© 2020 Imperial College London Page 1
MATH97083
BSc, MSci and MSc EXAMINATIONS (MATHEMATICS)
May-June 2020
This paper is also taken for the relevant examination for the
Associateship of the Royal College of Science
Applied Probability
BLACKBOARD (ONE FOR EACH QUESTION) WITH COMPLETED COVERSHEETS WITH
QUESTION.
.
Date: 13th May 2020
Time: 13.00pm - 15.30pm (BST)
Time Allowed: 2 Hours 30 Minutes
This paper has 5 Questions.
Candidates should start their solutions to each question on a new sheet of paper.
Each sheet of paper should have your CID, Question Number and Page Number on the
top.
Only use 1 side of the paper.
Allow margins for marking.
Any required additional material(s) will be provided.
Credit will be given for all questions attempted.
Each question carries equal weight.
We denote the natural numbers including 0 by N0 = N ∪ {0}.
1. (a) Consider a discrete-time homogeneous Markov chain (Xn)n∈N0 with state space E =
{1, 2, 3, 4, 5, 6, 7, 8} and transition matrix given by
P =

1
4 0
3
4 0 0 0 0 0
0 1 0 0 0 0 0 0
1
3
1
3
1
3 0 0 0 0 0
0 0 0 14
3
4 0 0 0
0 0 0 13
2
3 0 0 0
0 0 0 0 0 0 1 0
0 0 0 0 0 1 0 0
0 0 0 0 0 0 1 0

.
(i) Draw the transition diagram. (2 marks)
(ii) Specify the communicating classes and determine whether they are transient, null
(3 marks)
(iii) Find all stationary distributions. (5 marks)
(b) Suppose the influenza virus exists in K different strains, where K ≥ 2. Each year, the virus
either stays the same with probability 1 − a, for a ∈ (0, 1), or mutates to any of the other
strains with equal probability. Suppose you can model the virus mutation by a discrete-time
homogeneous Markov chain.
(i) We denote the state space by E = {1, . . . , K}. State the corresponding 1-step transition
probabilities of the Markov chain. (3 marks)
(ii) You decide to group the states: You consider the modified state space E˜ = {I, O} where
I stands for the initial state and O for the collection of the other K − 1 states.
(1.) State the corresponding 1-step transition probabilities of the Markov chain on E˜.
(2 marks)
(2.) Show that, for n ∈ N,
pII(n+ 1) = pII(n)
{
1− a− a
K − 1
}
+ a
K − 1 ,
and state all results from the lectures which you apply in your proof. (5 marks)
(Total: 20 marks)
MATH96052/MATH97083 Applied Probability (2020) Page 2
2. (a) Let T be a nonnegative discrete random variable on a probability space (Ω,F ,P) and let
A ∈ F be an event with P(A) > 0. Show that
E(T |A) =
∞∑
n=1
P(T ≥ n|A).
(2 marks)
(b) Consider a discrete-time homogeneous Markov chain on a countable state space E. Suppose
that the Markov chain is irreducible, has a stationary distribution denoted by pi and all states
are recurrent.
(i) Show that pii = µ−1i for all i ∈ E, where µi denotes the mean recurrence time for state
i. (5 marks)
(ii) Show that all states are positive recurrent. (3 marks)
(c) Consider a homogeneous Markov chain (Xn)n∈{0,1,2,...,1000} with state space E = {1, 2, 3, 4}
and transition matrix given by
P =

0.5 0 0.5 0
0 0.5 0 0.5
0 0.5 0 0.5
0.5 0 0.5 0

.
(i) Is this Markov chain irreducible? (2 marks)
(ii) How many stationary distributions does this Markov chain have? Find all stationary
distributions. (3 marks)
(iii) Is this Markov chain time-reversible? (5 marks)
(Total: 20 marks)
MATH96052/MATH97083 Applied Probability (2020) Page 3
3. (a) Consider two independent and Poisson distributed random variables X ∼ Poi(λ) and
Y ∼ Poi(µ) with λ, µ > 0. Show that
X
∣∣∣X + Y = n ∼ Bin(n, λ
λ+ µ
)
, for n ∈ N.
