t → ∞, w e require αµ = βλ + γ (λ + µ) t obreak even.

(b ) Th e a s s u m pt i on s re q u ir ed a r e t h a t t h e r a t e of b ecom i n g i ll a n d

ra te of recovery from illness are const a nt .

(c) This w ill certa inly not be true of a ny individual member but, i f

membership is large a nd t he a ge an d hea lth profiles of th e

members are constant by virtue of a constant influx of new

members, i t ma y be a reasonable approxima tion.

9 (i) I f V t is a ma rtinga le, then its expecta tion must be consta nt a nd equa l to

its initia l value e uy .

Therefore ( ) ( )t u y t B c u t e

+µ +α −E =

2 2( /2 ( ))u y u u c u t e

+ µ+ α − = e u y .

Thus w e must ha ve c (u ) = u µ + u 2α2 /2.

(ii) The optiona l s topping theorem sta tes tha t i f M t i s a m a r t i n g a l e, a n d T is a

random stopping time, then under some additional technical conditions

(such as M t ∧T being uniformly bounded) w e ha ve:

EM T

= M 0.

It is frequently used to evaluate the expectation of a function of T , such a s

the moment generating function (as in this instance).

(iii) Applying the optiona l s topping theorem to the mart ingale V t w e f in d t h a t

a T V E = e uy =

( ).a

u a c u T e

−E

The equa tion c (u ) = v h a s t w o r oot s u + , u − , one being nega tive and the

other positive (since v is positive).

Now a t T

V ∧ = ( ) t u v Y vt e

−a n d Y

t ≥ a for 0 ≤ t ≤ T

a . I f u (v ) < 0, then

0 < ( )

a

u v a

t T V e ∧ ≤ for a ll t , so that the technical condition is satisfied; thesam e ca nnot be said if u (v ) > 0.

Therefore the positive root is unacceptable and f (y , v ) = a vT

y e −

E = ( )( )u v y a

e − −

.

Comm ent: For the record , th er e were two ver y sl ight err or s in th is questi on, both

appeari ng as subscri pts. I n l i ne 4, f i r st form ul a: { }T α should have read { }a T , a n d i n pa r t

( i i i ) l i n e 1: { }u T shoul d have r ead { }a T . T hi s was taken in to account by the m ark ers, and

th e exami ner s ensur ed th at no m ar ks w er e lost by stud ent s because of eit her sm al l er r or.


32/189


Page 7

10 (i) P AA

(s , t ) =t s xd x e

−

µ=

2 2( ) /2t s e

−µ −

(ii) P[R s ≥ w ] = P

AA (s , s + w ) =

2 2(( ) ) /2s w s e

−µ + − =2 /2

.sw w

e −µ −µ Therefore

E[R s ] =

2/2

0 .sw w

e d w ∞ −µ −µ

Complete t he squa re a t the exponent to get

E[R s ] =

2 2/2 ( ) /2

0s s w

e e d w µ ∞ −µ +

=2 2

/2 /2s x

s

d x e e

µ ∞ −µ µ

= 1 ( )1

.( )

G s

g s

− µ

µ µ

(iii) From (ii) a nd the given bound

1 1[ ]s R

s ≤ µ µE = 1

s µ

3 3 /2

1 1 1[ ]s R

s s

≥ −

µµ µ

E =3 2

1 1.

s s −

µ µ

The first ineq ua lity y ields

µ ≤ 1

[ ]s s R E=

1

420= 0.00238 . (year)−2.

The second inequality can be written as

µ2E[R s ]

3

10,

s s

µ− + ≥

so µ must lie outside the interval:

2 3

4 [ ]1 1

2 [ ]

s

s

R

s s s

R

± − E

E=

4 [ ]1 1

2 [ ]

s

s

R

s

s R

± − E

E=

241 1

70

2 6 70

± −

× ×

= [0.00023, 0.00215]

In fa ct, since clea rly µ 1

[ ]s s R −

E

w e s e e t h a t µ must l ie in t he interval

[0.00215, 0.00238].

(i v) Us e t h e i n ve rs e t r a n s for m m e t h od , X = F −1(U ).

In t his ca se 1 − F (x ) = 70exp( [ ] )x

a bt d t −

+ = exp{−a (x − 70) − ½ b (x 2 − 702)}.


