General remarks: there is a problem with the symbol when uploading latex generated questions to moodle, thats why I left it out in all of the ques- tions in section A.

I will add these manually in the moodle implementation. The 0.10 penalty mentioned on each question is the default penalty for quizzes which allow for multiple attempts on each question. This option will be turned off during the exam (i.e. each student has only one attempt on all of the questions), so you can safely ignore this. Mathematical Finance 4: Degree exam part A 1. Force of interest numerical 3 points 0.10 penalty The force of interest is predicted to be the following function r of time t (in years) 0.02 , t 2 0.01t , 2 t 5 0.250.04t ,5t6 0.01 , t 6 r(t) = What is the constant annual rate of interest that, with monthly com- pounding, would give the same increase in value from t = 1 to t = 7? 2.75.1 0.0275.001 We are looking for the constant r satisfying r 612 7 r(t)dt Computation shows 7 r(t)dt = 0.165. Hence 1 1 r = 12(exp(0.165)72 1) 0.0275 2. Annuities multi 3 points 0.10 penalty Multiple Shuffle You are comparing mortgage offers from different building societies and are interested in the following two options for a loan of 120, 000: 1 1 + 12 = exp 1 (1) interest rate of 1.68% for the first 5 years and then 3.54% for the remaining term. No additional fees are charged. (2) fixed interest rate of 2.34% throughout the term but an additional fee of 1, 000 is added to the loan. Assuming that interest is converted at the end of every month, which of the following are true? (a) Assuming a monthly repayment of 760, the remaining debt after 5 years for offer (1) would be approximately 82, 972.51. (50%) (b) The monthly repayment to clear the debt after 20 years for offer (2) is A = 749.43 (50%) (c) Assuming offer (2) has a term of 20 years, the APR of (2) is 2.43% (50%) (d) Assuming a monthly repayment of 760, it would take at least 18 years to clear the entire debt for offer (1). (50%) (a) The remaining debt after k conversion periods is calculated by the formula r (1 + r )k 1 P(1+ )kA 12 n nr In this particular situation we calculate 120, 000(1 + 0.0168 )60 12 760(1+0.0168)601 82,972.51 0.0168 12 (b) We calculate r (1 + nr )T n 12 1.00195240 A = P n (1 + nr )T n 1 = 121, 000 0.00195 1.00195240 1 Hence A 631.79. (c) Using the result from (b) we can compute the APR a according to the formula 2012 a 120, 000 = 631.79(1 + 12 )k k=1 Using the substitution z = (1 + a )1 and the geometric sum on 12 the right hand side we can turn the equation into a polynomial. 2 The solutions of the polynomial equation can be found found using wolfram alpha or similar (I have shown students how to do this in class and there were plenty of similar exercises on the tutorial sheets). (d) According to (a) we have a debt of 82, 972.51 after 5 years. We are looking to determine the minimal T such that P (1 + r )12T 12 (1+ r )12T 1 A n r 0. After some basic algebraic manipulations this is n equivalent to In our case we get (1 + r )12T A 12 APnr 1.0029512T 1.475066250 which leads to T 10.9964. In particular, the entire debt would be already cleared after 11 + 5 = 16 years. 3. Bonds numerical 2 points 0.10 penalty Consider a government bond with nominal value 100 and a coupon rate of 1.2% maturing in 12 years time. Suppose you want to sell this bond on the secondary market. To attract buyers you want to offer a yield of 2.4%. What price should you ask for? Please round your answer to the second decimal place. 87.55.02 The price of the bond should equal the present value (with respect to the yield) of the cash flow sequence it generates, i.e. 24 P =0.6(1+ k=1 0.024 2 )k +100(1+ 0.024 2 )24. Setting z = 1 1.012 23 P =0.6zzk +100z24 k=0 1z24 = 0.6z 1z +100z24 87.55 3 4. Call/Put options general multi 2 points 0.10 penalty Single Shuffle Who of the following is potentially obligated to buy an asset at a pre- determined price? (a) A trader taking a short position in a call option (b) A trader taking a short position in a put option (c) A trader taking a long position in a call option (d) A trader taking a long position in a put option 5. Trading strategies multi 3 points 0.10 penalty Multiple Shuffle Select all the options you need to create a trading strategy with the following profit profile: (a) A short position in a call option with strike X2 (16.66667%) (b) A short position in a call option with strike X1 (16.66667%) (c) A short position in a put option with strike X2 (16.66667%) (d) A short position in a put option with strike X1 (16.66667%) (e) A long position in a put option with strike X1 (50%) (f) A long position in a put option with strike X2 (16.66667%) (g) A long position in a call option with strike X2 (50%) 4 (h) A long position in a call option with strike X1 (16.66667%) The depicted trading strategy is known as a long strangle. 