You may state and use without proof the distribution of X + Y . (4 marks)
(b) Let (Nt)t≥0 denote a Poisson process with rate λ > 0. For t1 < t2, show that
Nt1
∣∣∣Nt2 = n ∼ Bin(n, t1t2
)
, for n ∈ N.
(5 marks)
(c) Let (Nt)t≥0 denote a Poisson process with rate λ > 0. Let (Zi)i∈N denote independent and
identically distributed random variables with Bernoulli distribution with parameter p > 0.
Suppose that (Zi)i∈N and (Nt)t≥0 are independent. For t ≥ 0, define
Xt =
Nt∑
i=1
Zi, Yt = Nt −Xt.
(i) Show that (Xt)t≥0 and (Yt)t≥0 are Poisson processes with rates λp and λ(1 − p),
respectively. (4 marks)
(ii) Also show that for any t ≥ 0, Xt and Yt are independent. (2 marks)
(d) Consider the Cramér-Lundberg model in insurance mathematics.
(i) State the model for the total claim amount. (3 marks)
(ii) Derive the cumulative distribution function of the total claim amount at a fixed point in
time. (2 marks)
(Total: 20 marks)
MATH96052/MATH97083 Applied Probability (2020) Page 4
4. (a) Let N = (Nt)t≥0 and M = (Mt)t≥0 denote independent Poisson processes with rates λ > 0
and µ > 0, respectively. Let T = inf{t ≥ 0 : Mt = 1} denote the random time when the
first jump in M occurs. Determine P(NT/2 = k) for k ∈ N0 and name the distribution.
(7 marks)
(b) Consider a population of N individuals consisting at time 0 of one ‘infective’ and N − 1
‘susceptibles’. The process changes only by susceptibles becoming infective. If, at some time
t, there are i infectives, then, for each susceptible, there is a probability of i2λδ + o(δ) of
becoming infective in (t, t+ δ] for λ, δ > 0.
(i) If we consider the event of becoming an infective as a birth, what is the birth rate λi of
the process, when there are i infectives? (2 marks)
(ii) Let T denote the time to complete the epidemic, i.e. the first time when all N individuals
are infective.
(1.) Derive E(T ) (without using any type of generating functions). (2 marks)
(2.) Show that the Laplace transform of T is given by
E[e−sT ] =
N−1∏
i=1
(
λi
λi + s
)
, for s ≥ 0.
(2 marks)
(3.) Derive E(T ) by using the Laplace transform given in (2.). (2 marks)
You may leave your solution in (1.) and (3.) as a sum.
(c) Give an example of two non-identical continuous-time Markov chains which have the same
jump chain. (5 marks)
(Total: 20 marks)
MATH96052/MATH97083 Applied Probability (2020) Page 5
5. Let X = (Xt)t≥0 be a continuous-time Markov chain on a countable state space E with generator
G. We assume that the Markov chain is minimal.
(a) (i) Give a definition for the state i ∈ E to be recurrent. (2 marks)
(ii) Give a definition for the state i ∈ E to be transient. (1 mark)
(b) Suppose E = {1, 2, 3, 4} and
G =

−1 12 12 0
1
4 −12 0 14
1
6 0 −13 16
0 0 0 0

.
For each state in the state space, decide whether it is null/positive recurrent or transient and
(5 marks)
(c) Now let E = {1, 2} and
G =
 −1 1
2 −2
 .
(i) Find the stationary distribution of X and justify your answer. (4 marks)
(ii) Find the stationary distribution of the jump chain associated with X. (2 marks)
(iii) Formulate a general result which describes the relationship between the stationary
distributions of X and of its jump chain and show how the result applies in the context
of the Markov chain considered in this question (part (c)). (4 marks)
(iv) Find the transition matrix Pt = (pij(t))i,j∈E for all t ≥ 0. (2 marks)
Hint: You may use without a proof that G = ODO−1, where
O =
 1 −12
1 1
 , D =
 0 0
0 −3
 , O−1 =
 23 13−23 23
 .
(Total: 20 marks)
MATH96052/MATH97083 Applied Probability (2020) Page 6
BSc and MSci EXAMINATIONS (MATHEMATICS)
May-June 2020
This paper is also taken for the relevant examination for the Associateship.
