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Probability with Martingales

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To prove $b$ I tried:

$$A_n \ge A_{n-1}$$

$$\iff E[X_{n} - X_{n-1} | \mathscr F_{n-1}] \ge 0$$

$$\iff E[X_{n} | \mathscr F_{n-1}] \ge X_{n-1}$$

That suffices to show 'if' but what about 'only if'?

I think I can conclude that $P(A_n \ge A_{n+1}) = 1 \ \forall n$, but how do I conclude $P(A_n \le A_{n+1} \ \forall n$) = 1? I think I need right continuity or something? I remember indistinguishable vs modification being discussed in classes, but I think it wasn't discussed at this point in the book.

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This is discrete time, so right continuity is not an issue. It is simply the fact that if $P[D_n]=1$ for all $n$ then $P\left[\cap_n D_n\right]=1$ (and conversely).

John Dawkins
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  • John Dawkins, do we really have that $$P(\bigcap_{n=1}^{\infty} D_n) = 1 \iff P(D_n) = 1 \ \forall n \ge 1$$? I mean, 'only if' follows by monotonicity of probability but what about 'if' ? – BCLC Jun 11 '16 at 16:58
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    @BCLC to see why this is true, try to prove the complement has measure probability – clark Jun 11 '16 at 17:18
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    @BCLC, see Proposition 2.4(a) on page 25 of Probability with Martingales. – John Dawkins Jun 11 '16 at 17:50
  • @JohnDawkins Thanks! ^-^ – BCLC Jun 11 '16 at 18:15