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Taking a Limit of a Probability

  1. Feb 10, 2014 #1
    1. The problem statement, all variables and given/known data I want to show that if I have a consistent sequence of estimators [itex]W_n[/itex] for [itex]\theta[/itex], i.e. [itex]\lim_{n \rightarrow \infty} P(|W_n - \theta| < \epsilon) = 1[/itex], then [itex]U_n = a_nW_n + b_n[/itex] is also a consistent sequence of estimators for [itex]\theta[/itex] where [itex]\lim_{n \rightarrow \infty}a_n = 1 [/itex] and [itex]\lim_{n \rightarrow \infty}b_n = 0. [/itex]



    2. Relevant equations



    3. The attempt at a solution
    We are looking at [itex]\lim_{n \rightarrow \infty} P(|a_nW_n + b_n - \theta| < \epsilon)[/itex], which is equivalent to [itex]\lim_{n \rightarrow \infty} P(-\epsilon < a_nW_n + b_n - \theta < \epsilon)[/itex] or [itex]\lim_{n \rightarrow \infty} P(-\epsilon - b_n< a_nW_n - \theta < \epsilon - b_n) [/itex].
    My question is how I can apply what I know about the limits of [itex]W_n, a_n, b_n[/itex] in this expression. While I would like to be able to say that since the an's go to 1 and the bn's to 0 I can apply the consistency of Wn, but I don't know if that is acceptable.
     
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  3. Feb 10, 2014 #2

    haruspex

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    Use the definition of limit for the a and b sequences. Given ϵ > 0 there exists N such that...
     
  4. Feb 10, 2014 #3

    Ray Vickson

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    [tex] \{-\epsilon - b_n < a_n W_n - \theta < \epsilon -b_n \}
    = \left\{ \frac{-\epsilon - b_n + \theta (1 -a_n)}{a_n}
    < W_n - \theta < \frac{ \epsilon - b_n + \theta (1-a_n)}{a_n} \right\}[/tex]
     
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