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I would like to calculate stochastic differential of:

$$X_t=\left(\int_0^t(s^3+B_s) \,dB_s \right)(2t+tB_t)=Y_tZ_t$$

I would like to use: $d(Y_tZ_t)=Z_t \, dY_t +Y_t \, dZ_t+dY_t \, dZ_t\tag{$*$}$

$$Y_t:=\int_0^t(s^3+B_s) \, dB_s$$

$$Z_t:=2t+tB_t$$

$$dY_t=(t^3+B_t)\,dB_t$$

$$dZ_t=(2+B_t)\,dt+t\,dB_t$$

I put everything in $(*)$ and get a proper result, right?

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    $\begingroup$ If you compute $d\langle Y,Z\rangle$, yes, "right". Is this your question? $\endgroup$ – Did Feb 10 '15 at 20:11
  • $\begingroup$ $d\langle Y,Z\rangle = dY_t dZ_t$? $\endgroup$ – luka5z Feb 10 '15 at 20:14
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    $\begingroup$ Dunno, depends how you define $dY_tdZ_t$. $\endgroup$ – Did Feb 10 '15 at 20:43
  • $\begingroup$ Formal product of $dY_t$ and $dZ_t$, it yields $\sigma_t^Y \sigma_t^Z dt$ since $dtdB_t=0$ and $dtdt=0$ $\endgroup$ – luka5z Feb 10 '15 at 21:00
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    $\begingroup$ Looks more like a way to compute it than a definition, and the notation is $d\langle Y,Z\rangle$, not $dYdZ$, but ok. $\endgroup$ – Did Feb 10 '15 at 22:19

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