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Let $\gamma$ be a continuous function from $\mathbb{R}^+$ to $\mathbb{R}^+$ and consider the real valued inhomogeneous Ornstein-Uhlenbeck process satisfying $$ d X_t = -\gamma_t X_t d t + d W_t, \quad X_0 = x \in \mathbb{R}. $$ Assume that $\gamma_t \geq \nu > 0$. It seems reasonable to believe that $X_t$ has an invariant probability measure which should be gaussian. Indeed, a quick calculation gives $$ X_t = x \exp \left ( - \int_0^t \gamma_s d s\right ) + \int_0^t \exp \left ( - \int_s^t \gamma_u d u\right ) d W_s. $$ In order to identify the variance of the limiting gaussian variable, we shall need to calculate $$ \lim_{t \to \infty} \exp \left ( -2 \int_0^t \gamma_s d s \right ) \int_0^t \exp \left ( 2\int_0^s \gamma_u d u\right ) d s. $$ Question Is this limit easily computable? I am expecting something which looks like (I may be wrong)
$$ \frac{1}{2 \Gamma} $$ where $$ \Gamma \triangleq \lim_{t \to \infty} \frac{1}{t} \int_0^t \gamma_s d s. $$

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  • $\begingroup$ I think the $+2$ and $-2$ in the two exponents where you take the limit $t\rightarrow\infty$ should be interchanged, see the answer below. $\endgroup$ – Carlo Beenakker Mar 25 '17 at 13:56
  • $\begingroup$ Indeed, they have to be interchanged. thank you. it is updated. $\endgroup$ – megaproba Mar 26 '17 at 1:11
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If we define $$\Omega(t)= \exp \left (- 2 \int_0^t \gamma_s d s \right ) \int_0^t \exp \left ( 2\int_0^s \gamma_u d u\right ) d s,$$ then this function satisfies the differential equation $$\frac{d}{dt}\Omega(t)=1-2\gamma_t\,\Omega(t).$$ The limit $t\rightarrow\infty$ of $\Omega(t)$ exists if $\lim_{t\rightarrow\infty}\gamma_t=\Gamma>0$ exists, and is then equal to $$\lim_{t\rightarrow\infty}\Omega(t)=\frac{1}{2\Gamma}.$$

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