ST2133
Summer 2020 online assessment guidelines
ST2133 Advanced Statistics: Distribution Theory
The assessment will be an open-book take-home online assessment within
a 24-hour window. The requirements for this assessment remain the same as
the originally planned closed-book exam, with an expected time/effort of 2
hours.
Candidates should answer all FOUR questions: QUESTION 1 of Section A (40
marks) and all THREE questions from Section B (60 marks in total). Candidates
are strongly advised to divide their time accordingly.
You should complete your ST2133 paper using pen and paper. Please use BLACK
ink only.
Handwritten work then needs to be scanned, converted to PDF and then uploaded to
the VLE as ONE individual file including the coversheet. Each scanned sheet
should have your candidate number written clearly at the top or bottom. Please do
not write your name anywhere on any sheet.
The paper will be available at 12.00 midday (BST) on Friday 10 July 2020.
You have until 12.00 midday (BST) on Saturday 11 July 2020 to upload your file
into the VLE submission portal. However, you are advised not to leave your
submission to the last minute. We will deduct 5 marks if your submission is up to one
hour late, 10 marks if your submission is more than one hour late but less than two
hours late (etc.).
Workings should be submitted for all questions requiring calculations. Any necessary
assumptions introduced in answering a question are to be stated.
You may use any calculator for any appropriate calculations, but you may not use
any computer software to obtain solutions. Credit will only be given if all workings are
shown.
If you think there is any information missing or any error in any question, then you
should indicate this but proceed to answer the question stating any assumptions you
have made.
The assessment has been designed with a duration of 24 hours to provide a more
flexible window in which to complete the assessment and to appropriately test the
course learning outcomes. As an open-book assessment, the expected amount of
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effort required to complete all questions and upload your answers during this window
is no more than 2 hours. Organise your time well and avoid working all night.
You are assured that there will be no benefit in you going beyond the expected 2
hours of effort. Your assessment has been carefully designed to help you show what
you have learned in the hours allocated.
This is an open book assessment and as such you may have access to additional
materials including but not limited to subject guides and any recommended reading.
But the work you submit is expected to be 100% your own. Therefore, unless
instructed otherwise, you must not collaborate or confer with anyone during the
assessment. The University of London will carry out checks to ensure the academic
integrity of your work. Many students that break the University of London’s
assessment regulations did not intend to cheat but did not properly understand the
University of London’s regulations on referencing and plagiarism. The University of
London considers all forms of plagiarism, whether deliberate or otherwise, a very
serious matter and can apply severe penalties that might impact on your award. The
University of London 2019-20 Procedure for the Consideration of Allegations of
Assessment offences is available online at:
https://london.ac.uk/sites/default/files/governance/assessment-offence-procedure-
year-2019-2020.pdf
The University of London’s Rules for Taking Online Timed Assessments have been
included in an update to the University of London General Regulations and are
available at:
https://london.ac.uk/sites/default/files/regulations/progregs-general-2019-2020.pdf
© University of London 2020
Section A
Answer all three parts of question 1 (40 marks in total)
1. (a) The probability density function of a random variable X is given by
fX(x) =
 cx, 0 < x < 2;0, otherwise,
where c is a constant.
i. Find the value of c. [4 marks]
ii. Find E(|X − 1|). [4 marks]
iii. Let Y be an independent and identically distributed copy of X.
Find P (X + Y > 2). [5 marks]
(b) The probability mass function of X is given by
pX(x) =
µxe−µ
x!
, x = 0, 1, . . . ,
where µ > 0 is a constant.
i. Show that the moment generating function MX(t) of X is given by
MX(t) = exp(µe
t−µ), t ∈ R. [4 marks]
ii. Show that Var(X) = µ. [4 marks]
iii. Find E[X exp(X+Y )], where Y is an independent and identically distributed
copy of X. (Hint: M ′X(t) = E(X exp(tX)).) [5 marks]
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.
(c) In a factory, there are three machines producing identical products. The time it
takes for each machine to produce a product is exponentially distributed, such
that Xi ∼ Exp(λ), i = 1, 2, 3, and the machines are independent of each other.