6. One period model: Arbitrage/FTAP multi 2 points 0.10 penalty Multiple Shuffle Consider an arbitrage free one period market model (,F,P,,S). Which of the following are correct? (a) There might exist a portfolio such that 0, { | S() < 0} is a nullset, and { | S() > 0} is not a nullset (with respect to P). (50%) (b) Every probability measure Q [P ] is risk-neutral. (50%) (c) If two portfolios 1,2 have different prices (i.e. 1 = 2), then thereexistsasetAF withP(A)>0andS()=S() for all A. (50%) (d) If Q [P] denotes the set of risk-neutral probability measures equivalent to P, then Q either contains a single element, or it is uncountably infinite. (50%) 7. Multiperiod model numerical 3 points 0.10 penalty The current price of a stock is 32. It is believed that in each of the next two periods it will either increase or decrease by 2%. The risk-free interest rate is 0.6%. What is the probability (with respect to the risk- neutral probability measure Q) that a call option on the stock (written at time t = 0) with strike price 31 will be exercised? 0.8775.0001 87.75.01 After two periods we have the price 33.29 going up-up, 31.99 going up-down, or down-up; and 30.73 going down-down. We have q = ra = 0.006+0.02 = 0.65. Hence Q(1,1) = 0.652 = 0.4225, Q(1,1) = ba 0.02+0.02 Q(1, 1) = 0.65(1 0.65) = 0.2275 and Q(1, 1) = 0.352 = 0.1225. The call option will only be exercised if the market value is higher than the strike price, which happens precisely if there is at least one 1 in the path. Hence the probability is 0.4225 + 2 0.2275 = 0.8775 5 8. Black-Scholes multi 2 points 0.10 penalty Single Shuffle Suppose the volatility of a stock S is zero and C is a call option on S with strike price X. Then the Black-Scholes price for C is (a) (S0 XerT)+ (b) zero, since the option will not be exercised in any case (c) (X S0erT )+ (d) S0 (e) XerT The Black-Scholes price for a call option with volatility is computed as erT2 y2 2 (S0e Ty+rT1/2 T X)+e 2 dy erT rT + y2 rT rT Soif=0,weareleftwith2(S0e X) e 2 dy=e (S0e X)+ = (S0 XerT )+. 6 Friday, 1st January, 2021 gulogo_black.pdf EXAMINATION FOR THE DEGREES OF M.A., B.Sc. AND M.Sci. 4H HONOURS MATHEMATICS 4H: Mathematical Finance Course code: MATHS 4117 This paper consists of two sections: Section A and Section B. Candidates should answer all questions in Section A, and all questions in Section B. Unless otherwise stated, justification for your answers must be given. Indicative marks are shown for each question or question part. [CONTINUED OVERLEAF 2 Section A Attempt all questions from this section. Please complete the Moodle quiz on the Exam Moodle Page. Justification of answers is not required for the Moodle quiz. 20 Section B Attempt all questions from this section. B1. Consider a stock with current price . The risk reversal options trading strategy consists of buying a call option with strike price XC > , and selling a put option with strike price XP < on the stock with the same expiration date. (i) Find an expression for the payoff of a risk reversal trading strategy and sketch its profit pattern. 4 Solution. Let T be the expiration date of the options. The payoff of this trading strategy at time T is given as S T X P , S T X P C(ST)P(ST)= 0, XP ST XC STXC, STXC Wehaveprofit=C(ST)P(ST)C +P andsinceXP >XC weexpectthat the price for the total portfolio is C P > 0. Hence a qualitative sketch of the profit profile is as follows: Remark: similar examples of trading strategies and their profit profiles seen in class/on exercise sheets [CONTINUED OVERLEAF 3 (ii) Nowsuppose=48andthestockpriceatexpirationisbelievedtobeeither 46, 50, or 54 (each case being equally likely to occur). Moreover, there is a riskless bond available with interest rate 1.2%. (a) Specify the components of a one-period market model (, F, P, , S) mod- elling this situation. 2 Solution. Put = {1,2,3}, F = P() and P({i}) = 13 for all i . We further set = (1,48) and S = (S0,S1) with S0 = 1.012 and S1(1) = 46, S1(2) = 50, and S1(3) = 54. Remark: standard (b) Show that your model from (a) is arbitrage free and calculate the set of all risk-neutral probability measures equivalent to P. 4 Solution. A risk-neutral probability measure Q with qi := Q({i}) must satisfy 1.01248=(1+r)=EQ(S1)=46q1 +50q2 +54q3 and of course q1 + q2 + q3 = 1, or equivalently, q3 = 1 q1 q2. Substituting this into the first equation gives 5.424 = 8q1 4q2, or equivalently q2 =2q1 +1.356. Itfollowsthatalsoq3 =1q1 q2 =q1 0.356. Using these equalities it is straightforward to check that q3 (0,1) iff q1 (0.356, 1.356). Similarly, q2 (0, 1) iff q1 (0.356/2, 0.678). Since we require all qi to be contained in (0,1) we get that the set Q of risk- neutral probability measures equivalent to P (viewed as triples indicating their values for elementary outcomes) is {(q1, 2q1 + 1.356, q1 0.356) | q1 (0.356, 0.678)} Remark: similar example done in class (c) Calculate the set of arbitrage-free prices for a risk reversal trading strategy with strike prices XC = 49 and XP = 47. 