past_papers

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

## This note was uploaded on 02/01/2012 for the course MATH 5010 taught by Professor D during the Spring '11 term at HKU.

### Page1 / 12

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online