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Given that $\sigma e^{-ut}dB(t) = d(e^{-ut}X(t))$, where $X(t)$ is a stochastic process and $B(t)$ is a Wiener process, we have that:

$$ \int_0^t d(e^{-ut}X(s)) = X(0) + \sigma \int_0^t e^{-us}dB(s) $$

Why is it that there's an $X(0)$? I understand that it would be $X(0)$ if we had $\int_0^t dX(s)$ on the LHS, but I've never seen a LHS with $d(... X(s))$ where $...$ had other stuff in it.

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up vote 3 down vote accepted

There should not be. It seems there is a confusion with the identity $$ \sigma\int_0^t \mathrm e^{-us}\mathrm dB_s=\int_0^t \mathrm d(\mathrm e^{-ut}X(s)) =\mathrm e^{-ut}X(t)-X(0). $$

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