Course MFE/3F Practice Exam Solutions he chapter references below refer to the chapters of the ActuraialBrew.com Study Manual. Solution C Chapter 6, Sharpe Ratio If we (incorrectly) assume that the cost of the shares needed to delta-hedge the put option is $8.5, then we have: SD= 8.5 fi D is positive But the delta of a put option must be negative, so the equation above cannot be correct. Since (iv) tells us that the put option is delta-hedged by selling shares of stock, the original position must be a short position in the put option. he cost of the shares required to delta-hedge the put option is the number of shares required, D, times the cost of each share, S. Since shares are sold, the cost is negative. herefore, based on statement (iv) in the question, we have: D S =-8.5 he elasticity of the put option is: SD - W= = 8.5 V 7 he risk premium of the put option is equal to the elasticity of the put option times the risk premium of the underlying bond: g - r = W ( a -r) -8.5 g - 0.08 = (0. -0.08) 7 g = 0.008 Solution D Chapter 7, Black Formula he option expires in year, so =. he underlying bond matures year after the option expires, so s =. he bond forward price is: P(0, + s) 0.84 F = P0 (, + s) = = = 0.9099 P(0, ) 0.96 he volatility of the forward price is: s = 0.08 ActuarialBrew.com 04 Page
We have: d Ê F ˆ Ê0.9099ˆ ln Á + 0.5s ln + 0.5(0.08) () ËK Á Ë 0.93 = = = -0.454 s 0.08 d = d - s = -0.454-0.08 = -0.354 N( - d ) = N(0.454) = 0.59543 N( - d ) = N(0.354) = 0.660 he Black formula for the put price is: [ ] [ ] P = P(0, ) K N( -d ) - F N( -d ) = 0.96 0.930 0.660-0.9099 0.59543 = 0.03783 Solution 3 C Chapter 3, wo-period Binomial Model he factors u and d are constant in the model since: 0 44 96 u = = = =.0 00 0 80 80 96 64 d = = = = 0.80 00 0 80 he risk-neutral probability is: ( r-d ) h (0.05-0.00) e -d e -0.8 p* = = = 0.688 u-d.0-0.80 If the final stock price is $44, then the payoff of the call option is $50. If the final stock price is $96, then the payoff of the call option is $. If the final stock price is $64, then the payoff of the call option is $0. he value of the call option is: n -rhn ( ) ÈÊnˆ j n-j j n-j V( S0, K,0) = e  ÍÁ ( p*) ( p*) V( S0u d, K, hn) j - j = 0 ÎË -0.05() = e [(0.688) (50) + (0.688)( - 0.688) + 0] = 8.698 ActuarialBrew.com 04 Page
Solution 4 C Chapter 8, Elasticity All three statements are discussed in the last two paragraphs preceding Example.8 on page 363 of the third edition of the Derivatives Markets textbook. Statement I is false. he elasticity of a call option decreases as the option becomes more in the money. As the strike price decreases, the option becomes more in the money, and therefore the elasticity decreases. Statement II is false. he upper bound for a put option is 0, not -. he correct statement is W 0 Put. Statement III is true. he elasticity is equal to the percentage of the replicating portfolio invested in the stock. Since a call option is replicated by a leveraged investment in the stock, WCall. Solution 5 E Chapter, All-or-Nothing Options For the square of the final stock price to be greater than 64, the final stock price must be greater than 8: [ S ] > S > () 64 () 8 herefore, the option described in the question is 5 cash-or-nothing call options that have a strike price of 8. he current value of the option is: -r -0.03 5 CashCall( SK,, ) = 5 e Nd ( ) = 5 e Nd ( ) o find the delta of the option, we must find the derivative of the price with respect to the stock price: -0.03 ( 5 ( )) ( ( )) e N d Nd d -0.03-0.03 = 5e = 5 e N'( d ) S S S he derivative of d with respect to the stock price is: Ê Ê -d Se ˆ s ˆ Áln Á -r + Á ËKe Á -s s Á -r -d Ke e d ( d s ) Á - Ë -d -r Se Ke = = = = S S S s Ss = = 0.4667 8 0.30 ActuarialBrew.com 04 Page 3
