mid test solution 2015

STUDENT’S SURNAME ………………………………

OTHER NAMES ………………………………

STUDENT NUMBER ………………………………

2015 Second Semester

Class Examination

Unit : ACST307/817 – Quantitative Asset and Liability Modelling 2

Date : Monday 19th October 2015

Time Allowed : 100 minutes

No. of Questions: 3 (Answer ALL questions in the spaces provided).

Total Marks : 50 (Marks are shown at the end of each question)

Instructions : There is 1 question booklet which contains 3 questions. Write

your answers in the separate answer booklet provided.

Show ALL necessary working.

WRITE YOUR NAME AND STUDENT NUMBER ON

QUESTION/ANSWER BOOKLET. YOU MUST DO THIS

BEFORE YOU START THE TEST.

Candidates must cease writing immediately when instructed

to do so by the supervisor at the end of the examination.

Failure to do so will result in a mark of ZERO.

Calculators : You will be allowed to take a calculator that is silent and has

no text‐retrieval capacity.

Reference

Materials :

You will be allowed to take one A4 page into the exam

(handwritten or typed on one or two sides). The standard

normal table is provided on the last page.

Assessment : This test contributes to 10% of your final assessment.

Normal University Examination rules apply to the conduct of this test.

Student Name: ...................................

Student ID: ................................

ACST 307/817: Quantitative Asset-Liability Modelling 2

2015 Mid-Semester Examination

Question 1 [Total 22 marks]

Consider the following firm:

- XYZ mines copper, with fixed cost of $0.5/lb and variable cost of $0.4/lb.

Fixed cost is a cost which must be paid regardless of whether the prospected

copper is mined or not.

The 1-year forward price of copper is $1/lb. The continuously compounded

risk-free interest rate is 6% per annum. One-year option prices for copper are

shown in the table below.

Strike Call Put

0.950 $0.0649 $0.0178

0.975 $0.0500 $0.0265

1.000 $0.0376 $0.0376

1.025 $0.0274 $0.0509

1.034 $0.0243 $0.0563

1.050 $0.0194 $0.0665

In your answers, at a minimum consider copper prices in 1 year of $0.70, $0.80,

$0.90, $1.00, $1.10 and $1.20.

(i) If XYZ does nothing to manage copper price risk, what is its profit 1 year from

now, per pound of copper? If on the other hand XYZ sells forward its expected

copper production, what is its estimated profit 1 year from now? Construct

graphs illustrating both unhedged and hedged profit against copper price. [4

marks]

(ii) Suppose the 1-year copper forward price were $0.45 instead of $1. If XYZ

were to sell forward its expected copper production, what is its estimated profit

1 year from now? Should XYZ produce copper? [5 marks]

2

(iii) Compute estimated profit in 1 year if XYZ buys a put option with a strike

of $1.00 . Also draw a graph of profit against copper price for this case. [4

marks]

(iv) Compute estimated profit in 1 year if XYZ sells a call option with a strike

of $1.00 . Also draw a graph of profit against copper price for this case. [4

marks]

(v) Compute estimated profit in 1 year if XYZ buys a collars with the strike

$1.05 for the put and the strike $1.05 for the call (i.e. purchasing the put option

while simultaneously selling the call option). Also draw a graph of profit against

copper price for this case. [5 marks]

3


(i) Let be the time- price of a nondividend-paying stock. You are given that

- The stock price process is

= 0075+

where is a standard Brownian motion under the true probability distribution,

and 0 is a constant.

- Let ( ) be the time-t price of a derivative written on , when the time-

stock price is . Then ( ) satisfies

( )

+ 0045

( )

+ 00452

2 ( )

2= 0045 ( )

Find the market price of risk of the stock. [5 marks].

(ii) Assume the Black-Scholes framework. You are given:

- is the price of a nondividend-paying stock at time .

- The stock price process is given by

= 005+ 025

where is a standard Brownian motion under the true probability distribution.

- Under the risk-neutral probability distribution, the mean of 05 is −003.

Calculate the continuously compounded risk-free interest rate. [5 marks].

4


(i) Let { ≥ 0} be a standard Brownian motion and be the time- price

of a stock. It is given that

= 1000035+03

where 0 = 100.

(a) Calculate the mean and variance of 2 [2 marks]

(b) Find Pr(98 ≤ 2 ≤ 103). [2 marks]

(c) Find the expression for 2 in terms of 1, based on which find

Pr (98 ≤ 2 ≤ 103 | 1 = 102) .[4 marks]

(ii) Let be a standard Brownian motion and 0 be a constant. Find for

the following:

= 0

−2

2

+.

[4 marks]

(iii) Let be a standard Brownian motion. You are given that

- = 2− 3−,

- = √.

