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Suppose $B_t$ is standard Brownian motion and $X_t$ satisfies

$$dX_t = X_t ^2 dt + X_t B_t$$

Given $Y_t=\exp\left\{\int_0^t (X_s^2 +1 ) ds\right\}$, how to compute $d\langle Y\rangle_t$?


My try

$dY_t = Y_t (X_t^2 + 1) dt + Y_t (\int_0^t 2X_s ds) dX_t=(\ldots)dt + X_tY_t(\int_0^t 2X_s ds)dB_t$

Therefore, we have

$d\langle Y\rangle_t = \left(X_tY_t(\int_0^t 2X_s ds)\right)^2 dt$

Am I correct?

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  • $\begingroup$ Sorry but if $Y_t=\exp(Z_t)$ then $dY_t=Y_tdZ_t+\tfrac12Y_td\langle Z\rangle_t$. In your case, $Z_t=\int_0^t(X_s^2+1)ds$ hence $dZ_t=(X_t^2+1)dt$ and $d\langle Z\rangle_t=0$. Thus, $dY_t=Y_t(X_t^2+1)dt$ and $d\langle Y\rangle_t=0$. $\endgroup$ – Did Oct 30 '17 at 19:50

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