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Math Help - Convergence in probability

  1. #1
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    Convergence in probability

    Let be X_1, X_2, ..., X_n, Y_1, Y_2, ..., Y_n random variables on the same (\Omega, \mathbf{F}, \mathbb{P}) probabilty field. Assume that  X_n \buildrel P \over \rightarrow X and Y_n \buildrel P \over \rightarrow Y (denotes convergence in probabilty).
    Prove that:
    a) X_nY_n  \buildrel P \over \rightarrow XY.
    b) If Y_n \neq 0, Y \neq 0 (with probabilty 1), then X_n/Y_n  \buildrel P \over \rightarrow X/Y.

    I would really appreciate if you could help me.
    Thank you in advance!
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  2. #2
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    Re: Convergence in probability

    You might try one of the standard mathematical tricks: Do nothing, but do it in a smart way!

    In this case one such trick is to write

    0 = X_{n}Y-X_{n}Y

    and insert this in the salient difference

    |X_{n}Y_{n}-XY| \ .

    Invoking the Triangle inequality you can write

    |X_{n}Y_{n}-XY| \leq |X_{n}-X||Y| + |Y_{n}-Y||X_{n}| \ .

    If the left-hand side of this inequality is larger than some small positive number \varepsilon\ , then at least one of the two terms on the right-hand side of the inequality is greater than \varepsilon/2\ .
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  3. #3
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    Joined
    Oct 2010
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    Re: Convergence in probability

    Quote Originally Posted by Alibiki View Post
    You might try one of the standard mathematical tricks: Do nothing, but do it in a smart way!

    In this case one such trick is to write

    0 = X_{n}Y-X_{n}Y

    and insert this in the salient difference

    |X_{n}Y_{n}-XY| \ .

    Invoking the Triangle inequality you can write

    |X_{n}Y_{n}-XY| \leq |X_{n}-X||Y| + |Y_{n}-Y||X_{n}| \ .

    If the left-hand side of this inequality is larger than some small positive number \varepsilon\ , then at least one of the two terms on the right-hand side of the inequality is greater than \varepsilon/2\ .
    Thank you for your help!
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