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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 = t + W t X t can become negative. (2) Apply Itos lemma: (a) dY t = 0 + 3 X 2 t dX t + 1 2 6 X t ( dX t ) 2 = 3 X 2 t ( dt + dW t ) + 3 X t 2 dt = (3 Y 2 3 t + 3 Y 1 3 t 2 ) dt + 3 Y 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 )( dt + dW t ) + 8( Y t 10 t 2 ) 2 dt = (2 t + ( Y t 10 t 2 )(4 + 8 2 )) dt + 4 ( Y t 10 t 2 ) dW t Problem 3. Solution: Let f ( r t ,t ) = r t e t . So, by Its formula, we get: df ( r t ,t ) = r t e t dt + e t dr t = r t e t dt + e t ( ( r t r ) dt + dW t ) = e t r dt + e t dW t Integrating both sides from 0 to t yields: t integraldisplay df ( r t ,t ) = t integraldisplay e s r ds + t integraldisplay e s dW s r t e t r = r ( e t 1) + t integraldisplay e s dW s r t = r + e t t integraldisplay e s dW s Some additional questions: (2) The drift term is equal to ( r r t ). How can and r be interpreted? (2) here are many ways to interpret them. Lots of answers will be accepted. 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 e ( 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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 Spring '09
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