Solution of Derivative

# 64 of these options and short 2709 shares of stock

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: herefore we need to short 1.120 shares. The graph of our proﬁt is given in Figure 10. 8 x 10 -3 Figure 10 (Problem 13.18) 6 Overnight Profit (\$) 4 2 0 -2 -4 -6 -8 38 38.5 39 39.5 40 Stock Price (\$) 40.5 41 41.5 42 Question 13.20. We purchased a 91-day 40-strike call, denoted option 1. a) Using a 180-day 40-strike call (option 2) to delta-rho hedge we must write 50.64 of these options and short 27.09 shares of stock. Our one day proﬁt is given in Figure 13 on the next page. b) Using option 2 as well as a one year (365 day) 45-strike put (option 3) to delta-gamma-vega hedge, we have the following solution: n2 = −1.2259, n3 = −.2874, and nS = .0307. The one day proﬁt is given in Figure 14 on the next page. If we added another option, call it option 4, we can try to hedge all of the greeks (note will be taken care of by the Black Scholes Equation). Let Vega be noted by v + nS 2 n2 + 3 n3 + 4 n4 v2 n2 + v3 n3 + v4 n4 Rho2 n2 + Rho3 n3 + Rho4 n4 2 n2 + 3 n3 + 4 n4 103 = −.5824 = −.0652 = −.0780 = −.0511 (9) (10) (11) (12) Part 3 Options Figure 13 (Problem 13.20a) 0.6 0.5 Overnight Profit (\$) 0.4 0.3 0.2 0.1 0 -0.1 35 0.5 x 10 36 37 38 39 -3 40 41 Stock Price (\$) 42 43 44 45 Figure 14 (Problem 13.20b) 0 Overnight Profit (\$) -0.5 -1 -1.5 -2 -2.5 -3 -3.5 37 38 39 40 Stock Price (\$) 41 42 43 This is four equations and four unknowns (the coefﬁcients are from the Black Scholes model). Note we must try to solve the last three equations simultaneously, which give us the position of the three options, and then use the underlying asset to delta hedge. As an related note, occasionally you will ﬁnd strange things may happen we use options with the same maturity. For a given time to maturity, vega and gamma are proportional (i.e. vi = ki i ). If 104 Chapter 13 Market-Making and Delta-Hedging two options have the same time to maturity, then k1 = k2 . If we use option 2 to gamma hedge a position of option 1, 2 n2 = −n1 1 ; with the same maturity, we have v2 n2 = k2 2 n2 = −k1 n1 1 = −n1 v1 . (13) Hence gamma hedging takes care of vega hedging if the maturity matches. Similarly, if we use two options (call the 2 and 3) of the same maturity to hedge an option (call it 1) position with a different maturity we will have a problem for 2 n2 + 3 n3 = −n1 1 implies v2 n2 + v3 n3 = k2 ( 2 n2 + 3 n3 ) = −k2 n1 1 =− k2 n1 v1 k1 (14) If k1 = k2 (i.e. the option being hedged is different from the two traded options’ identical time to maturity), it will be impossible to both gamma and vega hedge. A simple algebraic way of looking at this is by trying to solve ax + by = c 2 (ax + by) = kc (15) (16) Unless k = 2 (in which case we have an inﬁnite number of solutions), there will be no solution. 105 Chapter 14 Exotic Options: I Question 14.2. The arithmetic average is 5 (three 5’s, one 4, and one 6) and the geometric average is (5 × 4 × 5 × 6 × 5)1/5 = 4.9593. For the next sequence, the arithmetic average does not change (= 5); however the geometric average, (3 × 4 × 5 × 6 × 7)1/5 = 4.7894 is much lower. As the standard deviation increases (holding arithmetic means constant), the geometric return decreases. As an example, suppose we have two observations, 1 + σ and 1 − σ . The arithmetic mean will be 1; however the √ √ geometric mean will be (1 + σ )(1 − σ ) = 1 − σ 2 < 1. Question 14.4. √ √ Using the forward tree speciﬁcation, u = exp(.08/2 + .3/ 2) = 1.2868, d = exp(.08/2 − .3/ 2) = .84187, and risk neutral probability p = (e.08/2 − d)/(u − d) = .44716. The two possible prices in 6 months are 128.68 and 84.19; the three possible 1 year prices are 165.58, 108.33, and 70.87. Using the 6m and 12m prices, the possible arithmetic averages are (in 1 year) are 147.13, 118.50, 96.26, and 77.53. The four possible geometric averages are 145.97, 118.07, 95.50, and 77.24. These are in the order: u-u, u-d, d-u, and d-d. a) The four intrinsic values will be 165.58 − 147.13 = 18.45 (u-u), 0 (u-d), 108.33 − 96.26 = 12.07 (d-u), and zero (d-d). This will give an up value of e−.04 p18.45 = 7.93, a down value of e−.04 p 12.07 = 5.19, and an initial value of e−.04 (p7.93 + (1 − p) 5.19) = 6.1602. b) The four intrinsic values will be 165.58 − 145.97 = 19.61 (u-u), zero (u-d), 108.33 − 95.50 = 12.83 (d-u), and zero (d-d). This will give an up value of e−.04 p19.61 = 8.43 and a down value of e−.04 p12.83 = 5.51 with an initial value of e−.04 (p8.43 + (1 − p) 5.51) = 6.55. Question 14.6. See Table Two for the prices and ratio. The longer the time to expiration, the greater the dispersion of ST . For the standard call option, this unambiguously increases the value (by standard convexity arguments). For the knock-out, there is a trade-off. The higher the dispersion, the greater chance for large payoffs; however, there will also be a higher chance for the barrier to be hit. 106 Chapter 14 Exotic Options: I Table Two (Problem 14.7) 6 Standard Knock-Out Ratio 0.25 0.9744 0.7323 1.3306 0.5 2.1304 1.2482 1.7067 1 4.1293 1.8...
View Full Document

## This document was uploaded on 03/11/2014 for the course FIN 402 at FPT University.

Ask a homework question - tutors are online