辅导案例-TERM 3 2019

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FAMILY NAME:
OTHER NAME(S):
STUDENT ID:
SIGNATURE:
SCHOOL OF RISK AND ACTUARIAL STUDIES
TERM 3 2019
ACTL3182: ASSET-LIABILITY AND DERIVATIVE MODELS
AND
ACTL5109: FINANCIAL ECONOMICS FOR INSURANCE AND
SUPERANNUATION
INSTRUCTIONS:
1. TIME ALLOWED2 HOURS
2. READING TIME10 MINUTES
3. THIS EXAMINATION PAPER HAS 27 PAGES.
4. TOTAL NUMBER OF QUESTIONS6
5. TOTAL MARKS AVAILABLE100
6. MARKS AVAILABLE FOR EACH QUESTION ARE SHOWN IN THE EXAMINATION PA-
PER (AND OVERLEAF). ALL QUESTIONS ARE NOT OF EQUAL VALUE.
7. ANSWER ALL QUESTIONS IN THE SPACE ALLOCATED TO THEM. IF MORE
SPACE IS REQUIRED, USE THE ADDITIONAL PAGES AT THE END.
8. CANDIDATES MAY BRING
a. THE TEXT FORMULÆ AND TABLES FOR ACTUARIAL EXAMINATIONS (ANY
EDITION) INTO THE EXAMINATION. ITMUST BEWHOLLY UNANNOTATED.
b. THEIR OWN UNSW APPROVED CALCULATOR
9. ALL ANSWERS MUST BE WRITTEN IN INK. EXCEPT WHERE THEY ARE EXPRESSLY
REQUIRED, PENCILS MAY BE USED ONLY FOR DRAWING, SKETCHING OR GRAPH-
ICAL WORKS.
10. THIS PAPER MAY NOT BE RETAINED BY THE CANDIDATE.
Question Mark
1 [18 marks]
2 [18 marks]
3 [12 marks]
4 [25 marks]
5 [11 marks]
6 [16 marks]
Total: 100
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Question 1. [18 marks]
(a) Maisiri is a small scale corn farmer in Northern New South Wales and he has to decide the
tonnage, X, to plant before the selling price, P , of corn is known towards the end of the
harvesting season. The selling price is Normally distributed with mean, µ and variance,
σ2.
The cost of growing X is 12hX
2
with h being a positive constant. Maisiri's profit, Y , is a
function of X and P expressed as
Y = PX − 1
2
hX2.
His utility-of-consequences function is
U(Y ) = − exp(−bY ),
where b is a positive constant which can be interpreted as the coefficient of absolute risk
aversion. Maisiri chooses X to maximize the expected utility.
Show that his optimal choice is given by
X =
µ
h+ bσ2
.
Hint: You may take advantage of your knowledge of moment generating functions for
normal distributions. [10]
Soln
The expression for Maisiri's expected utility as a function of X is
E[− exp(−bY )] = −E[exp{−b(PX − 1/2hX2)}]
= − exp(1/2bhX2)E[exp(−bXP )]
= − exp(1/2bhX2) exp{−bXµ+ 1/2(−bX)2σ2}
= − exp{−bµX + 1/2b(h+ bσ2)X2}.
where in the second line I have used the fact that exp(1/2bhX2) is non-random, and in the third
line I have used MGF of a normal distribution function
E[exp(kP )] = exp(kµ+ 1/2k2σ2).
Since exp is an increasing function and there is a negative sign in front, maximizing the final
expression for expected utility is equivalent to minimizing
−bµX + 1
2
b(h+ bσ2)X2
or maximizing
bµX − 1
2
b(h+ bσ2)X2
(You can as well work directly with the exponential and differentiate away. But that will be
messier and prone to errors.)
The first-order condition for the maximization is
µ− 1
2
(h+ bσ2)(2X) = 0,
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implying that
X =
µ
h+ bσ2
.
The second-order condition is
−(h+ bσ2) < 0.
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Page 4 of 27
(b) A year ago Jake, who is currency trader entered into a forward contract to buy £ 1 million
for $1.2 million. The contract now has six months to maturity. The daily volatility of
a six-month zero-coupon sterling bond (when its price is translated to dollars) is 0.055%
and the daily volatility of a six-month zero-coupon dollar bond is 0.041%. The correlation
between returns from the two bonds is 0.85. The current exchange rate is 1.21. Calculate
the standard deviation of the change in the dollar value of the forward contract in one day.
What is the 10-day 99% VaR? Assume that the six-month interest rate in both sterling
and dollars is 2% per annum with continuous compounding. [8]
Soln
(b) The contract is a long position in a sterling bond combined with a short position in a dollar
bond. The value of the sterling bond is 1.21×e−0.02×0.5 or $1.197960299 million. The value
of the dollar bond is 1.2× e−0.02×0.5 or $1.1880598 million. The variance of the change in
the value of the contract in one day is
1.1979602992 × 0.000552 + 1.18805982 × 0.000412
− 2× 0.85× 1.197960299× 0.00055× 1.1880598× 0.00041
= 1.258× 10−7
The standard deviation is therefore $0.000354667 million. The 10-day 99% VaR is $0.002613227
million.
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Question 2. [18 marks]
A financial market consists of three risky assets characterised by the following expected return
vector and covariance matrix:
z =
0.180.12
0.06
 , Σ =
0.4 0.3 0.20.3 0.20 0.25
0.2 0.25 0.10

