Consider the following ODE:

$$\frac{ d \gamma(x,t;\tau)}{d \tau} = R(\gamma(x,t;\tau)) ; \qquad \gamma(x,t,t)=x. $$

$R$ is smooth enough, bounded away from zero and a stationary process. Is there a chance that $\gamma(x,t;\tau)-x$ is stationary with respect to $x$ as well (i.e. $\gamma(x,t;\tau)-x$ has the same distribution as $\gamma(x+y,t;\tau)-x-y$ )

thanks for the any help

PS: one can solve the ODE as follows: let $C(x):=\int_0^x \frac{1}{R(y)} d y$ then $$\gamma(x,t;\tau)=C^{-1}(C(x)-t+\tau),$$

if that helps...


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.