2
102
a) 𝐴 = 1+0.02804 + (1+0.02636)2 = 98.7734
𝐵=
3
3
103
+
+
= 101.5295
2
(1 + 0.02458)3
1 + 0.02804 (1 + 0.02636)
b)
98.7734 =
2
102
+
1
(1 + 𝑌𝑇𝑀)
(1 + 𝑌𝑇𝑀)2
1
1+𝑌𝑇𝑀
Denote
=𝑥
2 + 102𝑥 2 − 98.7734 = 0
𝑥1 =
−2+√22 −4×102×(−98.7734)
2×102
=
𝑥2 =
−2−√22 −4×102×(−98.7734)
2×102
< 0 => Not accepted
1
1+𝑌𝑇𝑀
−2+200.75
204
= 0.974
1
= 0.974 => 𝑌𝑇𝑀 = 0.974 − 1 = 0.026377 = 2.6377%
𝑌𝑇𝑀𝐴 = 2.6377%
𝑌𝑇𝑀𝐵 =
c) 𝐷𝐴 =
𝐷𝐵 =
3
3
103
+
+
= 101.547
2
(1 + 0.0246484)3
1 + 0.0246484 (1 + 0.0246484)
2
102
+2∗(1+0.026377)2 ]
1+0.026377
[1∗
= 1.981
98.7734
3
[1 ∗ 1 + 0.0246484 + 2 ∗
3
103
+3∗
]
2
(1 + 0.0246484)
(1 + 0.0246484)3
= 2.914
101.5295
∆𝑃𝑟𝑖𝑐𝑒
+0.2%
d)
𝐴=
𝐵=
𝑃𝑟𝑖𝑐𝑒
∆𝑟
= − 1+𝑟 × 𝐷 = − 1+0.026 × 1.981 = −0.3862%
∆𝑃𝑟𝑖𝑐𝑒
∆𝑟
+0.2%
=−
×𝐷 =−
× 2.914 = −0.5688%
𝑃𝑟𝑖𝑐𝑒
1+𝑟
1 + 0.0246
New YTM of A= 2.6%+0.2%=2.8%
2
102
New Price of 𝐴 = 1+0.028 + (1+0.028)2 = 98.3934
Actual percentage change is
98.3934−98.7734
×
98.7734
100% = −0.3847%
New YTM of B= 2.46%+0.2%=2.66%
3
3
103
New Price of 𝐵 = 1+0.0266 + (1+0.0266)2 + (1+0.0266)3 = 100.9542
Actual percentage change is
e) 𝐶𝑜𝑛𝑣𝑒𝑥𝑖𝑡𝑦 =
100.9542−101.5295
×
101.5295
100% = −0.56662%
1
𝐶𝐹1
𝐶𝐹2
𝐶𝐹𝑛
[1∗2∗(1+𝑟)
1 +2∗3∗(1+𝑟)2 +⋯+𝑛∗(𝑛+1)∗(1+𝑟)𝑛 ]
(1+𝑟)2
𝐶𝐹1
𝐶𝐹2
𝐶𝐹𝑛
+
+⋯+(1+𝑟)
𝑛
(1+𝑟)1 (1+𝑟)2
1
2
102
+2∗3∗
[1 ∗ 2 ∗
]
(1 + 0.026)1
(1 + 0.026)2
(1 + 0.026)2
𝐶𝑜𝑛𝑣𝑒𝑥𝑖𝑡𝑦𝐴 =
= 5.62
98.7734
∆𝑃𝑟𝑖𝑐𝑒
𝑃𝑟𝑖𝑐𝑒
= −𝐷𝑚 ∗ ∆𝑟 +
1
2
∗ 𝐶 ∗ ∆𝑟2 = −1.981 ∗
1
1 + 0.026377
∗ 0.002 +
1
∗ 5.62 ∗ 0.0022
2
= −0.384%
𝐶𝑜𝑛𝑣𝑒𝑥𝑖𝑡𝑦𝐵 =
∆𝑃𝑟𝑖𝑐𝑒
𝑃𝑟𝑖𝑐𝑒
1
3
3
103
[1 ∗ 2 ∗
+2∗3∗
+3∗4∗
]
1 + 0.0246
(1 + 0.0246)2
(1 + 0.0246)2
(1 + 0.0246)3
101.5295
= −𝐷𝑚 ∗ ∆𝑟 +
1
2
∗ 𝐶 ∗ ∆𝑟2 = −2.914 ∗
1
∗ 0.002 + ∗ 10.99 ∗ 0.0022
1 + 0.0246484
2
= −0.5666%
1