MATH96052/MATH97083
Applied Probability (Solutions)
Setter’s signature Checker’s signature Editor’s signature
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
c© 2020 Imperial College London MATH96052/MATH97083 Page 1 of 10
meth seen ⇓
2,A
1. (a) (i) The transition diagram is given by
1 23 4 5 6 7 8
3
4
1
4 1
1
3
1
3
1
3
3
4
1
4
1
3
2
3
1
1
1
1
(ii) We have a finite state space which can be divided into five communicating
classes: The classes T1 = {1, 3}, T2 = {8} are not closed and hence transient. 1,A
The classes C1 = {2}, C2 = {4, 5}, C3 = {6, 7} are finite and closed and hence
positive recurrent. 2,A
(iii) This Markov chain does not have a unique stationary distribution pi since we
have three closed (essential) communicating classes. For the transient states
we know from the lectures that pii = 0 for i = 1, 3, 8. For the positive 1,A
recurrent states, we solve pi2 · 1 = pi2, (pi4, pi5) = (pi4, pi5)
(
1
4
3
4
1
3
2
3
)
and
(pi6, pi7) = (pi6, pi7)
(
0 1
1 0
)
, which leads to pi2 = pi2, pi5 =
9
4pi4 and pi6 = pi7.
2,A
There are various ways of representing all possible stationary distributions (only
one is needed!), e.g.:
∗ pi = (0, pi2, 0, pi4, 94pi4, pi6, pi6, 0) for all pi2, pi4, pi6 ≥ 0 with pi2 + 134 pi4 +
2pi6 = 1,
∗ pi = (0, pi2, 0, 49pi5, pi5, pi6, pi6, 0) for all pi2, pi5, pi6 ≥ 0 with pi2 + 139 pi5 +
2pi6 = 1,
∗ pi = a(0, 1, 0, 0, 0, 0, 0, 0) + b(0, 0, 0, 413 , 913 , 0, 0, 0) + c(0, 0, 0, 0, 0, 12 , 12 , 0)
for all a, b, c ≥ 0 with a+ b+ c = 1. 2,A
unseen ⇓
(b) (i) We have pii = 1 − a for all i ∈ E and we have that pij = b for all i 6= j for
some b ∈ (0, 1), hence 1 = ∑Kj=1 pij = 1− a+ (K − 1)b⇔ b = aK−1 , which
implies that pij =
a
K−1 for all i, j ∈ E, i 6= j. 3,B
(ii) (1.) Here the transition matrix corresponding to the Markov chain on E˜ =
{I,O} is given by
P =
(
1− a a
a
K−1 1− aK−1
)
.
2,B
(2.) Let n ∈ N. From the Chapman-Kolmogorov equations, we have that
Pn+1 = PnP, hence
pII(n+ 1) = pII(n) · pII + pIO(n) · pOI .
= pII(n) · (1− a) + pIO(n) · a
K − 1 .
2,C
Since Pn is a stochastic matrix, we have that pII(n) + pIO(n) = 1 ⇔
pIO(n) = 1− pII(n). 2,B
MATH96052/MATH97083 Applied Probability (Solutions) (2020) Page 2 of 10
Hence
pII(n+ 1) = pII(n) · pII + pIO(n) · pOI
= pII(n) · pII + (1− pII(n)) · pOI
= pII(n)(pII − pOI) + pOI
= pII(n)
(
1− a− a
K − 1
)
+
a
K − 1 .
[2 marks for applying and mentioning the Chapman- 1,B
Kolmogorov equations, 2 marks for using and mentioning that
Pn is a stochastic matrix, 1 mark for deriving the final formula.
If students made a mistake in the first part (1.) of the question
and plugged in the wrong 1-step transition probabilities here,
then they should not be penalised again, but can achieve full
marks in this part of the question.]
