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Finding the MVUE of a two-sided interval of a normal

  1. Apr 24, 2014 #1
    Our task is to determine if [itex]P(-c \le X \le c)[/itex] has a minimum variance unbiased estimator for a sample from a distribution that is [itex]N(\theta,1)[/itex]. The one-sided interval [itex]P(X \le c) = \Phi(x - \theta)[/itex] is unique, so constructing an MVUE is just a matter of applying Rao-Blackwell and Lehmann-Scheffe.

    However for our case, [itex]P(-c \le X \le c)[/itex] is the same for [itex]\theta[/itex] and [itex]-\theta[/itex]. So it seems like the MVUE isn't unique. I'm wondering if you can make a decision rule like choosing one unbiased estimator when [itex]\theta \ge 0[/itex] and the other when [itex]\theta < 0[/itex], but now instead of two non-unique unbiased estimators, we have three. Any thoughts? Is a MVUE just not possible?
     
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  3. Apr 25, 2014 #2

    Stephen Tashi

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    I assume this means that [itex] c [/itex] is given and [itex] \theta [/itex] is unknown. So you can't employ a rule that depends on knowing the sign of [itex] \theta [/itex].
     
  4. Apr 25, 2014 #3
    What if we construct two MVUE's, one for [itex]P(X \le c)[/itex], and one for [itex]P(X \le -c)[/itex], and then subtract one from the other? It still seems like we have the same problem, where the MVUE is not one-to-one...
     
  5. Apr 25, 2014 #4

    Stephen Tashi

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    There is ambiguity if you estimate [itex] P(-c \le X \e c) [/itex] first and try to estimate [itex] \theta [/itex] from that estimate. However the problem you stated doesn't insist we estimate [itex]\theta [/itex] in that manner. Wouldn't the simplest try be to estimate [itex] \theta [/itex] from the sample mean and then estimate [itex] P(-c \le X \le c) [/itex] from that estimate?
     
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