a) Risk-free asset payoff 𝐴𝑟𝑓 = (1) with price 𝑞𝑟𝑓 ≤ 1
1
b)
(1) 𝑞1 = 4𝑝1 + 2𝑝2 + 0𝑝3 = 2
(2) 𝑞2 = 0𝑝1 + 0𝑝2 + 4𝑝3 = 2
(3) 𝑞𝑟𝑓 = 𝑝1 + 𝑝2 + 𝑝3 ≤ 1
From (2) 𝑝3 = 0.5
From (1) 𝑝2 = 1 − 2𝑝1
Plug into (3): 𝑞𝑟𝑓 = 𝑝1 + 1 − 2𝑝1 + 0.5 ≤ 1
−𝑝1 + 1.5 ≤ 1
= 10.99
1
−𝑝1 ≤ −0.5
𝑝1 ≥ 0.5
From (1) 𝑝1 = 0.5 − 0.5𝑝2
0.5 − 0.5𝑝2 ≥ 0.5
−0.5𝑝2 ≥ 0
𝑝2 ≤ 0
Price 𝑝2 cannot be negative => 𝑝2 = 0
𝑝1 = 0.5
𝑝3 = 0.5
𝑞𝑟𝑓 = 0.5 + 0 + 0.5 = 1
Price of the second Arrow security if zero. This can provide arbitrage opportunity
If we short sell 𝐴1 and 𝐴2 , we would receive 𝑞1 + 𝑞2 = 4
We invest this into 4 risk-free assets and in the next period we receive 4 units no matter what ist he state of the
world
4
0
In the future we have liaibilities of 𝐴1 = (2) and 𝐴2 = (0)
0
4
In the states 1 and 3 we just have zero profit
In the state 2 we have 2 units of profit risklessly
c) 𝐴1 = 4𝑝1 + 2𝑝2 = 2
𝐴2 = 4𝑝3 = 1
𝐴3 = 1𝑝1 + 1𝑝2 + 1𝑝3 = 1
4𝑝3 = 1
𝑝3 =
1
= 0.25
4
1𝑝1 + 1𝑝2 + 0.25 = 1
1𝑝1 + 1𝑝2 = 0.75
4(0.75 − 𝑝2 ) + 2𝑝2 = 2 => 3 − 4𝑝2 + 2𝑝2 = 2
−2𝑝2 = −1
𝑝2 = 0.5
1𝑝1 + 0.5 = 0.75
𝑝1 = 0.25
Risk-free rate is 𝑟𝑓 =
1
1
𝑝1 +𝑝2 +𝑝3
− 1 = − 1 = 0%
1
bo
I
NPV
PV
a) c1= -(1+r)c0+w1
c1= -1.1c0 +110
Approximately should invest 57 in one project and get 90 in t=1
90
Value of the company after investment= 100 + (1.1 − 57) = 124.8181
b) Rate of return=
120
−
100
1 = 0.2 = 20%
120
Value of the company after investment=Current Value + NPV=100 + ( 1.1 − 100) = 109.0909
c) The two shareholders would agree on the amount of investment only under the conditions of the
perfect capital market. The shareholder who wants to consume more today can borrow at 10%, and the
shareholder who wants to consume more tomorrow can invest at 10%.
