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Asymptotically unbiased & consistent estimators

  1. Jul 8, 2011 #1
    Theorem: If "θ hat" is an unbiased estimator for θ AND Var(θ hat)->0 as n->∞, then it is a consistent estimator of θ.

    The textbook proved this theorem using Chebyshev's Inequality and Squeeze Theorem and I understand the proof.
    BUT then there is a remark that we can replace "unbiased" by "asymptotically unbiased" in the above theorem, and the result will still hold, but the textbook provided no proof. This is where I'm having a lot of trouble. I really don't see how we can prove this (i.e. asymptotically unbiased and variance->0 implies consistent). I tried to modify the original proof, but no way I can get it to work under the assumption of asymptotically unbiased.

    I'm frustrated and I hope someone can explain how to prove it. Thank you!
     
    Last edited: Jul 8, 2011
  2. jcsd
  3. Jul 8, 2011 #2

    micromass

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    Hi kingwinner! :smile:

    What about the following adjustment:

    [tex]P(|\hat{\theta}_n-\theta_0|\geq \varepsilon)\leq P(|\hat{\theta}_n-E(\hat{\theta}_n)|+|E(\hat{\theta}_n)-\theta_0|\geq \varepsilon)\leq \frac{Var(\hat{\theta}_n)}{(\varepsilon-|E(\hat{\theta}_n)-\theta_0|)^2}\rightarrow 0[/tex]
     
  4. Jul 8, 2011 #3
    Thanks for the help, but one of the assumptions of Chebyshev's inequality requires [tex]\varepsilon-|E(\hat{\theta}_n)-\theta_0|[/tex]>0 which is not necessarily true here?
     
  5. Jul 8, 2011 #4

    micromass

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    It's not necessarily true, but it is true for large n. We know that

    [tex]E(\hat{\theta}_n)\rightarrow \theta_0[/tex]

    So from a certain n0, we know that

    [tex]|E(\hat{\theta}_n)-\theta_0|<\varepsilon[/tex]

    So from that certain n0, we know that

    [tex]\varepsilon-|E(\hat{\theta}_n)-\theta_0|>0[/tex]
     
  6. Jul 8, 2011 #5
    Thanks for the help! :) You're a legend...
     
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