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I understand that result should be 0, but I don't understand how to get to it

Thanks a lot in advance

(ADDinfo)

I am actually trying to prove that the following process can be written as stochastic integrals with respect to B:

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The solutions then give the answer:

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And so however I try, I cannot understand how the equation for Yt with no dt arises

Alex
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    if you tell us what you're using this for, the question might be less likely to be closed – Taylor Jan 04 '18 at 15:52

1 Answers1

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If we write in shorthand $$ dX_t =- a^2/2\cos^2(at)dt -a \cos(at)dB_t $$ and we define $$ Y_t = f(X_t) = \exp(X_t), $$ then we can use Ito's Lemma to get \begin{align*} dY_t &= \left(- a^2/2\cos^2(at) -a^2/2 \cos^2(at)\right)Y_t d_t -a \cos(at)Y_tdB_t\\ &= - a^2\cos^2(at) Y_t d_t -a \cos(at)Y_tdB_t \end{align*} because

  1. $\frac{\partial f}{\partial t} = 0$

  2. $\frac{\partial f}{\partial x} = \exp(X_t) = Y_t$

  3. $\frac{\partial^2 f}{\partial x^2} = Y_t$.

Hopefully this is correct. I'm not totally sure what people do when they divide through by $dt$. It must be that $dB_t/dt$ has some meaning I don't know about.

Taylor
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