Skip to main content
1 of 2
user32372
  • 241
  • 1
  • 4

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 $(t_0+1,t_n+1)$. 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.

user32372
  • 241
  • 1
  • 4