The probability density function is given by
fX(x) = λe
−λx, x > 0.
i. Let T be the time until a product is produced by any of the 3 machines. Show
that
P (T > t) = exp(−3λt).
What is the distribution of T? (Hint: T > t if and only if Xi > t for all
i = 1, 2, 3). [4 marks]
ii. It was known that X2 > c. Show that
P (T > t|X2 > c) =
 exp(−2λt), 0 < t < c;exp(−3λt+ λc), t ≥ c.
[5 marks]
iii. Find E(T |X2 > c). You can use, for any positive random variable Y ,
E(Y ) =
∫ ∞
0
P (Y > y)dy.
[5 marks]
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.
Section B
Answer all three questions in this section (60 marks in total)
2. The joint probability density function of X and Y is given by
fX,Y (x, y) =
ax2y2√
x2 + y2
, 0 < x < y < 1.
Let
U =
X√
X2 + Y 2
, V = Y.
(a) Show that
X =
UV√
1− U2 and Y = V. [2 marks]
(b) Show that
0 < U <
1√
2
and 0 < V < 1. [3 marks]
(c) Show that
fU,V (u, v) =
au2v4
(1− u2)2 , 0 < u <
1√
2
, 0 < v < 1, and
fU(u) =
au2
5(1− u2)2 , 0 < u <
1√
2
. [10 marks]
(d) Find the value of a. Hint: use integration by parts on the left hand side of
∫ 1/√2
0
1
1− u2du = log(1 +
√
2). [5 marks]
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.
3. You are playing a game where you repeatedly throw a fair die (assume that throws
are independent). The game ends when you roll a 6. Let N be the total number
of throws. On each throw (including the last one), you win 1 point for an even
number, and lose 1 point for an odd number.
Let Xi be the points gained from the ith throw of the die, so that Xi = 1 or −1,
and the total points obtained at the end of the game is given by
W =
N∑
i=1
Xi.
(a) Write down the probability mass function of N . [3 marks]
(b) Show that P (Xi = 1|N > i) = 2/5 and P (Xi = −1|N > i) = 3/5. What is
P (Xi = 1|N = i)? [4 marks]
(c) Calculate E(etXi|N > i) and E(etXi|N = i). [3 marks]
(d) Show that the conditional moment generating function ofW given N is given
by
MW |N(t) = et
(
2et + 3e−t
5
)N−1
, t ∈ R.
[4 marks]
(e) Show that the moment generating function of W is given by
MW (t) =
et
6− 2et − 3e−t , 2e
t + 3e−t < 6.
[6 marks]
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.
4. In a particular emergency scenario, an ambulance arrives in time with probability
0.9. After arriving, the amount of time T the paramedics spent before leaving the
scene has probability density function (t > 0)
fT (t) =
 2e
−2t, ambulance arrives in time;
3e−3t, otherwise.
(a) Show that P (T > t) = (9e−2t + e−3t)/10. [2 marks]
(b) One observes that the paramedics spent less than 1 unit of time before
leaving. What is the probability they arrived in time? [5 marks]
(c) Let X be the time used for the paramedics to return to their hospital. The
probability density of X is given by (x > 0)
fX(x) =
 e
−x, T < c;
2e−2x, T ≥ c,
where c > 0. Let A be the event that the ambulance arrives in time.
i. Find P (X > c ∩ T < c ∩ A) and P (X > c ∩ T ≥ c ∩ A). [6 marks]
ii. By conditioning on T < c or T ≥ c only, show that
P (X > c) = e−c(1− 0.9e−2c + 0.8e−3c + 0.1e−4c). [4 marks]
iii. Using parts i and ii, show that
P (Ambulance arrives in time|X > c) = 9− 9e
−2c + 9e−3c
10− 9e−2c + 8e−3c + e−4c .
[3 marks]
UL20/0535
END OF PAPER
.
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.
Examiners’ commentaries 2020
Examiners’ commentaries 2020
ST2133 Advanced statistics: distribution theory
Important note
This commentary reflects the examination and assessment arrangements for this course in the
academic year 2019–20. The format and structure of the examination may change in future years,
and any such changes will be publicised on the virtual learning environment (VLE).
