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Can someone help me apply Ito's lemma to the function $f(t,x,k)$ where t is the time and x,k dimensions where x and k refer to dynamics

$dX(t)=\mu(t)dt+\sigma(t)dB(t)$

$dK(t)=\nu(t)dt+\theta(t)dW(t)$

What i have done so far $$ \begin{align}df(t,X(t),K(t))=\left(\frac{\partial f}{\partial t} + \mu(t)\frac{\partial f}{\partial x}+\nu(t)\frac{\partial f}{\partial k}+\frac{1}{2}\sigma(t)^{2}\frac{\partial ^{2}f}{\partial x^{2}}+\frac{1}{2}\theta(t)^{2}\frac{\partial ^{2}f}{\partial k^{2}}\right)dt + \sigma(t)\frac{\partial f}{\partial x}dB(t) + \theta(t)\frac{\partial f}{\partial k}dW(t) + \frac{\partial^2 f}{\partial x\partial k}dB(t)dW(t) \end{align} $$ Can someone correct me please ?

thanks for your time

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To clarify, you are trying to find the stochastic differential equation governing $f(t,X(t),K(t))$ where $X$ and $K$ satisfy the SDE's above, and $B$ and $W$ are independent brownian motions? –  Aaron Feb 3 '12 at 3:21
    
I've made some correction which are just notational, the most crucial one - you missed $2$ in $\partial^2 f$ for a mixed derivative. Seems to be correct for me. @Aaron: independence is not used as I see. –  Ilya Feb 3 '12 at 9:05
    
@Ilya Independence might not be necessary, but it would eliminate the $dB(t)dW(t)$ term, and knowing how they relate would simplify the term in any event. However, knowing for sure that they are brownian motions (and not some other process) is necessary to know that you only need up to quadratic terms, and that $dB^2=dt$, etc. The question confuses me, though, because I'm not really sure what it is asking, other than "is this right?" Without looking carefully at the individual terms, I have seen statements of Ito's lemma that look like what is written, so it doesn't seem an application. –  Aaron Feb 3 '12 at 9:54
    
First of all, thank you for you immediate responce. –  peter Feb 3 '12 at 12:21
    
@Ilya, i you are correct about the square :P –  peter Feb 3 '12 at 12:22

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