Solution of Derivative

37 then a2 2 e2 2 e2 1 2 04447212 96718 9 and 2

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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.

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