# tssolution2 - IEOR E4710 Term Structure Models Spring 2005...

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IEOR E4710 Term Structure Models: Spring 2005 Columbia University Instructor: Martin Haugh Solutions to Assignment 2 Question 1 Proof . By Itˆo’s Lemma, dY t = - X t exp - t 0 μ s ds μ t + exp - t 0 μ s ds X t μ t + 0 dt + exp - t 0 μ s ds X t σ t dB t = Y t σ t dB t . Then Y t = Y 0 + t 0 Y s σ s dB s = X 0 + t 0 Y s σ s dB s , which is a stochastic integral. By Theorem 2 (Martingale Property of Stochastic Integrals), Y t is a martingale. Thus for T > t , X t exp ( - t 0 μ s ds ) = E [ Y T | F t ] = E [ X T exp ( - T 0 μ s ds ) | F t ] = exp ( - t 0 μ s ds ) E [ X T exp ( - T t μ s ds ) | F t ] . Therefore, X t = E t [ X T exp ( - T t μ s ds )] . Question 2 Proof . By Itˆo’s Lemma, d ( B 3 t 3 ) = ( 0 + 0 + B t ) dt + B 2 t dB t = B t dt + B 2 t dB t . Then B 3 t 3 = 0 + t 0 B s ds + t 0 B 2 s dB s , from which we deduce t 0 B 2 s dB s = B 3 t 3 - t 0 B s ds . Question 3 (a) Proof . By Itˆo’s Lemma, d ( exp ( B t - t 2 )) = - 1 2 exp ( B t - t 2 )+ 0 + 1 2 exp ( B t - t 2 ) dt + exp ( B t - t 2 ) dB t = exp ( B t - t 2 ) dB t . Then exp ( B t - t 2 ) = 1 + t 0 exp ( B s - s 2 ) dB s , which is a stochastic integral. By Theorem 2 (Martingale Prop- erty of Stochastic Integrals), exp ( B t - t 2 ) is a martingale under probability measure P . (b) Solution . By (a) , E P [ exp ( B t - t 2 )] = 1 for all t [ 0 , T ] . Then Q ( A ) : = E P [ exp ( B T - T 2 ) 1 A ] is a probability measure since indeed Q ( Ω ) = E P [ exp ( B T - T 2 ) 1 Ω ] = 1. Under P - probability, B T Norm ( 0 , T ) . Then for some K > 0, P ( { exp ( B T ) K } ) = P ( { B T ln K } ) = 1 - Φ ( ln K T ) , where Φ ( x ) is the distribution function of standard normal random variable. And Q ( { exp ( B T ) K } ) = E P [ exp ( B T - T 2 ) 1 { B T ln K } ] = 1 2 π T ln K exp ( x - T 2 ) exp ( - x 2 2 T ) dx y = x - T T = 1 - Φ ( ln K - T T ) . 1
(c) Solution . Let η = - 1 in Girsanov’s Theorem, then ˜ B t = B t + t 0 ( - 1 ) ds = B t - t is a standard Q -Brownian motion. Thus, process A = t . Question 4 (Hedging Strategies) Solution . For any random variable G , E t [ G ] is a martingale. Then by applying Itˆo’s Lemma to d ( E t [ G ]) : = f ( t , B t ) , we must find out that the coefficient of the dt term is zero. Actually we will see in the following that for any random variable G which is a function of T and B T , the process θ ( s ) exactly equals the coefficient of the dB t term: f x ( t , B t ) . (a) M t : = E t [ G ] = E t [ 1 exp ( B T ) > K ] = E t [ 1 B T - B t > ln K - B t ] = P ( B T - B t > ln K - B t ) = 1 - Φ ( ln K - B t T - t ) . By Itˆo’s Lemma, d ( M t ) = 1 T - t Φ ( ln K - B t T - t ) dB t . Since M T = E T [ G ] = G and M 0 = E [ G ] , the process θ ( s ) = 1 T - s Φ ( ln K - B s T - s ) = 1 2 π ( T - s ) exp [ - ( ln K - B s ) 2 2 ( T - s ) ] .