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I am trying to solve this O-U process for which the long term mean is another O-U process.

$$ dX(t) = -\alpha_{x} X(t)dt + Y(t)dt + \sigma_{x} dW_{x}(t) $$ $$ dY(t) = -\alpha_{y}Y(t)dt + \sigma_{y}dW_{y}(t) $$

Solving for Y(t) is straightforward.

$$ Y(t) = e^{-\alpha_{y}*(t-s)}Y(s) + \sigma_{y} \int_{s}^{t}e^{-\alpha_{y}(t-u)}dW_{y}(u) $$

However, I struggle when plugging it in the first equation and trying to solve for X(t). Could anyone help me with this?

In general, I struggle when looking at problems in which one process depends on another. What should be the general way to approach these types of problems?

Thanks!

rodrigo
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    Solving the first SDE is also straightforward leaving the notation $Y(t)$ as an abbreviation in that solution. "Plugging in" the explicit form of $Y(t)$ that you found has no advantage whatsoever. – Kurt G. Nov 02 '24 at 02:39

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The solution to general linear SDE \begin{align*} \mathrm{d}X_t = (a(t)X_t+ b(t)) \mathrm{d}t + (g(t)X_t+ h(t))\mathrm{d}B_t \end{align*} is solved here Solution to General Linear SDE. In your case, you let $b(t)=e^{-a_y t}Y_0+ \sigma_{y} \int_{0}^{t}e^{-\alpha_{y}(t-u)}dW_{y}(u)$ (even though it is random, the proof using an integrating-factor still goes through) to get

\begin{align*} X_t = & X_0 e^{ -a_x t}+ e^{ -a_x t}\left( \int_0^t e^{ a_x s}b(s) \mathrm{d}s + \int_0^t e^{ a_x s}\sigma_x \mathrm{d}B_s\right). \end{align*}

Thomas Kojar
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  • This solution looks strange. In OP's case we have $g(t)=0$ so I would have thought with time dependent coefficients it is $$ X_t=e^{\int_0^ta(u),du}\left{X_0+\int_0^te^{-\int_0^sa(u),du}b(s),ds+\int_0^te^{-\int_0^sa(u),du}h(s),dB_s\right} $$ This is a special case of this. Since OP has constant $a(u)$ equal to $-a_x$ and constant $h(s)$ equal to $\sigma_x$ this can be simplified a bit but I cannot see how to get your expression. – Kurt G. Nov 05 '24 at 17:20
  • @KurtG. you are right, I used the wrong case. I will fix it. – Thomas Kojar Nov 05 '24 at 18:32
  • ok i fixed it to the right case. – Thomas Kojar Nov 05 '24 at 20:46