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past_papers - MAFS 501: Stochastic Calculus Some of the...

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Unformatted text preview: MAFS 501: Stochastic Calculus Some of the questions in previous final examinations ========================================= The following facts might be useful to you. • h Z t H s dX s , Z t K s dY s i = Z t H s K s d h X,Y i s . • Levy’s Characterization of Brownian motion: M t is a BM iff M t is a continuous martingale such that M = 0 and h M i t = t . • M t is Q-martingale ⇐⇒ Λ t M t is P-martingale. ========================================= Unless otherwise stated, B t denotes Brownian motion in R , B = 0. 1. Which of the following are martingales? (a) e t/ 2 cos( B t ), (b) t 2 B t- 2 Z t sB s ds , (c) ‡ a 1 / 3 + B t / 3 · 3 , (d) exp { σB t } , (e) B 4 t- 6 tB 2 t + 3 t 2 , (f) tB t- R t B s ds , (g) ( B t + t )exp {- B t- 1 2 t } , (h) X t = tB t e- B t , (i) X t = (1- t ) Z t dB s 1- s , where 0 ≤ t < 1, (j) X t = Φ( Y t ), where Y t = B t √ T- t , 0 ≤ t ≤ T , and Φ( x ) is the standard Normal d.f. Solution . One could the definition of martingales to check the results. Alternatively, one cold use the Ito’s Lemma and/or the integration by parts formula. (a) By integration by parts formula, dX t = d ‡ e t/ 2 cos( B t ) · = cos( B t ) de t/ 2 + e t/ 2 d (cos( B t )) = 1 2 e t/ 2 cos( B t ) dt + e t/ 2 d cos( B t ) = 1 2 e t/ 2 cos( B t ) dt + e t/ 2- sin( B t ) dB t- 1 2 cos( B t ) dt ¶ =- e t/ 2 sin( B t ) dB t So X t is a martingale. 1 Alternatively, one can use Ito formula by taking f ( t,x ) = e 1 2 t cos x , so f 1 = 1 2 f, f 2 =- e 1 2 t sin x, f 22 =- f. Therefore, dX t = f 1 dt + f 2 dB t + 1 2 f 22 d h B i t = 1 2 fdt- e 1 2 t sin B t dB t + 1 2 f 22 dt =- e 1 2 t sin B t dB t , which is a martingale. (b) dX t = d ( t 2 B t )- 2 tB t = t 2 dB t + B t dt 2- 2 tB t = ( t 2- 2 t ) dB t + 2 B t dt . So X t is NOT a martingale. (c) X t = a 1 / 3 + 1 3 B t ¶ 3 = Y 3 t , where Y t = a 1 / 3 + 1 3 B t . Hence, dX t = dY 3 t = 3 Y 2 t dY t + 1 2 6 Y t d h Y i t = Y 2 t dB t + 1 3 Y t dt. So X t is NOT a martingale. (d) d exp { σB t } = σ exp { σB t } dB t + 1 2 σ 2 exp { σB t } dt. It is NOT a martingale. (e) Now d ( B 4 t- 6 tB 2 t + 3 t 2 ) = (4 B 3 t dB t + 6 B 2 t dt )- (6 B 2 t dt + 6 tdB 2 t ) + 6 tdt = (4 B 3 t dB t + 6 B 2 t dt )- 6 B 2 t dt- 6 t (2 dB t + dt ) + 6 tdt = (4 B 3 t- 12 t ) dB t . It is a martingale. (f) d ( tB t- R t B s ds ) = tdB t + B t dt- B t dt = tdB t . It is a martingale. (g) Let X t = ( B t + t ), Y t = exp- B t- 1 2 t ¶ . First, dY t = exp- B t- 1 2 t ¶ d- B t- 1 2 t ¶ + 1 2 exp- B t- 1 2 t ¶ dt =- exp- B t- 1 2 t ¶ dB t Next, d ( X t Y t ) = X t dY t + Y t dX t + h X,Y i t = ( B t + t )(- 1)exp- B t- 1 2 t ¶ dB t + exp- B t- 1 2 t ¶ ( dB t + dt )- exp- B t- 1 2 t ¶ dt = (1- B t- t )exp- B t- 1 2 t ¶ dB t ....
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This note was uploaded on 02/01/2012 for the course MATH 5010 taught by Professor D during the Spring '11 term at HKU.

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past_papers - MAFS 501: Stochastic Calculus Some of the...

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