Faculty of Actuaries Institute of Actuaries

EXAMINATIONS

April 2004

Subject 103 Stochastic Modelling

EXAMINERS REPORT

Faculty of Actuaries
Institute of Actuaries
 Subject 103 (Stochastic Modelling) April 2004 Examiners Report

1 First method (direct method):

Let f(Xt) = ln Xt . Using Itô s Lemma we have

2
df = f X t dt X t dBt 1
2
f ( X t2 )dt

1 1 2 1
= Xt 1.
2
X t2 dt X t dBt
Xt X t2 Xt

1 2
= 2
dt dBt

Integrating from t0 to t we have

1 2
ln X t ln X t0 2
t t0 Bt Bt0

and using the initial condition we get

1 2
Xt x exp Bt Bt0 2
t t0 as required.

Alternative method (working backwards from the solution)

1 2
Let f (t , Bt ) x exp Bt Bt0 2
t t0 . Then
2
f 1 2 f f 2
f, f, 2
f , so that
t 2 B B
1 2 1 2
df t ft dt f t dBt f t dt f t dt ft dBt
2 2

This implies that Xt = f(t,Bt) satisfies the stochastic differential equation.

We need to verify that the initial condition holds for this solution. (It clearly does, but
the check needs to be performed.)

Considering that this stochastic differential equation is the most frequently used of all,
this question was on average very poorly answered.

2 (i) The solution to 1 0.63 z 0 is z = 1/0.63, which is greater than 1. Therefore

the model is stationary.

For invertibility, we should check that the roots of 1 0.45 z 0.34 z 2 0 are
outside the unit circle. They are ( 0.45 1.25)/( 2*0.34) = 2.5 or 1.18, both
OK.
 Subject 103 (Stochastic Modelling) April 2004 Examiners Report

(ii) Let 0 = Cov(Xt, et), 1 = Cov(Xt, et 1), 2 = Cov(Xt, et 2). Then

0 = 0.63 1 + 0 + 0.45 1 0.34 2

1 = 0.63 0 + 0.45 0 0.34 1
2 = 0.63 1 0.34 0
2
0= (this may be regarded as obvious and not stated explicitly)
2 = 1.08 2
1 = 0.63 0 + 0.45
2 = 0.3404 2
2 = 0.63 1 0.34

An alternative expression for 0 may be obtained using

0 Var(0.63 X t 1 et 0.45et 1 0.34 et 2 )

0.632 0 (1 0.45 2 0.34 2 ) 2
2 0.63[0.45 0 0.34 1 ]
which removes the need to calculate 2.

Having derived the equations, we need to solve them.

1 = 0.63 0 + 0.0828 2
2 2 + 1.422 2,
0 = 0.63(0.63 0 + 0.0828 ) + 1.370 = 0.3969 0

implying that

0 = 2.358 2, 1 = 1.5684 2, 2 = 0.6481 2.

Candidates were often unsure of the procedure required to derive the equations for the k but
did rather better at solving them. In particular, many candidates did not take correct
account of the relationship between Xt and past values of et. Marks were awarded for correct
methodology when deriving the solutions, even if the equations being solved were not the
right ones.

3 (i) It is a jump process because it remains in one state for a period of time and
then jumps to another state, or alternatively because it is a continuous-time
process with a discrete state space.
Given the entire past history of the process, the probability of a member
retiring and beginning to receive benefits in the next dt interval is dt, i.e.
independent of the past. The same applies to the probability of death between
times t and t + dt: it is dependent only on the state (the number of retired
members) at time t, and not on anything which happened before that. Thus the
Markov property holds.

(ii) n,n 1 , n,n 1 n

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 Subject 103 (Stochastic Modelling) April 2004 Examiners Report

(iii) From the Kolomogorov Forward equations we have that

dpn
( n,n 1 n,n 1 ) pn n 1,n pn 1 n 1,n pn 1
dt ,.
( n ) pn (n 1) pn 1 pn 1

(iv) The suggested form of pn does not depend on t, so its derivative is zero. The
RHS is

n n 1 n 1
( n )e (n 1) e e 0,
n! (n 1)! (n 1)!

where = / .

