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Let $$dr_t=(\alpha(t)-\beta r_t)dt+\sigma dW_t$$ where $\alpha$ is non stochastic process and $\beta$ and $\sigma$ are constant. Can we write process $r_t$ in the form $$r_t=x_t+y_t$$ where the process $x_t$ satisfies $$dx_t=-\beta x_t dt+\sigma dW_t$$ and $y_t$ be a deterministic function. I used Ito's lemma but was not useful.

Thanks in advanced.

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By the usual integrating factor method, \begin{align*} r_t = r_0e^{-\beta t} + \int_0^t \alpha(s) e^{-\beta(t-s)}ds +\sigma \int_0^t e^{-\beta(t-s)}dW_s. \end{align*} Let \begin{align*} x_t &=\sigma \int_0^t e^{-\beta(t-s)}dW_s, \textrm { and}\\ y_t &=r_0e^{-\beta t} + \int_0^t \alpha(s) e^{-\beta(t-s)}ds. \end{align*} Then $r_t = x_t + y_t$, moreover, \begin{align*} dx_t &= d\left(\sigma e^{-\beta t} \int_0^t e^{\beta s}dW_s \right)\\ &=-\beta \left(\sigma e^{-\beta t} \int_0^t e^{\beta s}dW_s\right)dt + \sigma dW_t\\ &=-\beta x_t dt + \sigma dW_t, \end{align*} and $y_t$ is a deterministic function.

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  • $\begingroup$ Thank you but $dx_t=-\beta x_t dt +\sigma dW_t$ $\endgroup$ – math Jul 8 '16 at 7:32

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