Information about the subject guide and the Essential reading
references
Unless otherwise stated, all cross-references will be to the latest version of the subject guide (2018).
You should always attempt to use the most recent edition of any Essential reading textbook, even if
the commentary and/or online reading list and/or subject guide refer to an earlier edition. If
different editions of Essential reading are listed, please check the VLE for reading supplements – if
none are available, please use the contents list and index of the new edition to find the relevant
section.
General remarks
Learning outcomes
At the end of this half course and having completed the Essential reading and activities, you should
be able to:
recall a large number of distributions and be a competent user of their mass/density and
distribution functions and moment generating functions
explain relationships between variables, conditioning, independence and correlation
relate the theory and method taught in the half course to solve practical problems.
Candidate performance
Candidates performed better in general compared to previous years, partly, inevitably, due to the
fact that they had more time to write their answers. Still, this examination successfully
differentiated candidates with a wide range of abilities.
Key steps to improvement
The ability to carefully calculate probabilities and moments are always assessed in each year’s
examination. Questions 1 (a) and 1 (b) (at least up to part ii.) are testing fundamental techniques in
probability and moment calculations, which are important to master in the course. The same goes
1
ST2133 Advanced statistics: distribution theory
for Question 2 when finding the region of integration after transforming the random variables, and
the calculation of joint densities for transformed random variables.
When calculating a probability or an expectation, especially when evaluating a double integral,
many candidates got the result wrong because of carelessly placing the wrong limits of integration.
Please practise more on how to find the limits correctly for a particular region of a joint density.
You should be ready to derive the moment generating functions of standard random variables, like
the normal, gamma, chi-squared, exponential (all continuous), or the geometric, binomial, Poisson
(all discrete), and ideally know the forms by heart. Examples include Question 1(b) i. and also
Question 3 (a). It is also important to know basic applications of these distributions, and apply the
correct formulae in probability questions.
Examination revision strategy
Many candidates are disappointed to find that their examination performance is poorer than they
expected. This may be due to a number of reasons, but one particular failing is ‘question
spotting’, that is, confining your examination preparation to a few questions and/or topics which
have come up in past papers for the course. This can have serious consequences.
We recognise that candidates might not cover all topics in the syllabus in the same depth, but you
need to be aware that examiners are free to set questions on any aspect of the syllabus. This
means that you need to study enough of the syllabus to enable you to answer the required number of
examination questions.
The syllabus can be found in the Course information sheet available on the VLE. You should read
the syllabus carefully and ensure that you cover sufficient material in preparation for the
examination. Examiners will vary the topics and questions from year to year and may well set
questions that have not appeared in past papers. Examination papers may legitimately include
questions on any topic in the syllabus. So, although past papers can be helpful during your revision,
you cannot assume that topics or specific questions that have come up in past examinations will
occur again.
If you rely on a question-spotting strategy, it is likely you will find yourself in difficulties
when you sit the examination. We strongly advise you not to adopt this strategy.
2
Examiners’ commentaries 2020
Examiners’ commentaries 2020
ST2133 Advanced statistics: distribution theory
Important note
This commentary reflects the examination and assessment arrangements for this course in the
academic year 2019–20. The format and structure of the examination may change in future years,
and any such changes will be publicised on the virtual learning environment (VLE).
Information about the subject guide and the Essential reading
references
Unless otherwise stated, all cross-references will be to the latest version of the subject guide (2018).
You should always attempt to use the most recent edition of any Essential reading textbook, even if
the commentary and/or online reading list and/or subject guide refer to an earlier edition. If
different editions of Essential reading are listed, please check the VLE for reading supplements – if
none are available, please use the contents list and index of the new edition to find the relevant
section.
Comments on specific questions
Candidates should answer all FOUR questions: QUESTION 1 of Section A (40 marks) and all
THREE questions from Section B (60 marks in total). Candidates are strongly advised to
divide their time accordingly.
Section A
Answer all three parts of question 1 (40 marks in total).