MATH96052/MATH97083 Applied Probability (Solutions) (2020) Page 3 of 10
sim. seen ⇓2. (a) We use the definition of the conditional expectation and the fact that m = ∑m−1n=0 1
to deduce that
E(T |A) =
∞∑
m=0
mP(T = m|A) =
∞∑
m=0
m−1∑
n=0
P(T = m|A) =
∞∑
n=0
∞∑
m=n+1
P(T = m|A)
=
∞∑
n=0
P(T ≥ n+ 1|A) =
∞∑
n=1
P(T ≥ n|A).
2,A
seen ⇓(b) (i) Suppose that X0 ∼ pi (i.e. P(X0 = i) = pii for each i). Let Ti = inf{n ≥
1 : Xn = i} denote the first hitting time for state i ∈ E, using part (a), we
get pijµj = P(X0 = j)E(Tj |X0 = j) =
∑∞
n=1 P(Tj ≥ n|X0 = j)P(X0 = j) =∑∞
n=1 P(Tj ≥ n,X0 = j). 2,D
Define an := P(Xm 6= j, 0 ≤ m ≤ n), for n ∈ N0.
Then P(Tj ≥ 1, X0 = j) = P(X0 = j) (since Tj ≥ 1 by definition) and for
n ≥ 2
P(Tj ≥ n,X0 = j) = P(X0 = j,Xm 6= j, 1 ≤ m ≤ n− 1)
= P(Xm 6= j, 1 ≤ m ≤ n− 1)− P(Xm 6= j, 0 ≤ m ≤ n− 1)
= P(Xm 6= j, 0 ≤ m ≤ n− 2)− P(Xm 6= j, 0 ≤ m ≤ n− 1)
= an−2 − an−1,
where we have used the homogeneity of the chain and the law of total
probability. 2,D
Then, summing over n (noting that we are dealing with a telescoping sum)
pijµj = P(X0 = j) +
∞∑
n=2
(an−2 − an−1)
= P(X0 = j) + P(X0 6= j)− lim
n→∞ an
= 1− lim
n→∞ an.
However, limn→∞ an = P(Xm 6= j, ∀m) = 0 by the recurrence of j. That is,
pi−1j = µj if pij > 0. 1,D
(ii) To see that pij > 0 for all j, suppose the converse; then
0 = pij =

i∈E
piipij(n) ≥ piipij(n)
for all i, n, yielding that pii = 0 whenever i → j. However, the chain is
irreducible, so that pii = 0 for each i - a contradiction to the fact that pi is a
stationary vector. Thus µi <∞ and all states are positive. 3,C
meth seen ⇓(c) (i) Yes, this Markov chain is irreducible since all states communicate with each
other.
2,A
MATH96052/MATH97083 Applied Probability (Solutions) (2020) Page 4 of 10
(ii) Since the Markov chain is irreducible and the state space is finite, there is
only one finite and closed communicating class, hence all states are positive
recurrent. From a theorem in lectures we can then deduce that there is a
unique stationary distribution. We observe that the transition matrix is doubly-
stochastic and hence we know from lectures/problem class that the uniform
distribution pi = (0.25, 0.25, 0.25, 0.25) is the unique stationary distribution of
the Markov chain. [Solving pi = piP for pii ≥ 0,
∑4
i=1 pii = 1 to derive
the stationary distribution is of course also a valid approach.] [1
mark for the correct answer and 2 marks for the justification.] 3,B
(iii) From the lectures we know that the Markov chain is time-reversible if and only if
the detailed-balance equations hold, i.e. piipij = pijpji for all i, j ∈ {1, 2, 3, 4}.
Here we have that pii = pij for all i, j, hence we need a symmetric transition
matrix for the detailed-balance equations to hold. However, here we have
e.g. p14 = 0 6= p41 = 0.5, hence the detailed-balance equations do not hold
and hence the Markov chain is not time-reversible. [2 marks for “not
time-reversible”, 3 marks for justification] 5,B
MATH96052/MATH97083 Applied Probability (Solutions) (2020) Page 5 of 10
unseen ⇓3. (a) We recall from lectures that X + Y ∼ Poi(λ+ µ).
