{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

stat2303_tutorial11

stat2303_tutorial11 - THE UNIVERSITY OF HONG KONG...

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

View Full Document Right Arrow Icon
Page 1 of 8 THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT2303 Probability Modelling (2010-2011) STAT2803 Stochastic Models (2010-2011) Tutorial 11 (Suggested Solutions) 1) (Random walk with drift) Let { 𝑋𝑋 ( 𝑡𝑡 ), 𝑡𝑡 = 0,1,2, … } be a stochastic process defined as 𝑋𝑋 (0) = 0 and 𝑋𝑋 ( 𝑡𝑡 ) = 𝑌𝑌 𝑖𝑖 𝑡𝑡 𝑖𝑖 =1 for 𝑡𝑡 ≥ 1 , where 𝑌𝑌 1 , 𝑌𝑌 2 , … is a sequence of independently and identically distributed random variables with the distribution 𝑁𝑁 ( 𝜇𝜇 , 𝜎𝜎 2 ) . a) Show that { 𝑋𝑋 ( 𝑡𝑡 ), 𝑡𝑡 = 0,1,2, … } has independent increments For any 0 ≤ 𝑡𝑡 1 ≤ 𝑡𝑡 2 ≤ 𝑡𝑡 3 < 𝑡𝑡 4 , 𝑋𝑋 ( 𝑡𝑡 2 ) − 𝑋𝑋 ( 𝑡𝑡 1 ) = 𝑌𝑌 𝑖𝑖 𝑡𝑡 2 𝑖𝑖 = 𝑡𝑡 1 +1 𝑋𝑋 ( 𝑡𝑡 4 ) − 𝑋𝑋 ( 𝑡𝑡 3 ) = 𝑌𝑌 𝑖𝑖 𝑡𝑡 4 𝑖𝑖 = 𝑡𝑡 3 +1 Since these increments are the sum of non-overlapping subsequences of { 𝑌𝑌 1 , 𝑌𝑌 2 , … } . They are independent and hence the process { 𝑋𝑋 ( 𝑡𝑡 ), 𝑡𝑡 = 0,1,2, … } has independent increments. b) Find the distribution of 𝑋𝑋 ( 𝑡𝑡 + 𝑠𝑠 ) − 𝑋𝑋 ( 𝑠𝑠 ) for 𝑠𝑠 ≥ 0 and 𝑡𝑡 > 0 . Does { 𝑋𝑋 ( 𝑡𝑡 ), 𝑡𝑡 = 0,1,2, … } has stationary increment? For 𝑡𝑡 > 0 , 𝑋𝑋 ( 𝑡𝑡 + 𝑠𝑠 ) − 𝑋𝑋 ( 𝑠𝑠 ) = 𝑌𝑌 𝑖𝑖 𝑠𝑠 + 𝑡𝑡 𝑖𝑖 = 𝑠𝑠 +1 ~ 𝑁𝑁 ( 𝜇𝜇𝑡𝑡 , 𝜎𝜎 2 𝑡𝑡 ) which does not depend on 𝑠𝑠 . Therefore, { 𝑋𝑋 ( 𝑡𝑡 ), 𝑡𝑡 = 0,1,2, … } has stationary increments. c) Find 𝐶𝐶𝐶𝐶𝐶𝐶 [ 𝑋𝑋 ( 𝑠𝑠 ), 𝑋𝑋 ( 𝑡𝑡 )] for 0 ≤ 𝑠𝑠 ≤ 𝑡𝑡 For 0 ≤ 𝑠𝑠 ≤ 𝑡𝑡 , 𝐶𝐶𝐶𝐶𝐶𝐶 [ 𝑋𝑋 ( 𝑠𝑠 ), 𝑋𝑋 ( 𝑡𝑡 )] = 𝐶𝐶𝐶𝐶𝐶𝐶 [ 𝑋𝑋 ( 𝑠𝑠 ), 𝑋𝑋 ( 𝑡𝑡 ) + 𝑋𝑋 ( 𝑠𝑠 ) − 𝑋𝑋 ( 𝑠𝑠 )] = 𝐶𝐶𝐶𝐶𝐶𝐶 [ 𝑋𝑋 ( 𝑠𝑠 ), 𝑋𝑋 ( 𝑠𝑠 )] + 𝐶𝐶𝐶𝐶𝐶𝐶 ( 𝑋𝑋 ( 𝑠𝑠 ), 𝑋𝑋 ( 𝑡𝑡 ) − 𝑋𝑋 ( 𝑠𝑠 )] = 𝑉𝑉𝑉𝑉𝑉𝑉 [ 𝑋𝑋 ( 𝑠𝑠 )] = 𝜎𝜎 2 𝑠𝑠 as a result of independent increments
Background image of page 1

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

View Full Document Right Arrow Icon
STAT2303/2803 Probability Modelling/Stochastic Models Semester 1, 2010-11 Page 2 of 8 2) Let { 𝐵𝐵 ( 𝑡𝑡 ), 𝑡𝑡 ≥ 0} be a standard Brownian motion. What is the distribution of 𝐵𝐵 ( 𝑠𝑠 ) + 𝐵𝐵 ( 𝑡𝑡 ) where 𝑠𝑠 ≤ 𝑡𝑡 ? 𝐵𝐵 ( 𝑠𝑠 ) + 𝐵𝐵 ( 𝑡𝑡 ) = 𝐵𝐵 ( 𝑠𝑠 ) + 𝐵𝐵 ( 𝑡𝑡 ) + 𝐵𝐵 ( 𝑠𝑠 ) − 𝐵𝐵 ( 𝑠𝑠 ) = 2 𝐵𝐵 ( 𝑠𝑠 ) + [ 𝐵𝐵 ( 𝑡𝑡 ) − 𝐵𝐵 ( 𝑠𝑠 )] Since a Brownian motion has independent increments, 𝐵𝐵 ( 𝑠𝑠 ) and 𝐵𝐵 ( 𝑡𝑡 ) − 𝐵𝐵 ( 𝑠𝑠 ) are independent. As 𝐵𝐵 ( 𝑠𝑠 ) + 𝐵𝐵 ( 𝑡𝑡 ) is a linear combination of independent normal random variables and hence it is also distributed as normal with mean and variance 𝐸𝐸 [ 𝐵𝐵 ( 𝑠𝑠 ) + 𝐵𝐵 ( 𝑡𝑡 )] = 0 and 𝑉𝑉𝑉𝑉𝑉𝑉 [ 𝐵𝐵 ( 𝑠𝑠 ) + 𝐵𝐵 ( 𝑡𝑡 )] = 𝑉𝑉𝑉𝑉𝑉𝑉 [2 𝐵𝐵 ( 𝑠𝑠 )] + 𝑉𝑉𝑉𝑉𝑉𝑉 [ 𝐵𝐵 ( 𝑡𝑡 ) − 𝐵𝐵 ( 𝑠𝑠 )] = 4 𝑠𝑠 + (
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}