he current value of d is: d Ê -d ˆ - s Ê 0.0ˆ Se 8e 0.30 ln Á - - - ln r Á -0.03 ËKe Ë8e = = = -0.667 0.30 s he density function for the standard normal random variable is: N'( x) = e p -0.5x We can now calculate the delta of the option: -0.03 d -0.03-0.5d 5 e N'( d ) = 5e e 0.4667 S p -0.03-0.5(-0.667) = 5e e 0.4667 p = 5 0.60 = 8.03 herefore, 8.03 shares must be purchased to delta-hedge the option. Solution 6 C Chapter 5, Forward Price of S a he stock price follows geometric Brownian motion, and the claim pays S ( ) a, where a = 5. he forward price on the claim is therefore: ( r) 0.5 ( ) F a 0, S( ) a a a a S(0) e a ( r) 0.5 ( ) S(0) e aa e a 0.5 ( ) F0, ( S) aa e (.5) 0.65 5 0.5(5)(5)(0.5) () e Solution 7 C Chapter 9, Delta-Gamma Hedging he gamma of the position to be hedged is: -,000 0.04 = - 4.0 ActuarialBrew.com 04 Page 4
We can solve for the quantity, Q, of the other put option that must be purchased to bring the hedged portfolio s gamma to zero: - 4. + 0.08Q = 0.00 Q =,496.5 he delta of the position becomes: -,000 (- 0.5875) +,496.5 (- 0.339) = 79.9 he quantity of underlying stock that must be purchased, of the position being hedged: Q S =- 79.9 Q S, is the opposite of the delta herefore, in order to delta-hedge and gamma-hedge the position, we must sell 79.9 units of stock and purchase,496.5 units of Put-II. Solution 8 B Chapter 5, Geometric Brownian Motion Equivalencies A lognormal stock price implies that changes in the stock price follow geometric Brownian motion: - - + ( a d 0.5 s ) t sz( t) St S e dst Stdt StdZt () = (0) () = ( a - d) () + s () () Substituting 0.4 for s, we have: ( 0.500.4 ) t0.4 Z( t) St () S(0) e dst () ( ) Stdt () 0.4 StdZt () () ( 0.08) t0.4 Z( t) St () S(0) e dst () ( ) Stdt () 0.4 StdZt () () he expression for ds( t ) can be rewritten to match Choice B: dst () = ( a - d) Stdt () + 0.4 StdZt () () = ( a -d - 0.08) S( t) dt + 0.4 S( t) dz( t) + 0.08 S( t) dt Solution 9 E Chapter 7, Black-Scholes Call Price he first step is to calculate d and d : ln( S/ K) + ( r - d + 0.5 s ) ln(30 / 35) + (0.06-0.0 + 0.5 0.3 ) 0.5 d = = s 0.3 0.5 =-0.8094 d = d - s = -0.8094-0.3 0.5 = -0.98094 ActuarialBrew.com 04 Page 5
We have: Nd ( ) N ( 0.8094) 0.0584 Nd ( ) N ( 0.98094) 0.633 he value of one European call option is: r C Se N( d ) Ke N( d ) Eur 0.0(0.5) 0.06(0.5) 30e 0.0584 35e 0.633 0.53649 he value of 00 of the European call options is: 00 0.53649 = 5.36 Solution 0 D Chapter 5, Stock Price Probability he stock price follows geometric Brownian motion with: a - d = 0.5-0.03 = 0. s =-0.30 s =- 0.30 = 0.30 o answer this question, we use: ( ) Prob S < = (- ˆ K N d ) When calculating ˆd, we use the absolute value of the volatility coefficient, s, so below we have s = 0.3 : dˆ ( ) ÊS ln t ˆ Ê S ( 0.5 )( ) ln t ˆ Á + a -d - s -t + 0.5-0.03-0.5(0.3) 0.5 Ë K Á Ë.0S t = = s - t 0.3 0.5 =-0.75 he probability that the stock price does not increase by more than 0% over the next 6 months is: ( ) Prob S.0 S N( d ˆ ) N(0.75) 0.60739 t+ 0.5 < t = - = = ActuarialBrew.com 04 Page 6
Solution E Chapter 3, Historical Volatility he Derivatives Markets textbook provides two formulas for estimating the volatility of the stock:. Chapter Formula:  ( ri - r) = sˆ i = h k -. Chapter 4 Formula: k k  ( ri ) ˆ i = sh = h k - If we were to use the Chapter formula, then we would need to obtain r : k   4 r i i i= i= r = = k 4 r Since we are not given the first 9 prices, we are not able to obtain r. herefore we use the Chapter 4 formula instead. he original volatility estimate is based on 0 prices. here are 9 returns, so k = 9 : k  ( ri ) ˆ i= s h = h k - 0.8 = / 5 9  ( ri ) = 0.000857 i= 9  ( ri ) i= 9- ActuarialBrew.com 04 Page 7
