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程序代写案例-ST2133

时间：2021-05-05

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

UL20/0535

Page 1 of 7

UL20/0535

Page 2 of 7

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]

UL20/0535 .

Page 3 of 7

.

(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]

UL20/0535 .

Page 4 of 7

.

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]

UL20/0535 .

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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]

UL20/0535 .

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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

.

Page 7 of 7

.

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 .

13

学霸联盟

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

UL20/0535

Page 1 of 7

UL20/0535

Page 2 of 7

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]

UL20/0535 .

Page 3 of 7

.

(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]

UL20/0535 .

Page 4 of 7

.

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]

UL20/0535 .

Page 5 of 7

.

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]

UL20/0535 .

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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

.

Page 7 of 7

.

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 .

13

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