Question 1
(a) The probability density function of a random variable X is given by:
fX(x) =
{
cx for 0 < x < 2
0 otherwise
where c is a constant.
i. Find the value of c.
(4 marks)
ii. Find E(|X − 1|).
(4 marks)
iii. Let Y be an independent and identically distributed copy of X.
Find P (X + Y > 2).
(5 marks)
3
ST2133 Advanced statistics: distribution theory
Reading for this question
Chapter 3, especially Section 3.3.2 for parts i. and ii. For part iii., read Chapter 4, especially
Section 4.2.2.
Approaching the question
Part i. was well-answered. Mistakes were made with very careless calculations by a few
candidates. Many more mistakes were made in part ii., with many not knowing how to deal
with the absolute value of a random variable. The key is to first remove the absolute sign.
To do this, you need to differentiate two cases, one is when X ≥ 1, when |X − 1| = X − 1.
The other case is X < 1, when |X − 1| = −(X − 1). Hence using first principles of
expectation calculation:
E(|X − 1|) =
∫ 1
0
−(x− 1) cxdx+
∫ 2
1
(x− 1) cxdx.
Several candidates got the answer wrong in part iii. because of incorrect limits of integration
in the double integral. Since X + Y > 2, we must have X > 2− Y , so that, with the
imposed upper limit of X, we must have 2− Y < X ≤ 2. These will be the limits of
integration of the inner integral. Then in the outer integral, Y can vary freely from 0 to 2
(more rigorously, 0 < Y ≤ 2, because Y cannot be 0 since then P (X > 2) = 0).
i. We must have:
1 =
∫ 2
0
cxdx = c
[
x2
2
]2
0
= 2c
so that c = 1/2.
ii. We have:
E(|X − 1|) =
∫ 1
0
−(x− 1) cxdx+
∫ 2
1
(x− 1) cxdx
=
[
cx2
2
− cx
3
3
]1
0
+
[
cx3
3
− cx
2
2
]2
1
=
( c
2
− c
3
)
+
(
8c
3
− 2c
)
−
( c
3
− c
2
)
= c
=
1
2
.
iii. We have:
P (X + Y > 2) =
∫ 2
0
∫ 2
2−y
c2xy dx dy
= c2
∫ 2
0
y
[
x2
2
]2
2−y
dy
= c2
∫ 2
0
(
2y − y
3 − 4y2 + 4y
2
)
dy
= c2
[
−y
4
8
+
2y3
3
]2
0
=
10c2
3
=
5
6
.
4
Examiners’ commentaries 2020
(b) The probability mass function of X is given by:
pX(x) =
µxe−µ
x!
for x = 0, 1, . . .
where µ > 0 is a constant.
i. Show that the moment generating function MX(t) of X is given by:
MX(t) = exp(µe
t − µ) for t ∈ R.
(4 marks)
ii. Show that Var(X) = µ.
(4 marks)
iii. Find E(X exp(X + Y )), where Y is an independent and identically
distributed copy of X. (Hint: M ′X(t) = E(X exp(tX)).)
(5 marks)
Reading for this question
Chapter 3, especially Section 3.5.
Approaching the question
Part i. was well-answered since it is just the derivation of a moment generating function of
an elementary distribution, with which you should be familiar. Part ii. was a standard
application of the moment generating function from part i., and many candidates were able
to get their answers right. Some derived the variance from first principles by calculating
E(X) =
∑
x x pX(x) and E(X(X − 1)) =
∑
x(x− 1)x pX(x), and then:
Var(X) = E(X(X − 1)) + E(X)− (E(X))2
which is fine. Some candidates also calculated the cumulant generating function and
differentiated it twice, which should be the fastest solution. For part iii., the key is to note
that:
X exp(X + Y ) = X exp(X) exp(Y )
where X exp(X) and exp(Y ) are independent since X and Y are. Hence:
E(X exp(X + Y )) = E(X exp(X)) E(exp(Y )).
