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Unformatted text preview: STAT/ACTSC 446/846 Assignment #1 (Solution) Problem 1. Solution: Cov ( Z t ,W u ) = Cov (3 t + 2 W t ,W u ) = 2 min( t,u ) = 2 t Problem 2. Solution: (1) X t = μt + σW t X t can become negative. (2) Apply Ito’s lemma: (a) dY t = 0 + 3 X 2 t dX t + 1 2 6 X t ( dX t ) 2 = 3 X 2 t ( μdt + σdW t ) + 3 X t σ 2 dt = (3 Y 2 3 t μ + 3 Y 1 3 t σ 2 ) dt + 3 σY 2 3 t dW t (b) dY t = 2 tdt + 4 e 4 X t dX t + 1 2 16 e 4 X t ( dX t ) 2 = 2 tdt + 4( Y t 10 t 2 )( μdt + σdW t ) + 8( Y t 10 t 2 ) σ 2 dt = (2 t + ( Y t 10 t 2 )(4 μ + 8 σ 2 )) dt + 4 σ ( Y t 10 t 2 ) dW t Problem 3. Solution: Let f ( r t ,t ) = r t e αt . So, by It’s formula, we get: df ( r t ,t ) = αr t e αt dt + e αt dr t = αr t e αt dt + e αt ( α ( r t r ) dt + σdW t ) = e αt αr dt + σe αt dW t Integrating both sides from 0 to t yields: t integraldisplay df ( r t ,t ) = t integraldisplay e αs αr ds + t integraldisplay σe αs dW s r t e αt r = r ( e αt 1) + t integraldisplay σe αs dW s r t = r + e αt t integraldisplay σe αs dW s Some additional questions: (2) The drift term is equal to α ( r r t ). How can α and r be interpreted? (2) here are many ways to interpret them. Lot’s of answers will be accepted. α can be interpreted as the rate at which the process reverts to its mean, and r can be interpreted as the mean to which the process reverts. If you take the expectation of r t it is equal to r since the expectation of integraltext t σe α ( s t ) dW s is equal to 0 (one easy way to prove it is to say that it is a martingale, so it has constant expectation and thus equal to the expectation at time 0, and the process is...
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This note was uploaded on 01/16/2011 for the course ACTSC actsc 446 taught by Professor Idk.. during the Spring '09 term at Waterloo.
 Spring '09
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