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gentinex
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Joined: June 8th, 2006, 1:16 pm

SDE question

November 5th, 2006, 11:19 pm

I am trying to solve the SDE where r and alpha are real numbers, and B_t is standard Brownian motion. In fact, this question has been posted on the forum recently, and I even explained to the original poster of that thread how to do it (namely, multiply both sides by the integrating factor exp(-alpha * B_t + 0.5 * alpha^2 * t) and then use Ito's formula in reverse), but looking back at it, I was wondering what is wrong with the following reasoning:Multiply both sides by the factor exp(-alpha * B_t):Now apply Ito's formula to the function f(t,Y_t,B_t) = Y_t * exp(-alpha * B_t):Then the product of the two dB_t's is just dt, and the product of dB_t and dY_t is alpha * Y_t dt, so the latter two terms cancel out, and the sum of the first two simplifies by using our expression for exp(-alpha * B_t) dY_t, so we would haveI know this isn't the right answer, so I think there's a hole in my reasoning, but where?
Last edited by gentinex on November 5th, 2006, 11:00 pm, edited 1 time in total.
 
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Normal
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Joined: August 29th, 2006, 2:08 pm

SDE question

November 25th, 2006, 2:56 pm

I think the hole is when you multiply both sides by . Recall that is short hand for The trick here is to multiply an exponential term to kill the dB_t term and then convert the SDE into an ODE. The right multiplying term is the exponential martingale I can provide the details if needed.
Last edited by Normal on November 24th, 2006, 11:00 pm, edited 1 time in total.
 
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gentinex
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SDE question

November 26th, 2006, 3:58 pm

Multiplying both sides by \exp(-\alpha B_t) is not in and of itself an incorrect thing to do. One could multiply by any factor, including the one you mentioned (which I believe was the same one I told you to use when you asked this question a while ago!). However, I was trying to find out where I---if you, say, looked at my calculations step by step---had made an incorrect calculation in my reasoning, using this different multiplying factor of \exp(-\alpha B_t).But looking back at things, I do see where I made an error: in my calculation for Ito's formula for f(t,Y_t,B_t) = Y_t \exp(-\alpha B_t). Since I am counting terms dY_t \cdot dB_t and dB_t \cdot dY_t, the last term should have a factor of -1, not -1/2, which makes the corresponding integral more complicated since there's still a Y_t term on the right hand side. (I also see that there's a typo in this term in my original derivation, namely that the -1/2 term has a Y_t factor in it when there shouldn't be one, but that's less serious than the fact that the -1/2 should be -1).