{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# tu12-sol - ACTL2003&5103 Tutorial 12 Solutions 1 Ross(8th...

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

ACTL2003&5103 Tutorial 12- Solutions 1. Ross (8th edition): Chapter 19, Q25 (a) E [ Y ] = E Z 1 0 tdB t = E B 1 - Z 1 0 B t dt = E [ B 1 ] - Z 1 0 E [ B t ] dt = 0 and the variance is V ar [ Y ] = V ar Z 1 0 tdB t = Z 1 0 t 2 dt = 1 3 . (b) E [ Z ] = E Z 1 0 t 2 dB t = E B 1 - Z 1 0 2 B t tdt = E [ B 1 ] - Z 1 0 2 E [ B t ] tdt = 0 and the variance is V ar [ Z ] = V ar Z 1 0 t 2 dB t = Z 1 0 t 4 dt = 1 5 . 1

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

View Full Document
2. Z 2 . 5 1 . 5 f ( t ) dB t = Z 2 1 . 5 3 dB t + Z 2 . 5 2 4 dB t = 3( B 2 - B 1 . 5 ) + 4( B 2 . 5 - B 2 ) Since we know that 3( B 2 - B 1 . 5 ) + 4( B 2 . 5 - B 2 ) is the sum of two inde- pendent normal distributions N (0 , 4 . 5) and N (0 , 8) respectively, then Z 2 . 5 1 . 5 f ( t ) dB t is N (0 , 13 . 5) 3. (a) First, note that we can re-write the stochastic process Y t as Y t = log ( X t ) = log ( X 0 ) + μt + σB t . Let F ( x, t ) = log( X 0 ) + μt + σx . We apply Ito’s lemma on Y t = F ( B t , t ) . Since ∂x F ( x, t ) = σ , 2 ∂x 2 F ( x, t ) = 0 and ∂t F ( x, t ) = μ , We have dY t = [ ∂x F ( x, t )] x = B t dB t + [ ∂t F ( x, t )] x = B t dt + 1 2 σ 2 t [ 2 ∂x 2 F ( x, t )] x = B t dt = σdB t + μdt or = μdt + σdB t .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}