Exam MFE May 007 SOA Exam MFE Solutions: May 007 Solution 1 B Chapter 1, Put-Call Parity Let each dividend amount be D. The first dividend occurs at the end of months, and the second dividend occurs at the end of 5 months. We can use put-call parity to find D: Eur (, ) + = 0-0, T ( ) + Eur (, ) -0.06(0.5) -0.06( /1) -0.06(5 /1) C K T Ke S PV Div P K T 4.50 + 50e = 5 - De - De +.45-0.06( /1) -0.06(5 /1) -0.06(0.5) De [ + e ] = 5 +.45-4.50-50e D = 0.764 Solution A Chapter 3, Realistic Probability We can solve for p, the true probability of the stock price going up, using the following formula: ( a -d) h (0.10-0) 1 e -d e -0.756 p = = = 0.51576 u-d 1.433-0.756 Solution 3 C Chapter 7, Black-Scholes Formula The first step is to calculate d 1 and d : We have: ln( S/ K) + ( r - d + 0.5 s ) T d1 = s T ln(100 / 98) + (0.055-0.01 + 0.5 0.5 ) 0.5 = = 0.9756 0.5 0.5 d = d - s T = 0.9756-0.5 0.5 = -0.05600 1 N( d1 ) N( 0.9756) N( 0.30) 1 N(0.30) 1 0.6179 0.381 N( d ) N(0.05600) N(0.06) 0.539 ActuarialBrew.com 014 Page 1
Exam MFE May 007 The value of the European put option is: rt T PEur Ke N( d) Se N( d1) 0.055(0.5) 0.01(0.5) 98e 0.539 100e 0.381 11.93 Solution 4 E Chapter 1, Put-Call Parity For each put option, the choice is between having the exercise value now or having a 1-year European put option. Therefore, the decision depends on whether the exercise value is greater than the value of the European put option. The value of each European put option is found using put-call parity: -d T CEur ( K, T) + Ke = S0e + PEur ( K, T) -d T PEur ( K, T) = CEur ( K, T) + Ke -S0e The values of each of the 1-year European put options are: -0.04(1) -0.08(1) P (40,1) = 9.1 + 40e - 50e = 1.40 Eur -0.04(1) -0.08(1) P (50,1) = 4.91 + 50e - 50e = 6.79 Eur -0.04(1) -0.08(1) P (60,1) = 0.71 + 60e - 50e = 1.0 Eur -0.04(1) -0.08(1) P (70,1) = 0.00 + 70e - 50e = 1.10 Eur In the third and fourth columns of the table below, we compare the exercise value with the value of the European put options. The exercise value is Max( K - S 0,0). Exercise European K C Value Put 40.00 9.1 0.00 1.40 50.00 4.91 0.00 6.79 60.00 0.71 10.00 1.0 70.00 0.00 0.00 1.10 In each case, the exercise value is less than the value of the European put option. Therefore, it is not optimal to exercise any of these put options. ActuarialBrew.com 014 Page
Exam MFE May 007 Solution 5 D Chapter 8, Volatility of an Option The first step is to calculate d 1 and d : We have: ln( S/ K) + ( r - d + 0.5 s ) T d1 = s T ln(85 / 80) + (0.055-0.00 + 0.5 0.5 ) 1 = = 0.4815 0.5 1 d = d - s T = 0.4815-0.5 1 = -0.01875 1 Nd ( 1 ) N(0.4815) N(0.48) 0.6844 Nd ( ) N( 0.01875) N( 0.0) 1 N(0.0) 1 0.5080 0.490 The value of the call option is: -d T CEur ( S, K, s, r, T, d) = Se N( d1) - Ke N( d) -0(1) -0.055(1) = 85e 0.6844-80e 0.490 = 0.903 The value of delta is: -d T -0.0(1) D Call = e N( d1 ) = e 0.6844 = 0.6844 The elasticity of the option is: SD 85 0.6844 W= = =.7807 V 0.903 The volatility of the call option is: s = s W = 0.5.7807 = 1.3904 Option Stock Option ActuarialBrew.com 014 Page 3
