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Consider the SDE $$ dX = \alpha (\beta - X)dt + \gamma dW $$ with initial condition $X(0) = X_0$. Let $ Y(t) = X(t)e^{\alpha t}$. One way to write the formal solution of $X(t)$ is $X(t) = X_0 + \int_0^t \alpha (\beta - X(s))ds + \int_0^t \gamma dW(s) $. Write down the SDE $Y(t)$ satisfies and its formal solution. From there, write down an expression of the solution of $X(t)$ that does not involve $X$ inside the integral.

So far this is what I'm thinking: $$Y(t) = X(t)e^{\alpha t} = e^{\alpha t}X_0 + e^{\alpha t}\int_0^t \alpha (\beta - X(s))ds + e^{\alpha t}\int_0^t \gamma dW(s) $$ as the formal solution, but I'm not sure.

Then the SDE for $Y(t)$ would be: $$ dY = e^{\alpha t} + e^{\alpha t}\alpha (\beta - X)dt + e^{\alpha t}\gamma dW $$

But it's asking for a solution where $X$ is not inside the integral. I'm confused by what that means. Thank you.

MelHA
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  • It means that you have to apply Itô's formula in order to get rid of the $X$ inside the integral. Let $(X_t)_t$ be a solution to the given SDE, then what does Itô's formula tell you about $X(t) e^{\alpha t}$...? – saz Mar 04 '18 at 10:13
  • I don't fully understand what you're saying. I thought we could just multiply $e^{\alpha t}$ across...? – MelHA Mar 04 '18 at 21:56
  • Sure you can do that, but it doesn't help you to get rid of the $X$ under the integral. The SDE which you derived for $Y_t$ is wrong. In order to get the proper differntial of $Y_t$ you have to apply Itô's formula. – saz Mar 05 '18 at 06:43
  • Okay, so this is what I did: $$dY = \alpha X e^{\alpha t} + e^{\alpha t} dX$$ $$ = \alpha \beta e^{\alpha t} dt + \gamma e^{\alpha t} dW$$

    Then integrating: $$ Y(t) = Y_0 + \alpha \int_0^t \beta e^{\alpha s} ds + \gamma \int_0^t e^{\alpha s} dW(s)$$

    Our X(t) is then: $$ X(t) = X_0 e^{\alpha t} + \beta (1 - e^{- \alpha t} ) + \gamma \int_0^t e^{\alpha (s - t)} dW(s)$$

     – MelHA Mar 05 '18 at 07:09
  • The first term in the very last line should be $X_0 e^{-\alpha t}$, I think, but apart from that it looks okay. – saz Mar 05 '18 at 07:48
  • Yes! I forgot to add the negative sign. Okay, phew. Thank you so much! – MelHA Mar 05 '18 at 07:58

1 Answers1

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Indeed, this is just the integrating-factor approach (see Solution to General Linear SDE for general linear SDE).

By Ito-product-rule we have

$$dY=d(e^{at}X)=ae^{at}X_tdt+e^{at}dX_t=ae^{at}X_tdt+e^{at}a\beta dt-ae^{at}X_tdt+e^{at}\gamma dW_{t}$$ $$=e^{at}a\beta dt+e^{at}\gamma dW_{t}$$

and so

$$X_{t}=e^{-at}(x_0+\int e^{as}a\beta ds+\int e^{as}\gamma dW_{s}).$$

Thomas Kojar
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