1,A
Let n ∈ N and k ∈ {0, 1, . . . , n}, then, using Bayes’ rule, we have
P(X = k|X + Y = n) = P(X + Y = n|X = k)P(X = k)
P(X + Y = n)
=
P(Y = n− k|X = k)P(X = k)
P(X + Y = n)
by independence of X and Y
=
P(Y = n− k)P(X = k)
P(X + Y = n)
=
µn−k
(n− k)!e
−µλk
k!
e−λ
(
(λ+ µ)n
n!
e−(µ+λ)
)−1
=
(
n
k
)(
λ
λ+ µ
)k ( µ
λ+ µ
)n−k
,
which is indeed the probability mass function of a Bin
(
n, λλ+µ
)
random variable. 3,A
unseen ⇓(b) Since (0, t1], (t1, t2] are disjoint, the increments Nt1 − N0 = Nt1 , Nt2 − Nt1 are
independent. Also, we have that Nt1 ∼ Poi(λt1), Nt2 ∼ Poi(λt2), Nt2 − Nt1 ∼
Poi(λ(t2 − t1)). Hence we can apply part (a) with X = Nt1 and Y = Nt2 −Nt1
to conclude that
Nt1
∣∣Nt2 = n ∼ Bin(n, λt1λt1 + λ(t2 − t1)
)
⇔ Nt1
∣∣Nt2 = n ∼ Bin(n, t1t2
)
.
5,A
sim. seen ⇓(c) We answer (i) and (ii) jointly: Both X = (Xt)t≥0 and Y = (Yt)t≥0 are stochastic
processes. We need to check the defining properties of the Poisson process:
∗ X0 = 0 and Y0 = 0 almost surely, since N0 = 0 almost surely. 1,A
∗ Independent increments: (Nt)t≥0 has independent increments. By
construction (the Zis are all independent of each other and of (Nt)) (Xt)t≥0
and (Yt)t≥0 inherit the independent increments property from (Nt)t≥0.
∗ Stationary increments: (Nt)t≥0 has stationary increments. By construction
(the Zis are all independent of each other and of (Nt) and identically
distributed) (Xt)t≥0 and (Yt)t≥0 inherit the stationary increments property
from (Nt)t≥0. 1,A
∗ Let n,m ∈ N0, t ≥ 0, then
P(Xt = m,Yt = n) = P(Xt = m,Nt = m+ n)
= P(Xt = m|Nt = m+ n)P(Nt = m+ n)
=
(
m+ n
m
)
pm(1− p)n (λt)
m+n
(m+ n)!
e−λt
=
(λtp)m
m!
e−λtp
(λt(1− p))n
n!
e−λt(1−p) = P(Xt = m)P(Yt = n).
So, we found that Xt and Yt are independent with Xt ∼ Poi(λtp) and 2,A; 2,B
Yt ∼ Poi(λt(1 − p)) and X and Y are Poisson processes with rates λp and
λ(1−p), respectively. [The final 4 marks are split as follows: 2 marks
for the marginal distributions (possibly derived separately), 2
marks to show the independence.]
MATH96052/MATH97083 Applied Probability (Solutions) (2020) Page 6 of 10
seen ⇓(d) (i) The total claim amount in the Crame´r-Lundberg model is defined as a
compound Poisson process (St)t≥0 satisfying
St =
{ ∑Nt
i=1 Yi, Nt > 0,
0, Nt = 0,
where N = (Nt)t≥0 is a Poisson process of rate λ > 0 and the claim size
process is denoted by Y = (Yi)i∈N, where the Yi denote positive i.i.d. random
variables with finite mean and finite variance. Also, N and Y are assumed to
be independent of each other. [Any alternative correct definition of 3,A
the (compound) Poisson process should be awarded full marks.]
(ii) Let t ≥ 0. Then FSt(x) = P(St ≤ x) = 0 for x < 0. Let x ≥ 0, then
FSt(x) = P(St ≤ x) =
∞∑
n=0
P(St ≤ x,Nt = n) =
∞∑
n=0
P
(
n∑
i=1
Yi ≤ x,Nt = n
)
=
∞∑
n=0
P
(
n∑
i=1
Yi ≤ x
)
P(Nt = n) =
∞∑
n=0
e−λt
(λt)n
n!
P
(
n∑
i=1
Yi ≤ x
)
.
