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Given a standard Brownian motion $(B_t)_{t\geq 0}$, define $$ X_t=tB_t,\qquad Y_t=e^{B_t},\qquad Z_t=X_t/Y_t. $$ I want to compute the quadratic variation of the latter, i.e. $\langle Z\rangle_t$. It is an easy exercise to obtain $$ \langle X\rangle_t=\frac{1}{3}t^3,\qquad \langle Y\rangle_t=\int_0^tY_s^2\;ds. $$ Moreover, $dZ_t=X_td(Y_t^{-1})+Y_t^{-1}dX_t+d\langle X_t,Y^{-1}_t\rangle$ by Ito's product rule. This can be massaged further to obtain $$ dZ_t=\left(\frac{t}{Y_t}-Z_t\right)dB_t+B_tdt+d\langle X_t,Y_t^{-1}\rangle. $$ Consequently, we infer $$ \langle Z\rangle_t=\int_0^t\frac{s}{Y_s}-Z_s\;ds. $$ I was wondering whether there is an expression for $d\langle X_t,Y_t^{-1}\rangle$ even if this does not alternate the final result.

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