d) Shareholder on the left wants to consume approximately 85 today. He must borrow 85-43 = 42 000
EUR at 10%
Shareholder on the right wants to consume approximately 40 today. He can lend 43-40 = 3 000 EUR at
10%
e) Yes, both would agree but the shareholder on the right most likely would change his optimal
consumption
a) Variances: 𝜎21 = 0.42 = 0.16, 𝜎22 = 0.22 = 0.04
Covariances: 𝜎1,2 = 𝜌1,1 × 𝜎1 × 𝜎2 = −0.5 ∗ 0.4 ∗ 0.2 = −0.04
(
0.16
−0.04
−0.04
)
0.04
b) Portfolio:
Expected return: 𝜇𝑝 = 𝑤1 𝜇1 + 𝑤2 𝜇2 + (1 − 𝑤1 − 𝑤2 )𝜇𝑟𝑓
Variance of portfolio return: 𝜎𝑝2 = 𝑤12 𝜎12 + 𝑤22 𝜎22 + 2𝑤1 𝑤2 𝜎1,2
Min 𝜎𝑝2 = 𝑤12 𝜎12 + 𝑤22 𝜎22 + 2𝑤1 𝑤2 𝜎1,2
s.t ̅̅̅̅
𝜇𝑝 = 𝑤1 𝜇1 + 𝑤2 𝜇2 + (1 − 𝑤1 − 𝑤2 )𝜇𝑟𝑓
Where ̅̅̅̅
𝜇𝑝 is the required return (investor wants to receive this return, e.g. 10% or 15%)
FOCs:
𝑑𝜎𝑝2
𝑑𝑤1
𝑑𝜎𝑝2
𝑑𝑤2
= 2𝑤1 𝜎12 + 2𝑤2 𝜎1,2 = 0
= 2𝑤2 𝜎22 + 2𝑤1 𝜎1,2 = 0
From (1):
𝑤1 = −
𝑤2 𝜎1,2
𝜎12
Plug into the constraint:
̅̅̅̅
𝜇𝑝 = −
̅̅̅̅
𝜇𝑝 = −
𝑤2 𝜎1,2
𝜎12
𝑤2 𝜎1,2
𝜎12
̅̅̅̅
𝜇𝑝 − 𝜇𝑟𝑓 = −
𝜇1 + 𝑤2 𝜇2 + (1 +
𝑤2 𝜎1,2
𝜎12
𝜇1 + 𝑤2 𝜇2 + 𝜇𝑟𝑓
𝜎1,2
𝜎12
𝜇1 + 𝜇2 +
̅̅̅̅̅−𝜇
𝜇𝑝 𝑟𝑓
(𝜇1 −𝜇𝑟𝑓 )+(𝜇2 −𝜇𝑟𝑓 )
𝜎2
1
𝑤1 = −
𝜎1,2
𝜎12
𝜎12
𝜇1 + 𝑤2 𝜇2 + 𝜇𝑟𝑓 + 𝜇𝑟𝑓
̅̅̅̅
𝜇𝑝 − 𝜇𝑟𝑓 = 𝑤2 (−
𝑤2 = − 𝜎1,2
𝑤2 𝜎1,2
(− 𝜎1,2
̅̅̅̅̅−𝜇
𝜇𝑝 𝑟𝑓
=−
𝑤2 𝜎1,2
𝜎12
𝑤2 𝜎1,2
𝜇𝑟𝑓 𝜎1,2
𝜎12
−
− 𝑤2 ) 𝜇𝑟𝑓
𝜎12
− 𝜇𝑟𝑓 𝑤2
− 𝜇𝑟𝑓 𝑤2