Finally, note that MX(t) = E(e
tX) and M ′X(t) = E(X exp(tX)), and hence
MX(1) = E(exp(X)) and M
′
X(1) = E(X exp(X)). Many candidates made the mistake of
thinking E(X exp(X + Y )) = E(X exp(2X)), which is completely wrong since X and Y are
independent, and hence cannot be equal in general!
i. We have:
MX(t) = E(e
tX) =
∞∑
x=0
(µet)xe−µ
x!
= e−µ exp(µet) = exp(µet − µ)
for t ∈ R.
ii. Consider:
M ′X(t) = exp(µe
t − µ)µet = µetMX(t)
and so:
M ′′X(t) = µe
tMX(t) + µe
tM ′X(t).
Now M ′X(0) = µ, and so:
Var(X) = M ′′X(0)− µ2 = µ+ µ2 − µ2 = µ.
5
ST2133 Advanced statistics: distribution theory
iii. Note that:
M ′X(t) = E
(
d
dt
etX
)
= E(XetX).
Hence:
E(X exp(X + Y )) = E(X exp(X)) E(exp(Y ))
= M ′X(1)MY (1)
= M ′X(1)MX(1)
= µe(MX(1))
2
= µ exp(1 + 2µe− 2µ).
(c) In a factory, there are three machines producing identical products. The time it
takes for each machine to produce a product is exponentially distributed, such
that Xi ∼ Exp(λ), for i = 1, 2, 3, and the machines are independent of each
other. The probability density function is given by:
fX(x) = λe
−λx for x > 0.
i. Let T be the time until a product is produced by any of the 3 machines.
Show that:
P (T > t) = exp(−3λt).
What is the distribution of T? (Hint: T > t if and only if Xi > t for all
i = 1, 2, 3.)
(4 marks)
ii. It was known that X2 > c. Show that:
P (T > t |X2 > c) =
{
exp(−2λt) for 0 < t < c
exp(−3λt+ λc) for t ≥ c.
(5 marks)
iii. Find E(T |X2 > c). You can use, for any positive random variable Y :
E(Y ) =
∫ ∞
0
P (Y > y) dy.
(5 marks)
Reading for this question
Chapter 3 for part i., Chapter 5 for parts ii. and iii., especially Section 5.4.
Approaching the question
Part i. requires you to realise that the event {T > t} is equivalent to Xi > t for all i = 1, 2, 3,
which is already given in the hint. Hence using the independence of X1, X2 and X3, we have:
P (T > t) = P (X1 > t,X2 > t,X3 > t) = P (X1 > t)P (X2 > t)P (X3 > t)
and you should know how to calculate P (X1 > t) from first principles, and P (X2 > t) and
P (X3 > t) are exactly the same. Part ii. was one of the worst answered in the whole paper.
You need to use Bayes’ theorem and know how to simplify {T > t} ∩ {X2 > c}. Note that:
{T > t} ∩ {X2 > c} = {X1 > t ∩X2 > t ∩X3 > t ∩X2 > c}
= {X1 > t ∩X3 > t ∩ (X2 > t and X2 > c)}.
6
Examiners’ commentaries 2020
Also, we have:
(X2 > t and X2 > c) =
{
{X2 > c} for t < c
{X2 > t} for t ≥ c.
The majority of candidates did not know how to deal with the above, and attempted to
make up some as-if-correct derivations. Unfortunately, it is not difficult to see when a
candidate does not have an idea on how to derive the solution in the first place. Part iii. just
needed you to replace P (Y > y) given in the hint with P (T > t |X2 > c). Many candidates
got this right. Some mistakes were also made when calculating the integral involved.
i. We have:
P (T > t) = P (Xi > t, i = 1, 2, 3)
= P (X1 > t)P (X2 > t)P (X3 > t)
= (P (X1 > t))
3
=
(∫ ∞
t
λe−λx dx
)3
= e−3λt.