(v) The implication is that Poisson( ) is a stationary distribution for the Markov
jump process. (In this case it is also a limiting distribution, but that is not
deducible from the Core Reading.)

Very few candidates suggested that the rate at which deaths take place should be
proportional to the number of retired scheme members.

In (iv) examiners awarded marks for correct attempts to carry out the verification procedure
even when the differential equation in (iii) was wrong.

In (v), terms such as limiting distribution or equilibrium distribution were given full
credit.

4 (i) Model fitting involves first choosing a suitable family of models, then a
suitable state space and finally estimating the values of the model parameters
to isolate a single member of the family.

Having performed the model fitting process, we have the best-fitting model
from the given family. Model validation involves checking whether this is an
adequate explanation of the observations.

Simulation can be used in model validation to test that the paths typically
followed by a simulated version of the process are broadly similar to the paths
observed in practice.

(ii) (a) Under the given model, the successive increments St form a
sequence of i.i.d. N( , 2) random variables, so the parameters can be
estimated using the sample mean and sample variance of the observed
increments.

(b) The first thing to do is to take the log of the observations, obtaining x1,
, xn say. Then apply the same technique as in (a).
 Subject 103 (Stochastic Modelling) April 2004 Examiners Report

(c) This is a time series model, so a time series technique (such as Box-
Jenkins) would be appropriate.

Parameters are estimated by Method of Moments or Maximum

Likelihood, or simply by applying a computer package to do the
estimation for you.

(iii) Any of the tests described here are equally valid.

(a) The model implies that the increments are Normally distributed, so a
standard test of normality, such as the Anderson-Darling test, the
Kolmogorov-Smirnov test or a chi-squared goodness-of-fit test can be
employed to test this.

Another testable property of the model is independence of increments:

it is possible to investigate the lag-1 sample ACF and determine
whether it is significantly different from zero, or use the Ljung-Box
chi-square statistics to perform a portmanteau test on all lags of the
sample ACF simultaneously.

A similar test is to inspect the sample ACF and PACF of the residuals
to see whether the observed values fall outside the 5% significance
band.

We could test whether there was a relationship between the increment

st and st; this is a test for homoscedasticity, though in the situation
described in the question it is unlikely to provide useful results.

A turning points test, a run test or a sign change test could also be
used, in each case comparing a calculated statistic with tables of a
reference distribution. These are tests for serial independence (usually
applied to a sequence of residuals).

(b) Apply the same tests as in (a) to the logarithm of the data.

(c) Tests can be applied to the sequence of residuals arising from the
fitting process. Since, if the model is accurate, these should form a
sequence of uncorrelated Normal observations, the same tests as in (a)
can be applied.

A procedure which might be applied to the model in part (c) but not to
the models in (a) or (b) is to inspect the sample partial ACF of the
increment process St; if the process really is a first-order
autoregression as stated in the model, then the sample PACF should
have no significant values at lags higher than 1.

This question differentiated well between candidates, with some very good answers indicating
that the candidates understood the principles and practice of modelling, and others
highlighting deficiencies of understanding.

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 Subject 103 (Stochastic Modelling) April 2004 Examiners Report

5 (i) (a) If X = B(½) and Y = B(1), then the marginal distribution of X is N(0,0.5)
and the conditional distribution of Y given X = x is N(x, 0.5).

This means that f X ,Y ( x, y ) is given by

1 x2 1 ( y x)2
exp . exp
2 (0.5) 2(0.5) 2 (0.5) 2(0.5)

1
exp( ( y 2 2 yx 2 x 2 ))

(b) The conditional density function of X given Y is

1
f X ,Y ( x, b) exp( (b 2 2bx 2 x 2 ))
f X |Y ( x | b)
fY (b) 1
exp ½b 2
2

2
exp 2( x ½b) 2

This is the density function of the Normal distribution with mean ½b

and variance ¼.

(ii) (a) Here we seek P(Y(1) > 0 | Y(½) = 5) = P(B(1) > 0 | B(½) = 1)

= P B( 12 ) 1

= P 2 B( 12 ) 0 2

= 1 2 = 0.9213.

