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UNIVERSITY OF LONDON
IMPERIAL COLLEGE OF SCIENCE, TECHNOLOGY
AND MEDICINE
Course:M3S4/M4S4
Setter:
Hand
Checker:
Crowder
Editor:
McCoy
External:
Kent
Date:
December 2003
BSc and MSci EXAMINATIONS (MATHEMATICS)
MAY-JUNE 2004
This paper is also taken for the relevant examination for the Associateship
SOLUTIONS
M3S4/M4S4 APPLIED PROBABILITY
© 2000 University of London
M3S4/M4S4 Page of 10.
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1. SOLUTION
(a) [Seen]
(i) [2]
pn  t  h   pn  t  . 1   h  o  h  
 pn 1  t  .   h  o  h  
o  h 
(ii) [2]
Simplifying and rearranging the above, we obtain
pn  t  h   pn  t 
o h
  pn1  t   pn  t   
h
h
so that, in the limit as h  0 ,
dpn  t 
  pn 1  t   pn  t   (1a)
dt
In the special case of n  0 the equation is
dp0  t 
(1b)
  p0  t  
dt
(iii) [6]
Multiplying the equations in (1) by s n and summing over n  0,1, 2,... yields



d
d
p0  t    pn  t  s n   p0  t     pn  t  s n    pn 1  t  s n
dt
n 1 dt
n 1
n 1

Then, using   s, t    pn  t  s n , this becomes
n 0

  1  s  
t
(iv) [3]
Integrating this equation yields
log    1  s  t  A  s 
so that
  B  s  exp   1  s  t 
with A  s   ln B  s  , and using the condition given in the question that   s,0  1
we obtain B  s   1.
(v) [2]
This is the pgf of a Poisson distribution with parameter t
(b) [5] [Seen similar]
The number of casualties which have occurred by time t is a compound Poisson
process. In the lectures we derived that the expected number of casualties in such a
process is E  S  t    t , where  is the mean of the number of casualties at each
accident (full marks for either remembering this or deriving it). Since the mean of a
© 2000 University of London
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G0  q  distribution is   q p , we have that the mean number of casualties by time t
is qt p .
© 2000 University of London
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2. SOLUTION
[Part seen]
(i) [5]
P  X  x 

 P  X  x | T  t  P t  dt

e  t   t 

 e  t dt
x!
0



x
 x
x!

e
    t x
t dt
0
 x
x!
x !     x 1

  

     
x
for x  0,1, 2,... .
This is a geometric distribution with parameter      . In the lectures we called it
a G0        distribution.
(ii) [5]
We have a geometric distribution with p  


0
0
    so that
E  x    xp  x    xqp x  qp
 qp
d  x
p
dp 0
d  1  qp p 
 


dp  1  p  q 2 q 
Likewise, the pgf is


0
0
  s    s x qp x  q  s x p x  q 1  sp          s 
(iii) [5]
From the lectures, the probability that the population will die out is given by the
smallest positive root of      . Now, from   q 1  p  we obtain
 2 p   q  0
so that  p  q  1  0 with solutions   q p and 1.
If q  p the smaller of these is 1, and if q  p the smaller is q p . In terms of  and
 , these are, respectively, 1 and, for    ,  
(iv) [5]
From the lectures, we know that the expected number born in generation n is  n   n ,
where  is the mean of the offspring probability distribution. Hence,
© 2000 University of London
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 n   n      . If    the expected number of births up to and including the
nth generation is thus
   2  ...   n
1