Exam MFE May 007 Solution 6 C Chapter 1, Exchange Options The first page of Study Note MFE-7-07 tells us that for each share of the stock the amount of dividends paid between time t and time t + dt is assumed to be S(t ) d dt. Therefore, the continuously compounded dividend rate for Stock 1 is 5%, and the continuously compounded dividend rate for Stock is 10%. The claim has the following payoff at time 3: Max ÈÎS 1(3), S (3) A portfolio consisting of a share of Stock and the option to exchange Stock for Stock 1 effectively gives its owner the stock with this maximum value. If the value of Stock is greater than the value of Stock 1 at time 3, then the owner keeps Stock and allows the exchange option to expire unexercised. If the value of Stock 1 is greater than the value of Stock, then the owner exercises the option, giving up Stock for Stock 1. Since Stock has a continuously compounded dividend rate of 10%, the cost now of a share of Stock at time 3 is: P -d T F0, T ( S) = e S0 P -0.10(3) F0,3( S ) = e 00 = 148.16 The cost of an exchange option allowing its owner to exchange Stock for Stock 1 at time 3 is $10. Adding the costs together, we obtain the cost of the claim: 148.16 + 10 = 158.16 Solution 7 E Chapter 17, Black Formula The option expires in 1 year, so T = 1. The underlying bond matures 1 year after the option expires, so s = 1. The bond forward price is: P(0, T + s) 0.8817 F = P0 ( T, T + s) = = = 0.93460 P(0, T) 0.9434 The volatility of the forward price is: s = 0.05 ActuarialBrew.com 014 Page 4
Exam MFE May 007 We have: d 1 Ê F ˆ Ê0.93460ˆ ln Á + 0.5s T ln Á + 0.5(0.05) (1) ËK Ë 0.959 = = = 0.101 s T 0.05 1 d = d - s T = 0.101-0.05 1 = 0.1601 1 Nd ( ) = N(0.101) ª N(0.1) = 0.583 1 Nd ( ) = N(0.1601) ª N(0.16) = 0.5636 The Black formula for the call price is: C = P(0, T) ÈÎF N( d1) - K N( d) = 0.9434 0.9346 0.583-0.959 0.5636 = 0.019 Solution 8 C Chapter 7, Black-Scholes Formula The volatility of the stock is: Var ln S( t) 0.4t s = = = t t 0.4 We can use the version of the Black-Scholes formula that is based on prepaid forward prices to find the value of the call option: Ê P F ˆ 0, T ( S) ln + 0.5s T Ê100ˆ Á P Á + ËF ln 0.5 0.4 10 0, T ( K) Ë100 d1 = = = 1 s T 0.4 10 d = d1 - s T = 1-0.4 10 = -1 Nd ( 1 ) = N(1) = 0.8413 Nd ( ) = N( - 1) = 1 - N(1) = 0.1587 The price of the call option is: ( ) CEur P P F0, T ( S), F0, T ( K), s, T P P = F0, T ( SNd ) ( 1 )- F0, T ( KNd ) ( ) = 100 0.8413-100 0.1587 = 68.6 ActuarialBrew.com 014 Page 5
Exam MFE May 007 Solution 9 A Chapter 18, Interest Rate Cap The tree of interest rates is: 9.89% 7.704% 6.000% 6.000% 4.673% 3.639% Although the cap payments are made at the end of each year, we will follow the textbook s convention of valuing them at the beginning of each year using the following formula: Max ÈÎ0, RT -1 - KR T-year caplet payoff at time ( T - 1) = Notional 1 + RT -1 The interest rate cap consists of a 1-year caplet, a -year caplet, and a 3-year caplet. The payoff for the 1-year caplet is zero since 6% is less than 7.5%. The payoff for the -year caplet is positive only when the short-term interest rate increases to 7.704%, since 4.673% is less than 7.5%: 0.07704-0.075 R 1 = 0.07704 fi 100 = 0.1894 1.07704 The payoff for the 3-year caplet is positive only when the short-term rate increases to 9.89%, since the other short-term interest rates are less than 7.5%: 0.0989 0.075 R 0.0989 100.1767 1.0989 The possible payments from the cap are illustrated in the tree below:.1767 0.1894 0.0000 0.0000 0.0000 0.0000 The value of the -year caplet is: 0.1894 Value of -year caplet 0.5 0.0893 1.06 The value of the 3-year caplet is:.1767 Value of the 3-year caplet (0.5) 0.4766 1.06 1.07704 ActuarialBrew.com 014 Page 6