33/189


Page 8

Therefore ½ bx 2 + ax = −log(1 − F (x )) + ½ 702b + 70a . Replace F (x ) by u t oget

x = F −1(u ) =

2 22 [ log(1 ) ½ 70 70 ]a a b u b a

b

− + + − − + +

= −r + 2 2 170 140 2 log(1 ),r r b u −+ + − −

w h e r e r = ab −1. I f u is an observation of a uniform pseudo-random variate,

then x is an observation from the required distribution.

11 (i) C on su m er pr ices d o tend t o exhibit regular sea sona l va riat ion, t hough n ot

a great deal these days. And, s ince prices tend to go up rat her more tha n

they come down, it is probably worth including a trend term in any model.

It is certa inly possible to test w hether th e trend t erm is equal t o zero.

(ii) (a ) X n + 1 − x n = α(x n − x n −1) + e n + 1 + βe n .

(b ) Th e pa r a m e t er s a r e α, β a n d 2 .e σ The t rend remova l process w ouldha ve a ccounted for a ny µ p a r a m e t er .

(iii) ˆ (1)n x = E(X n + 1x n , . .., x 1) = x n + α(x n − x n −1) + E(e n + 1 + βe n x n , . .., x 1). Now e

n + 1 ha s mea n 0 a nd is conventiona lly supposed independent of everyt hing

tha t ha ppens before n .

On the other hand, e n

ca n be deduced from pa st da ta ,

e.g. e n

= x n − x

n −1 − α(x n −1 − x n −2) − βe n −1 , which ma y be iterat ed back to gete

n in t erms of the known x a n d t h e kn ow n e

0.

Thus

ˆ (1)n x = x n + α(x n − x n −1) + βe n .

Similarly,

ˆ (2)n x = E(X n + 2Fn ) = E(X n + 1 + α(X n + 1 − x n ) + e n + 2 + βe n + 1Fn )

= (1 + α) ˆ (1)n x − αx n .

We see tha t X n + 1 − ˆ (1)n x = e n + 1 , so that the prediction variance is just

Var(e n + 1

) = 2 .e σ


34/189


Page 9

(iv ) S in ce e n

= x n − 1ˆ (1)n x − , w e h a v e

ˆ (1)n x = x n + α(x n − x n −1) + β(x n − 1ˆ (1));n x −

if we set α = 0 a n d β = −ξ, the equa tion is identica l to the updat ingequa tion for exponent ial smoothing.

(v ) An AR IM A(p , d , q ) model is I (d ); in t his case, x is I (1).

A st a t i on a r y (I (0)) model ha s a n equilibrium distribut ion: t he distr ibution

of the forecast of X n + k would converge t o equilibrium for lar ge k . An I (1)

process is the partial sum of an I (0) process, so would have increasing

va riance, even if th e mean ha ppened t o be stable.

(v i) Tw o s er ies {x } a n d {y } are cointegrat ed if both a re I (1) but th ere a re some

constants a a n d b such t ha t {a x + by } is s tat iona ry.

Two processes are likely to be cointegrated if one drives the other, or if

both a re driven by the sa me underlying process. In t he given insta nce the

suggestion is certainly worth investigating.


35/189


EXAMINATIONS

10 April 2001 (pm)


Time allowed: Three hours

INSTRUCTIONS TO THE CANDIDATE

1. Write your surname in full, the initials of your other names and your

Candidate’s Number on the front of the answer booklet.

2. Mark allocations are shown in brackets.

3. Attempt all 10 questions, beginning your answer to each question on a

separate sheet.

Graph paper is not required for this paper.

AT THE END OF THE EXAMINATION

Hand in BOTH your answer booklet and this question paper.

In addition to this paper you should have available

Actuarial Tables and an electronic calculator.


103—A2001 Institute of Actuaries


36/189

103 A2001—2

1 {N (t) : t ≥ 0} is a Poisson process with rate λ and { t : t ≥ 0} is the filtrationassociated with N .

(i) Write down the conditional distribution of N (t + s) − N (t) given t , where

s > 0 and use your answer to find E (θN (t+s) t). [3]

(ii) Find a process of the form M (t) = η(t)θN (t) which is a martingale. [2][Total 5]

2 An insurance company wishes to test the assumption that claims of a particulartype arrive according to a Poisson process model. The times of arrival of the next

20 incoming claims of this type are to be recorded, giving a sequence T 1 , …, T 20.