As n   this converges to
1 

1 
1    
n
© 2000 University of London
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3. SOLUTION
(a) [Seen]
(i) [2]    P ,  i  0 ,

i
1
(ii) [1] A set of states, C, which it is impossible to leave: pij  0 if i  C , j  C
(iii) [2] Two states communicate if there is a non-zero probability of moving (perhaps in
more than one step) from either state to the other. A communicating class is a set of
communicating states.
Since, by definition, each state communicates with itself, each state does lie in a
communicating class. To prove the assertion we need to show that no state lies in more
than one such class. This follows from the fact that if a state s lies in two such classes,
A and B, then it is possible to get from an state in A to any state in B, and vice versa, via
s. Thus A and B form a single communicating class. An informal argument like this is
acceptable, or a more formal one if they give it.
(b) [Unseen]
(i) [3]
(ii) [2] Communicating classes: {0,1}, {2}, {3}. Classes {2} and {3} are closed since,
once entered, they cannot be left.
(iii) [2] Any distribution of the form  0,0, p,1  p  will be a stationary distribution.
(c) [Unseen]
(i) [6] Using the equations describing the first two elements and the fourth in    Q we
obtain the equations
0.5 0  0.25 1   0
0.5 0  0.5 1  0.5 2   1
 3  0.5 3
and combining these with
 i  1
yields  0  0.25, 1  0.5,  2  0.25,  3  0 .
(ii) [2] The mean time between returns is 1  0  4
© 2000 University of London
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4. SOLUTION
(i) [2] [Seen] Traffic intensity is the ratio of the expected value of the service time
distribution to that of the inter-arrival time distribution. The traffic intensity of an
M    / M   /1 queue is     .
(ii) [5] [Seen] For a steady state, we have
0   px1  px1      px
and
0   p1   p0 .
From these, p1   p0 , p2   2 p1 , and in general px   x p0 .
(iii) [5] [Seen] For the px , x  0,1, 2,... to form a proper distribution, we must have

p
x
 1 . This is only possible if 0    1 from which p 0  1    . The steady
0
state queue length distribution is thus geometric. In the lectures, we called it G0   
(iv) [8] [Unseen] The mean queue length is  1          , either from memory
or by deriving the mean of the geometric distribution in (iii). Thus a       . By
the memoryless property of the exponential distribution, the mean of the service time
distribution is 1  . Hence b  1  . From these two equations we obtain
  a b 1  a  , so that the mean inter-arrival time is b 1  a  a .
© 2000 University of London
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5. SOLUTION
(a) [8] [Unseen] The equation is in Lagrange form with
f  2st
g  s
h  2t
The auxiliary equations are
ds dt d 


.
2 st  s 2t 
The first equation gives  ds   2tdt from which c1  s  t 2 .
ds
d

from which c2   s .
s

so that   s  s  t 2  .
The second equation gives
Hence  s    s  t 2 

Now using the condition that   se s when t  0 we obtain ses  s  s  or
  s   es .
Hence the solution is   s, t   se s t .
2
(b) [Seen similar]
(i) [2] The bacterial population is not wiped out at time t if no drug treatment has occurred
by that time. Since the rate at which doses of treatment are administered is  , the time
interval between treatments is distributed as M   . Hence, P(no treatment by time t) =
exp  t , so that PY t   0  exp  t  .
(ii) [2] Either: since no treatment has occurred by time t, the population behaves according
to a simple birth process, so that, from the lectures, its size at time t has a geometric
distribution G1 e  t . Or, from the pgf given in the question, the distribution is a
geometric distribution G1 e  t . So,
 
 

PY t   y | Y t   0  e  t 1  e  t

y 1
y  1,2,...
(iii) [5] The unconditional probability distribution of Y t  is given by
PY t   0  1  PY t   0  1  e t
and when y  0 ,
PY t   y   PY t   y | Y t   0PY t   0

 e  t 1  e  t


y 1
 e    t 1  e  t
The pgf of Y t  is given by
© 2000 University of London
e t

y 1
M3S4/M4S4 Page of 10.
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
  s, t    P Y  t   y  s y
y 0

 1  e  t   e
    t
y 1
 1  e  t 
 1
1  e 
t
y 1
sy
se  
1  s 1  e   t 
   t
e  t  s  1
1  s 1  e   t 
(iv) [3] Differentiating s, t  with respect to s yields

s, t 
e    t
se    t 1  e  t


2
s
1  s 1  e  t
1  s 1  e  t

  
So that
  s, t 

s
 e
s 1

e
     t
e  t

e
     t


1  e 
t
e2  t
   t
© 2000 University of London
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