Exam MFE May 007 The value of the 3-year cap is: (1-year caplet) (-year caplet) (3-year caplet) 0.0000 0.0893 0.4766 0.5660 Solution 10 B Chapter 9, Delta-Gamma Hedging The gamma of the position to be hedged is: - 1,000 0.0651 = -65.10 We can solve for the quantity, Q, of the other call option that must be purchased to bring the hedged portfolio s gamma to zero: - 65.10 + 0.0746Q = 0.00 Q = 87.7 Since only choice B specifies the purchase of 87.7 units of Call-II, we already have enough information to see that Choice B must be the correct answer. The delta of the position becomes: - 1,000 0.585 + 87.7 0.7773 = 95.8 The quantity of underlying stock that must be purchased, delta of the position being hedged: Q S =-95.8 Q S, is the opposite of the Therefore, in order to delta-hedge and gamma-hedge the position, we must sell 95.8 units of stock and purchase 87.7 units of Call-II. Solution 11 D Chapter 4, Two-Period Binomial Tree The risk-neutral probability of an upward movement is: ( r-d ) h 0.05(0.5) e -d e -0.890 p* = = = 0.46500 u-d 1.181-0.890 The stock price tree is: 97.6333 8.6700 70.0000 73.5763 6.3000 55.4470 ActuarialBrew.com 014 Page 7
Exam MFE May 007 The tree of prices for the American put option is: 0.0000 3.3518 10.7558 6.437 17.7000 4.5530 If the stock price reaches $8.67 in 6 months, then the value of option is: -0.05(0.5) e (0.46500)0.0000 + (1-0.46500)6.437 = 3.3518 If the stock price reaches $6.30 in 6 months, then early exercise is optimal because the value of holding the option is only: -0.05(0.5) e (0.46500)6.437 + (1-0.46500)4.5530 = 15.748 Since the exercise value is $17.70, the option is exercised early if the stock price reaches $6.30. The value of the option is: e -0.05(0.5) (0.46500)3.3518 + (1-0.46500)17.7000 = 10.7558 Solution 1 A Chapter 14, Itô s Lemma Since we are given the risk-free rate in both the U.S. and Great Britain, we have: r - r * = 0.08-0.10 = -0.0 The expression for Gt ( ) is therefore: -0.0( T -t) Gt () = Ste () The partial derivatives are: -0.0( T -t) GS = e GSS = 0 È -0.0T 0.0t Ste () e Î -0.0T 0.0t -0.0( T -t) Gt = = S( t) e (0.0) e = S( t)(0.0) e t ActuarialBrew.com 014 Page 8
Exam MFE May 007 From Itô s Lemma, we have: 1 dg() t = GSdS() t + GSS[ ds()] t + Gtdt -0.0( T -t) 1-0.0( T -t) = e ds() t + (0)[ ds()] t + S()(0.0) t e dt -0.0( T -t) = e ds() t + G()(0.0) t dt T t -0.0( - ) = e S()0.1 t dt + 0.4 dz() t + G()(0.0) t dt = G( t) 0.1dt + 0.4 dz( t) + G( t)(0.0) dt = Gt[ ()0.1dt + 0.4 dz( t) ] Solution 13 E Chapter 19, Vasicek Model In the Vasicek Model, we have: -B(, tt) r PrtT (,, ) = AtTe (, ) We use the following two facts about the Vasicek model: A(, tt ) and B( tt, ) do not depend on r. A(, tt) = A(0, T-t ) and B(, tt) = B(0, T-t ). This implies: A(0,) = A(1,3) = A(,4) B(0,) = B(1,3) = B(,4) We have two equations and two unknowns: -B(0,)(0.04) A(0,) e = 0.9445 -B(0,)(0.05) A(0,) e = 0.931 Dividing the second equation into the first equation allows us to find B (0,) : - B(0,)(0.04) + B(0,)(0.05) 0.9445 e = 0.931 Ê0.9445ˆ B(0,)(0.01) = ln Á Ë 0.931 B(0,) = 1.3156 We can now solve for the value of A (0,) : -B(0,)(0.04) A(0,) e = 0.9445 B(0,)(0.04) A(0,) = 0.9445e (1.3156)(0.04) A(0,) = 0.9445e A(0,) = 0.99577 ActuarialBrew.com 014 Page 9
Exam MFE May 007 We can now solve for r * : -B(0,)( r*) A(0,) e = 0.8960-1.3156 r* 0.99577e = 0.8960 Ê 0.8960 ˆ - 1.3156 r* = ln Á Ë 0.99577 r* = 0.07989 Solution 14 E Chapter 3, One-Period Binomial Tree If the stock price moves up, then the straddle pays $0. If the stock price moves down, then the straddle pays $5: 70 0 = 50-70 60 V 45 5 = 50-45 In this case, u 70 / 60 and d 45 / 60. The risk-neutral probability of an upward movement is: ( -d ) (0.08-0.00)1 - - 45 e r h d 60 p* = = e = 0.79989 u- d 70-45 60 60 The value of the straddle is: -rh -0.08(1) V = e ÈÎ( p*) Vu + (1 - p*) Vd = e 0.79989(0) + (1-0.79989)(5) = 15.69 Solution 15 C Chapter 7, Black-Scholes Formula The volatility parameter s is not defined in the question, but let s assume that it is the volatility of the prepaid forward. The prepaid forward prices of the stock and the strike price are: P 0, T P 0, T -0.05(4 /1) = - = -0.05(0.5) = = F ( S) 50 1.50e 48.548 F ( K) 50e 48.7655 The volatility of the prepaid forward price is: s = 0.30 PF ActuarialBrew.com 014 Page 10