After a week passes, the analyst can add 5 more returns into the sum of squared returns: 4 9 4 4 Â( ri) = Â( ri) + Â ( ri) = 0.000857 + Â ( ri) i= i= i= 0 i= 0 È Ê 98 ˆ È Ê96ˆ È Ê9ˆ È Ê90ˆ È Ê94ˆ = 0.000857 + Íln Á + Íln Á + Íln Á + Íln Á + Ë Ë Ë Ë Íln Á Ë Î 00 Î 98 Î 96 Î 9 Î 90 = 0.000857 + 0.0050865 = 0.006047 Now we can calculate the volatility using all 5 prices. here are 4 returns, so k = 4 : k 4 Â( ri) Â( ri) s i i h = = 0.006047 ˆ = = = = 34.4% h k - / 5 4- / 5 4- We can use the I-30XS Multiview calculator to obtain the sum of the 5 additional squared returns above: [data] [data] 4 (to clear the data table) (enter the data below) L L L3 00 98 ----------- 98 96 96 9 9 90 90 94 (place cursor in the L3 column) [data] Æ (to highlight FORMULA) [ln] [data] / [data] ) [enter] [ nd ] [quit] [ nd ] [stat] DAA: (highlight L3) FRQ: (highlight one) (select CALC) [enter] Statistic number 6 is 0.0050865. Solution D Chapter 8, Black-Derman-oy Model In each column of rates, each rate is greater than the rate below it by a factor of: i h e s ActuarialBrew.com 04 Page 8
herefore, the missing rate in the third column is: -s 0.086 0.086e i = 0.086 = 0.56 0.785 he missing rate in the fourth column is: -s 0.473 0.479e i = 0.479 = 0.95 0.085 he tree of short-term rates is then: 4.79% 7.85% 7.47% 9.5% 4.00% 0.86% 4.59% 0.85% 5.6% 4.73% he caplet pays off only if the interest rate at the end of the fourth year is greater than 4.00%. he payoff table is:.5467 0.0000 0.0000 4.69 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 he payments have been converted to their equivalents payable at the end of 3 years. he calculations are shown below: 00 (0.479-0.400) =.5467.479 00 (0.95-0.400) = 4.645.95 he expected present value of these payments is the value of the 3-year caplet: È - * V0 = E ÍV Í ( + r ) Î i= 0 i 3 Ê.5467 4.645 = 0.5 Á + Ë (.4)(.747)(.785) (.4)(.747)(.785) 4.645 4.645 ˆ + + (.4)(.747)(.086) (.4)(.459)(.086) =.89 ActuarialBrew.com 04 Page 9
Alternatively, the value of the caplet can be found recursively, as shown in the tree below:.5467 6.575 3.5466 4.645.895.7609 0.7684 0.0000 0.0000 0.0000 Solution 3 D Chapter, Currency Options he first option gives its owner the right to: Give up $.00 Get 0.4 pounds he value of this option is: 0.03 0.03 pounds = dollars = $0.07895 0.38 he second option gives its owner the right to: Give up $.50 Get.0 pound he payoff is.5 times the payoff of the first option, so the value of the option is.5 times the value of the first option:.5 $0.07895 = $0.974 Solution 4 E Chapter 4, Options on Futures Contracts he values of u F and d F are: s h 0.5 0.5 uf = e = e =.9336 -s h -0.5 0.5 df = e = e = 0.83797 he futures price tree and the call option tree are below. here is no need to find the current value of the call option: F 0 F European Call,3.700 3.700,00.0000 9.7636 0.0000 ActuarialBrew.com 04 Page 0
