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Math Help - random variable convergence

  1. #1
    Junior Member
    Joined
    Jan 2010
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    random variable convergence

    Hello,

    question is about about one point in proposition saying that convergence of random variable almost surely implies convergence in probability.

    A sequence (X_n),n\in  \mathbb{N} of random variables is said to converge almost surely to the random variable X if P(\omega\mid X_n\to X, n\to \infty)=1, which is equivalent to \forall \varepsilon \forall  \delta \exists N : P(\bigcap_{n=N}^\infty\[\omega: \mid {X_n - X}\mid< \varepsilon])\geq 1-\delta.

    In proof of this proposition (in book of John B.Thomas) it is said that:
    P(\bigcap_{n=N}^\infty\[\omega: \mid {X_n - X}\mid< \varepsilon])\geq  1-\delta is equivalent to P(\bigcap_{n=N}^\infty\[\omega: \mid {X_n - X}\mid\geq \varepsilon])< \delta

    Although, using De Morgan's law and property of pobability measure I get that
    P(\bigcap_{n=N}^\infty\[\omega: \mid {X_n - X}\mid<  \varepsilon])\geq  1-\delta is equivalent to P(\bigcup_{n=N}^\infty\[\omega: \mid {X_n - X}\mid\geq  \varepsilon])< \delta,
    which is essentially different statement.

    Can anybody comment this?
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  2. #2
    Senior Member
    Joined
    Oct 2009
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    It looks like a typo to me.

    For me, the easiest thing to remember is that X_n \to X almost surely if and only if for every \epsilon > 0 we have P([\omega: |X_n - X| \ge \epsilon \mbox{ for infinitely many $n$ }]) = P(\limsup [\omega: |X_n -X| \ge \epsilon]) =  0 (the first equality being by definition). Then the result follows from

    \limsup P([\omega: |X_n - X| \ge \epsilon]) \le P(\limsup [\omega: |X_n - X| \ge \epsilon]).
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