Let $$ [X,Y]=\int_{0}^{\tau} (dX_t)(dY_t)$$ where $X_t$ is cadlag process and $Y_t$ is a Brownian motion. Can we say $$ \frac{\partial}{\partial\tau}\int_{0}^{\tau} (dX_t)(dY_t)= 0 $$
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Are you sure you have both $dX_t$ and $dY_t$? Also, if the integral is a continuous process, then sure. – William M. Apr 14 '17 at 01:26
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I made a mistake! Just edited! – Dr. Kandy Junior Apr 14 '17 at 01:27
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What does $dX_t dY_t$ mean? Did you mean $d[X,Y]_t$? – William M. Apr 14 '17 at 01:33
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No. Take $X_t=Y_t=B_t$ to be a Brownian motion started from zero. Then $[X,Y]_t=t$, so in this case $\frac{d}{dt}[X,Y]_t=1$, not zero.
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