(b) We require P Y ( 12 ) 0 | Y (1) 5 = P B ( 12 ) 0 | B(1) 1

0 ½
Using part (i)(b), this is 1 (1) 0.8413.
¼

This question was designed in such a way that candidates could tackle part (ii) even if part (i)
had not worked out right, but many allowed themselves to become discouraged by difficulties
in the early part of the question and gave up.

6 (i) Use a linear congruential generator (LCG). Specify three positive integers
a c m with m a m c and an initial value x0 , then generate a sequence of
x1 x2 xn in the range {0 1 2 m 1} by the recursive rule
 Subject 103 (Stochastic Modelling) April 2004 Examiners Report

xn (axn 1 c) (mod m) n 12 N

Then un xn m is a sequence of pseudo-random numbers in [0 1] .

To generate a sequence over the range (a b) , a linear transformation

vn (b a )un a is needed. Thus the procedure is: generate a variable
U U (0 1) using a LCG as above; return V (b a)U a .

(ii) The main advantage of using pseudo-random numbers is reproducibility.

When testing a model, this almost certainly depends on a number of
parameters and assumptions, so it is very often desirable to examine the
sensitivity of the model to these assumptions and the values of the parameters
used. It is thus necessary to reduce the random element to a minimum and
rerun the experiment using the same data .

(iii) (a) We use the inverse transform method. The cdf of the Pareto is

x a 1
F ( x) a 1
1
0 (1 x) (1 x) a

so the inverse is

F 1 ( y ) (1 y ) 1a
1

Thus the procedure is

Generate an observation U from the Uniform (0,1) distribution.

1a 1a
Return X (1 U ) 1 (or, equally good, X U 1 ) as an
observation from the Pareto.

(b) Since the number of possible values for X is n 1 , we need to split

the interval (0 1) into n 1 subintervals, the length of which has to be
proportional to the corresponding probabilities. Given a uniform
variable U , one such procedure is:

For i 1 2 n , if (i 1) (n 2) U i (n 2) , then set X i

If U n (n 2) , then set X 0 .

(iv) From (iii), we see that the tail of the Pareto distribution is 1 F ( x) (1 x) a ,
which decreases to zero much more slowly than e.g. the exponential or normal
distributions.
Alternatively, it is called fat-tailed because of the relatively high probability
of producing values which are a long way from the mean/median.

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 Subject 103 (Stochastic Modelling) April 2004 Examiners Report

In general, the Pareto offers a suitable choice for portfolios where there is a
non-negligible chance of very large claims, which is reasonable in various
forms of general insurance, in particular insurance that deals with natural
catastrophes (floods, earthquakes, hurricanes etc).

This question was generally well answered and caused few difficulties.

7 (i) (a) We will use the matrix form of the Kolmogorov DE, which states that
P '(t ) AP (t ) .
d T T dP T
It follows that P(t ) AP(t ) 0T P(t ) 0T .
dt dt
[The other Kolmogorov DE, P '(t ) P(t ) A , cannot be used to complete the proof]

(b) If X0 is random with distribution , then the distribution of Xt is given

by TP(t).

The fact that the differential of this is equal to zero implies that TP(t)
= TP(0) = T for all t, in other words Xt has the same distribution for
all t.

For an irreducible Markov chain on a finite state space, the stationary

distribution is also the limiting (equilibrium, long-term, steady state)
distribution.

(ii) (a) The states can be labelled as 0, 1, 2:0, 2:1, 2:2, where 0 and 1 represent
the number of operators occupied and 2:j means that both operators are
occupied and j calls are on hold.

Candidates with a different collection of states can earn 0.5 marks, as

long as their states are states. For example, Call arrives is not a
state, but an event triggering a transition from one state to another.

(b)
1/2 1/2 1/2 1/2
0 1 2:0 2:1 2:2
1/3 2/3 2/3 2/3

The labels on the arcs represent the transition rates.

(c) From the transition diagram, the generator matrix is as below.

 Subject 103 (Stochastic Modelling) April 2004 Examiners Report

1/ 2 1/ 2 0 0 0
1/ 3 5/6 1/ 2 0 0
A 0 2/3 7/6 1/ 2 0
0 0 2/3 7/6 1/ 2
0 0 0 2/3 2/3

(d) We have

1 1 3
0 1 1 0
2 3 2
5 1 2 9
1 0 2:0 2:0 0
6 2 3 8
7 1 2 27
2:0 1 2:1 2:1 0
6 2 3 32
7 1 2 81
2:1 2:0 2:2 2:2 0
6 2 3 128

To solve this we add the condition that the i must sum to 1.

