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I have an SDE that looks that Ornstein-Uhlenbeck SDE :

$$dX_t=(-\alpha +\beta X_t)dt+\sigma dB_t.$$

I know that $$d(X_te^{\beta t})=e^{-\beta t}dX_t-e^{\beta t}X_t\beta dt=e^{-\beta t}(-\alpha +\beta X_t)dt-e^{\beta t}\sigma dB_t+e^{\beta t}X_t\beta dt$$ $$=-\alpha e^{-\beta t} dt+\sigma e^{-\beta t}dB_t$$ does it work ?

nmasanta
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John
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    You made an error, the second term in the first equality should be $X_t \beta e^{\beta t} dt$. But you'll see that the terms involving $X_t dt$ don't cancel after fixing that error, which was the whole point of the substitution. That should suggest how to fix the substitution to allow that cancellation. – Ian Jun 20 '19 at 15:06
  • Your error is still there after the second equals sign. – Ian Jun 20 '19 at 15:08
  • @Ian: thanks I corrected the typo. But How can I do now ? I can see that nothing cancel, that's why I ask here :) – John Jun 20 '19 at 15:08
  • Think about the ODE situation. If you have $y'=y$, does it really help to look at $u=e^t y$? No, because then $u'=e^t y + e^t y' = 2u$, which is just as complicated as before. But $u=e^{-t} y$ helps. The goal of making this substitution in the SDE is exactly the same, to remove the $X_t dt$ term. – Ian Jun 20 '19 at 15:09
  • I see. I edited. Does it work now ? @Ian – John Jun 20 '19 at 15:12
  • Minus the persistent minor errors, you're on the right track now, and all that's left to do is to actually integrate, which you can now do because now there's no dependence on an unknown function in the integrals. The actual SDE part is just computing $\int_0^t e^{-\beta s} dB_s$. – Ian Jun 20 '19 at 15:13

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Indeed using the integrating factor approach as here Solution to General Linear SDE for $dX_t = \big( a(t) X_t + b(t) \big) dt + \big( g(t) X_t + h(t) \big) dB_t,$ we get

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

Thomas Kojar
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