This preview shows pages 1–4. Sign up to view the full content.
Chapter 12
Financial Engineering and Security Design
±
Question 12.1
Let
06
.
R
e
.
=
The present value of the dividends is
12
3 4
5
(1 50)
2
(2 50)
3
8 1317
RR
R
−−
− −
−
+.
+
+
=
.
.
The note originally sells for 100 8 1317
91 868.
−.
= .
With the 50 cent permanent increase, the present value
of dividends rises by
12345
2.0957
2
RRRRR
−−−−−
++++
=
to 10.2274, leading the note value to fall to100 10 2274
89 773.
=.
±
Question 12.2
For this problem, let
1
103
.
B
.
=
1.
The prepaid forward price is
015 3
0
1200
1147 20.
T
Se
e
δ
.
×
==
.
2.
We have to solve the coupon,
c
, that solves
6
1
52 8
1147 20
1200
9 7467
5 4172
i
i
cB
c
⎛⎞
⎜⎟
=
⎝⎠
.
=
⇒
.
.
.
∑
3.
The prepaid forward price for 1 share at time
t
is
015
1200
;
P
t
t
Fe
−.
=
for each semiannual share, we
can write the relevant prepaid forward price as 1200
D
i
where
015 2
.
De
−.
/
=
With this formulation we
have a similar analysis for the fractional shares,
c
*
:
6
**
1
52 8
1200 1147 20
1200
007528 shares.
1200 5 845
i
i
cD
c
=
.
=
⇒
.
×.
∑
To interpret, we will receive 007528
.
units of the index every six months. This has a current value
of 1200 ( 007528)
9 0336.
.=
.
We could quote
c
*
in dollars ($9.0336) instead of units.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document 146
McDonald •
Fundamentals of Derivatives Markets
±
Question 12.3
1.
This is the twoyear prepaid forward price:
015(2)
0
1200
1164 5.
T
Se
e
δ
−−
.
==
.
2.
As in Equation (12.5),
015(2)
0
8
1
1200(1
)
4762
7 4475
i
P
T
t
i
SF
e
c
P
−.
=
−
−
=
.
.
.
∑
3.
As in the Problem 12.2c, letting
015 4
,
De
−.
/
=
8
8
*8
*
8
1
1
1
1200 1200
1200
003757 shares,
i
i
i
i
D
cD
D
c
D
⎛⎞
⎜⎟
=
⎝⎠
=
−
+=
⇒
.
∑
∑
which is currently worth
()
003757 $1200
$4 5084.
.=
.
±
Question 12.4
The relevant twoyear interest rate is ln(1 8763) 2
6 6023%.
/.
/ = .
1.
The embedded option is worth 247.91. The prepaid forward is worth
015(2)
1200
1164 53.
e
−.
=.
The bond price is worth the sum 1164 53
247 91 1412 44
.+ .=
..
2.
λ
must solve 1164 53
247 91 1200
35 47 247 91
1431.
λλ
.+
⇒
=./ .=
.
±
Question 12.5
As in the previous question, we use
6 6023%.
r
1.
The embedded option is worth 247.91. The bond price is worth1200 ( 8763)
247 91 1299 47
.
=
.
.
2.
must solve 1200 ( 8763)
247 88 1200
59884.
.+.
=
⇒
±
Question 12.6
We continue to use 6.6023% as the relevant twoyear interest rate.
1.
The outofthemoney option (i.e.,
K
=
1500) is worth 141.56, making the bond have a value
of 1164 53
247 91 141 56
1270 89.
.+ .− .=
.
2.
We must solve 1164 53
(247 91 141 56)
1200
λ
.− . =
for a solution of
3335.
3.
If
1,
=
we have to adjust the strike (from Part (1), we know we have to lower
K
from 1500) to make
the outofthemoney option worth
( )
1164 53
247 91 1200
212 44
CK
+
.
−
=
.
⇒
1284 50.
K
Chapter 12
Financial Engineering and Security Design
147
±
Question 12.7
Let
06 5 5
55
7189
BP e
−.
× .
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 01/16/2011 for the course FIN 512 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Staff

Click to edit the document details