Therefore the stationary distribution is

128 192 144 108 81

653 653 653 653 653

(in decimal form this is

0.1960 0.2940 0.2205 0.1654 0.1240 )

Many candidates made a good attempt at part (i), though not many scored full marks. In part
(ii) a surprising number of candidates were unable to distinguish between states in which
the process stays for a certain length of time and events, which occur instantaneously and
trigger transitions from one state to another.

8 (i) Suppose at time t X has just arrived in state i. The probability that X remains
in state i until time t + k and then leaves, giving a duration of k + 1 steps in
state i, is piik (1 pii ) . In other words, P ( Di ,n d ) piid 1 (1 pii ) , which is
the probability function of a geometric random variable.

The fact that Di,n is independent of previous durations follows from the
Markov property: What happens to X after arriving in state i is independent of
anything that happened before that moment.

(ii) A sequence of successive observations of a geometric random variable is

likely to produce 1 more often than any other value. The durations of the
restorer s stays in the various locations are always at least 2 days and often
much more. There is an indication that the geometric distribution is
inappropriate for the data set provided.

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 Subject 103 (Stochastic Modelling) April 2004 Examiners Report

An alternative indication that the Markov property is dubious is the fact that
the restorer appears to return to Bath at regular intervals, i.e. regardless of the
time spent in each state, the path taken appears to lack randomness.

(iii) pij nij / ni , where nij is the number of transitions from state i to state j and
ni+ is the total number of transitions out of state i. For example, of the 8 days
spent in Warwick, one is followed by a trip to Caernarvon, one by a trip to
Bath and the remaining 6 are followed by another day in Warwick, so that nWC
= 1, nWB = 1 and nWW = 6. We therefore have (using the order B, C, S, W)

9 2 1
0
12 12 12
2 13
0 0
15 15
P
1 8 1
0
10 10 10
1 1 6
0
8 8 8

(iv) The model is irreducible. Starting from Bath it is possible to visit Stratford,
Warwick, Caernarvon and return to Bath, completing the circuit.

It is also aperiodic, since pii > 0 for some (in fact for all) state i.

(v) We need the entry of P3 corresponding to the row and column of Warwick
(4th row, 4th column in this case).

.5729 .0271 .2583 .1417

.2156 .7511 .0222 .0111
Now P 2 , so the required answer is
.0258 .1792 .6400 .1550
.2042 .2021 .0208 .5729

1 1 0 6
.1417 .0111 .1550 .5729 = 0.4488.
8 8 8 8

Each part of this question attracted some unexpected answers as well as some good
ones. The answers to part (i) were on the whole disappointing, but the final parts were
rather better answered in general.
 Subject 103 (Stochastic Modelling) April 2004 Examiners Report

9 (i) (a) In one sense it is reasonable, because in the absence of rainfall

gardeners might increase their demand for water. But the relationship
does not work in reverse: in particularly rainy weather the demand is
likely to be no less than in normally rainy weather.

(b) Yes. It reflects the seasonal pattern of rainfall.

(ii) No. It satisfies an equation which is explicitly dependent on t, so it will not be

stationary unless all the t are equal to one another. (Arguments based on
whether | | > 1 are only relevant if all the t are equal.)

(iii) (a) Normality: by induction. If Rt 1 and et are Normal, then Rt must be

Normal, as it is a linear combination of these two. Since R0 is Normal,
normality follows for all t.

E ( Rt ) t [ E ( Rt 1 ) t 1] . This can be iterated backwards to

zero, showing that E ( Rt ) t .

An alternative approach is to solve explicitly for Rt and read off the

answers from there.
t 1
t k
Rt t ( R0 0) et k
k 0
Now we can see that Rt is a linear combination of Normal random
variables with some constants, hence Normally distributed, and that the
mean is t.