Hence T ∼ Exp(3λ).
ii. We have:
P (T > t |X2 > c) = P (X1 > t,X2 > t,X3 > t,X2 > c)
P (X2 > c)
=

P (X1 > t)P (X2 > c)P (X3 > t)
P (X2 > c)
for 0 < t < c
P (X1 > t)P (X2 > t)P (X3 > t)
P (X2 > c)
for t ≥ c
=

e−2λt for 0 < t < c
e−2λte−λt
e−λc
for t ≥ c
=
{
e−2λt for 0 < t < c
e−3λt+λc for t ≥ c.
iii. We have:
E(T |X2 > c) =
∫ ∞
0
P (T > t |X2 > c) dt
=
∫ c
0
e−2λt dt+
∫ ∞
c
e−3λt+λc dt
=
[
e−2λt
−2λ
]c
0
+
[
e−3λt+λc
−3λ
]∞
c
=
1− e−2λc
2λ
+
e−2λc
3λ
=
1
2λ
− 1
6λ
e−2λc.
7
ST2133 Advanced statistics: distribution theory
Section B
Answer all three questions in this section (60 marks in total).
Question 2
The joint probability density function of X and Y is given by:
fX,Y (x, y) =
ax2y2√
x2 + y2
for 0 < x < y < 1.
Let:
U =
X√
X2 + Y 2
and V = Y.
(a) Show that:
X =
UV√
1− U2 and Y = V.
(2 marks)
(b) Show that:
0 < U <
1√
2
and 0 < V < 1.
(3 marks)
(c) Show that:
fU,V (u, v) =
au2v4
(1− u2)2 for 0 < u <
1√
2
, 0 < v < 1
and:
fU(u) =
au2
5(1− u2)2 for 0 < u <
1√
2
.
(10 marks)
(d) Find the value of a. Hint: use integration by parts on the left-hand side of:∫ 1/√2
0
1
1− u2 du = log(1 +
√
2).
(5 marks)
Reading for this question
Chapter 4, especially Section 4.6.
Approaching the question
Part (a) was in general well-answered, which is just simple algebra. Part (b) was also
well-answered, where you needed to use 0 < X < Y < 1, and replace X and Y with the
corresponding functions of U and V , respectively, then solve for U and V . Part (c) was again
well-answered in general, with the transformation formula well-derived and the Jacobian
calculation quite accurate in general. Part (d) proved more challenging for many. The key is to
realise that you need to calculate:∫ 1/√2
0
fU (u) du =
a
5
∫ 1/√2
0
u2
(1− u2)2 du
which can be obtained if you use integration by parts correctly using the hint.
8
Examiners’ commentaries 2020
(a) We have:
U =
X√
X2 + V 2
⇒ U2X2 −X2 = −U2V 2
hence:
X =
UV√
1− U2
since X is positive.
(b) We have:
0 <
UV√
1− U2 < V < 1.
Solving for UV/
√
1− U2 < V , we have:
0 < U <
1√
2
since U and V are positive. Finally, the right-hand inequality means 0 < V < 1.
(c) The joint density of U and V is:
fU,V (u, v) = fX,Y
(
uv√
1− u2 , v
) ∣∣∣∣∣∣∣
∣∣∣∣∣∣∣
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
∣∣∣∣∣∣∣
∣∣∣∣∣∣∣
= auv2
uv√
1− u2
∣∣∣∣∣∣
∣∣∣∣∣∣
v√
1− u2 +
u2v
(1− u2)3/2
u√
1− u2
0 1
∣∣∣∣∣∣
∣∣∣∣∣∣
=
au2v3
(1− u2)1/2
v
(1− u2)3/2
=
au2v4
(1− u2)2 for 0 < u <
1√
2
and 0 < v < 1.
Hence:
fU (u) =
au2
(1− u2)2
∫ 1
0
v4 dv =
au2
5(1− u2)2 for 0 < u <
1√
2
and 0 otherwise.
(d) Using integration by parts on the left-hand side of the given integral:
log(1 +
√
2) =
[
u
1− u2
]1/√2
0
−
∫ 1/√2
0
u
2u
(1− u2)2 du
hence: ∫ 1/√2
0
u2
(1− u2)2 du =
√
2− log(1 +√2)
2
.
Therefore:
1 =
∫ 1/√2
0
fU (u) du = a
√
2− log(1 +√2)
10
⇒ a = 10√
2− log(1 +√2) .