Exam MFE May 007 We use the prepaid forward volatility in the Black-Scholes Formula: d1 We have: Ê P F ˆ 0, T ( S) Á ln + 0.5s Ê48.548 PFT ˆ Á P Á + ln 0.5 0.30 0.5 ËF0, T ( K) Ë48.7655 = = s T 0.30 0.5 = 0.087 d = d - s T = 0.087-0.30 0.5 = -0.194 1 1 PF PF N( d ) N( 0.087) N( 0.08) 1 N(0.08) 1 0.5319 0.4681 N( d ) N(0.194) N(0.13) 0.5517 The value of the European put option is: P P rt 0, 0, T T 0 0, T 1 P F ( S ), F ( K ),, T Ke N ( d ) S PV ( Div ) N ( d ) Eur 48.7655 0.5517 48.548 0.4681 4.1895 Solution 16 D Perpetual Options This question is based on Section 1.6 of the textbook, which is no longer assigned for Exam MFE/3F. The values of h 1 and h are: h h 1 r -d Êr -d 1ˆ r = - + Á - + s Ë s s 1 1 r -d Êr -d 1ˆ r = - - Á - + s Ë s s Therefore: ( r - d ) h1 + h = 1 - s ActuarialBrew.com 014 Page 11
Exam MFE May 007 We can solve for d : ( r - d ) h1 + h = 1 - s 7 (0.05 - d ) = 1-9 0.30 (0.05 - d ) = 9 0.30 0.01 = 0.05 -d d = 0.04 We now have the information necessary to solve for h 1 : h 1 r -d Êr -d 1ˆ r = - + Á - + s Ë s s 1 1 0.05-0.04 Ê0.05-0.04 1 ˆ (0.05) = - + Á - + = 1.514 Ë 0.30 0.30 0.30 Solution 17 B Chapter 10, Gap Options For the gap option, we have: Strike Price: K1 = 90 Trigger Price: K = 100 The delta of the regular call option is: -d T D Call = e N( d 1 ) = 0. For the regular European call option, we have: -d T 4 = Se Nd ( 1) - Ke Nd ( ) 4 = 80(0.0) -100 e N( d ) e N( d ) = 0.1 Since the regular call option and the gap call option have the same values for d 1 and d, we can substitute the final line above into the equation for the value of the gap call option: -d T Gap call price = Se N( d1) - K1e N( d) = 80(0.) -90(0.1) = 5.0 ActuarialBrew.com 014 Page 1
Exam MFE May 007 Solution 18 A Chapter 15, Sharpe Ratio When the price follows geometric Brownian motion, the natural log of the price follows arithmetic Brownian motion: ( ) dst () = astdt () + sstdzt () () dln St () = a - 0.5s dt+ sdz Therefore: dy() t = Gdt + HdZ() t Yt () d[ ln Y() t ] = ( G - 0.5H ) dt + HdZ The arithmetic Brownian motion provided in the question for d[ ln Y( t )] allows us to find an expression for H and G: [ ln ( )] 0.06 s ( ) and [ ln ( )] ( 0.5 ) d Y t = dt + dz t d Y t = G - H dt + HdZ H = s G - 0.5H = 0.06 G = 0.06 + 0.5H Since X and Y have the same source of randomness, dz( t ), they must have the same Sharpe ratio. 0.07-0.04 G -0.04 = 0.1 H G - 0.04 0.5 = H 0.06 + 0.5H -0.04 0.5 = H 0.5H = 0.0 + 0.5H 0.5H - 0.5H + 0.0 = 0 We can use the quadratic formula to solve for H: H 0.5 ± 0.5-4(0.5)(0.0) = = (0.5) 0.1 or 0.4 Since we are given that s < 0.5 and we know that H = 0.1 H = s, it must be the case that: ActuarialBrew.com 014 Page 13
Exam MFE May 007 We can now find the value of G: G - 0.04 0.5 = H G - 0.04 0.5 = 0.1 G = 0.065 Solution 19 D Chapter 9, Delta-Gamma Approximation The delta-gamma approximation for the new price is: Vt h Vt t t The change in the stock price is: = S - S = 31.50-30.00 = 1.50 e t+ h t The delta-gamma approximation is: Vt h Vt t t 1.50 4.00 (1.50)( 0.8) (0.10) 3.695 ActuarialBrew.com 014 Page 14