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

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