(a) Suppose an investor wishes to construct a feasible portfolio on the minimum variance set
with an expected return of 16%. What are the corresponding weights of each security and
the associated standard deviation of this portfolio? [10]
(b) Compute the expected return and standard deviation of the minimum variance point. [4]
(c) Draw a rough sketch of the minimum variance point in mean-standard deviation space,
indicating the coordinates of the minimum variance point as well as the coordinates of the
three securities. [4]
Hint: you may utilise the following inverse of the variance covariance matrix:
Σ−1 =
10.625 −5 −8.75−5 0 10
−8.75 10 2.5

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Page 6 of 27
Soln:
(a) The trick here is to realize that inverting a diagonal matrix (of full rank) is trivial, you just
invert all non-zero entries componentwise. First compute A, B, C and ∆ where
Σ−1 =
10.625 −5 −8.75−5 0 10
−8.75 10 2.5

This implies that
A = 1

Σ−11 = 5.625
B = z

Σ−11 = 0.2625
C = z

Σ−1z = 0.09225
∆ = AC −B2 = 0.45.
Next we compute the Lagrange multiplies:
λ =
C −BµP

= 0.111667, γ =
AµP −B

= 1.416667
The weights on the efficient frontier can be computed from the expression
w = λΣ−11+ γΣ−1z =
(
23
30
,
2
15
,
1
10
)′
.
Variance of a portfolio can be expressed as
σ2P =
Aµ2P − 2BµP + C

= 0.338333,
σP ≈ 0.581664
(b) The expected return and standard deviation of returns for the minimum variance point as
computed as follows
µG =
B
A
= 0.046667, σG =

