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FINM345A09Homework4 - b ± b 2 p X t ² dP t where E P t =...

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FINM345/STAT390 Stochastic Calculus Hanson Autumn 2009 Lecture 4 Homework: Stochastic Jump and Jump-Diffusion Processes (Due by Lecture 5 in Chalk FINM345 Assignment Submenu) { Note dropping the Digital Dropbox } You must show your work, code and/or worksheet for full credit. There are 10 points per question if correct answer. 1. Inverse Problem for Poisson Integral. Find X ( t ) = F ( P ( t )) if Z t 0 X ( s ) dP ( s ) = e cP ( t ) ln( aP ( t ) + b ) - ln( b ) , (1) where a , b and c are real constants. 2. Solution to a Jump SDE. Solve the following jump SDE for X ( t ), E[ X ( t )], and Var[ X ( t )], dX ( t ) = aX 2 ( t ) dt - bX 2 ( t ) 1 + bX ( t ) dP ( t ) , (2) where a < 0, b > 0 and c > 0 are constants such that E[ P ( t )] = ct , while X (0) = x 0 > 0, with probability one. { Hint: seek a transformation Y ( t ) = F ( X ( t )) for some F such that Y ( t ) satisfies a constant coefficient SDE. The answer may be left as a Poisson distribution sum. } 3. Solution to another Jump SDE.. Solve the following Poisson jump SDE for X ( t ) and E[ X ( t )]: dX ( t ) = a p X ( t ) dt
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Unformatted text preview: + b ± b + 2 p X ( t ) ² dP ( t ) , where E[ P ( t )] = λ t and X (0) = x > 0, with probability one, while λ , a and b are real constants. ( Hint: Find a power transformation to convert the SDE to a constant coefficient SDE. ) 4. For Jump-Diffusion SDE, Find Coefficients Transformable to Constant Co-efficient SDE. Show that the (Itˆ o) jump-diffusion SDE for X ( t ), dX ( t ) = f ( X ( t )) dt + bX a ( t ) dW ( t ) + h ( X ( t )) dP ( t ) , (3) can be transformed by Y ( t ) = F ( X ( t )) to a constant coefficient SDE , where b and a 6 = 1 are real constants, and X (0) = x > 0, with probability one. In a proper answer, derive the power forms of f ( X ( t )) and h ( X ( t )) from the constant coefficient SDE conditions. Also, what is the answer when a = 1? 1...
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