In the course of a proof, I used the following principle, which seems so intuitive that it should have a name:

Suppose one has a stochastic process $X_t$, for $t \in \omega$, on a (possibly infinite) state space $S$. Suppose one has $n$ Boolean functions $A_i, B_i: S \rightarrow \{ 0, 1 \}$. Suppose that, for all $s \in S$ and all $t \in \omega$, one has the condition that $P(B_i(X_{t+1}) = 1 \mid X_t = s) \leq p_i$.

Now consider the following event $E$:

  1. For all $i = 1, ..., n$, there is a unique time $t_i$ such that $A(X_{t_i}) = 1$;

  2. For all $i \neq i'$ we have $t_i \neq t_{i'}$

  3. For all $i = 1, ..., n$ we have $B(X_{t_i+1}) = 1$

Then the probability of the event $E$ is at most $p_1 p_2 \dots p_n$.

Is the principle already known? Would one consider it "obvious" ?

  • 1
    $\begingroup$ What is the meaning of the condition $B_i=s$ if $B_i$ is Boolean? $\endgroup$ – Algernon Mar 20 '13 at 13:01
  • $\begingroup$ Sorry, that was a typo, it is fixed $\endgroup$ – David Harris Mar 20 '13 at 13:14

I don't think this is true.

Consider one dimensional Brownian motion with $X_0 = 1$ and let $B_i$ be the indicator of the event that the $k$th decimal place is a $0$ (so all of our $B_i$'s are the same). Let $k$ be large enough that we may choose $p_i = p< 1$.

Now let $A_i$ be the indicator of the interval $\left(0,\frac {i\cdot\varepsilon}n\right)$. Abuse notation and let $t_0$ be the first hitting time of $0$.

Now let $E'$ be the event that $B_i = 1$ for the entire time interval $(t_n+1,t_0+1)$. Then $P(E')$ is strictly positive and we have $E'\supset E$.

But as $t_n$ is the first hitting time of $\varepsilon$ the event $E'$ does not depend on $n$, hence we may choose $n$ large enough that $P(E') > p^n$.

So I would say that you're missing at least one assumption.

  • $\begingroup$ 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. $\endgroup$ – David Harris Apr 11 '13 at 16:05
  • $\begingroup$ $p_i$ 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. $\endgroup$ – user32372 Apr 12 '13 at 13:12
  • $\begingroup$ 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. $\endgroup$ – user32372 Apr 12 '13 at 13:24

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