Solution of Derivative

# Question 1912 see table one for a typical simulation

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Unformatted text preview: , 9.51, and .000135 respectively. Question 19.8. By log normality P (St < 95) = P 100 exp P z< √ .1 − .22 /2 t + .2 tz < 95 ln (95/100) − .1 − .22 /2 t √ .2 t with t = 1/365 this is N (−4.9207) = 4 × 10−7 . This magnitude negative return should, on average, occur once every 2.5 million days. With t = 1/252 (i.e. one trading day) this becomes N (−4.0965) = 2.097 × 10−5 ; making such a drop similarly unlikely. 126 Chapter 19 Monte Carlo Valuation Question 19.10. The simulations should be done by generating 1000 standard normals ε1 and another . 1000 (independent) standard normals ε2 . Then let S1 = 40 exp .08−.3/2 + √3 ε1 and let S2 = 12 12 √ .08−.5/2 .5 2 ε . The means, standard deviations, and 100 exp + √ z where z = .45ε1 + 1 − .45 2 12 12 correlation from Monte Carlo should approximate their theoretical counterparts. Question 19.12. See Table One for a typical simulation. With weekly data, the lognormality (as opposed to normal) of the stock price isn’t strong (i.e. there is little skewness or kurtosis). This implies the simple return is not that different from the continuous (normal) return. However, when we look at yearly distributions, there is now signiﬁcant kurtosis and skewness. TABLE ONE (Problem 19.12) Returns Week Stock Price Year Week Year Mean 0.0030 0.1551 100.3862 122.1316 SD 0.0415 0.2995 4.1708 37.4255 Skewness 0.0013 0.0013 0.1267 0.9511 Kurtosis 3.0133 3.0133 3.0390 4.6230 Question 19.14. Using simple gross returns (i.e. P ayoff/Cost ), the mean should be around 1.62, the standard deviation around 1.49, skewness around 1.79, and kurtosis around 7.59. 127 Chapter 20 Brownian Motion and Itô’s Lemma Question 20.2. √ If y = S 2 then S = y and dy = 2Sα (S, t) + σ (S, t)2 d t + 2Sσ (S, t) dZt where α (S, t) is the drift of S and σ (S, t) is the volatility of S . For the three speciﬁcations: a) √ √ dy = 2α y + σ 2 d t + 2 yσ dZt b) √ √ √ dy = 2 yλ a − y + σ 2 d t + 2 yσ dZt √ √ = 2λa y − 2λy + σ 2 d t + 2 yσ dZt c) (1) (2) dy = 2α + σ 2 y dt + 2σydZt Question 20.4. √ If y = S then S = y 2 and 1 1 1 −1/2 α (S, t) − S −3/2 σ (S, t)2 d t + S −1/2 σ (S, t) dZt S 2 8 2 1 1 1 α (S, t) − 3 σ (S, t)2 d t + σ (S, t) dZt 2y 8y 2y dy = = a) dy = α 2y − σ2 8y 3 b) dy = λa 2y − λ 2y 2 c) dy = α 2 − σ2 8 dt + dt + (4) σ 2y dZt σ2 8y 3 (3) − σ 2y dZt y dt + σ ydZt . 2 Question 20.6. If y = ln (SQ) = ln (S) + ln (Q) then dy = d ln (S) + d ln (Q) (5) 2 2 = αS − δS − σS /2 + αQ − δQ − σQ /2 d t + σS dZS + σQ dZQ . 128 (6) Chapter 20 Brownian Motion and Itô’s Lemma Question 20.8. Since the process y = S a Qb follows geometric Brownian motion, i.e. dy = αy ydt + σy ydZy the price of the claims will be e−r E ∗ (y1 ) = y0 e(αy −r ) . We use Ito’s lemma, as in equation (20.38), with δ = 0 and αS = αQ = r to arrive at the drift 1 1 2 2 αy = ar + br + a (a − 1) σS + b (b − 1) σQ + abρσS σQ 2 2 .42 .22 = .06 (a + b) + a (a − 1) + b (b − 1) − .3 (.4) (.2) ab. 2 2 (7) (8) a) Since a = b = 1, y0 = 10000 and αy = .12 − .024 = .096 hence the claim is worth .096−.06 = 10366.56. 10000e b) Since a = 1 and b = −1, y0 = 1 and αy = .22 + .024 = .064 hence the claim is worth e.064−.06 = 1.004. c) Since a = 1/2 and b = 1/2, y0 = 100 and αy = .029 hence the claim is worth 100e.029−.06 = 96.948. d) Since a = −1 and b = −1, y0 = 1/10000 and αy = .056 hence the claim is worth .056−.06 /10000 = 9.960 1 × 10−5 . e e) Since a = 2 and b = 1, y0 = 1000000 and αy = .292 hence the claim is worth 1000000e.292−.06 = 1.2612 million. Question 20.10. Note that if V (S) satisﬁes the given equation, then 1 E ∗ (dV ) = (r − δ) SVS + σ 2 S 2 VSS d t = rV dt. 2 (9) Since V (S) = kS h1 where a is constant, showing y = S a satisﬁes E ∗ (dy) = rydt when a = h1 is sufﬁcient (i.e. the constant term is irrelevant). Using Ito’s lemma, 1 E ∗ (dy) = aS a −1 (r − δ) S + a (a − 1) S a −2 σ 2 S 2 2 σ2 = a (r − δ) y + a (a − 1) y d t. 2 129 (10) (11) Part 5 Advanced Pricing Theory If E ∗ (dy) = rydt then a must satisfy a (r − δ) + a (a − 1) σ2 =r 2 (12) The two solutions are h1 and h2 as given (12.11) and (12.12) which one can verify directly. Question 20.12. We must try to ﬁnd a position in S and Q that eliminates risk. Let us buy one unit of S and let θ be the position in Q. Let It be our bond investment. We have Vt = St + θt Qt + It with V0 = 0. Since this strategy must be self ﬁnancing, dV = αS S + θαQ Q + rI d t + (σS S − ηθQ) dZ (13) hence we will set θ = σS S/ (ηQ). This will make our zero cost, self ﬁnancing strategy riskless. Hence the drift and the value must be zero. Mathematically, if Vt = 0 then I = −S − σS S Q. The ηQ drift being zero implies αS S + σS S σS S αQ Q − r S + ηQ η = 0. (14) Dividing both sides by S and simplifying leads to αQ = r − αS − r η. σS (15) Since Q is negatively related to Z , i...
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## This document was uploaded on 03/11/2014 for the course FIN 402 at FPT University.

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