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Thread: Convergence in probability

  1. #1
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    Convergence in probability

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

    $\displaystyle 0 = X_{n}Y-X_{n}Y$

    and insert this in the salient difference

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

    Invoking the Triangle inequality you can write

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

    If the left-hand side of this inequality is larger than some small positive number $\displaystyle \varepsilon\ ,$ then at least one of the two terms on the right-hand side of the inequality is greater than $\displaystyle \varepsilon/2\ .$
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  3. #3
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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

    $\displaystyle 0 = X_{n}Y-X_{n}Y$

    and insert this in the salient difference

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

    Invoking the Triangle inequality you can write

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

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