M3S4/M4S4: Applied probability: 2007-8
Assessed Coursework 2: SOLUTIONS
1. Matrices with non-negative entries in which the rows sum to 1 are sometimes called
stochastic matrices. The transition matrices of Markov chains are stochastic matrices.
(a) Show t
M3/M4 j
54
Imperial College -
London
BSc, MSci and MSc EXAMINATIONS (MATHEMATICS)
May - June 2011
This paperl is also taken for the relevant examination for the Associateship of the
Royal College of Science.
Applied Probability
Date: Friday, 20 May 2011.
(i) (a) I: P(exactly 1 event occurs in (t,t + 6t]) = Mt + 0(6t). seen u
[o(6t)/§t -) 0 as 6t —> 0].
II: P(2 or more events occur in (t,t+ 6t]) = 0(6t).
III: Occurrence of events after time t is independent of occurrence of events
before t. E
(b) Let puff.
'.'._.d._._ _._ ._L._._.'._
Marks 84
seen / unseen
(i) Either of the following deﬁnitions, are appropriate for full marks. A counting
Process, {N;}120, is a Poisson Process of rate A > 0 if
(a) N0 = D.
(b) The increments are independent, that is,
BSc and MSci EXAMINATIONS (MATHEMATICS)
May-June 2009
This paper is also taken for the relevant examination for the Associateship.
M3S4/M4S4
Applied Probability
Date:
Wednesday, 3rd June 2009
Time:
10 am 12 pm
Credit will be given for all questions attemp
M3S4/M4S4: Applied probability: 2007-8
Solutions 1: Introduction
1.
E(X ) =
0
[1 F (x)] dx
=
0
=
f (y ) dy dx =
x
0
y
f (y )
0
dx dy =
0
y
f (y ) dx dy
0
f (y )y dy = E (X ).
0
2. T > t if and only if Ti > t for i = 1, . . . , n. Thus the event [T > t] is
M3S4/M4S4: Applied probability: 2007-8
Solutions 2: Point Processes
1. (a) Ti is the sum of three independent Poisson variables, each with rate . Hence the
distribution of Ti is Gamma(3, ).
(b) Wn = T1 + . . . + Tn Gamma(3n, ).
2. The distribution between
M3S4/M4S4: Applied probability: 2007-8
Solutions 3: pgfs and branching processes
1.
Y (s) = E(sY ) = E( saX +b ) = sb E(saX ) = sb X (sa ).
2.
(s) =
e x x
s = e
x!
x
p( x) s =
x=0
(s)x
= e es = exp(1 s).
x!
3. This is a G1 (p) distribution, with p(x) = q
M3S4/M4S4: Applied probability: 2007-8
Solutions 4: Random Walks
1. We need to handle cases p = q and p = q separately. Let qA,j and qB,aj be the
probability of A and B losing all their money when they start with $j and $(a j )
respectively.
To show that
4
Branching Processes
Organise by generations: Discrete time.
If P(no offspring) = 0 there is a probability that the process will die out.
Let X = number of offspring of an individual
p(x) = P(X = x) =
offspring prob. function
Assume:
(i) p same for all i
6
Markov Chains
A stochastic process cfw_Xn ; n = 0, 1, . . . in discrete time with finite or infinite state
space S is a Markov Chain with stationary transition probabilities if it satisfies:
(1) For each n 1, if A is an event depending only on any subse
7
Continuous time Markov processes
X (t) develops in continuous time (t 0) (state space still discrete).
Markov Property
P(X (t) = j | X (t1 ) = i1 , X (t2 ) = i2 , . . . , X (tn ) = in ) = P(X (t) = j | X (tn ) = in )
for any n > 1 and 0 t1 < t2 < . . .
M3S4/M4S4: Applied probability: 2007-8
Assessed Coursework 1: SOLUTIONS
1. (a) If exactly one event of a Poisson process took place in an interval [0 , t] find and
name the distribution of the time at which the event occurred.
Let N (t1 , t2 ) = number of
(3) Either of the following definitions. are appropriate for full marks. A counting
Process. {Nthzm is a Poisson Process of rate A > 0 if
1. N9 = 0.
2. The increments are independent, that is. for any khkg 6 2+. 0 < s < t
P({N¢ — N5 =Iki}l{N, = k2,0 g r S