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A mean reverting Ornstein-Uhlenbeck SDE is given by $$=(−)+;_0=,$$ where m and are positive constants and is a standard Brownian motion in 1 dimension. I have obtained the solution of this equation, $$X_t = xe^{-t} + m(1-e^{-t}) + \sigma \int_0^t e^{-(t-s)} dW_s.$$ My questions are: Is my solution correct? What is the long time behaviour of the solution? How to obtain the equation of the second moment?

Sybil
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  • What do you mean by equation of the second moment ? – Surb Dec 09 '20 at 12:54
  • I think the second moment is also known as the variance. https://en.wikipedia.org/wiki/Moment_(mathematics)#Variance – Sybil Dec 09 '20 at 13:15
  • I know what is that the second moment (and is not the variance. The variance is the centred second moment). But what is the "equation of the second moment" ? If you look for the variance, then my answer below allow you to compute the variance of $X_t$. – Surb Dec 09 '20 at 14:49
  • Oh I see, thank you for your explanation. What I mean by "the equation of the second moment" should be the equation obtained by evaluating $M_n(t)=E(X_t^n)$ where $n=2$. – Sybil Dec 10 '20 at 16:27

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Hint

I didn't check into detail the solution you found, but at least, the solution is something of this form. Now, one can prove that if $f\in L^2[0,t]$, then $$\int_0^tf(s)\,\mathrm d W_s\sim \mathcal N\left(0,\int_0^t f(s)^2\,\mathrm d s\right).$$

Surb
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