1
A
= 0.421637.
0
0.05
0.1
0.15
0.2
0.25
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
MVS
(c)
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Question 3. [12 marks]
(a) State four key differences in assumptions between the single factor Arbitrage Pricing Theory
(APT) and the Capital Asset Pricing Model (CAPM). [4]
Soln
ˆ CAPM is based on utility theory, and in particular on mean-variance portfolio opti-
mizers.
ˆ CAPM uses equilibrium.
ˆ APT is based on the law of one price, that is the no arbitrage assumption.
ˆ APT says nothing about the optimality of the market portfolio.
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Page 10 of 27
(b) Assume the following two-factor model describes security returns in the market:
ri = ai + bi,1f1 + bi,2f2 + ξi,
and suppose the following three well-diversified portfolios are observed:
Portfolio µi βi,1 βi,2
W 12% 1.0 0.5
X 13.4% 3.0 0.2
Y 12% 3.0 -0.5
i. Determine the equation of the plane that describes equilibrium returns under the
assumptions of the arbitrage pricing theory (APT). [3]
ii. Show that an arbitrage opportunity arises if a portfolio Z exists, with µZ = 10%,
bZ,1 = 2.0 and bZ,2 = 0.0. [2]
iii. Demonstrate how you would exploit the opportunity in Part ii. above. [3]
Soln
i. In equilibrium, all expected returns are given by an equation of the form
µi = λ0 + bi,1λ1 + bi,2λ2.
Substituting the given data into this equation yields the following three simultaneous equa-
tions:
λ0 + λ1 + 0.5λ2 = 0.12
λ0 + 3λ1 + 0.2λ2 = 0.134
λ0 + 3λ1 − 0.5λ2 = 0.12.
Solving these equations, we obtain the following risk-free rate of return and factor prices:
λ0 = 0.1, λ1 = 0.01 and λ2 = 0.02. The equation of the plane describing equilibrium
returns is thus
µi = 0.1 + 0.01bi,1 + 0.02bi,2.
ii. Substituting bZ,1 = 2 and bZ,1 = 0 into the above equation indicates that the equilibrium
expected return of portfolio Z should be 0.12. Since this is greater than the observed
expected return of Z, which is µZ = 0.10, we conclude that an arbitrage opportunity exists
as Z is being overvalued by the market.
iii. Let wW , wX and wY = 1−wW −wX be the weights of a portfolio P containing W , X and
Y . This implies that bP,1 = 3− 2wW and bP,2 = wW + 0.7wX − 0.5.
To exploit the arbitrage, we must solve for bP,1 = bZ,1 = 2 and bP,2 = bZ,2 = 0. Doing so
yields wW = 0.5, wX = 0 and wY = 0.5. We make an investment in P which is completely
funded by short-selling Z, that is
Investment Weight
W 0.5
X 0.0
Y 0.5
Z -1
Notice the following about this portfolio:
Page 11 of 27
ˆ Its weights sum to zero, implying that it requires zero initial capital;
ˆ Its factor loadings are all zero, which implies that it is riskless; and
ˆ It produces a positive payoff.
So we have constructed a portfolio that requires zero initial investment, but delivers a
riskfree payoff. This is an arbitrage opportunity.
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Question 4. [25 marks]
(a) Derive the lower and upper bounds of a European put option. [5]
Soln Create two portfolios, A and B where A consists of a long put option and a long
position in the stock whose initial price is S0. Portfolio B is a cash only position with
Ke−rT .
Upon maturity of the put option, the value of portfolio A is either ST if ST > K or K if
ST < K.
The value of Portfolio B is always equal to K regardless of the stock price. Hence Portfolio
is at at least as greater that the cash only position, this yields the lower bound of a European
put option, that is
p0 + S0 ≥ Ke−rT or pt + St ≥ Ke−r(T−t).
For the upper bound, the maximum attainable value of a put option is its strike which only
happens when the stock price is ruined
pt ≤ Ke−r(T−t).
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Page 15 of 27
(b) Define what is meant by a self-financing strategy under the general multistep Binomial
model. [2]
(c) Consider a two-step Binomial market model with four scenarios Ω = {ω1, ω2, ω3, ω4} and
a risk-free security whose prices are
B(0) = 100, B(1) = 110, B(2) = 121,
and a risky asset whose price can follow any of the following four possible scenarios:
Scenario S(0) S(1) S(2)
ω1 70 91 109.2
ω2 70 91 95.55
ω3 70 73.5 88.2
ω4 70 73.5 69.825.
Also assume that there is a European call option in the market whose strike price is K = 80
and expires in two periods as detected by the two-step Binomial model.
i. Find the risk neutral probabilities for each of the market scenarios {ω1, ω2, ω3, ω4}. [8]
ii. Demonstrate whether there is an arbitrage opportunity in this market. [2]
iii. Find the call option prices C(n), for n = 0, 1, 2 for each of the market scenarios
{ω1, ω2, ω3, ω4}. [8]
Soln:
(b) In a discrete market model with stock prices S(n) and bond prices B(n) for n = 0, 1, · · ·N
a self-financing strategy φ(n), ψ(n) for n = 1, · · · , N is defined by the condition φ(n)S(n)+
ψ(n)B(n) = φ(n+ 1)S(n) + ψ(n+ 1)B(n) for each n = 1, · · · , N − 1
(c) i. Let E∗ be the risk-neutral expectation.
First step.
E∗
(
S(1)
B(1)
)
=
S(0)
B(0)
.
Denote the risk-neutral probability of S(1) = 91 by p∗. Then
p∗
91
110
+ (1− p∗)73.5
110
=
70
100
17.5p∗ = 77− 73.5 = 3.5
p∗ =
3.5
17.5
= 0.2.
Second step.
E∗
(
S(2)
B(2)
|S(1) = 91
)
=
S(1, ω1)
B(1)
.
Denote the risk-neutral probability of S(2) = 109.2 given that S(1) = 91 by q∗. Then
q∗
109.2
121
+ (1− q∗)95.55
121
=
91
110
13.65q∗ =
91
110
· 121− 95.55 =
q∗ =
4.55
13.65
=
1
3
' 0.3333.
E∗
(
S(2)
B(2)
|S(1) = 73.5
)
=
S(1, ω3)
B(1)
.
Page 16 of 27
Denote the risk-neutral probability of S(2) = 88.2 given that S(1) = 73.5 by r∗. Then
r∗
88.2
121
+ (1− r∗)69.825
121
=
73.5
110
18.375r∗ =
73.5
110
· 121− 69.825 = 11.025
r∗ =
11.025
18.375
= 0.6
We have
P∗(ω1) = p∗q∗ = 0.2× 0.3333 = 0.0667
P∗(ω2) = p∗(1− q∗) = 0.2× 0.6667 = 0.1333
P∗(ω3) = (1− p∗)r∗ = 0.8× 0.6 = 0.48
P∗(ω4) = (1− p∗)(1− r∗) = 0.8× 0.4 = 0.32
ii. There is no arbitrage because there exists a risk-neutral probability P∗ on the space
of market scenarios.
iii.
C(2, ω1) = max (S(2, ω1)−X, 0) = max(109.2− 80, 0) = 29.2
C(2, ω2) = max (S(2, ω2)−X, 0) = max(95.55− 80, 0) = 15.55
C(2, ω3) = max (S(2, ω3)−X, 0) = max(88.2− 80, 0) = 8.2
C(2, ω4) = max (S(2, ω4)−X, 0) = max(69.825− 80, 0) = 0
For the risk-neutral expectation we have
E∗
(
C(2)
B(2)
|S(1) = 91
)
=
C(1, ω1)
B(1)
.
C(1, ω1) =
A(1)
A(2)
[q∗C(2, ω1) + (1− q∗)C(2, ω2)]
=
0.3333× 29.2 + 0.6667× 15.55
1.1
= 18. 2723,
and C(1, ω2) = C(1, ω1) = 18. 2723.
Also,
E∗
(
C(2)
B(2)
|S(1) = 73.5
)
=
C(1, ω3)
B(1)
.
C(1, ω3) =
A(1)
A(2)
[r∗C(2, ω3) + (1− r∗)C(2, ω4)]
=
0.6× 8.2 + 0.4× 0
1.1
= 4. 4727,
and C(1, ω4) = C(1, ω3) = 4. 4727.
E∗
(
C(1)
B(1)
)
=
C(0)
B(0)
C(0) =
A(0)
A(1)
[p∗C(1, ω1) + (1− p∗)C(1, ω3)]
=
0.2× 18. 2723 + 0.8× 4. 4727
1.1
= 6. 5751.
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Question 5. [11 marks]
(a) State the stochastic differential equation for the stock price dynamics and the ordinary
differential equation for the money market account under the Black-Scholes framework. [3]
Soln:
The stock price dynamics can be represented as
dS(t) = µS(t)dt+ σS(t)dW (t),
where µ is the instantaneous expected return and σ is the volatility of the stock price
process. The corresponding money market account dynamics is
dB(t) = rB(t)dt,
with r being the risk-free interest rate.
(b) Derive the explicit solutions of the stock price and money market account in Part (a). [8]
Soln:
For the stock price dynamics, let X(t) = lnS(t) and application of Ito's Lemma yields
dX(t) =
(
µ− 1
2
σ2
)
dt+ σdW (t).
Integrating both sides yields∫ t
0
dX(s) =
∫ t
0
(
µ− 1
2
σ2
)
ds+
∫ t
0
σdW (s),
X(t)−X(0) =
(
µ− 1
2
σ2
)
t+ σ[W (t)−W (0)],
Implying that
X(t) = X(0) +
(
µ− 1
2
σ2
)
t+ σW (t).
Recall that X(t) = lnS(t), substituting in the above equation yields
lnS(t) = lnS(0) +
(
µ− 1
2
σ2
)
t+ σW (t),
Page 19 of 27
which can be expressed as
S(t) = S(0) exp
(
(µ− 1
2
σ2)t+ σW (t)
)
.
The solution of the money market account ordinary differential equation is
B(t) = B(0)ert.
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Question 6. [16 marks]
Given that the price of a European call option under the Black-Scholes framework is
C(t) = S(t)N(d1)−Ke−r(T−t)N(d2),
where
d1 =
ln S(t)K +
(
r ± 12σ2
)
(T − t)
σ

