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Derivatives: Week 1 Harjoat S. Bhamra Table 1: US Government Bond Prices on December 31, 1993 maturity 19940630 19941231 19950630 19951231 19960630 19961231 19970630 19971231 19980630 19981231 coupon 0.000 7.625 4.125 4.250 7.875 6.125 6.375 6.000 5.125 5.125 bid price 98.3911 103.8125 100.1875 100.0000 108.6875 104.5313 105.1563 103.7813 100.0625 99.6250 ask price 98.4012 103.8750 100.2500 100.0625 108.7500 104.5938 105.2188 103.8438 100.1250 99.6875 1. Using the data in Table 1, find both the bid and ask discount factors via bootstrapping. Assume the current date t is December 31, 1993. Observe that all coupon payments are made 6 months apart. You can use MATLAB, Octave, R, Python, or any other programming language of your choice. Now answer the following questions (a) The 5 year ask discount factor Zask(0, 5) is equal to A 0.7708 B 0.7709 C 0.7703 D 0.7944 E 0.7950 (b) The 2-year bid implied USD risk-free rate rbid(0, 2) is equal to A 4.23% B 4.20% C 4.21% D 4.26% E 4.27% (c) The bid price of a 3 year US Government bond paying a coupon of 5% semiannually is equal to A 101.4507 B 101.3897 C 101.3899 D 101.4506 E 101.4508 (d) You are working on the swaps trading desk of the well known bank Silberfrau Schwert. A salesperson is interested in selling a 3 year USD LIBOR swap to a client and wants to know what the 3 year USD LIBOR swap rate would be. Which rate would you like to quote? A a rate less than or equal to 4.48075% B a rate greater than 4.48075% C a rate less than or equal to 4.50227% D 4.50227% E 4.48075% 1 abdulrahman abdulrahman 2. You now decide to estimate discount factors using the Nelson-Siegel model. You can use any programming language of your choice. (a) Using your new set of DF’s, what is the the bid price of a 3 year US Government bond paying a coupon of 5% semiannually? A 101.4507 B 101.3897 C 101.3899 D 101.2657 E 101.3281 (b) The head of your desk has noticed that you used bootstrapping to estimate DF’s to price a 3 year USD LIBOR swap for a client. She asks you to use the Nelson-Siegel approach to find the 3 year USD LIBOR swap rate. Based on the Nelson-Siegel model, which rate would you like to quote? A a rate less than or equal to 4.48075% B a rate greater than 4.48075% C a rate less than or equal to 4.52446% D 4.52446% E 4.48075% 3. You will have learned how to use Ito’s Lemma to derive df(xt), given some function f(·) and and a stochastic process xt. You should be able to do the following questions: (a) Suppose dxt = µdt+ σdZt, where Z = (Zt)t∈[0,∞) is a standard Brownian motion under the probability measure P. i. For α > 2, find d(xαt ) A αxα−2t [( µxt + 1 2σ 2(α− 1) ) dt+ xtσdZt ] B αxα−2t [( µxt + 1 2σ 2α ) dt+ xtσdZt ] C αxα−2t [( µxt + σ 2(α− 1) ) dt+ xtσdZt ] D αxα−2t [( µxt + 1 2σ 2(α− 1) ) dt+ σdZt ] E αxα−2t [( µxt + σ 2(α− 1) ) dt+ x2tσdZt ] ii. Find dyt, where yt = e xt A dytyt = yt ( µ+ 12σ 2 ) dt+ σdZt B dytyt = ( µ− 12σ 2 ) dt+ σdZt C dytyt = ( µ+ 12σ 2 ) dt+ σdZt D dytyt = µdt+ σdZt E dyt = ( µ+ 12σ 2 ) ytdt+ σdZt (b) Suppose dSt St = µSdt+ σSdZt. i. Find d lnSt A d lnSt = ( µS − 12Stσ 2 S ) dt+ σSdZt B d lnSt = ( µS − 12Stσ 