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I have a question regarding the Ornstein -Uhlenbeck process. We have a simplified version with Stochastic Integral Equation: $X_t=-a\int^t_0 X_s\,ds +B_t$. B is the Brownian motion.

And its analytic solution is $X_t=e^{-at}\int^t_0 e^{as}\,dB_s$.

How do I prove that this is the case? I know that $e^{-at}=-a\int^t_0 e^{-as}\,ds$ and that I'm somehow supposed to make use of this result but I am still unable to get from the analytic solution to the integral equation.

Thanks!

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If you want to check if solution is correct, why not apply Ito formula? In the case you wonder how the analytic solution was obtained from the integral equation - that's a bit more elaborate. –  Ilya Dec 3 '11 at 13:22
    
Could you elaborate on how to apply Ito's formula? Sorry I'm not very good with this subject. I know that Ito formula is $f(B_t)-f(B_0)=\int^t_0 f'(B_u)\,dB_u+{1/2}\int^t_0 f''(B_u)\,du$ but I don't know what function should f(x) be. –  Sharon Reed Dec 3 '11 at 14:42
    
No, its in my lecture notes. I just can't understand that bit. –  Sharon Reed Dec 3 '11 at 15:02
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1 Answer

up vote 6 down vote accepted

The Ito formula, for $\mathrm{d} X_t = - a X_t \mathrm{d} t + \mathrm{d} B_t$, you need is: $$ \mathrm{d}\left( f(t, X_t) \right) = \left( \partial_t f(t,X_t) - a \partial_x f(t,X_t) + \frac{1}{2} \partial_{xx} f(t,X_t) \right) \mathrm{d} t + \left( \partial_x f(t,X_t) \right) \mathrm{d} B_t $$

HINT: use $f(t,X_t) = \mathrm{e}^{a t} X_t$.

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Okay thanks! I'll go work on it. –  Sharon Reed Dec 3 '11 at 15:07
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