Zheng Hong HW4 - CNETzhengh HW4 Zheng Hong Contents 1...

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CNETzhengh HW4 Zheng Hong October 26, 2009 Contents 1 Inverse Problem for Poisson Integral 2 2 Solution to a Jump SDE 2 3 Solution to another Jump SDE 2 4 Find Coefficients Transformable 3 1
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1 Inverse Problem for Poisson Integral If R t 0 X ( s ) dP ( s ) = e cP ( t ) ln ( aP ( t ) + b ) - ln ( b ), then X ( t ) dP ( t ) = de cP ( t ) ln ( aP ( t ) + b ); set Y ( P ( t )) = e cP ( t ) ln ( aP ( t ) + b ), dY ( P ( t )) = Y ( P ( t ) + dP ( t )) - Y ( P ( t )) = zol ( Y ( P ( t ) + 1) - Y ( P ( t ))) dP ( t ) = [ e c ( P ( t )+1) ln ( a ( P ( t ) + 1) + b ) - e cP ( t ) ln ( aP ( t ) + b )] dP ( t ) = X ( t ) dP ( t ) (1) So, X ( t ) = F ( P ( t )) = e c ( P ( t )+1) ln ( a ( P ( t ) + 1) + b ) - e cP ( t ) ln ( aP ( t ) + b ) 2 Solution to a Jump SDE Set Y ( t ) = F ( X ( t )) = 1 X ( t ) , Y (0) = 1 X (0) = 1 x 0 ; F x ( X ( t )) = - 1 X 2 ( t ) , F t ( X ( t )) 0; dX ( t ) = aX 2 ( t ) dt - bX 2 ( t ) 1 + bX ( t ) dP ( t ) dY ( t ) = dF ( X ( t )) = ( F t + F x f )( X ( t ) , t ) dt + [ F ]( X
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This note was uploaded on 04/19/2010 for the course FIN 390 taught by Professor Hansen during the Fall '09 term at University of Illinois, Urbana Champaign.

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Zheng Hong HW4 - CNETzhengh HW4 Zheng Hong Contents 1...

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