Timeline for Is this probabilistic principle for stochastic processes known?
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| Apr 12, 2013 at 13:24 | comment | added | user32372 | I'm not sure if that helps, I don't quite understand your comment. The point is that neither the probability $p_i$, nor the event $E'$ depend on $n$, but they have nontrivial probabilities. The idea of the counterexample is to force the $t_i$ close together without allowing them to be equal. (you can set $\varepsilon \ll 10^{-k}$) So in continuous time you would need to strengthen condition 2 to make it work. | |
| Apr 12, 2013 at 13:12 | comment | added | user32372 | $p_i$ is the probability that the $k$th digit of $X_{t+1}$ is a zero given $X_t = s$, maximized over $s$. For reasonably large $k$ this will be pretty much $0.1$. The time $t_n$ is the first hitting time of $\varepsilon$, so does not depend on $n$. $E'$ is the event that the $k$th digit is $0$ for the entire time interval $(t_n_1,t_0+1)$ this means that the Brownian motion can't leave the interval $\left(10N\times 10^{-k},10N+1\times 10^{-k}\right)$. The probability that a Brownian motion stays within a fixed interval for a given time is strictly positive. | |
| Apr 12, 2013 at 12:43 | history | edited | user32372 | CC BY-SA 3.0 |
Typo: changed the time interval $(t_0+1,t_n+1)$ to $(t_n+1,t_0+1)$ because $t_n<t_0$.
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| Apr 11, 2013 at 16:05 | comment | added | David Harris | I have been puzzling over this answer for a few weeks. How can $P(E')$ be strictly positive and yet still $p_i < 1$? The latter condition means that each stage of the Brownian motion has an independent probability of flipping the $k$th decimal place. As $n \rightarrow \infty$, the time interval between states decreases, so that $p_i$ should approach to one. | |
| Mar 20, 2013 at 16:18 | history | answered | user32372 | CC BY-SA 3.0 |