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CaptainAwesome
05-16-2007, 10:31 PM
For part (i)
Can anyone tell me if this sounds right?

U(t) = 2Z(t) - 2

1) the partial of U with respect to Z(t) is 2
2) the second partial of U with respect to Z(t) is 0
3) the partial of U with respect to t is 2dZ(t)

so, applying ito's lemma gives:
dU(t) = 2dZ(t) + 2dZ(t)dt = 2dZ(t), since dZ(t)dt = 0

Is that right or do I not understand this?

nkt
05-16-2007, 11:12 PM
I think you are wrong, because you are confused with z(t), that z as a function of t. This is a Taylor expansion, not the partial rule. Let u(t) to be y and z(t) to be x. so you have y = 2x -2. What is dy? dy=2dx, right. If you had y=x^2, then dy = 2xdx + 0.5*2*(dx)^2. There is a .5 which your method doesn't explain. Just compare 14 i) and ii).
I don't want to confuse. It's late, god luck tomorrow.

jason.
05-17-2007, 12:08 AM
For part (i)
Can anyone tell me if this sounds right?

U(t) = 2Z(t) - 2

1) the partial of U with respect to Z(t) is 2
2) the second partial of U with respect to Z(t) is 0
3) the partial of U with respect to t is 2dZ(t)

so, applying ito's lemma gives:
dU(t) = 2dZ(t) + 2dZ(t)dt = 2dZ(t), since dZ(t)dt = 0

Is that right or do I not understand this?

Almost. Your misstep was 3). Let's collect some facts:

1. {\partial U \over \partial Z} = 2
2. {\partial^2 U \over \partial Z^2} = 0
3. {\partial U \over \partial t} = 0!

Then, by Ito's Lemma, dU = {\partial U \over \partial Z} \,dZ + {1 \over 2} {\partial^2 U \over \partial Z^2} \,(dZ)^2 + {\partial U \over \partial t} \, dt = 0 \,dt + 2\, dZ. The coefficient of dt is the drift. Here that coefficient is zero so the drift is zero.