− 𝜇𝑟𝑓 ) = 𝑤2 (−
𝜎1,2
𝜎12
̅̅̅̅̅−0.02
𝜇𝑝
0.04
(0.08−0.02)+(0.05−0.02)
0.16
)=
(𝜇1 −𝜇𝑟𝑓 )+(𝜇2 −𝜇𝑟𝑓 )
𝜎2
1
(𝜇1 − 𝜇𝑟𝑓 ) + (𝜇2 − 𝜇𝑟𝑓 ))
=
̅̅̅̅̅−𝜇
𝜇𝑝 𝑟𝑓
𝜎2
(𝜇1 −𝜇𝑟𝑓 )+(𝜇2 −𝜇𝑟𝑓 )𝜎 1
1,2
̅̅̅̅̅−0.02
𝜇𝑝
0.015
=
̅̅̅̅̅−0.02
𝜇𝑝
(0.08−0.02)+(0.05−0.02)
0.16
0.04
=
̅̅̅̅̅−0.02
𝜇𝑝
0.18
For example, if ̅̅̅̅
𝜇𝑝 = 10%, then 𝑤1 = 0.44 and 𝑤2 = 5.33 and 𝑤𝑟𝑓 = 1 − 0.44 − 5.33 = −4.77
if ̅̅̅̅
𝜇𝑝 = 6%, then 𝑤1 = 0.22 and 𝑤2 = 2.66 and 𝑤𝑟𝑓 = 1 − 0.22 − 2.66 = −1.88
(c) Equation for the portfolio frontier:
2
̅̅̅̅̅−0.02
𝜇𝑝
Min 𝜎𝑝2 = (
0.18
a)
(100 − 𝑐0 ) × 1.06 = 𝑐1
𝑐1 = 106 − 1.06𝑐0
Slope of CML is -1.06
c1= -(1+r)c0+w1
c1= -1.06c0 +106
CML=-1.06
2
̅̅̅̅̅−0.02
𝜇𝑝
) 𝜎12 + (
0.015
) 𝜎22 + 2
̅̅̅̅̅−0.02
𝜇𝑝
0.015
×
̅̅̅̅̅−0.02
𝜇𝑝
0.015
𝜎1,2
1
𝑈(𝑐0 , 𝑐1 ) = ln(1 + 𝑐0 ) + ln(1 + 𝑐1 )
4
𝑑𝑈
1
=
𝑑𝑐0 𝑐0 + 1
𝑑𝑈
1
=
𝑑𝑐1 4𝑐1 + 4
-
𝜕𝑈
𝜕𝑐0
𝜕𝑈
𝜕𝑐1
=−
1
𝑐0 +1
1
4𝑐1 +4
=−
4𝑐1 +4
𝑐0 +1
At the optimum the slope of the indifference curve must be equal to the slope of CML:
−
4𝑐1 +4
𝑐0 +1
= −1.06
4𝑐1 + 4 = 1.06(𝑐0 + 1)
𝑐1 =
1.06𝑐0
4
+
1.06−4
4
= 0.265𝑐0 − 0.735
Plug into the CML:
0.265𝑐0 − 0.735 = 106 − 1.06𝑐0
1.325𝑐0 = 106.735
𝑐0∗ = 80.55
𝑐1∗ = 20.61
Check:
I have 100 and consume 80.55, the rest (100-80.55 = 19.45) is invested at 6%. Next year I receive 20.61
exactly what I want to consume.