(i) Give reasons why tests for the goodness of fit should be based on the inter-

arrival times X i = T i − T i−1 rather than on the arrival times T i . [1]

(ii) Write down the distribution of the inter-arrival times if the Poissonprocess model is correct and state one statistical test which could be

applied to determine whether this distribution is realised in practice. [2]

(iii) State the relationship between successive values of the inter-arrival times

if the Poisson process model is correct and state one method which could

be applied to determine whether this relationship holds in practice. [2]

[Total 5]

3 (i) Give a definition of the spectral density of a stationary time series,

expressed in terms of the autocovariance function {γ k : k ∈ Z } of theprocess. Use this definition to derive the spectral density of a first-ordermoving average process and of a first-order autoregression. [5]

(ii) Suppose the “inverse” of a time series model with spectral density f (ω) is

defined to be the model with spectral density1

( ) f ω. Using part (i), state

the form of the inverse of a first-order moving average and state the way

in which the inverse of an invertible MA(1) differs from the inverse of a

non-invertible MA(1). [2]

[Total 7]


37/189

103 A2001—3 PLEASE TURN OVER

4 Let X n denote an autoregressive high frequency time series modelled by:

X n+1 = (1 − α) X n + θ + τen ,

where en = ±1 with equal probabilities and α, θ, τ are constant parameters. Ananalyst wishes to investigate whether this series may be approximated by some

continuous time diffusion, i.e. X n ≈ Y nh , where Y t satisfies a stochastic differential

equation

dY t = µ(Y t) dt + dBt

and Bt denotes standard Brownian motion.

(i) State the expectation and variance of dY t = Y t+h − Y t , the increment of theprocess Y t over a small interval of size h, conditional on Y t = y. [2]

(ii) Calculate the expectation and variance of the increment X n+1 − X n of the

autoregression, conditional on X n = y. [2]

(iii) Find, by equating the first and second moments of the increments in (i)

and (ii) above, an expression for the drift µ( y) of the approximatingdiffusion in a form which does not involve the time increment h. [2]

(iv) State a condition under which the approximating process in (iii) is a

Brownian motion with drift. [1]

(v) State a condition under which the approximating process in (iii) is an

Ornstein-Uhlenbeck process. [1]

[Total 8]

5 (i) Derive expressions for ρ1 and ρ2 , the autocorrelation function of X at lags1 and 2, in the case that X is a stationary process satisfying the recursion:

X t = α X t−1 + et + βet−1 ,

where {et : t = 1, 2, …} is a sequence of uncorrelated random variables with

mean 0, variance σ2. [5]

(ii) A company’s monthly sales figures, corrected for trend and seasonalfactors, exhibit sample autocorrelation function at lags 1 and 2 of r1 = 0.5,

r2 = 0.4. Find method of moments estimators of α and β for the modelin (i). [3]

[Total 8]


38/189

103 A2001—4

6 A motor insurance company has 80,000 policy holders, paying an average annualpremium of £400. The company receives claims at a rate of 2000 per month, the

sizes of the claims having mean £1,200, standard deviation σ = £200.

Let S (t) denote the company’s total surplus at time t, with S (0) equal to the

initial reserve, set at £20,000,000.

(i) Calculate the expectation of the total amount paid out in claims in a given

month and the safety loading employed by the company. [2]

(ii) State, with reasons, whether it would be appropriate to use a diffusion

approximation to calculate the probability of ruin, that is the probability

that the process {S (t) : t ≥ 0} ever hits 0. [3]

(iii) Describe a simulation-based method for estimating the probability of ruin.

Indicate why it would be important to use a method of generating pseudo-

random variables which gives rise to reproducible sequences. [4]

[Total 9]

7 The evolution of a stock price S t is modelled by

S t = ,tt Be

µ +σ

where Bt represents a standard Brownian motion, µ and σ are fixed parametersand the initial value of the stock is S 0 = 1.

(i) Derive an expression for P {S t ≤ x }. [2]

(ii) Derive expressions for the median of S t and the expectation of S t . [4]

(iii) (a) Determine an expression for the conditional expectation E (S tF s),where s < t and where {F s : s ≥ 0} denotes the filtration associatedwith the process S .

(b) Find conditions on µ and σ under which the process {S t : t ≥ 0} is amartingale.

(c) State, with reasons, whether or not the stock would be a good long

term investment in this case.

[5]

[Total 11]


39/189

103 A2001—5 PLEASE TURN OVER

8 A company assesses the credit-worthiness of various firms every quarter; theratings are, in order of decreasing merit, A, B, C and D (default). Historical data

support the view that the credit rating of a typical firm evolves as a Markov

chain with transition matrix

P =

2 2

2 2

2 2

1 0

1 2

1 20 0 0 1

− α − α α α

α − α − α α α

α α − α − α α

for some parameter α

(i) Draw the transition graph of the chain. [2]

(ii) Determine the range of values of α for which the matrix P is a validtransition matrix. [2]

(iii) State, with reasons, whether the chain is irreducible and aperiodic.