Question 3
You are playing a game where you repeatedly throw a fair die (assume that throws
are independent). The game ends when you roll a 6. Let N be the total number of
9
ST2133 Advanced statistics: distribution theory
throws. On each throw (including the last one), you win 1 point for an even
number, and lose 1 point for an odd number.
Let Xi be the points gained from the ith throw of the die, so that Xi = 1 or −1,
and the total points obtained at the end of the game is given by:
W =
N∑
i=1
Xi.
(a) Write down the probability mass function of N .
(3 marks)
(b) Show that P (Xi = 1 |N > i) = 2/5 and P (Xi = −1 |N > i) = 3/5. What is
P (Xi = 1 |N = i)?
(4 marks)
(c) Calculate E(etXi |N > i) and E(etXi |N = i).
(3 marks)
(d) Show that the conditional moment generating function of W given N is given
by:
MW |N(t) = et
(
2et + 3e−t
5
)N−1
for t ∈ R.
(4 marks)
(e) Show that the moment generating function of W is given by:
MW (t) =
et
6− 2et − 3e−t for 2e
t + 3e−t < 6.
(6 marks)
Reading for this question
Chapter 5, especially Section 5.6.
Approaching the question
Part (a) is a standard question which needs you to realise N has a geometric distribution under
this specific application. Then you need to specify the probability of success, which many
candidates answered well. Part (b) was well-answered in general. The key to answer this lies in
the fact that as long as N > i, then Xi cannot be a 6. That means given N > i, Xi can only be
1, 2, 3, 4 or 5. Those who could answer part (b) answered part (c) correctly in general, which
required you to use part (b) and then use first principles to calculate the conditional moment
generating function of Xi given N > i or N = i.
Part (d) needed one to utilise the answers in part (c) and still realise that the Xis are
independent given N > i, and when N = i the moment generating function is just et. You also
need to realise the calculation is assuming N is known, which can then be treated as a constant.
Finally, part (e) was well-answered for those candidates who in general got the previous part
right. There were still careless mistakes but most got the general idea of how to proceed, which is
to take the expectation of the answer in part (d), and realise that such an expectation can be
seen as a moment generating function of N with a particular value of t.
(a) It follows a geometric distribution with probability of success 1/6. Hence:
pN (n) =
(
5
6
)n−1
1
6
for n = 1, 2, . . .
and 0 otherwise.
10
Examiners’ commentaries 2020
(b) If it is a 6, the game ends. Hence when N > i, Xi can only take values from 1 to 5. Hence:
P (Xi = 1 |N > i) = P (Outcome = 2 or 4 | 6 excluded) = 2
5
and:
P (Xi = −1 |N > i) = P (Outcome = 1, 3 or 5 | 6 excluded) = 3
5
.
Also P (Xi = 1 |N = i) = 1 since the last throw has to be 6, so that XN = 1.
(c) We have:
E(etXi |N > i) = et × 2
5
+ et(−1) × 3
5
=
2et + 3e−t
5
and:
E(etXi |N = i) = et × 1 = et.
(d) We have:
MW |N (t) = E(etW |N) = E
(
exp
(
t
N∑
i=1
Xi
)∣∣∣N)
=
N∏
i=1
E(etXi)
=
(
N−1∏
i=1
E(etXi)
)
E(etXN )
= et
(
2et + 3e−t
5
)N−1
for t ∈ R.
(e) The moment generating function MN (t) of N is:
MN (t) =
∞∑
n=1
(
5et
6
)n
1
5
=
et
6− 5et for t < log
(
6
5
)
.
Hence:
MW (t) = E(e
tW ) = E(E(etW |N))
= E(MW |N (t))
= E
(
et
(
2et + 3e−t
5
)N−1)
=
5et
2et + 3e−t
E
(
exp
(
N log
(
2et + 3e−t
5
)))
=
5et
2et + 3e−t
MN
(
log
(
2et + 3e−t
5
))
=
5et
2et + 3e−t
2et + 3e−t
5(6− 2et − 3e−t) for log
(
2et + 3e−t
5
)
< log
(
6
5
)
=
et
6− 2et − 3e−t for 2e
t + 3e−t < 6.
