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Unformatted text preview: s a European call option with strike H . You can therefore value a shout call option by using the
binomial method, where at every node you check the optimality of shouting by comparing the value
if you don’t shout to the value of shouting. From the point of view of methodology, this is exactly
like valuing an American call, except that instead of receiving S − K at exercise, when you shout
you get P V (H − K) + C(S, H, t), where t is the remaining life of the option. 139 Chapter 23
Interest Rate Models
Question 23.2.
The two year forward price is F = P (0, 3) /P (0, 2) = .7722/.8495 = .90901.
a)
Since F P (0, 2) = P (0, 3) the ﬁrst input into the formula will be .7722. The present value
of the strike price is .9P (0, 2) = .9 × .8495 = .76455. We can use this as the strike with no interest
rate; we could also use a strike of .9 with an interest rate equal to the 2 year yield. Either way the
option is worth
BSCall (.7722, .76455, .105, 0, 2, 0) = $0.0494
b) (1) Using put call parity for futures options,
p = c + KP (0, 2) − F P (0, 2) = .0494 + .76455 − .7722 = $0.4175 (2) c)
The caplet is worth 1.11 two year put options with strike 1/1.11 = .9009. The no interest
formula will use (.9009) (.8495) = .7653 as the strike. The caplet has a value of
1.11BSP ut (.7722, .7653, .105, 0, 2, 0) = $0.0468. (3) Question 23.4.
A ﬂat yield curve implies the two bond prices are P1 = e−.08(3) = .78663 and P2 = e−.08(6) =
.61878. If we have purchased the three year bond, the duration hedge is a position of
N =− 1
1 P1
= − e3(.08) = −.63562.
2 P2
2 (4) in the six year bond. Notice the total cost of this strategy is
V8% = .78663 − .63562 (.61878) = .39332. (5) which implies we will owe .39332e.08/365 = .39341 in one day. If yields rise to 8.25%, our portfolio
will have a value
V8.25% = e−.0825(3−1/365) − .63562e−.0825(6−1/365) = .39338.
140 (6) Chapter 23 Interest Rate Models If yields fall to 7.75%, the value will be
V7.75% = e−.0775(3−1/365) − .63562e−.0775(6−1/365) = .39338. (7) Either way we lose .00003. This is a binomial version of the impossibility of a no arbitrage ﬂat
(stochastic) yield curve.
Question 23.6.
Note that the interest rate risk premium of zero implies φ = 0.
a) Beginning with the CIR model and using equation (23.37),
γ= a 2 + 2σ 2 = .22 + 2 (.44721)2 = .66332. (8) Let A2 and B2 be the 2 year bond’s A and B term in equation (23.37). Then
A2 = 2γ e.2+γ
(.2 + γ ) e2γ − 1 + 2γ .04/.447212 = .96718 (9) and
2 e2γ − 1 = 1.4897. (10) P (0, 2) = .96718e−1.4897(.05) = .89776. (11) B2 = (.2 + γ ) e2γ − 1 + 2γ This gives a price of the two year bond equal to The delta is Pr = −B2 P (0, 2) = −1.4897(.89776) = −1.3374 and the gamma is Prr =
2
B2 P (0, 2) = 1.48972 (.89776) = 1.9923. Similar analysis for the ten year bond will yield a price
of P (0, 10) = .6107, a delta of −1.4119, and a gamma of 3.2643. The “true” duration of the bonds
should be −Pr /P which equal 1.49 and 2.31 (respectively) quite different from 2 and 10 years.
The “true” convexity should be Prr /P which equals 2.22 and 5.35; the traditional convexities are
Pyy /P = 4 and 100.
Using similar notation for the Vasicek model and equation (23.24) the two year bond price is derived
from the components
r = .1 − 0.5 .12 /.22 = −.025, (12) B2 = 1 − e−2(.2) /.2 = 1.6484, (13) 141 Part 5 Advanced Pricing Theory and
A2 = e−.025(1.6484−2)−.16484 2 /.8 = .97514. (14) The two year bond will be worth P (0, 2) = .97514e−1.6484(.05) = .89799. As in the CIR analysis, the delta will be Pr = −B2 P (0, 2) = −1.6484 (.89799) = −1.4802 and a gamma of Prr =
2
B2 P (0, 2) = 1.64842 (.89799) = 2.44. Similarly, the price of the 10 year bond is .735, the delta is
−3.1776, and the gamma is 13.74. The “true” durations −Pr /P are 1.65 and 4.32 are substantially
different from 2 and 10. The convexity measures Prr /P which equal 2.72 and 18.694 are also quite
different from 4 and 100.
b) The duration hedge will use a position of
Nduration = − 2 P (0, 2)
= −.2 (.89799) /.735 = −.24435.
10 P (0, 10) (15) The delta hedge is
Ndelta = − 1.4802
Pr (0, 2)
= −.4658.
=−
3.177
Pr (0, 10) (16) The duration hedged portfolio has a cost of .89799 − .24435 (.735) = .71839 and the delta √
hedge
costs .89799 − .4658 (.735) = .55563. The one day standard deviation for r will be .05 ± .1/ 365.
In the “up” scenario the bond prices will become P2 = .8904 and P10 = .7186. The return in the
up scenario for the two hedges are
returnduration = .8904 − .24435 (.7186) − .71839e.05/365 = −.00368 (17) returndelta = .8904 − .4658 (.7186) − .55563e.05/365 = −.00003. (18) and In the “down” scenario the bond prices will be P2 = .905895 and P10 = .751818. The return in the
down scenario for the two hedges are
returnduration = .905895 − .24435 (.751818) − .71839e.05/365 = .0037 (19) returndelta = .905895 − .4658 (.751818) − .55563e.05/365 = −.00009. (20) and The delta...
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This document was uploaded on 03/11/2014 for the course FIN 402 at FPT University.
 Spring '14
 NguyenThiMaiLan
 Derivatives

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