1
(b) 𝑈(80.55. 20.61) = ln(1 + 80.55) + 4 ln(1 + 20.61) = 5.1695
CML 𝑐1 = 106 − 1.06𝑐0
𝑐1 = 50
1) Find 𝑐0 from CML
50 = 106 − 1.06𝑐0
1.06𝑐0 = 56 => 𝑐0 = 52.8301
2) Find 𝑐0 from utility function
1
ln(1 + 𝑐0 ) + ln(1 + 50) = 5.1695
4
ln(1 + 𝑐0 ) + 0.982956 = 5.1695
ln(1 + 𝑐0 ) = 5.1695 − 0.982956 = 4.1865
1 + 𝑐0 = 𝑒 4.1865 = 65.7921
𝑐0 = 64.7921
The optimality condition is −
4𝑐1∗ +4
𝑐0∗ +1
= −1.06
Check both cases
a) In two years we have 𝐴 = (
0
420
) and 𝐵 = (
)
420
0
Prices 𝑝𝐴 = 180 and 𝑝𝐵 = 220
Pure discount bond brings the payoff of 𝐷𝐵 = (
100
)
100
Price of DB must be such that there is no arbitrage opportunity (based on the law of one price)
𝑝𝐷𝐵 =
100
420
× 220 +
100
420
× 180 = 52.38 + 42.86 = 95.2381
b)
i) Annually: 95.2381 × (1 + 𝑟)2 = 100 => 𝑟 = (
100
)
0.5
95.2381
− 1 = 2.4695%
4
𝑟
ii) Semi-annually: 95.2381 × (1 + ) = 100 => 𝑟 = 2 × [(
2
iii) Monthly: 95.2381 × (1 +
𝑟
12
)
24
= 100 => 𝑟 = 12 × [(
iv) Continuously: 95.2381 × 𝑒 2𝑟 = 1 00 => 𝑒 2𝑟 =
lim (1 +
𝑚→∞
𝑟
𝑚
)
𝑚×𝑛
100
95.2381
1
4
100
95.2381
) − 1] = 2.4544%
1
24
100
95.2381
) − 1] = 2.4420%
=> 2𝑟 = ln
100
95.2381
=> 𝑟 = 2.4395%
= 𝑒 𝑟𝑛
c) Assume there are three asset prices: 𝑝1 , 𝑝2 and 𝑝3
Discrete returns: 𝑟2 =
𝑝2 −𝑝1
𝑝1
, 𝑟3 =
𝑝3 −𝑝2
𝑝2
, average return 𝑟̅ = √𝑟2 × 𝑟3 , discrete returns are not additive
𝑝
𝑝
𝑟2 +𝑟3
𝑝1
𝑝2
2
Cont. returns: 𝑟2 = ln ( 2), 𝑟3 = ln ( 3) , average return 𝑟̅ =
, cont returns are additive
With the help of additivity we can compute variances, covariances and other statistical characteristics of the asset
returns
a) 𝐶𝐵1 =
3
1+0.02
103
+ (1+0.02)2 = 2.94 + 99 = 101.94
𝐶𝐵2 =
100
= 82.03
1.0210
3
103
+2∗
2.94
99
1 + 0.02
(1 + 0.02)2
= 1×
+2×
= 0.02884 + 1.942319 = 1.97
101.94
101.94
101.94
1∗
𝐷1 =
𝐷2 =
100
1.0210
10∗
82.03
= 10
b) 𝑃 = 20 × 101.94 + 30 × 82.03 = 2038.8 + 2460.9 = 4499.7
𝐷𝑃 = 𝑤1 × 𝐷1 + 𝑤2 × 𝐷2 =
2038.8
4499.7
× 1.97 +
2460.9
4499.7
× 10 = 0.8926 + 5.4690 = 6.3616
c) At t = 10
𝐹𝑉 = 20 × (3 × 1.029 + 103 × 1.028 ) + 30 × 100 = 5485.32
or 𝐹𝑉 = 4499.7 × 1.0210
At t = 6.36
𝐹𝑉 = 20 × (3 × 1.025.36 + 103 × 1.024.36 ) + 30 ×
100
1.0210−6.36
= 5103.85
or 𝐹𝑉 = 4499.7 × 1.026.36
d) Sketch discuss later
e)
3
103
+2∗
1 + 0.025
(1 + 0.025)2
= 1.95
101.94
1∗
𝐷1 =
𝐷2 =
100
1.02510
10∗
82.03
= 9.52
Price change 𝐷𝑃 = 𝑤1 × 𝐷1 + 𝑤2 × 𝐷2 =
2038.8
4499.7
× 1.95 +
2460.9
4499.7