he number of futures contracts that the investor must be long is: Vu - Vd 3.700-0.0000 D= = = 0.7999 Fu ( F -df ),3.700-9.7636 Solution 5 B Chapter 3, Realistic Probability In the Cox-Ross-Rubinstein model, the values of u and d are: s h 0.3 0.5 u = e = e =.363 -s h -0.3 0.5 d = e = e = 0.80886 We can solve for p, the true probability of the stock price going up, using the following formula: ( a -d) h (0.5-0.05)0.5 e -d e -0.80886 p = = = 0.567 u-d.363-0.80886 Solution 6 A Chapter 7, Options on Futures he 6-month futures price is: ( r-d )( -0) F0, = F S F te (0.09-0.05)(0.5) F0,0.5 = 50e = 5.00 he values of d and d are: We have: F0, K ln / 0.5 ln 5.00/ 55 0.5(0.30) (0.5) d F 0.4895 0.30 0.5 d d 0.4895 0.30 0.5 0.4608 Nd ( ) N ( 0.4895) 0.4070 Nd ( ) N ( 0.4608) 0.337 he value of the call option is: Eur r r 0, F 0, F 0.09(0.5) e 0.09(0.5) e C ( F, K,, r,, r) F e N( d ) Ke N( d ) 5.00 0.4070 55 0.337.6389 ActuarialBrew.com 04 Page
Solution 7 A Chapter 9, Frequency of Re-Hedging he gamma the 00 calls is: 00 0.034 = 3.4 he return on the delta-hedged position is: Rh = S s G( - z ) h We can solve for the random values that produce a profit greater than $.40:.40 75 0.30 3.4 ( z ) 365.40.3775( z ) 0.5908 z z 0.4097 z 0.4097 and z 0.4097 z 0.64009 and z 0.64009 Since z is a standard normal random variable, the probability that both of the inequalities above are satisfied is: [ ] N(0.64009) - N( - 0.64009) = N(0.64009) - - N (0.64009) We have: N (0.64009) 0.73894 he probability of the profit exceeding $.40 is therefore: N(0.64009) N (0.64009) 0.73894 ( 0.73894) 0.47788 Solution 8 A Chapter, Exchange Options Since we need to find the value of the option in yen, we use yen as the base currency. Let s define the underlying asset to be euro and the strike asset to be.45 Canadian dollars. his makes the option an exchange put option with: S = 65 K =.45 5 = 66.75 he volatility of ln( S/ K ) is: S K S K 0.8 0.3 (0.59)(0.8)(0.3) 0.036448 0.909 ActuarialBrew.com 04 Page
he values of d and d are: d Ê -ds ˆ - s Ê 0.05() ˆ Se 65e 0.909 () ln Á + + -d ln Á -0.07() Ë K Ke Ë66.75e = = = 0.4495 s 0.909 d = d - s = 0.4495-0.909 = -0.04596 We have: N( - d ) = N( - 0.4495) = 0.4438 N - d = N = ( ) (0.04596) 0.5833 he value of the exchange put is: ExchangePutPrice K ( ) S Ke N d Se N( d) 0.07() 0.05() 66.75 e (0.5833) 65 e (0.4438).55 Solution 9 E Chapter, Put-Call Parity he dividends paid before the expiration of the options occur at time month, 4 months, and 7 months. We can use put-call parity to find the value of the put option: r Eur(, ) 0 0, ( ) Eur (, ) 0.(0.75) 0.(/) 0.(4 /) 0.(7 /) C K Ke S PV Div P K 6.38 75e 73 e e e P (75,0.75) P Eur (75,0.75) 7.693 Eur Solution 0 D Chapter 9, Risk-Neutral Cox-Ingersoll-Ross Model Ann uses the CIR model for the short rate. We can use the Sharpe ratio in Ann s model to solve for the parameter f : f(0.6,0) =.0 r f =.0 s 0.6 f =.0 0.05 f = 0.5 ActuarialBrew.com 04 Page 3
he risk-neutral version of Ann s model is: dr = [ a( r) + s( r) f( r, t) ] dt + s( r) dz È r = Ía( r) + s rf dt + s( r) dz ÍÎ s = È Îa ( r) + fr dt + s( r) dz = [ 0.4(0.09375 - r) + 0.5r] dt + 0.05 rdz = 0.5(0.5 - rdt ) + 0.05 rdz Mike s model will produce the same prices (and therefore the same yields) as Ann s model if the risk-neutral version of his model is the same as the risk-neutral version of Ann s model. We can rule out Choices A and B, because they are based on the Vasicek model. Ann s model is a CIR model. We can use the Sharpe ratio in Choice C to solve for the parameterf : 0.6 f = 0.05 0.05 f = 0.0065 he risk-neutral version of Choice C is: dr = È Îa( r) + fr dt + s( r) dz = [ 0.5(0.5 - r) + 0.0065r] dt + 0.05 rdz = 0.4375(0.53846 - rdt ) + 0.05 rdz he risk-neutral version of Choice C is not the same as the risk-neutral version of Ann s model. We can use the Sharpe ratio in Choice D to solve for the parameterf : 0.6 f = 0.40 0.05 f = 0.05 he risk-neutral version of Choice D is: dr a( r) r dt ( r) dz 0.3(0.5 r) 0.05rdt0.05 rdz 0.5(0.5 rdt ) 0.05 rdz Choice D has the same risk-neutral process as Ann s model, and therefore it will produce the same prices and yields as Ann s model. herefore, Choice D is the correct answer. ActuarialBrew.com 04 Page 4