(b) Var(Rt) is equal to 2 Var(Rt 1) + 2. If Rr has constant variance R

2

then R2 = 2 R2 + 2, implying that R2 = 2 / (1 2).

If instead the above expression for Rt is used, we can see that

2 2t 2
t 1
2k 2 2 1 1 2t 2
R R , so that R
k 0 1 2t 1 2
(iv) t represents the mean rainfall in one particular month of the year. Estimate
this by calculating the average rainfall in that month over the 10-year period,
1
t (rt rt 12 rt 24 rt 108 ) .
10

is best estimated using the lag-1 autocorrelation of R ,

120
(rt t )(rt 1 t 1 )
t 2
120
.
2
(rt t )
t 1

Noticing that Cov(Dt, Rt) = Var(Rt), we suggest minus the ratio of the
sample covariance of D with R divided by the sample variance of the R, i.e.

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 Subject 103 (Stochastic Modelling) April 2004 Examiners Report

120
(rt r )(dt d)
t 1
120
, although in fact a better estimate would be obtained
2
(rt r)
t 1
if account was taken of the variable mean of the Rt.

2
(Less appropriate is using the fact that Var( Dt ) Var( Rt ) to suggest
2
sD / sR2 , since the sign of is lost in this operation.)

Now we have an estimate for , we can use Rt Dt .

(v) X121 X120 R121 D121 . In addition, R121 1 ( R120 0) e121 and
D121 R121 .

Therefore X121 X120 (1 )[ 1 ( R120 0 )] (1 )e121 , which

implies that x120 (1) X120 (1 )[ 1 ( R120 0 )]
and Var( X121 x120 (1)) (1 )2 2
.

Most candidates made encouraging progress with this question, though part (iv) proved
difficult for most. In part (v), as well, the relationships between the variables were not
always well understood.

10 (i) The increment St+s St is equal to cs (Nt+s Nt), which is independent of

anything that happened before time t and has the same distribution as
cs (Ns N0), using the independent increment property of the Poisson
process.

Alternatively, it is possible to use the Decomposition Theorem, to state that a

Lévy process is any sum of three components: a deterministic linear function
of time, a purely continuous random component (a multiple of Brownian
motion) and a purely discontinuous random component (a compound Poisson
process). In this case the coefficient of the Brownian component is zero and
the jumps of the compound Poisson process all have size 1.

(ii) By the independent increments property, Nt s Nt ~ Po( s ) given Nt.

[u c (t s )] Nt [u c (t s ) N t ] s[ e 1]
(iii) (a) E[Yt s | Ft ] e E[e s | Ft ] e .

This is equal to Yt if c [1 e ] .

(b) If c < , the line y = c increases slowly from 0 as increases,

whereas [1 e ] increases more rapidly to start with, but never
 Subject 103 (Stochastic Modelling) April 2004 Examiners Report

exceeds . Therefore there must be a crossing point at a positive value

of .

If, on the other hand, c > , then there is no positive crossing point; the
line y = c goes slowly to as , whereas [1 e ] tends to
exponentially fast as .

(c) Yes; this is the condition for the company to remain solvent.

(iv) (a) If S hits K first, then ST = K exactly, since increases in S occur

continuously. But if S becomes negative before hitting K, then the
negative movement must have been caused by a downward jump;
jumps are of magnitude 1, so in this case ST can be no less than 1.

(b) Since St is between 1 and K for all 0 t T, it follows that Ymin(t ,T ) is

bounded above and below, from which we deduce that the optional
stopping theorem applies.

(c) ST = K with probability 1 , or with probability , ST takes a value

somewhere between 1 and 0. Therefore YT takes the value e K with
probability 1 , or with probability it takes some value between 1
and e . Hence the result.

K u K
E[YT ] e e e
(d) Clearly, K K
.
1 e 1 e

As K becomes larger, the fact that < 0 means that e K 0, so in the

limit we have e u.

There was an error in the final part of the question; where the question asked for a lower
bound, the quantity which could be derived from the previous part of the question was in fact
an upper bound. Candidates who reached the last part and were confused by the error were
treated generously.

It appeared that many candidates tried this question when they were short on time. Those
candidates who attempted part (iv) often did quite well on it, even if they had omitted earlier
parts of the question.

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