× 9.52 = 0.8835 + 5.2065 = 6.09
NPV
I
PV
a) 𝑃𝑟𝑜𝑗𝑒𝑐𝑡1 =
42
30
− 1 = 0.4 = 40%
𝑃𝑟𝑜𝑗𝑒𝑐𝑡2 =
21
− 1 = 0.05 = 5%
20
𝑃𝑟𝑜𝑗𝑒𝑐𝑡3 =
52
− 1 = 0.4 = 4%
50
b) 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑟𝑒𝑡𝑢𝑟𝑛 =
115
100
− 1 = 0.15 = 15%
max amount can consume today: 100
max amount of consumption in t=1: 115
c) Should invest in project 1 (invest 30 and get 42) since rate of return higher than interest rate
NPV = 42/1.1 - 30
PV = 42/1.1
d) (𝑤0 − 𝑐0 ) × (1 + 𝑟) = 𝑐1
(𝑤0 − 𝑐0 ) × (1 + 𝑟) = 𝑐1
𝑤0 = 100 + 𝑁𝑃𝑉 = 100 + 𝑃𝑉 − 𝐼
(𝑤0 − 𝑐0 ) × (1 + 0.1) = 𝑐1
Assume we do not know what is 𝑤0
We know that the point 𝑐0 = 70 and 𝑐1 = 42 belongs to the CML
42
(𝑤0 − 70) × 1.1 = 42 => 𝑤0 = + 70 = 108.18
1.1
𝑤1 = 108.18 × 1.1 = 119
𝑐1 = 119 − 1.1𝑐0
Max amount today: If 𝑐1 = 0, 𝑐0 =
119
1.1
= 108.18
Max amount tomorrow: If 𝑐0 = 0, 𝑐1 = 119
e) 119 − 1.1𝑐0 = 𝑐1
𝑑𝑈
1
=
𝑑𝑐0 𝑐0 + 1
𝑑𝑈
1
=
𝑑𝑐1 10𝑐1 + 10
-
𝜕𝑈
𝜕𝑐0
𝜕𝑈
𝜕𝑐1
=−
1
𝑐0 +1
1
10𝑐1 +10
=−
10𝑐1 +10
𝑐0 +1
= −1.1
10𝑐1∗ + 10 = 1.1(𝑐0∗ + 1)
10(𝑐1∗ + 1) = 1.1(𝑐0∗ + 1)
(𝑐1∗ + 1) =
𝑐1∗ =
1.1(𝑐0∗ +1)
10
1.1(𝑐0∗ +1)
−
10
1
𝑐1∗ = 0.11𝑐0∗ + 0.11 − 1 = 0.11𝑐0∗ − 0.89
Plug into CML:
119 − 1.1𝑐0∗ = 0.11𝑐0∗ − 0.89
1.1𝑐0∗ + 0.11𝑐0∗ = 119 + 0.89
1.21𝑐0∗ = 119.89
𝑐0∗ =
119.89
1.21
= 99.08
𝑐1∗ = 0.11 × 99.08 − 0.89 = 10
Need to borrow 99.08 – (100 – 30) = 29.08 in period 0
Investor has 100 today
Invest 30 into the project that brings tomorrow 42
The rest amount (100- 30) he consumes today
Additionally he borrows 29.08 today and in total consumes 70 + 29.08 = 99.08
Tomorrows comes and he receives 42. But he must repay 29.08*1.1 = 32. He can consume tomorrow
only 42 – 32 = 10
a) 1 year forward rate t=1
(1 + 𝑟1 )(1 + 𝑓1,1 ) = (1 + 𝑟2 )2
𝑓1,1 =
(1 + 0.00323)2
− 1 = 0.0059 = 0.59%
1 − 0.00053
5-year forward rate t=2
5
(1 + 𝑟2 )2 (1 + 𝑓5,2 ) = (1 + 𝑟7 )7
1
𝑓5,2
(1 + 0.01153)7 5
=(
) − 1 = 0.0149 = 1.49%
(1 + 0.00323)2
t=10
5
(1 + 𝑟10 )10 (1 + 𝑓5,10 ) = (1 + 𝑟15 )15
1
𝑓5,10
(1 + 0.01678)15 5
=(
) − 1 = 0.0213 = 2.13%
(1 + 0.01452)10
t=15
5
(1 + 𝑟15 )15 (1 + 𝑓5,15 ) = (1 + 𝑟20 )20
1
𝑓5,15
(1 + 0.01745)20 5
=(
) − 1 = 0.1011 = 10.11%
(1 + 0.01678)15
t=20
5
(1 + 𝑟20 )20 (1 + 𝑓5,20 ) = (1 + 𝑟25 )25
1
𝑓5,20
(1 + 0.01637)25 5
=(
) − 1 = 0.1577 = 15.77%
(1 + 0.01745)15
t=25
5
(1 + 𝑟25 )25 (1 + 𝑓5,25 ) = (1 + 𝑟30 )30
1
𝑓5,25
(1 + 0.01673)30 5
=(
) − 1 = 0.0961 = 9.61%
(1 + 0.01637)25
b) ?
