HO4-sol-09

HO4-sol-09 - Homework 4 Solutions Problem 1 Using Ito’s formula on XX T d XX T =(d X X T X d X T d X(d X T = AX d t σ d B X T X X T A T d t d B

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Homework 4 Solutions March 11, 2009 Problem 1. Using Ito’s formula on XX T d( XX T ) = (d X ) X T + X d X T + d X (d X ) T = ( AX d t + σ d B ) X T + X ( X T A T d t + d B T σ T ) + σ d B d B T σ T = AXX T d t + σ d BX T + XX T A T d t + d B T σ T + σσ T d t where we’ve used d B d B T = I d t 1 . Integrating from 0 to t and taking expectation, E [ XX T ] = xx T + E bracketleftbiggintegraldisplay t ( AXX T + XX T A T )d s bracketrightbigg + E bracketleftbiggintegraldisplay t σσ T d s bracketrightbigg + E bracketleftbiggintegraldisplay t ( σ d BX T + X d B T σ T ) bracketrightbigg C ( t ) = C (0) + integraldisplay t ( AC ( s ) + C ( s ) A T )d s + integraldisplay t σσ T d s Differentiating, we get d C ( t ) d t = AC ( t ) + C ( t ) A T + σσ T C (0) = xx T It then follows from the notes that the OU process can be expressed by: X ( t ) = e At x + integraldisplay t e A ( t − s ) σ d B ( s ) Taking expectation E [ XX T ] = e At xx T e A T t + E bracketleftbigg ( integraldisplay t e A ( t − s ) σ d B ( s ))( integraldisplay t e A ( t − s ) σ d B ( s )) T bracketrightbigg C ( t ) = e At C (0) e A T t + integraldisplay t e A ( t − s ) σσ T e A T ( t − s ) d s by Ito-Isometry. Starting at some time t instead of zero, X ( t ) = e A ( t − t ) X ( t ) + integraldisplay t t e A ( t − s ) σ d B ( s ) We now want to take the limit of X ( t ) as t → -∞ . Using the fact that all of A ’s eigenvalues have negative real parts, we can see from expressing A in Jordan canonical form that lim t →−∞ e A ( t − t ) X ( t ) = 0 Thus X e ( t ) = integraldisplay t −∞ e A ( t − s ) σ d B ( s ) 1 C ( t ) is the second order moment of X ( t ) not the variance of X ( t ) as stated in the original problem 1 Note that E [ X e ( t )] = 0 and that X e ( t ) doesn’t depend on the starting point x . Direct calculation shows that C e ( t ) = E [ X e ( t ) X e ( t ) T ] = E bracketleftbigg ( integraldisplay t −∞ e A ( t − s ) σ d B ( s ))( integraldisplay t −∞ e A ( t − s ) σ d B ( s )) T bracketrightbigg = integraldisplay t −∞ e A ( t − s ) σσ T e A T ( t − s ) d s Making the substitution t- s = s ′ , we see C e ( t ) = integraldisplay ∞ e As ′ σσ T e A T s ′ d s ′ Notice that C e ( t ) is independent of t as it should be since it is the covariance of the equilibrium process. Since it is independent of t then clearly d C e ( t ) d t = 0 and thus it follows that C e ( t ) solves the following Lyapunov equation AC e + C e A T + σσ T = 0 This equation would be very difficult to solve without knowing the OU process was behind this....
View Full Document

This note was uploaded on 11/08/2009 for the course MATH 236 taught by Professor Papanicolaou during the Winter '08 term at Stanford.

Page1 / 6

HO4-sol-09 - Homework 4 Solutions Problem 1 Using Ito’s formula on XX T d XX T =(d X X T X d X T d X(d X T = AX d t σ d B X T X X T A T d t d B

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online