Suppose that

= ( ) + ( )

Find (1 2)

(1 2)

[6 marks]

The End

5

z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.53590.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.57530.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.61410.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.65170.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879

0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.72240.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.75490.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.78520.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.81330.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389

1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.86211.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.88301.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.90151.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.91771.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319

1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.94411.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.95451.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.96331.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.97061.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767

2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.98172.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.98572.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.98902.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.99162.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936

2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.99522.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.99642.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.99742.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.99812.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986

3.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.99903.1 0.9990 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.99933.2 0.9993 0.9993 0.9994 0.9994 0.9994 0.9994 0.9994 0.9995 0.9995 0.99953.3 0.9995 0.9995 0.9995 0.9996 0.9996 0.9996 0.9996 0.9996 0.9996 0.99973.4 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9997 0.9998

Standard Normal Cumulative Probability Table

Cumulative probabilities for POSITIVE z-values are shown in the following table:

2015 Class Examination

Model solutions


(i) The following table summarizes the unhedged and hedged profit calculations:

Copper price

in one yearTotal cost Unhedged profit Profit on short forward

Net income on

hedged profit

$070 $090 −$020 $030 ($1− $070) $010 ($1− $090)$080 $090 −$010 $020 ($1− $080) $010

$090 $090 $000 $010 $010

$100 $090 $010 $000 $010

$110 $090 $020 −$010 $010

$120 $090 $030 −$020 $010

[3 marks]

We obtain the following profit diagram:

1

[1 mark]

(ii) If the forward price were $0.45 instead of $1, we would get the following table:

Copper price

in one yearTotal cost Unhedged profit Profit on short forward

Net income on

hedged profit

$070 $090 −$020 −$025 ($045− $070) −$045 ($045− $090)$080 $090 −$010 −$035 ($045− $080) −$045$090 $090 −$000 −$045 −$045$100 $090 $010 −$055 −$045$110 $090 $020 −$055 −$045$120 $090 $030 −$075 −$045

[3 marks]

Although the copper forward price of $0.45 is below our total costs of $0.90, it

is higher than the variable cost of $0.40. It still makes sense to produce copper

because even at a price of $0.45 in one year, we will be able to partially cover

our fixed costs. [2 marks]

(iii) In this exercise, we need to find the future value of the put premia. For the

$1-strike put, it is: $00376× 006 = 3 992 5× 10−2 ≈ $004. The table on thefollowing page shows the profit calculations for the $1.00-strike put. The figure

below compares the profit diagrams of all other possible hedging strategies.

Copper

price in

one year

Total costUnhedged

profit

Profit on long

$1-strike put

option

Put

premium

paid

Net income

on hedged profit

$070 $090 −$020 $030 ($1− $070) $004$006

($1− $090− $004)$080 $090 −$010 $020 ($1− $070) $004

$006

($1− $090− $004)$090 $090 $000 $010 $004

$006

($1− $090− $004)$100 $090 $010 $000 $004

$006

($1− $090− $004)$110 $090 $020 $000 $004

$016

($110− $090− $004)$120 $090 $030 $000 $004

$026

($120− $090− $004)

[3 marks]

Profit diagram of the different put strategies:

2

[1 mark]

(iv) The future value of the $1.00-strike call premium amounts to: $00376 ×006 = $004.

Copper

price in

one year

Total costUnhedged

profit

Profit on short

$1-strike call

option

Call

premium

received

Net income

on hedged profit

$070 $090 −$020 $000 $004−$006($070 + 004− $090)

$080 $090 −$010 $000 $004−$006($080 + 004− $090)

$090 $090 $000 $000 $004$004

($090 + 004− $090)$100 $090 $010 $000 $004

$014

($1 + $004− $090)$110 $090 $020 −$010 ($1− $110) $004

$014

($1 + $004− $090)$120 $090 $030 −$020 ($1− $120) $004

$014

($1 + $004− $090)

3

[3 marks]

We obtain the following payoff graphs:

[1 mark]

(v) XYZ will buy collars, which means that they buy the put leg and sell the call

leg. We have to compute the net option premium position, and find its future

value, i.e.

($00665− $00194)× 006 = 5 001 3× 10−2 ≈ $005

[1 mark]

4

Copper

price in

one year

Total cost

Profit on long

$1.05-strike

put option

Profit on short

$1.05-strike

call option

Net

premiumHedged profit

$070 $090$035

($105− $070) $000 $005$01

($105− 090− $005)$080 $090

$025

($105− $080) $000 $005$01

($105− 090− $005)$090 $090

$015

($105− $090) $000 $005$01

($105− 090− $005)$100 $090

$005

($105− $100) $000 $005$01

($105− 090− $005)$110 $090 $000

−$005($105− $110) $005

$01

($105− 090− $005)$120 $090 $000

−$015($105− $120) $005

$01

($105− 090− $005)

We see that we are completely and perfectly hedged. Buying a collar where the

put and call leg have equal strike prices perfectly offsets the copper price risk.