For the sake of thoroughness, let s consider Choice E as well. he Sharpe ratio for Choice E is the same as the Sharpe ratio we found for Choice C: 0.6 f = 0.05 0.05 f = 0.0065 he risk-neutral version of Choice E is: dr = È Îa( r) + fr dt + s( r) dz = [ 0.3(0.5 - r) + 0.0065r] dt + 0.05 rdz = 0.9375(0.539 - rdt ) + 0.05 rdz he risk-neutral version of Choice E is not the same as the risk-neutral version of Ann s model. Solution C Chapter 5, Estimating Parameters from Observed Data he quickest way to do this problem is to input the data into a calculator. Using the I-30X IIS, the procedure is: [nd][sa] (Select -VAR) [DAA] [ENER] X= ln ( 95 /00) ØØ (Hit the down arrow twice) X= ln ( 05 / 95 ) ØØ X3= ln ( /05 ) ØØ X4= ln ( 03 /) [ENER] [SAVAR] Æ Æ (Arrow over to Sx) [ENER] (he result is 0.90608) [SAVAR] Æ (Arrow over to x ) [ENER] (he result is 0.088676407) o exit the statistics mode: [nd] [EXISA] [ENER] Alternatively, using the BA II Plus calculator, the procedure is: [nd][daa] [nd][clr WORK] X0= 95/00 = LN [ENER] ØØ (Hit the down arrow twice) X0= 05/95 = LN [ENER] ØØ ActuarialBrew.com 04 Page 5
X03= /05 = LN [ENER] ØØ X04= 03/ = LN [ENER] [nd][sa] ØØØ = (he result is 0.90608) [nd][sa] ØØ = (he result is 0.0886764) o exit the statistics mode: [nd][qui] Alternatively, using the I-30XS MultiView, the procedure is: [data] [data] 4 (enter the data below) (to clear the data table) L L L3 00 95 ----------- 95 05 05 03 (place cursor in the L3 column) [data] Æ (to highlight FORMULA) [ln] [data] / [data] ) [enter] [ nd ] [quit] [ nd ] [stat] DAA: (highlight L3) FRQ: (highlight one) (select CALC) [enter] (to obtain x ) x [enter] he result is 0.0886764 [ nd ] [stat] 3 3 (to obtain Sx) Sx [enter] he result is 0.906008 herefore, the annualized estimates for the mean and standard deviation of the normal distribution are: ˆ ˆ a -d - 0.5s = 0.088676 sˆ = 0.9060 he estimate for the annualized expected return is: r ˆ a = + d + sˆ 0.5 = 0.088676 + 0 + 0.5 (0.9060) = 0.3090 h ActuarialBrew.com 04 Page 6
We know that the stock price at the end of month 4 is $03. he expected value of the stock at the end of month is: ( a -d)( -t) ES [ ] = Se t (0.3090-0)( - 4 /) 0.3090 / 3 ES [ ] = S 4 e = 03e =.394 Solution A Chapter 9, Market-Maker Profit Question 47 of the Sample Exam Questions uses a constant risk-free interest rate, but in the solution provided by the Society of Actuaries a remark describes how the question could still be answered if the risk-free interest rate is deterministic but not necessarily constant. Let s assume that several months ago was time 0. Further, let s assume that the options expire at time and that the current time is time t. he interest rate is not necessarily constant, so it can vary over time. herefore, we replace the usual discount factors as described below: e e -r -r ( -t) is replaced by e is replaced by e -Ú rsds ( ) 0 -Ú rsds ( ) t We can use put-call parity to obtain a system of equations: rsds ( ) 7.6 Ke 0 00 6.84 rsds ( ) 8.4 Ke t 90 0.49 his can be solved to find t Ú rsds ( ) e : t r( s) ds r( s) ds r( s) ds 0 89.58 0 t Ke Ke e t 89.58 rsds ( ) 9.08 e 0 rsds ( ) rsds ( ) 89.58 Ke t 9.08 Ke t 9.08 he market-maker sold 00 of the put options. From the market-maker s perspective, the value of this position was equal to the quantity owned times the price: 00 6.84 684.00 he delta of the call option can be used to determine the delta of the put option: -d - 0 D Put = DCall - e = 0.697 - e = 0.697 - = -0.303 he delta of the position, from the perspective of the market-maker, was: Delta of a short position in 00 puts 00 ( 0.303) 30.3 ActuarialBrew.com 04 Page 7