[2]

(iv) Derive a stationary probability distribution for the chain and establish

whether it is unique. [4]

(v) For the value α = 0.1, calculate the probability that the company’s ratingin the third quarter, X 3, is in the default state D:

(a) in the case where the company’s rating in the first quarter, X 1, is

equal to A

(b) in the case X 1 = B

(c) in the case X 1 = C

(d) in the case X 1 = D

[3]

[Total 13]


40/189

103 A2001—6

9 A continuous-time Markov sickness and death model has four states: H (healthy),S (sick), T (terminally ill) and D (dead). From a healthy state transitions are

possible to states S and D, each at rate 0.05 per year. A sick person recovers his

health at rate 1.0 per year; other possible transitions are to D and T , each with

rate 0.1 per year. Only one transition is possible from the terminally ill state,

and that is to state D with transition rate 0.4 per year.

(i) Draw the transition graph for this process. [2]

(ii) Define P (t) = { pij(t) : i, j ∈ H, S, T, D} where pij(t) denotes the probability of being in state j at time t given that the individual was in state i at time 0.

State the Kolmogorov forward equation satisfied by the matrix P (t),

making sure that you specify the entries of the matrix A which appears.

[3]

(iii) Calculate the probability of being healthy for at least 10 uninterrupted

years given that you are healthy now. [1]

(iv) Let d j denote the probability that a life which is currently in state j will

never suffer a terminal illness. By considering the first transition from

state H , show that dH =1 12 2

+ dS and deduce similarly that dS =51

12 6 H d+ .

Hence evaluate dH and dS . [5]

(v) Write down the expected duration of a terminal illness, starting from the

moment of the first transition into state T . Use the result of (iv) to deduce

the expectation of the future time spent terminally ill by an individual

who is currently healthy. [4]

[Total 15]

10 A family agrees an expenditure target, Y n , for year n, in such a way that theannual increase in the expenditure target is proportional to the increase in the

family income over the previous year. The actual expenditure during the year,

X n , is assumed to be related to the expenditure target, but incorporating an

element of randomness and a factor accounting for the family’s propensity to

overspend. The family income, I n , is assumed to grow at a constant annual rate,

before randomness is taken into account.

The head of the household believes that the following three equations form an

appropriate representation of the above information:

Y n = Y n−1 + β(I n−1 − I n−2)

X n = (1 + π) Y n +(1)ne

I n = (1 + α) I n−1 +(2 )ne

where (1) (2){( , ) :n ne e n = 1, 2, …} is a sequence of zero-mean bivariate Normal

random variables and α, β and π are positive parameters (with β < 1).


41/189

103 A2001—7

(i) Express the first of the equations in terms of the backshift operator, B,

and deduce that a linear relationship exists between Y n and I n−1. [3]

(ii) Show that the process Zn = ( X n , I n) is a first-order multivariate

autoregressive process. [2]

(iii) State, with reasons, whether {I n : n ≥ 1} is a stationary time series, andhence determine whether

{Z

n : n

≥ 1} is I (0), I (1) or neither. [3]

(iv) Find an estimator for the parameter α by minimising the quantity

=Σ(2) 2

2( )nt te . [3]

(v) The head of the household wishes to perform a simulation to investigate

whether the propensity to overspend will result in negative net savings.

It is assumed that (1) Var( )ne =2 (2)1 , Var( )neσ =

22σ and

(1) (2)Cov( , )n ne e = ρσ1σ2 , where −1 < ρ < 1.

(a) Describe a method of simulating an observation of the pair(1) (2)( , )n ne e starting from two uniformly distributed pseudo-random

variables U 1 , U 2 .

(b) Describe the role of sensitivity analysis in drawing conclusions

from the simulation. [6]

(vi) An alternative model is proposed, involving the logarithms of the

quantities I n , X n and Y n :

ln Y n = ln Y n−1 + ln I n−1 − ln I n−2

ln X n = θ + ln Y n +(1)ne

ln I n = φ + ln I n−1 +(2 )ne

Discuss whether this model is more suitable than the original model. [2]

[Total 19]


42/189


EXAMINATIONS

April 2001


EXAMINERS’ REPORT


Institute of Actuaries


43/189


Page 2

1 (i) Given t we know that N (t + s) − N (t) ~ Poisson(λ s).

Hence E (θN (t+s) t) = θN (t) e(θ−1)λ s.