[3 marks]

Profit diagram:

[1 mark]

5


(i) Comparing

( )

+ 0045

( )

+ 00452

2 ( )

2= 0045 ( )

with the Black-Scholes partial differential equation (PDE), i.e.

+

+

2

22

2

2=

we have

= 0045 and2

2= 0045

which gives = 03 [3 marks]

Moreover, from

= 0075+

we know that

= 0075

[1 mark]

Hence the market price of risk of the stock is

−

=0075− 0045

03= 01

[1 mark]

(ii) From

= 005+ 025

the market price of risk of the stock is given by

=−

=005−

025= 4 (005− )

6

[1 mark]

It is known that

¡

¢= − and

¡

¢=

i.e. under world, is normally distributed with mean − and variance

, i.e.

˜ (− ) under Q world.

[3 marks]

Proof (See 2008 mid-session test Q3)

Using the given Radon-Nikodym derivative,

we have

³

´=

µ

¶=

³−

− 1

22

´= −

122 ×

n(−)

o= −

122 ×

12(−)2

= −+122

where since ˜ (0 ) under world we have

n

o=

() =

122

==================================================

Hence we have

¡05

¢= − × 05 = −4 (005− )× 05 = −003

and = 0035 [1 mark]

7


(i) (a)

(2) = ¡1000035×2+032

¢= 1000035×2

¡032

¢= 100× 0035×2 ×

12×2×(03)2 = 1173511

where

() = () =

122

[1 mark]

( 22 ) =

n¡1000035×2+032

¢2o=

©1002 × 2(0035×2+032)

ª= 10022×0035×2

¡062

¢= 1002 × 2×0035×2 ×

12×2×(06)2 = 16487

so

(2) = ( 22 )− {(2)}2 = 16487− (1173511)2

= 10022×0035×2 × 12×2×(06)2 −

³100× 0035×2 ×

12×2×(03)2

´2= 27159351

[1 mark]

(b)

ln2 = ln 100 + 0035× 2 + 032

So we have

(ln2) = ln 100 + 0035× 2 = 467517

and

(ln2) = (03)2 × 2 = 018

Hence

ln2 ˜ (467517 018) .

8

[1 mark]

Now

Pr (98 ≤ 2 ≤ 103) = Pr (ln 98 ≤ ln2 ≤ ln 103)= Pr

µln 98− 467517√

018≤ ≤ ln 103− 467517√

018

¶= Pr (−02126 ≤ ≤ −00953)= (1− 05398)− (1− 05832)= 00434

[1 mark]

(c) We have

2

1=1000035×2+032

1000035×1+031= 0035+03(2−1)

which imples

2 = 10035+03(2−1)

By independent increments, the increment 2 − 1 does not depend on the

given condition 1 = 103. Hence

2 −1 | 1 = 103 ˜ (0 1)

[2 marks]

Now

Pr (98 ≤ 2 ≤ 103 | 1 = 102)= Pr

¡98 ≤ 1

0035+03(2−1) ≤ 103 | 1 = 102¢

= Pr¡98 ≤ 1020035+03(2−1) ≤ 103¢

= Pr

µ1

03

µln98

102− 0035

¶≤ 2 − 1 ≤ 1

03

µln103

102− 0035

¶¶= Pr (−02500 ≤ ≤ −00841)= 05987− 05319 = 00668.

[2 marks]

(ii) Since

9

= ( ) = 0

−2

2

+

from Itô’s Lemma, we have

( ) = = + +1

2 ()

2

= 0

µ− 2

2

¶

−2

2

++ 0

−2

2

+ + 0

2

2

−2

2

+ ()

2

= 0

−2

2

++ 0

−2

2

+

Hence

= +

[4 marks]

(iii) We have

= ( ) = 05

Using Itô’s Lemma

= + +1

2 ()

2

= 05 +

205

+1

2

Ã−

432

!()

2

Hence with ()2= 9−22

,

=p+

2√

¡2− 3−

¢+9−22

832

=

µp +

p − 9

8−2

p

¶− 3

2−

p

=

µ

+ − 9

8−2

¶− 3

2−

10

[4 marks]

So we have ( )

( )=

+ − 9

8−2

−32−

= −1+ 1− 9

8−2

32−

Hence

(1 2)

(1 2)= −1 + 1−

98−2

32−1

= −162 − 912

= −33485

[2 marks]

11

mid test solution 2015

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