o delta-hedge the position, the market-maker sold 30.3 shares of stock. he value of this position was the quantity owned times the price: 30.3 00 3,030.00 he proceeds from selling the put options and the stock were available to be lent at the risk-free rate. he value of this cash position at time 0 was: 684.00 + 3,030.00 = 3,74.00 he initial position, from the perspective of the market-maker, was: Component Value Options 684.00 Shares 3,030.00 Risk-Free Asset 3,74.00 Net 0.00 After t years elapsed, the value of the options changed by: 00 (0.49 6.84) 365.00 After t years elapsed, the value of the shares of stock changed by: 30.3 (90.00 00.00) 303.00 After t years elapsed, the value of the funds that were lent at the risk-free rate changed by: t 0 rsds 3,74.00 e ( ) 3,74.00 9.08 03.65 89.58 he sum of these changes is the profit. Component Change Gain on Options 365.00 Gain on Stock 303.00 Interest 03.65 Overnight Profit 4.65 he change in the value of the position is $4.65, and this is the profit. Solution 3 E Chapter, Forward Price with Monte Carlo Valuation o find the value of the option at time 0, we use the risk-neutral distribution. he Monte Carlo estimate for the current price of the contingent claim is: V = -r ( -t) n e ÂVi ( ) n i= - 0.08 e È,000,000,000,000,000,000,000 7 Í 3.9 83.3 3.97.05.56 3.54 36.50 = + + + + + + Î ActuarialBrew.com 04 Page 8
he forward price for an asset that does not pay dividends is its current price accumulated forward at the risk-free rate of return. herefore, the forward price of the contingent claim is the expression above, accumulated at 8%: - 0.08 e È,000,000,000,000,000,000,000 0.08 7 Í + + + + + + 3.9 83.3 3.97.05.56 3.54 36.50 e Î È,000,000,000,000,000,000,000 = 7 Í + + + + + + 3.9 83.3 3.97.05.56 3.54 36.50 Î = 33.75 he Monte Carlo estimate for the forward price of the contingent claim is 33.75. he quickest way to answer this question is to use the I-30XS MultiView Calculator: [data] [data] 4 (to clear the data table) (enter the data below) L L L3 3.9 ----------- ----------- 83.3 3.97.05.56 3.54 36.50 (place cursor in the L column) [data] Æ (to highlight FORMULA),000 / [data] ) [enter] [ nd ] [quit] [ nd ] [stat] DAA: (highlight L) FRQ: (highlight one) (select CALC) [enter] x is 33.747389. Solution 4 D Chapter 8, Elasticity he elasticity of the contingent claim is: S 43 V.33 o find the delta of this contingent claim, we break it down into its component parts. he contingent claim's payoff is the same as a straddle's payoff, except that the contingent claim's payoff is shifted down by $0 compared to the straddle's payoff. A straddle consists of a long position in a call and a long position in a put. he payoff of the straddle can be shifted down by $0 by borrowing the present value of $0 at time 0. ActuarialBrew.com 04 Page 9