(ii) Now E (η(t + s) θN (t+s)

t) = η(t + s) θN (t)

e(θ−1)λ s

, which needs to be equal toM (t) = η(t) θN (t). It follows that η(t) = e−(θ−1)λ t.

2 (i) The inter-arrival times are much more suitable because they areindependent.

(ii) They should be exponentially distributed with the same mean.

Kolmogorov-Smirnov, Anderson-Darling or χ2 goodness-of-fit test can allbe used.

(iii) Successive values should be independent.

Regress X i on X i−1 using ordinary least squares, or fit an AR(1) and test

the α1 parameter for significance (equivalent to Durbin-Watson test).

3 (i) Spectral density f (ω) = 12

∞−∞π Σ γ ke

ikω , or equivalent.

For MA(1), therefore, we have

f (ω) =2

2

eσπ

(1 + β2 + 2β cos ω).

And for AR(1),

f (ω) =2

2

1

2 1 2 cos

eσπ + α − α ω

.

(ii) Clearly from (i) the inverse of the MA(1) is an AR(1), with α = −β and with

a different value of2eσ .

The word “invertible” attached to a MA(1) indicates that the inverse is a

stationary AR(1), whereas a non-“invertible” MA(1) has as inverse an

AR(1) model which cannot be stationary, such as X t = 2 X t−1 + et.


44/189


Page 3

4 (i) ( | ) ( ) ( )t tE dY Y y y h o h= = µ + , Var( | ) ( )t tdY Y y h o h= = +

(ii) 1( | )n n nE X X X y y+ − = = θ − α ,2

1 Var( | )n n n X X X y+ − = = τ .

(iii) µ( y)h = (θ − α y) and h = τ2

, so µ( y) = (θ − α y) / τ2

.

(iv) The increments of a brownian motion do not depend on its current value,

i.e. α = 0.

(v) An OU process drifts towards zero, so that θ = 0.

5 (i) Let γ k denote the autocovariance function of X . Then

Cov ( X t , et) = 0 + σ2 + 0 = σ2;

Cov ( X t , et−1) = αγ 0 + 0 + βσ2;

γ 2 = αγ 1

γ 1 = αγ 0 + 0 + β Cov ( X t−1 , et−1) = αγ 0 + βσ2

γ 0 = αγ 1 + Cov ( X t , et) + β Cov ( X t , et−1) = α2γ 0 + (1 + 2αβ + β

2) σ2,

implying that

γ 0 =2

2

2(1 2 )

1

σ+ αβ + β

− α,

ρ1 = 2( ) (1 )

1 2

α + β + αβ+ αβ + β

, ρ2 = αρ1.

(ii) Estimate of α is r2 / r1 = 0.8; estimate of β is given by12

(1 + 2αβ + β2) = (α + β) (1 + αβ), or 0.3β2 + 0.84β + 0.3 = 0, with solution

β =

−1.4

± 0.96 .

In this case we take the positive square root to ensure invertibility.


45/189


Page 4

6 (i) Expectation is £2.4m. Safety loading is

ρ =C − αµ

αµ =

(400 /12) 80,000 2,000 1,200

2,000 1,200

× − ××

= 11.11%

where α denotes the mean arrival rate of claims and µ the mean claimsize.

(ii) Conditions for validity of diffusion approximation are uµ large, ρ small,

µρu moderate, where u is the reserve.

In this caseu

µ is large, ρ is not particularly small and

uµρ =

7 22 10 11.1 10

1,200

−× × × = 1,852, clearly too large. We conclude that the

diffusion approximation is not appropriate.

(iii) Decide on a quantum of time, which may be a month or may be smaller.

For each time period generate a Poisson variate to indicate the number of

claims received and, conditional on this, a Normal variate with

appropriate mean and variance to represent the total sum claimed.

Subtract this from the total premium income over the period

(deterministic), using the resulting quantity as the increment of the

surplus process. Run the simulation for an extended period of time,

stopping if/when it goes below zero. A large number of simulations should

be performed, with the probability of ruin being estimated using standard

techniques based on the Binomial distribution.

The importance of reproducibility is for sensitivity analysis. The

estimated probability may depend heavily on the values assumed for

mean and standard deviation of the claim size, or on other numerical

parameters. It is necessary to vary the initial assumptions and run the

simulation again, just to ensure that conclusions are not substantially

changed if the parameter values used do not adequately reflect the actual

conditions experienced.

