Given the Ornstein-Uhlenbeck transition pdf (where $t_2\geq t_1 \geq 0$ and $x_2 \geq x_1 \geq 0$ and $\gamma >0$):
$$p(x_2,t_2;x_1,t_1) = \frac{1}{\sqrt{2\pi(1-e^{-2\gamma(t_2-t_1)})}}\exp \left( -\frac{(x_2-x_1e^{-\gamma (t_2-t_1)})^2}{2(1-e^{-2\gamma(t_2-t_1)})}\right)$$
And the continuity condition:
$$\lim_{\Delta t \rightarrow 0} \frac{1}{\Delta t} \int_{|x_2-x_1|>\epsilon} p(x_2,t_2;x_1,t_1) dx_2 = 0$$
How do I show that it is true in the most simple way?
I tried to use the Markov inequality here, but it does not do the trick. Shall I treat the integral like so: $\int_{|x_2-x_1|>\epsilon} \equiv 1-\int_{-\epsilon}^{\epsilon}$ and then do the Taylor series expansion of the transition pdf?
And then what is the simple way to obtain the drift and diffusion terms: $\alpha(x_1,t_1)$ and $\beta^2(x_1,t_1)$, using the following conditions:
$$\lim_{\Delta t \rightarrow 0} \frac{1}{\Delta t} \int_{|x_2-x_1|<\epsilon} (x_2-x_1) p(x_2,t_2;x_1,t_1) dx_2 = \alpha(x_1,t_1)$$ and
$$\lim_{\Delta t \rightarrow 0} \frac{1}{\Delta t} \int_{|x_2-x_1|<\epsilon} (x_2-x_1)^2 p(x_2,t_2;x_1,t_1) dx_2 = \beta^2(x_1,t_1)$$
It looks like the above can be treated as an expectation, but I am unsure as to how. Thank you!