с)
d)
3
1−0.00053
3
1+0.00595
3
103
3
103
+ (1+0.00323)2 + (1+0.00603)3 = 107.14
+ (1+0.00595)2 + (1+0.00595)3 = 107.13
a) A4 can be a call option on A1 with strike 130
A4 can be a call option on A3 with strike 80
A5 can be a put option on A1 with strike 130
A5 can be a put option on A3 with strike 80
b) On the left hand side of the equation we have price of call (c) and price of bond with the face value of K
Their future payoffs are: max{S1 – K, 0} + K = max{S1, K}
On the right hand side we have price of put (p) and price of stock (S0)
Their future payoffs are: max{K – S1 ,0} + S1 = max {K, S1}
Since the payoffs are equal, prices of two portfolios must also be equal (call + bond and stock + put)
a)
𝑐0 = 2 − 𝐼 => 𝐼 = 2 − 𝑐0
1
1
1
5
5
5
𝑐1 = (2 − 𝑐0 )(4 + 2 − (2 − 𝑐0 )) = (2 − 𝑐0 )(4 + 𝑐0 ) = (8 − 2𝑐0 − 𝑐02 )
a) 𝐴 =
4
1+𝑟1
104
+ (1+𝑟
2
2)
= 103.845
𝐵=
2
102
+
= 99.981
1 + 𝑟1 (1 + 𝑟2 )2
𝐶=
5
105
105
+
+
= 100.363
2
(1 + 𝑟3 )3
1 + 𝑟1 (1 + 𝑟2 )
𝐷=
100
= 62.275
(1 + 𝑟7 )7
Subtract (B) from (A):
103.845 − 99.981 =
3.864 =
4
2
104
102
−
+
−
1 + 𝑟1 1 + 𝑟1 (1 + 𝑟2 )2 (1 + 𝑟2 )2
2
2
+
1 + 𝑟1 (1 + 𝑟2 )2
2
2
= 3.864 −
(1 + 𝑟2 )2
1 + 𝑟1
Plug into B
3.864 −
2
102
+
= 99.981
(1 + 𝑟2 )2 (1 + 𝑟2 )2
3.864 +
102
= 99.981
(1 + 𝑟2 )2
100
= 96.117
(1 + 𝑟2 )2
1
(1 + 𝑟2
)2
100 2
=(
) − 1 = 0.019999
96.117
𝑟2 = 2%
2
2
= 3.864 −
(1 + 0.02)2
1 + 𝑟1
2
= 1.922
1 + 𝑟1
1 + 𝑟1 = 1.04 => 0.04
𝑟1 = 4%
5
5
105
+
+
= 100.363
1 + 0.04 (1 + 0.02)2 (1 + 𝑟3 )3
105
= 100.363 − 9.613
(1 + 𝑟3 )3
1
105 3
𝑟3 = (
) − 1 = 0.05 = 5%
90.75
1
100
100 7
62.275 =
=> 𝑟7 = (
) − 1 = 0.07 = 7%
7
(1 + 𝑟7 )
62.275
𝑟1 = 4%, 𝑟2 = 2% 𝑟3 = 5% 𝑟7 = 7%
Forward rate from t=3 to t=7
4
(1 + 0.05)3 ∗ (1 + 𝑓1,4 ) = (1 + 0.07)7
4
(1 + 𝑓1,2 ) =
(1 + 0.07)7
(1 + 0.05)3
1
