2
$\begingroup$

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.

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you but $dx_t=-\beta x_t dt +\sigma dW_t$ $\endgroup$
    – math
    Jul 8, 2016 at 7:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.