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Let $Y$ is a ${\mathbb R}$-valued random variable, $(X_{t})_{t \geq 0}$ is one-dimensional stochastic process and $(W_{t})_{t \geq 0}$ is a one-dimensional Brownian motion. Then is the following formula correct? $$ Y \int_{0}^{t}X_{s}{\rm d}W_{s}=\int_{0}^{t}YX_{s}{\rm d}W_{s}, \quad t \geq 0. $$ Here, these integrands $(X_{t})_{t \geq 0}$ and $(YX_{t})_{t \geq 0}$ have some conditions that allow the stochastic integrals to be defined.

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  • $\begingroup$ Does $Y$ depend on $X$ or $W$? $\endgroup$ – user658409 Nov 1 '19 at 2:06
  • $\begingroup$ This statement is explained in detail on page 148 of the following book: I. Karatzas and S. E. Shreve. Brownian Motion and Stochastic Calculus, 2nd ed. Springer-Verlag New York, Inc. (1998). $\endgroup$ – 720773 Feb 14 at 17:33
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Yes, because in terms of an outcome $\omega$, $\begin{eqnarray*} &&Y(\omega) \int_{0}^{t}X_{s}(\omega){\rm d}W_{s}(\omega)\\ &=& Y(\omega) \lim_n\sum_{k=0}^{tn} X_{k/n}(\omega)(W_{k/n+\Delta}-W_{k/n})(\omega)\\ &=& \lim_n\sum_{k=0}^{tn} Y(\omega)X_{k/n}(\omega)(W_{k/n+\Delta}-W_{k/n})(\omega)\\ &=& \int_{0}^{t}Y(\omega)X_{s}(\omega){\rm d}W_{s}(\omega) \end{eqnarray*} $

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