2,A
MATH96052/MATH97083 Applied Probability (Solutions) (2020) Page 7 of 10
unseen ⇓4. (a) Since N and M are independent, we deduce that N and T are independent. From
lectures we know that T ∼ Exp(µ). 2,D
Then, using the continuous version of the law of total probability, we have for
k ∈ N0:
P(NT/2 = k) =
∫ ∞
−∞
P(NT/2 = k|T = t)fT (t)dt =
∫ ∞
0
P(Nt/2 = k|T = t)µe−µtdt
N,T independent
=
∫ ∞
0
P(Nt/2 = k)µe−µtdt =
∫ ∞
0
(λt/2)k
k!
e−λt/2µe−µtdt.
2,D
We change the variables z = (λ/2 + µ)t and use the fact that
∫∞
0 z
ke−zdz =
Γ(k + 1) = k! to deduce that
P(NT/2 = k) =
µ
k!
(
λ
2
)k (λ
2
+ µ
)−(k+1) ∫ ∞
0
zke−zdz
= µ
(
λ
2
)k (λ
2
+ µ
)−(k+1)
=
(
λ
λ+ 2µ
)k ( 2µ
λ+ 2µ
)
.
2,D
Hence NT/2 has a geometric distribution with parameter

λ+2µ . 1,D
sim. seen ⇓
(b) (i) If there are i infectives, then there are N − i susceptibles and hence the birth
rate is given by λi = (N − i)i2λ if i = 1, . . . , N − 1 and 0 otherwise.
2,B(ii) (1.) Let Xi be the time spent in state i (where i denotes the number of
infectives), then we have that the time to complete the epidemic is
T = X1 + · · ·+XN−1,
where the Xi are independent of each other with Xi ∼ Exp(λi). By the
linearity of the expectation,
E[T ] =
N−1∑
i=1
E(Xi) =
N−1∑
i=1
1
λi
=
N−1∑
i=1
1
(N − i)i2λ.
2,C
(2.) Let s ≥ 0. Using the notation from (1.), the Laplace transform of Xi is
given by
E(e−sXi) =
∫ ∞
0
e−sxλie−λixdx = λi
∫ ∞
0
e−(s+λi)xdx =
λi
λi + s
.
Hence
E[e−sT ] = E[e−s
∑N−1
i=1 Xi ]
independence of Xis
=
N−1∏
i=1
E[e−sXi ] =
N−1∏
i=1
(
λi
λi + s
)
.
2,C
MATH96052/MATH97083 Applied Probability (Solutions) (2020) Page 8 of 10
(3.) To compute the expectation, we calculate the logarithm of the Laplace
transform and use the fact that
E[T ] = − d
ds
[
log
{
E[e−sT ]
}]∣∣∣∣
s=0
.
Here log
{
E[e−sT ]
}
=
∑N−1
i=1 log
(
λi
λi+s
)
and 1,C
d
ds
log
{
E[e−sT ]
}
=
N−1∑
i=1
(λi + s)
λi
· (λi + s) · 0− λi · 1
(λi + s)2
= −
N−1∑
i=1
1
(λi + s)
.
Hence,
E[T ] =
N−1∑
i=1
1
λi
=
N−1∑
i=1
1
(N − i)i2λ.
1,C
meth seen ⇓(c) Many examples are possible: For instance, take a non-explosive birth process
starting at 0 with rates λi > 0 for i ∈ N0 and a Poisson process of rate λ > 0.
Both processes are continuous-time Markov chains on N0. 3,D
Then the jump chain for both processes, denoted by (Zn)n∈N0 , is given by Zn = n
for n ∈ N0 and its transition probabilities are given by pZij = 1 when j = i + 1
and 0 otherwise (for i, j ∈ N0), since both continuous-time Markov chains are 1,D
non-decreasing processes which can only jump up by one step at a time.
1,C
Alternatively, one could consider Poisson processes of different rates etc.
