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I have that $P(a < X_n | F_n) >p$, if I want to find $P(a > X_n | F_n)$, can I just reverse the (direction) probability and does $P(a>X_n | F_n) < 1-p$ or $P(a>X_n | F_n) > 1-p$? Note, $F_n$ is a filtration.

Also, if $P(a < X_n | F_n) >p$, can I say that $P(a < X_{cn} | F_{cn}) >p$ for a constant $c$?

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First part:

$$ \begin{align} \mathbb{P}(a>X_n|\mathcal{F}_n)&= \mathbb{E}(\mathbb{1}(a>X_n)|\mathcal{F}_n)\\ &=1-\mathbb{E}(1-\mathbb{1}(a>X_n)|\mathcal{F}_n)\\ &=1-\mathbb{E}(\mathbb{1}(a\leq X_n)|\mathcal{F}_n)\\ &=1-\mathbb{P}(a\leq X_n|\mathcal{F}_n) \end{align} $$

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  • $\begingroup$ Thanks! Any ideas for the 2nd part? $\endgroup$ – Murer Mar 28 '12 at 22:06
  • $\begingroup$ @Murer Yes. Perhaps you could edit your question to include your attempt at the second part? $\endgroup$ – Ben Derrett Mar 29 '12 at 9:46

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