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MTHE/STAT455, STAT855 -- Final Exam, 2014
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QUEEN’S UNIVERSITY
DEPARTMENT OF MATHEMATICS AND STATISTICS
FACULTY OF ARTS AND SCIENCE
MTHE/STAT455, STAT855 FALL 2014, FINAL EXAM
7:00PM, DECEMBER 16, 2014
GLEN TAKAHARA
Instructions:
• “Proctors are unable to respond to queries about the interpretation of exam questions.
Do your best to answer exam questions as written.”
• This material is copyrighted and is for the sole use of students registered in MTHE/STAT455,
STAT855 and writing this examination. This material shall not be distributed or disseminated. Failure to abide by these conditions is a breach of copyright and may
also constitute a breach of academic integrity under the University Senates Academic
Integrity Policy Statement.
• This examination is THREE HOURS in length. It is closed-book – no books, notes,
or any other resource material is allowed except as indicated in the next item.
• Simple non-communicating calculators without text storage capabilities (Casio 991,
blue or gold sticker) and any notes on a single 8 12 × 11 inch sheet of paper (both sides)
are allowed.
• Please answer all questions in the booklets provided. Put your student number on the
front of all answer booklets and number the answer booklets if more than one answer
booklet is used.
• There are 5 questions. STAT855 students must do all 5 questions. MTHE/STAT455
students must to 4 of the 5 questions. For MTHE/STAT455 students, if you answer
all 5 questions please indicate which 4 questions you want marked. By default, unless
otherwise specified, the first 4 questions will be marked.
• Each question is worth 15 marks for a possible total of 60 for MTHE/STAT455 students
and 75 for STAT855 students.
• Show all your work. You may receive partial credit if you get an answer wrong but
show your work, whereas you will receive no credit if you get an answer wrong and do
not show your work.
MTHE/STAT455, STAT855 -- Final Exam, 2014
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1. (Total 15 marks) Consider the following searching problem on the nonnegative integers.
Suppose A is initially at position 0 and B is initially at position k, where k ≥ 0. At the
end of each time unit, A moves one position to the right while B moves one position
to the right with probability α and remains at his current position with probability
1 − α. Whenever A and B are at the same position, we say that A has found B and
they stay in that position forever.
(a) (6 marks) Let Mk be the expected number of time units until A finds B when B
is initially at position k. Find M0 and M1 , and by conditioning on the action of
B in the first time unit, find Mk for general k.
(b) (9 marks) Let B start at position k > 0. Define the discrete time Markov chain
{Xn : n ≥ 0} by Xn = B’s position minus A’s position just after the nth time
unit.
i. Give the state space for this Markov chain.
ii. Write down the transition matrix for this Markov chain.
iii. Classify each state of this Markov chain according to its period and whether
the state is recurrent or transient.
iv. Is this Markov chain irreducible. Why or why not?
2. (Total 15 marks) Let X = {Xn : n = 0, 1, 2, . . .} be an irreducible, positive recurrent
discrete-time Markov chain with state space S and transition probabilities pij , i, j ∈ S.
State whether each statement below is true, in which case prove it, or false, in which
case give a counterexample.
(a) (4 marks) If X is time-homogeneous then it is stationary.
(b) (4 marks) If X is stationary then it is time-homogeneous.
(c) (4 marks) If X is time reversible then pij > 0 if and only if pji > 0 for all i, j ∈ S.
(d) (3 marks) If pij > 0 if and only if pji > 0 for all i, j ∈ S then X is time reversible.
3. (Total 15 marks) Consider a branching process in which there are two possible types of
individuals in any given generation, those with a particular gene mutation and those
without. An individual without the mutation will have 0 offspring with probability
1/4, 2 offspring without the mutation with probability 2/3, or 2 offspring with the
mutation with probability 1/12. An individual with the mutation will have 0 offspring
with probability 1/4 or 2 offspring with the mutation with probability 3/4. Let Xn
be the number of individuals in generation n without the mutation and let Yn be the
number of individuals in generation n with the mutation. Assume X0 = 1 and Y0 = 0.
(a) (7 marks) Find the probability that there will eventually be no individuals without
the mutation. Hint: Follow just the unmutated individuals.
(b) (8 marks) Find E[Xn ] and E[Yn ]. Hint: For E[Yn ] condition on Xn−1 and Yn−1 .
MTHE/STAT455, STAT855 -- Final Exam, 2014
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4. (Total 15 marks) Let {Xn : n ≥ 0} be a discrete time Markov chain with state space
S = {1, 2, 3} and transition matrix

 1
P=
1
3
0
3
2
3
2
3
2
3
0
1
3
0


.
(a) (9 marks) Find the stationary distribution of the Markov chain {Xn : n ≥ 0}.
(b) (6 marks) Show that there is no time reversible continuous time Markov chain
such that {Xn : n ≥ 0} is the embedded discrete time chain. Hint: Write the
local balance equations for the continuous time chain and show that they cannot
be satisfied.
5. (Total 15 marks) Let {X(t) : t ≥ 0} be a continuous time Markov chain with state
P
space S and transition rates (qij )i6=j∈S and holding rates vi = j6=i qij . Let A be a
given subset of states of S and define TA to be the first time the Markov chain enters
the set A.
(a) (9 marks) Let ηi = P (TA < ∞ | X(0) = i) be the probability that the chain ever
visits A given that it starts in state i. Note that ηi = 1 for i ∈ A. Show that
vi ηi =
X
qij ηj
j6=i
for i ∈
/ A. Hint: Obtain similar equations for the embedded jump chain by
conditioning on the first jump, and differentiate between whether the first jump
is to a state in A or to a state not in A.
(b) (6 marks) Let µi = E[TA | X(0) = i] be the expected time for the chain to enter
the set A given that it starts in state i. Note that µi = 0 for i ∈ A. Show that
vi µi = 1 +
X
qij µj
j6=i
for i ∈
/ A. Hint: Condition again on the first jump of the chain.