2 S ) dt− σSdZt C d lnSt = µSdt+ σSdZt D d lnSt = ( µS + 1 2σ 2 S ) dt+ σSdZt E d lnSt = ( µS − 12σ 2 S ) dt+ σSdZt 2 ii. Find d(Sαt )/S α t A α [( µS − 12 (1− α)σ 2 S ) dt+ ασSdZt ] B ( µS − 12 (1− α)σ 2 S ) dt+ StσSdZt C α [( µS + 1 2 (1− α)σ 2 S ) dt+ σSdZt ] D α [( µS − 12 (1− α)σ 2 S ) dt+ σSdZt ] E α [( µS − 12σ 2 S ) dt+ σSdZt ] 4. You may not be fully comfortable with indicator functions and may find the use of a martingale to define a new probability measure to be somewhat exotic, so let’s explore these ideas in some examples. (a) Consider a fair coin which can land on either heads or tails. Label the event of heads via H and the event of tails via T . Now define an indicator function 1H via 1H = { 1 event H occurs 0 event H does not occur i. What is E[1H ]? A 0.4 B 1 C 0 D 0.52 E 0.5 ii. Now suppose someone tampers with the coin so that the probability of it landing on heads is now 0.51. What is E[1H ]? A 0.4 B 1 C 0 D 0.51 E 0.5 (b) Now consider an economy with two dates t ∈ {0, 1} = T . The state of the economy at date-1 is not known at date-0. What is known is that there are K possible states, where K is some strictly positive integer. We denote the K-states by ω1, . . . , ωK . When we collect them all together in one set, we have the state space (sometimes called the sample space) Ω = {ω1, . . . , ωK}. The physical probability of being in the state ωk at date-1 is denoted by Pk, where k ∈ {1, . . . ,K}. Of course, ∀ k ∈ {1, . . . ,K}, 0 ≤ Pk < 1="" and="" ∑k="" k="1" pk="1." define="" a="" set="" of="" k="" securities,="" where="" security="" k="" has="" date-1="" payoff="" given="" by="" the="" indicator="" function="" 1ωk="" .="" i.="" find="" e0[1ωk="" ].="" a="" pk−1="" b="" pk="" c="" 1="" d="" pk="" +="" pk+1="" e="" pk+pk+12="" 3="" ii.="" define="" the="" discrete-time="" stochastic="" process="" m="{Mt}t∈T" =="" {m0,m1}.="" at="" date-0="" we="" know="" that="" m0="1," but="" m1="" is="" a="" random="" variable="" which="" takes="" the="" value="" m1(ωk)="" if="" the="" economy="" ends="" up="" in="" state="" ωk="" at="" date-1.="" we="" shall="" also="" impose="" the="" conditions="" that="" ∀="" k="" ∈="" {1,="" .="" .="" .="" ,k},="" m1(ωk)=""> 0 and E0[M1] = M0. Hopefully, it is obvious that M is a martingale with respect to the physical probabilities and the state space we have described above. M is also strictly positive. We now use M1 to define Q1, Q2, . . . , QK via Qk = E0[M11ωk ], ∀ k ∈ {1, . . . ,K}. A. What is ∑K k=1Qk? A Q2 B 0 C 1 D 0.5 E Pk+Pk+12 When you think about this question, you will see that we used a martingale which was strictly positive to define a new set of numbers, Q1, Q2, . . . , QK , which represent probabilities – do you see why?. These probabilities are different from the physical probabilities – they are a new set of probabilities. iii. Which one of the following statements is true? A M1(ωk) = Pk Qk B M1(ωk) = Qk Pk C M1(ωk) = Q2k Pk D M1(ωk) = 2 Qk P 2k E M1(ωk) = Qk P 2k iv. Where EQ0 [·] denotes a date-0 conditional expectation using the new probabilities, Q1, . . . , QK , find E Q 0 [M1] A ∑K k=1 Qk 2Pk B 1 C ∑K k=1 Q2k Pk D ∑K k=1 Q2k P 2k E ∑K k=1 Q2k 2Pk 4