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I am currently working on a stochastic calculus exercise at the moment and I am slightly confused when it comes to finding cross variation.

We are given that the process $X_t = W_t^3$ ($W_t$ is standard Brownian Motion), and that $Y_t = sin(W_t)$.

We are then asked to find the cross variation between $Z_t$ and $U_t$, where $Z_t = Y_t^2$ and $U_t = e^{X_t}$.

I've found (I hope correctly) the Ito decomposition of $Z_t$ and $U_t$ to be given by

$dZ_t = cos(2W_t) dt + sin(2W_t) dW_t$

$dU_t = \frac{1}{2} e^{W_t} dt + e^{W_t} dWt$

I'm unsure though how to proceed and find $<U, Z>$.

Any help would be appreciated - many thanks.

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  • $\begingroup$ If I'm not mistaken the only cross term contributing to the cross-variation process are the $dW_t$ terms, then (assuming your differentials are correct, and no correlation) $d[U,Z](t)=sin(2W_t)e^{W_t}dt$ $\endgroup$ – zebullon Jun 14 '15 at 3:21
  • $\begingroup$ Thanks @zebullon - that makes sense to me. I forgot to put another part of the question in my post - is the quadratic variation of $Z_t$ equal to $(sin(2W_t))^2 dt$ ? $\endgroup$ – Shaz Jun 14 '15 at 3:50
  • $\begingroup$ @Shaz Which definition of (cross)variation do you use? $\endgroup$ – saz Jun 14 '15 at 5:50
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    $\begingroup$ Recall that if $dU=AdW+Bdt$ and $dV=CdW+Ddt$ then by definition $d[U,V]=ACdt$. For example, $d[Z,Z]=\sin^2(2W)dt$. (But watch out for your $dU$, you computed $d(e^W)$, not $d(e^{W^3})$.) $\endgroup$ – Did Jun 14 '15 at 20:40

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