he lowest payoff for a straddle occurs at its strike price, so the straddle has strike price of $40. he contingent claim can be replicated with a portfolio consisting of a long call with a strike price of $40, a long put with a strike price of $40, and borrowing the present value of $0. he current value of the contingent claim is: -0.09 C(40) + P(40) -0e We need to find the call option's delta and the put option's delta. he first step is to calculate d : d We have: ln( S/ K) ( r 0.5 ) ln(43 / 40) (0.09 0.04 0.5 0.35 ) 0.35 0.5449 Nd ( ) = N(0.5449) = 0.70003 he delta of the call option is: 0.04 Call e N( d ) e 0.70003 0.6758 he delta of the put option is: -d - 0.04 D Put = DCall - e = 0.6758 - e = - 0.88 he delta of the contingent claim is the delta of the call plus the delta of the put minus the delta of the loan (i.e., zero): 0.6758 ( 0.88) 0.0000 0.38437 he elasticity of the contingent claim is: S 43 (0.38437) 7.0936 V.33 Solution 5 E Chapter, Strike Price Grows Over ime he prices of the options decrease as time to maturity increases. herefore, if the strike price increases at a rate that is greater than the risk-free rate, then arbitrage is available. Option C expires 0.5 years after Option B, so let s accumulate Option B s strike price for 0.5 years at the risk-free rate: 0.3 0.5 53e = 56.5594 ActuarialBrew.com 04 Page 0
Since the strike price of Option C is $57, the strike price grows from time.5 to time.0 at a rate that is greater than the risk-free rate of return. Consequently, arbitrage can be earned by purchasing Option C and selling Option B. he arbitrageur buys the -year option for $3.00 and sells the.5-year option for $3.50 he difference of $0.50 is lent at the risk-free rate of return. he.5-year option After.5 years, the stock price is $5.00. herefore, the.5-year option is exercised against the arbitrageur. he arbitrageur borrows $53 and uses it to buy a share of stock. As a result, at the end of years the arbitrageur owns the share of stock and owes the accumulated value of the $53. his position results in the following cash flow at the end of years: 0.3 0.5 58.00-53e =.4406 he -year option he stock price of $58.00 at the end of years is greater than the strike price of the -year option, which is $57. herefore, the -year put option expires worthless, and the resulting cash flow is zero. he net cash flow he net cash flow at the end of years is the sum of the accumulated value of the $0.50 that was obtained by establishing the position, the $.4406 resulting from the.5-year option, and the $0.00 resulting from the -year option: 0.3 0.50e +.4406 + 0.00 =.0890 Solution 6 B Chapter 3, Expected Return he up and down factors are constant throughout the tree, and therefore the risk-neutral probability of an upward movement is also constant and can be calculated using any node. Below we use the first node: (0.00.00)() 56 ( r ) h e e d p* 80 0.6758 u d 04 56 80 80 ActuarialBrew.com 04 Page
Working from right to left, we create the tree of prices for the European put option: Stock European Put 35.0 0.0000 04.00 7.998 80.00 7.80 5.050 7.000 56.00 34.4837 39.0 60.8000 he realistic probability is also constant throughout the tree, so as with the risk-neutral probability, we can use the values at the first node to calculate it: (0.50.00)() 56 ( ) h e e d p 80 0.7697 u d 04 56 80 80 When the stock price is $80.00, we have: [( ) u + (- ) d] g h pv pv e = V g (0.7697)(7.998) + ( -0.7697)(34.4837) e = 5.05 Ê(0.7697)(7.998) + ( -0.7697)(34.4837) ˆ g = ln Á = -0.0634 Ë 5.05 When the stock price is $04.00, we have: [( ) u + (- ) d] g h pv pv e = V g (0.7697)(0.00) + ( -0.7697)(7.0) e = 7.998 Ê(0.7697)(0.00) + ( -0.7697)(7.0) ˆ g = ln Á = -0.437 Ë 7.998 When the stock price is $56.00, we have: [( ) u + (- ) d] g h pv pv e = V g (0.7697)(7.0) + ( -0.7697)(60.80) e = 34.4837 Ê(0.7697)(7.0) + ( -0.7697)(60.80) ˆ g = ln Á = 0.03 Ë 34.4837 ActuarialBrew.com 04 Page