11
ST2133 Advanced statistics: distribution theory
Question 4
In a particular emergency scenario, an ambulance arrives in time with probability
0.9. After arriving, the amount of time T the paramedics spent before leaving the
scene has probability density function (t > 0):
fT (t) =
{
2e−2t ambulance arrives in time
3e−3t otherwise.
(a) Show that P (T > t) = (9e−2t + e−3t)/10.
(2 marks)
(b) One observes that the paramedics spent less than 1 unit of time before leaving.
What is the probability they arrived in time?
(5 marks)
(c) Let X be the time used for the paramedics to return to their hospital. The
probability density of X is given by (x > 0):
fX(x) =
{
e−x for T < c
2e−2x for T ≥ c
where c > 0. Let A be the event that the ambulance arrives in time.
i. Find P (X > c ∩ T < c ∩A) and P (X > c ∩ T ≥ c ∩A).
(6 marks)
ii. By conditioning on T < c or T ≥ c only, show that:
P (X > c) = e−c(1− 0.9e−2c + 0.8e−3c + 0.1e−4c).
(4 marks)
iii. Using parts i. and ii., show that:
P (Ambulance arrives in time |X > c) = 9− 9e
−2c + 9e−3c
10− 9e−2c + 8e−3c + e−4c .
(3 marks)
Reading for this question
Chapter 2, especially Section 2.4 for Bayes’ theorem. Chapter 3.
Approaching the question
Part (a) requires you to use the law of total probability and condition T > t on whether an
ambulance arrives in time or not. Many candidates got this right. Part (b) is a standard
application for using Bayes’ theorem, and many got this correct too. Part (c) i. requires you to
write the probabilities into a product of those you can calculate. For example:
P (X > c ∩ T < c ∩A) = P (X > c |T < c ∩A)P (T < c ∩A)
= P (X > c |T < c)P (T < c |A)P (A)
where we used what we are given in the question to know that the probability of X, given T < c
or T ≥ c, is independent of the event A, and hence P (X > c |T < c ∩A) = P (X > c |T < c).
Finally, part (c) ii. is just a use of the law of total probability conditioning on T < c or T ≥ c,
which was answered well. Part (c) iii. is just a simple use of Bayes’ theorem using part i and ii.
12
Examiners’ commentaries 2020
(a) Let A denote the event ‘Ambulance arrives in time’. Hence:
P (T > t) = P (T > t |A)P (A) + P (T > t |Ac)P (Ac)
= 0.9×
∫ ∞
t
2e−2x dx+ 0.1×
∫ ∞
t
3e−3x dx
=
9e−2t + e−3t
10
.
(b) We have:
P (A |T < 1) = P (T < 1 |A)P (A)
P (T < 1)
=
0.9(1− e−2)
1− 0.9e−2 − 0.1e−3 .
(c) i. We have:
P (X > c, T < c,A) = P (X > c |T < c,A)P (T < c |A)P (A)
= P (X > c |T < c)P (T < c |A)P (A)
= e−c(1− e−2c)0.9
= 0.9e−c(1− e−2c).
Similarly:
P (X > c, T ≥ c, A) = P (X > c |T ≥ c, A)P (T ≥ c |A)P (A)
= e−2ce−2c × 0.9
= 0.9e−4c.
ii. We have:
P (X > c) = P (X > c |T < c)P (T < c) + P (X > c |T ≥ c)P (T ≥ c)
= e−c(1− 0.9e−2c − 0.1e−3c) + e−2c(0.9e−2c + 0.1e−3c)
= e−c(1− 0.9e−2c + 0.8e−3c + 0.1e−4c).
iii. We have:
P (A |X > c) = P (X > c, T < c,A) + P (X > c, T ≥ c, A)
P (X > c)
=
0.9e−c(1− e−2c) + 0.9e−4c
e−c(1− 0.9e−2c + 0.8e−3c + 0.1e−4c)
=
9− 9e−2c + 9e−3c
10− 9e−2c + 8e−3c + e−4c .
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