# Can we take out an integrand random variable which does not depend on time from the stochastic integral?

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.

• Does $Y$ depend on $X$ or $W$? – user658409 Nov 1 '19 at 2:06
• 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). – 720773 Feb 14 at 17:33

Yes, because in terms of an outcome $$\omega$$, $$\begin{eqnarray*} &&Y(\omega) \int_{0}^{t}X_{s}(\omega){\rm d}W_{s}(\omega)\\ &=& Y(\omega) \lim_n\sum_{k=0}^{tn} X_{k/n}(\omega)(W_{k/n+\Delta}-W_{k/n})(\omega)\\ &=& \lim_n\sum_{k=0}^{tn} Y(\omega)X_{k/n}(\omega)(W_{k/n+\Delta}-W_{k/n})(\omega)\\ &=& \int_{0}^{t}Y(\omega)X_{s}(\omega){\rm d}W_{s}(\omega) \end{eqnarray*}$$