MATH96052/MATH97083 Applied Probability (Solutions) (2020) Page 9 of 10
5. (a) (i) We say that state i ∈ E is recurrent if P({t ≥ 0 : Xt = i} is unbounded |X0 =
i) = 1. 2
(ii) We say that state i ∈ E is transient if P({t ≥ 0 : Xt = i} is unbounded |X0 =
i) = 0. 1
(b) We derive the transition matrix of the corresponding jump chain:
P =

0 12
1
2 0
1
2 0 0
1
2
1
2 0 0
1
2
0 0 0 1
 .
We observe that the jump chain has two communicating classes: T = {1, 2, 3}, C =
{4}. T is not closed, hence all states in T are transient. C is finite and closed,
hence state 4 is (positive) recurrent. 3
We know that if a state is recurrent (transient) for the jump chain, then it is
recurrent (transient) for the continuous-time Markov chain. So we conclude that
states 1, 2, 3 are transient and state 4 is recurrent for the continuous-time Markov
chain. Moreover, since g44 = 0, we have that state 4 is positive recurrent. 2
(c) (i) We denote the stationary distribution of X by λ = (λ1, λ2). We solve λG = 0,
for λ1, λ2 ≥ 0, λ1 + λ2 = 1. Here we have −λ1 + 2λ2 = 0⇔ λ1 = 2λ2. Then
1 = λ1 + λ2 ⇔ 1 = 3λ2 ⇔ λ1 = 23 , λ2 = 13 . 2
Since G is irreducible and recurrent, we get that λG = 0⇔ λPt = λ for all
t ≥ 0, where (Pt)t≥0 denotes the matrix of transition probabilities associated
with X. 2
(ii) The transition matrix of the associated jump chain is given by
P =
(
0 1
1 0
)
.
We denote the stationary distribution of the jump chain by pi = (pi1, pi2). We
solve piP = pi, for pi1, pi2 ≥ 0, pi1 + pi2 = 1. Here we have pi1 = pi2. Then
1 = pi1 + pi2 ⇔ pi1 = 12 , pi2 = 12 . 2
(iii) Let G be a generator and let P denote the transition matrix of the associated
jump chain. Let λ be a measure. Then λ satisfies λG = 0 if and only if
µP = µ where µi = λi · (−gii) for all i ∈ E. 3
In our case, we can obtain µ as follows: µ1 = λ1 · (−g11) = 23 · 1 = 23 , µ2 =
λ2 · (−g22) = 13 · 2 = 23 , i.e. µ1 = µ2. When we normalise, we obtain that
pi = (0.5, 0.5) is the stationary distribution of the jump chain. 1
(iv) From lectures, we know that, for t ≥ 0, we have Pt = etG =∑∞
n=0
tn
n!OD
nO−1 = O
∑∞
n=0
tn
n!D
nO−1. Hence
Pt = O
(
et·0 0
0 e−3t
)
O−1 =
(
2
3 +
1
3e
−3t 1
3 − 13e−3t
2
3 − 23e−3t 13 + 23e−3t
)
.
2
MATH96052/MATH97083 Applied Probability (Solutions) (2020) Page 10 of 10
MATH97083 1
Overall students did well on this question. The two common mistakes were: first, failure to
represent all possible stationary distributions correctly in iii) by omitting to mention the non‐
negativity constraints. Second, not justifying each stop of the derivation in b)ii)2) properly.
MATH97083 2
A lot of students struggled with the proof of b) ii), and a handful of students struggled with b) i).
This was surprising since b) i) has a much more involved proof than b) ii).
MATH97083 3
Parts a and b were done well; some students unnecessarily repeated calculations in part b which
were identical those in part a. Many students addressed part c using generating functions and
other expectations, which is not the correct approach. Part d was book work and so most
students answered these questions easily, although some did not state enough technical detail
MATH97083 4
There were some mixed efforts in trying to solve Q4, which was possibly the hardest question in
this exam. Very few students gave a detailed derivation with sufficient justifications of all the
required steps in 4a). Q4b) was generally done better, but many students lost marks for using
sloppy notation and/or insufficient justifications. The results in Q4c) were mixed, but it was nice
to see that many students got the general idea right.
MATH97083 5
The Mastery question was generally done well. Some students did not provide sufficient
justifications for the results in 5b) and 5ci). Some struggled with some more detailed
computations in 5ciii) and 5civ).
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