he tree of values for g is filled in below: N/A 0.437 0.0634 N/A 0.03 N/A he highest of the three values above is.3%. Solution 7 E Chapter 9, Rendleman-Bartter Model In the Rendleman-Bartter Model, the short-term rate follows geometric Brownian motion. When a stock's price follows geometric Brownian motion, we can find its expected value as shown below at right: ( - ) [ ] t dst () = ( a - d) Stdt () + sstdzt () () fi ESt () = S(0) e a d In this case, the short rate follows geometric Brownian motion, so we have: [ ] 0. drt () = 0.() rtdt+ 0.0() rtdzt () fi Ert () = r(0) e he expected value of r () is: 0. 0. [ ] Er() = r(0) e = 0.08e = 0.090 t Solution 8 C Chapter 7, Black-Scholes Formula Using Prepaid Forward Prices he values of d and d for Stock Y are: Ê P F0, ( S) ˆ Ê - 4 0 ln 0.5 80e ˆ Á + s P ln + 0.5 0.4 4 ÁF 0.07 4 0, ( K) Á - Ë ËKe d = = s 0.4 4 Ê 80 ˆ ln Á + 0.5 Ë -0.8 Ke = 0.48 d = d - s = d - 0.4 = d -0.48 ActuarialBrew.com 04 Page 3
he values of d and d for Stock Z are: Ê P F0, ( S) ˆ Ê - 0 6 ln 0.5 40e ˆ Á + s P ln + 0.5 0. 6 ÁF 0.84 0.07 6 0, ( K) Á - Ë Ë0.5Ke e d = = s 0. 6 Ê 40 ˆ Ê 80 ˆ ln Á + 0.5 ln + 0.5 Ë -0.8-0.8 0.5Ke Á Ë Ke = = 0.48 0.48 d = d - s 6 = d - 0. 4 = d -0.48 Note that the call on Stock Y and the call on Stock Z have the same values for d and d. he Black-Scholes formula for the call option on Stock Y is: -d -r d = - - 0 4-0.07 4-0.8 C ( S, K, s, r,, ) Se N( d ) Ke N( d ) Eur 6.30 = 80 e N( d ) - Ke N( d ) 6.30 = 80 Nd ( ) - Ke Nd ( ) We can use the Black-Scholes formula for the call option on Stock Z to see that the call option on stock Z has a value that is half the value of the call option on Stock Y: -d -r - 0 6 0.84-0.07 6 e N d Ke e N d -0.8 Nd Ke Nd È -0.8 Nd Ke Nd C ( S, K, s, r,, d) = Se N( d )- Ke N( d ) Eur = 40 ( ) -0.5 ( ) = 40 ( ) -0.5 ( ) = 0.5 80 ( ) - ( ) Î = 0.5 6.30 = 8.5 Solution 9 C Chapter 0, Compound Options Let S be the ex-dividend price of the stock. Since it doesn't makes sense to exercise the 6-strike American call, the exercise value must be less than the value of a 6-strike - year European call option. Below we use put-call parity to substitute for the value of the European call option: 0 S 6 CEur(6,) 0 S 6 6 S PEur(6,).5 7.60 PEur(6,) Since the value of the European put option is greater than 7.60, it makes sense to exercise Option A and not exercise Option B. herefore Option A is worth more than zero. ActuarialBrew.com 04 Page 4
Since it makes sense to exercise the 44-strike American call, the exercise value must be less than the value of a 44-strike -year European call option. Below we use put-call parity to substitute for the value of the European call option: 0 S44 CEur(44,) 44 0 S44 S PEur(44,) 4.6 P (44,) Eur.5 Since the value of the European put option is less than 4.6, it makes sense to exercise Option D and not exercise Option C. herefore Option D is worth more than zero. Options A and D are worth more than zero, and options B and C expire worthless. Solution 30 B Chapter 5, Drift We can use the theta of the claim to determine b : Vt () = 0 t 0 t {[ St ()] 3 bt} + t 0+ b = 0 b = 0 = 0 he value of the claim is therefore: We have: Vt () = [ St ()] 3 VS = 3S VSS = 6S V t = 0 From Itô s Lemma: S 0.5 SS( ) t dv = V ds + V ds + V dt ( ) ( )( ) = 3S 0.08Sdt + 0.0SdZ + 0.5 6S 0.08Sdt + 0.0SdZ + 0 ( 0.4 0.60 ) 3 ( 0.04 ) 3 = S dt + dz + S S dt 3 3 = 0.36S dt + 0.60S dz = 0.36Vdt + 0.60VdZ he drift is the expected change per unit of time in the price of the claim. From the stochastic differential equation above, we observe that the drift is 0.36 Vt. ( ) ActuarialBrew.com 04 Page 5