7 (i) P {S t ≤ x } = P {exp (µt + σ Bt) ≤ x } = P {µt + tN σ ≤ ln(x )} = ( )ln( )x tt− µσΦ , whereΦ denotes the standard Normal distribution function.

(ii) We have to find m so that P {S t ≤ m} = { }tt B P e mµ +σ ≤ = P {µt + σ Bt ≤ ln (m)}

= P {σ Bt ≤ ln (m) −µt} = 12 . Since σ Bt is a symmetric normal variable, itsmedian is 0 and the last equation can only be satisfied if ln(m) − µt = 0and m = eµt .

The expectation is ES t = Ee(µt+σ B(t)) = Eeµt tN eσ =

2

2 tte e

σµ =( )22 t

eσµ+

.


46/189


Page 5

(iii) By the same token, E (S tF s) = S (s) E (eµ(t−s)+σ( B(t)− B(s))) = S (s)

( )22 ( )t se

σµ+ −.

If S is to be a martingale, the conditional expectation must be equal to

S (s).

This will happen if µ = 212

− σ .

From part (ii) we see that for this stock with initial value 1, the median of

the distribution at time t goes to 0 exponentially fast for large t hence, a

very bad investment!

8 (i) Transition Graph

(ii) All transition probabilities must lie in [0,1].

Now 1 − 2α − α2 ≤ 1 − α − α2 ≤ 1 for α ≥ 0, so it suffices to ensurethat 1 − 2α − α2 ≥ 0 i.e. α ≤ 2 − 1. So the range of possible values of α is [0, 2 − 1].

(iii) The chain is not irreducible since D is a trap state.

The chain is aperiodic by inspection.

(iv) A stationary probability distribution, if it exists, must obey

(1 − α − α2) π A + απ B + α2πC = π A

απ A + (1 − 2α − α2) π B + απC = π B

α2π A + απ B + (1 − 2α − α2) πC = πC

α2π B + απC + π D = π D

The last equation implies π B = πC = 0, and this in turn shows that π A = 0.

Hence the stationary probability distribution is π = (0, 0, 0, 1)T

.

1 − 2α − α2

α

α B

C D

α

α

1

1 − α − α2

1 − 2α − α2

α2α2

α

α2


47/189


Page 6

It is unique: there is just one recurrent class and it is aperiodic. (Or point

out that there is no other solutions to the equations.)

(v) With α = 0.1, the transition matrix is

0.89 0.1 0.01 00.1 0.79 0.1 0.01

0.01 0.1 0.79 0.1

0 0 0 1

Its square is

0.8022 0.169 0.0268 0.002

0.169 0.6441 0.159 0.0279

0.0268 0.159 0.6342 0.18

0 0 0 1

the relevant entries being the last column.

9 (i) Transition Graph:

(ii) KFE: ( ) P t′ = P (t) A,

A =

0.1 0.05 0 0.05

1.0 1.2 0.1 0.1

0 0 0.4 0.4

0 0 0 0

− − −

.

(iii) The probability of staying in state H for 10 years is 10∞ 0.1e−0.1x dx = e−1.

H T

D

S

0.05 0.1

1.0

0.1 0.40.05


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(iv) First transition from H must be to S or D, each equally likely. If to D, then

it is certain that no terminal illness will occur; otherwise, the probability

of avoiding a terminal illness is S d .

From S similarly, except that the transition probabilities are to H with

prob. 1.0 51.2 6= , to D or to T , each with prob. 0.1 11.2 12= . Once in T it is notpossible to avoid terminal illness.

Solving the above equations, dS =51 1

12 6 2(1 ),S d+ × + implying that dS =

67

,

dH =1314

.

(v) The Markov property implies that the time spent in state T has

exponential distribution. The rate is 0.4 per year, so the expectation is 2.5

years.

The expected time spent in terminal illness given current health is (ever

hit T X 0 = H ) × 2.5 years =2.514

years.

10 (i) (1 − B) Y = β B (1 − B) I ,with solution Y = β BI + const

(ii) We have the vector equation

(1)1

(2)1

0 (1 ) const

= ,0 1 0

n n n

n n n

X X e

I I e

−

−

+ π β

+ + + α

which clearly represents a vector AR(1).

(iii) I is not stationary: the condition for an AR(1) to be stationary is that the

autoregressive parameter is less than 1 in absolute value.

I is not I (1), either, since ∇I = e(2) + α BI which, as already stated, is notstationary.

Z is therefore neither I (0) nor I (1).

(iv) The equation for the sum of squares is

SS = −= =

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Stochastic Modelling 2000-2004