5 Solution. We first calculate the prices of the put and call options sepa- rately. An arbitrage-free price for C is given by 1 EQ(C), where Q is a 1.012 risk-neutral probability measure equivalent to P. Note first that the call option will not be exercised in case 1, i.e. C(1). Using this and our results from (b) we get 1111 1.012EQ(C) = 1.012(C(2)q2+C(3)q3) = 1.012(q2+5q3) = 1.012(3q10.424) Similarly, an arbitrage-free price for the put option is given by 111 1.012EQ(P) = 1.012P(1)q1 = 1.012q1 Since the total price for the strategy is the difference between the prices for the call option and the put option, we get for the set of arbitrage free prices (CP)={ 1 (2q1 0.424)|q1 (0.356,0.678)} 1.012 [CONTINUED OVERLEAF Applying the bounds for q1 from (b) this (up to rounding error) coincides with the interval (0.28,0.92). Remark: similar to examples in class/on exercise sheets B2. Consider a 2-period binomial model with a single risky stock whose price process is depicted below. 72 76.32 67.68 80.90 71.74 4 63.62 (i) Find the possible values of the risk-free interest rate r for which this model is arbitrage free. 1 Solution. We have 76.32/72 = 1.06 and 67.68/72 = 0.94 = 10.06. Quick checks show that the stock price either increases or decreases by 6% in each period. I.e. a = 0.06 and b = 0.06. Hence any r (0.06, 0.06) will provide an arbitrage-free 2-period binomial model. Remark: standard example (ii) An Asian call option with strike price X is an option that depends on the average price Sav = 13 (S0 + S1 + S2) of the underlying asset (where St : R denotes the price of the asset at time t). The payoff of such an option can be modelled using the random variable Ccall := (S X)+ av av Assuming that r = 0.02, calculate the arbitrage-free price of an Asian call option with strike price 71. 4 Solution. We first compute the average prices for each possible path in the diagram. Let 1 denote going up and 1 denote going down in the tree depicted above. Then we compute Sav(1,1) = 72+76.32+80.90 76.41. Similarly, we get 3 Sav(1, 1) 73.35, Sav(1, 1) 70.47, and Sav(1, 1) 67.77. Now q = ra = ba 23 and Q() = qk(1q)2k where k is the number of 1s in the path . According to the lecture notes we can calculate the arbitrage-free price of the asian call 1 14 2 = 1.022 (5.41Q(1, 1) + 2.35Q(1, 1)) 1.022 5.41 9 + 2.35 9 2.81 1 call 1call = (1+r)2EQ(Cav ) = (1+r)2 Subbing our known data we compute Cav ()Q() Remark: an example like this is on the exercise sheets [CONTINUED OVERLEAF 5 (iii) We would now like to describe the behaviour of the stock above by the Black- Scholes model. (a) Based on the data for 2 periods given above, suggest a reasonable value for the volatility of the stock, and hence provide a description of the ran- dom variable S1 describing the stock price in terms of a standard normally distributed random variable Z. 2 Solution. When approximating Black-Scholes one assumes aN = exp( ) 1andbN =exp( )1.InourcaseN=2weget N 0.06=b2 =exp( )1. 2 Solving this for gives 0.082. Hence in the Black-Scholes model we get 2 S1 = 72 exp(0.082Z + r 2 ) = 72 exp(0.082Z + 0.016638), N where Z N (0, 1). Remark: standard example (b) Use the Black-Scholes model to determine the probability that a European call option with strike price 70 expiring in 1 year is exercised. 3 Solution. The call option is exercised if equivalently if i.e. if 70 S1 = 72 exp(0.082Z + 0.016638) 0.082Z + 0.016638 ln(70) Z 72 1 (ln(70) 0.016638) 0.5464 0.082 72 According to the normal distribution table we have P(Z 0.5464) P(Z < 0.55) = 0.7088, i.e. the chance that the option is exercised is roughly 70.88%. Remark: a similar example was calculated in class and on exercise sheets (iv) Let (,F,(Ft)t,P,(St)) be an arbitrage-free T-period market model and let , be two self-financing trading strategies with associated value processes (V ) and (V) . Prove the law of one price in this setting, i.e. show that if V = V holds tt TT P-almostsurely,thenV =V holdsP-almostsurelyalsoforallt{0,...,T1}. tt 5 Solution. Recall that V = X where (X ) is the discounted price process ttttt Xti = Sti . We have that VT VT = 0 holds P -almost surely. It follows that if Q St0 is a martingale measure equivalent to P , then VT VT = 0 holds also Q-almost [CONTINUED OVERLEAF tt 6 surely. Using that and are self-financing in line (2) and that Q is a martingale measure in line (4) below we compute: V V = ( ) X (1) T1 T1 T1 T1 T1 =(T T)XT1 (2) =(i i)Xi (3) T T T1 i = ( Ti Ti ) E Q ( X Ti | F T 1 ) (4) i =E (i i)Xi |F (5) Q TTTT1 i =EQ VTVT|FT1 =0 (6) Continuing inductively in the same fashion proves the result for all t {0, . . . T }. Remark: most difficult part of exam. Proof for one-period models seen in class, but the students havent seen the proof in a multiperiod setting. END]