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Unformatted text preview: past. If the stock price rises in one month to (say) $120, the simple insurance
option will be less effective whereas the rollover will provide a new insurance option with a strike
of .95 (120) = 114. 109 Chapter 15
Financial Engineering and Security Design
Question 15.2.
For this problem let B = 1
1.03 . a) The prepaid forward price is 1200e−.015(3) = 1147.20. b) We have to solve the coupon, c, that solves
6 Bi c + 1147.20 = 1200 =⇒ c = 52.8 i =1 1−B
B − B7 = 9.7467. (1) c)
The prepaid forward price for 1 share at time t is FtP = 1200e−.015t ; for each semiannual
share, we can write the relevant prepaid forward price as 1200D i where D = e−.015/2 . With this
formulation we have a similar analysis for the fractional shares, c∗ : c 6 ∗ D i 1200 + 1147.20 = 1200 =⇒ c∗ = i =1 52.8
1200 1−D
D − D7 = .007528 shares. (2) Note this is interpreted as 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.
a) S0 e−δT = 1200e−.015(2) = 1164.5. b) As in equation (15.5),
c= c) P
S0 − FT
8
i =1 Pti = 1200 1 − e−.015(2)
= 4.762.
7.4475 (3) As in the problem 15.2c, letting D = e−.015/4 , c ∗ 8 D i 1200 + 1200D 8 = 1200 =⇒ c∗ = i =1 which is currently worth .003757 $1200 = $4.5084.
110 1 − D8
8
i
i =1 D = .003757 shares. (4) Chapter 15 Financial Engineering and Security Design Question 15.4.
As in the previous question, we use r = 6.6%.
a)
The embedded option is worth 247.88. The bond price is worth 1200 (.8763) + 247.88 =
1299.44.
b) λ must solve 1200 (.8763) + λ247.88 = 1200 =⇒ λ = .59884. Question 15.6.
Let B = e−.06×5.5 be the relevant discount factor. The equity linked CD is worth 1300B + .7C . Let
the 2 year forward price be F0 = 1300e(r −q)2 . By putcall parity, C − P = (F0 − 1300) B . This
implies the CD is equivalent to a long forward position on .7 on the index (zero cost), a long position
of .7 at the money puts, and an investment of .3 (1300B) + .7F0 (B) dollars in the risk free bond
(i.e. .3 + .7F0 bonds). The ﬁnal payoff will be
.7 (max (1300 − S5.5 , 0) + S5.5 − F0 ) + .7F0 + .3 (1300) (5) If S5.5 < 1300 this equals .7 (1300 − F0 ) + .7F0 + .3 (1300) = 1300 and if S5.5 ≥ 1300 it is equal
to
.7 (S5.5 − F0 ) + .7F0 + .3 (1300) = 1300 + .7 (S5.5 − 1300) (6) which is the same as the CD.
Question 15.8.
For notation clarity, let the six month discount factor be B = e−.06/2 . The ATM put option will be
worth 178.99. We need to ﬁnd the cash payment, c, to solve
1300 = 1300e −.06×5.5 B − B 12
− 178.99 + c
1−B for a solution of c = 58.984.
Question 15.10.
λ must solve 1300 = 1200e−.06×5.5 + λ (441.44) for a solution of λ = .9906. 111 (7) Part 4 Financial Engineering and Applications Question 15.12.
Using the information from the Problems Table:
a) The prepaid forward price is Pt8 Ft8 = .8763 (19.8) = 17.351. b) The cash payment solves
c (.9388 + .8763) = 20.90 − 17.351 =⇒ c = 1.9553 (8) Question 15.14.
We can value the option as a futures option (or use a spot option with a dividend yield equal to the
lease rate). Either way, the option is worth 1.073 (use 19.8e−.066×2 = 17.352 in the nondividend
BSCall ). Hence λ must solve 20.9 = 17.352 + λ1.073 for a solution of λ = 3.3066.
Question 15.16.
The contract has the party receiving a K2 call and writing a K1 put. We will assume K2 = K1 + 2
(otherwise there are an inﬁnite number of solutions).
a)
The strikes must solve C (K1 + 2) = P (K1 ). K1 = 19.577 satisﬁes this condition. If K2 −
K1 = 2, there are inﬁnite solutions to this. One solution is K2 = 22 and K1 = 19.229, another
being K1 = 18 and K2 = 23.607.
b)
We must solve 8=1 C (K1 + 2, ti ) =
i
19.456 and K2 = 21.456. 8
i =1 P (K1 , ti ). We can solve numerically for K1 = Question 15.18.
An investor who needs to hold the stock (perhaps tracking an industry) but is pessimistic about the
prospects of the company. If the stock goes down, the PEPS outperforms the stock, while if the
stock rises sufﬁciently, the stock outperforms the PEPS. 112 Chapter 16
Corporate Applications
Question 16.2.
Let K be the maturity value of the debt. Equity is valued as a call option with strike K on the assets
A = 100, with the given interest and volatility. By Black Scholes, the four values of equity will
be E = 5.9977, 11.4278, 23.3041, and 37.0374; Debt will be D = 100 − E = 94.0023, 88.5722,
76.6959, and 62.9626. Yields are given by DeyT = K which implies y = ln (K/D) /T = 0.3042,
0.2118, 0.1495, and 0.1245. The debt to equity ratios are D/E = 15.6732, 7.7506, 3.2911, and
1.7000.
a) See Table One. Table One (Problem 16.3a)
Yields on the four classes of debt
Time to Maturity
1
2
5
10
Senior
8.00% 8.01% 8.08% 8.15%
IntermA
8.74% 9.43% 9.62% 9.21%
IntermB
24.65% 19.40% 13.63% 10.88%
Junior
80.03%
40.44% 19.03% 12.69%
b)
See Table Two. The higher the volatility on assets and risk free rate, the higher the yield
on the debt issues. The only exce...
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 Spring '14
 NguyenThiMaiLan
 Derivatives

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