𝑓1,4 = 1.387134 − 1 = 0.08524 ≈ 8.5%
b) 𝐹𝑎𝑖𝑟 𝑝𝑟𝑖𝑐𝑒 =
4
1+0.04
+
4
(1+0.02)2
+
104
(1+0.05)3
= 3.8462 + 3.8447 + 89.84 = 97.5299
This bond is overpriced by the market
We can short sell this bond at 100 now
Split this amount into three parts:
Invest 3.8462 at 4% for 1 year 3.8447 at 2% for 2 years 89.84 at 5% for 3 years
After 1 year we have 4 and we can pay this amount as an obligation for the bond
After 2 years we have 4 and we can pay
After 3 years we have 104.001 and we an pay
There is a risk-free profit 0.001 from each bond
c) 𝐴 =
4
104
1+𝑌𝑇𝑀
Denote
+ (1+𝑌𝑇𝑀)2 = 103.845
1
1+𝑌𝑇𝑀
=𝑥
4𝑥 + 104𝑥 2 − 103.845 = 0
𝑥1 =
−4+√4 2 −4×104×(−103.845)
2×104
= 0.9802
𝑥2 =
−4−√4 2 −4×104×(−103.845)
2×104
< 0 => Not accepted
1
1+𝑌𝑇𝑀
= 0.9802 => 𝑌𝑇𝑀 =
1
−
0.9802
4
𝐷=
[1 ∗ 1 + 0.0202 + 2 ∗
𝐷𝑚 =
104
(1 + 0.0202)2
103.845
4
104
+2∗
]
1+0.0202
(1+0.0202)2
[1∗
d) ) 𝐷𝐷 =
103.845
100
(1+0.07)7
7∗
62.275
1 = 2.02%
]
= 1.9622
1
1
∗ (1+0.0202) = 1.9622 ∗ (1+0.0202) = 1.9233
= 6.9
1
100
[6 ∗ 7 ∗
]
(1 + 0.07)2
(1 + 0.07)7
𝐶𝑜𝑛𝑣𝑒𝑥𝑖𝑡𝑦 =
= 36.684
62.275
∆𝑟
𝐷𝑜𝑙𝑙𝑎𝑟 𝑑𝑢𝑟𝑎𝑡𝑖𝑜𝑛 = 𝐷 ∗ (
)∗𝑃
1+𝑟
𝐷𝑜𝑙𝑙𝑎𝑟𝐷𝑢𝑟𝑎𝑡𝑖𝑜𝑛 = 6.9 ∗
0.005
∗ 62.275 = 2.008
1 + 0.07
Price change with dollar duration?
𝐷𝑜𝑙𝑙𝑎𝑟𝐶𝑜𝑛𝑣𝑒𝑥𝑖𝑡𝑦 = 𝐶 ∗ 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑃
1
𝐷𝑜𝑙𝑙𝑎𝑟 𝑝𝑟𝑖𝑐𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 = ∗ 𝐷𝑜𝑙𝑙𝑎𝑟𝐶𝑜𝑛𝑣𝑒𝑥𝑖𝑡𝑦 ∗ ∆𝑟 2
2
𝐷𝑜𝑙𝑙𝑎𝑟 𝑝𝑟𝑖𝑐𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 =
1
∗ 36.684 ∗ 0.0052 = 0.000459 ≈ 0.0459%
2
Sharpe’s ratio:
SR for A:
SR for B:
SR for C:
(𝜇𝑘 −𝑟𝑓 )
𝜎𝑘
10%−2%
8%
20%−2%
10%
8%−2%
5%
=1
= 1.8
= 1.4
The best portfolio would consist or both investors from the risk-free asset and asset B
If there is a risk-free asset, investors are looking for the investment where
(𝜇𝑘 −𝑟𝑓 )
𝜎𝑘
is maximized