T − t , and d2 = d1 − σ

T − t,
with
N(x) =
1√
2pi
∫ x
−∞
e−
y2
2 dy =
1√
2pi
∫ +∞
−x
e−
y2
2 dy
being the cumulative distribution function of the standard normal distribution N(0, 1) with
mean 0 and variance 1.
(a) Derive the replicating strategy (φ(t), ψ(t)) for the European call option under Black-Scholes
framework where φ(t) and ψ(t) denote the units of stocks and bonds at current time, t.
[10]
(b) Compute the stock and bond position in the strategy (φ(t), ψ(t)) at time t = 0 of a
European call with strike K = 100, risk free rate r = 2%, volatility σ = 15% that will
mature in T = 1 year and the initial stock price being S = 100. [6]
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Page 23 of 27
Soln
(a) The replication strategy proceeds as follows
φ(t) = f ′x(t, x) =
∂C(t)
∂S(t)
= N(d1).
ψ(t) = (f(t, S(t))− φ(t)S(t))ert = −Ke−rTN(d2).
In fact, with n(x) = 1√
2pi
e−
x2
2
and let τ = T − t. We know that
∂d1
∂S
=
∂d2
∂S
=
1


τ
;
n(d1) =
1√
2pi
e−
d21
2 =
1√
2pi
e−
(d2+σ

τ)2
2 =
1√
2pi
e−
d22+2d2σ

τ+σ2τ
2
= n(d2)e
−d2σ√τ− 12σ2τ = n(d2)e

(
ln
S(t)
K
+(r− 12σ2)τ
)
− 1
2
σ2τ
= n(d2)
K
S
e−rτ ,
So we have
Sn(d1) = Ke
−rτn(d2).
Hence,
φ(t) =
∂C(t)
∂S(t)
= N(d1) + S · n(d1)∂d1
∂S
−Ke−rτ · n(d1)∂d1
∂S
= N(d1) +
(
Sn(d1)−Ke−rτn(d2)
) 1


τ
= N(d1).
(b) Based on the data, t = 0, T = 1, (r − 0.5σ2)T = 0.0088, (r + 0.5σ2)T = 0.0313 and
σ

T = 0.15 and also
d1,2 =
ln S(t)K +
(
r ± 12σ2
)
(T − t)
σ

T − t
Hence, read from the table, when S = 100, we have d1 = 0.2083 and
φ(0) = N(d1) = 0.5832.
We also have d2 = d1 − σ

T = 0.2083 − 0.15 = 0.0583, hence N(d2) = 0.5239 and the
position on bond will be
ψ(0) = −Ke−rTN(d2) = −51.35.
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