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How to show that the Method of Moment Estimators for the Normal

  1. Jul 22, 2013 #1
    So I'm trying to show that the estimators for the normal distribution by the method of moments are consistent. So to show consistency, I have to :

    1) Show E(θ(estimator) = θ (parameter)

    2) lim n-->∞ Var(θ(estimator) = 0

    So Since there are two estimators in the normal distribution ( [itex]\mu[/itex],[itex]\sigma[/itex]2) I have to prove that they are each consistent:

    To prove [itex]\mu[/itex](estimator):

    E([itex]\mu[/itex](estimate) = E(([itex]\sum[/itex]Xi)/n)

    Working along I get that [itex]\mu[/itex] (estimate) is unbias as well as [itex]\sigma[/itex]2, but now to show the second condition I get tripped up for both estimators.

    for [itex]\mu[/itex](estimate):

    V([itex]\mu[/itex](estimate) = V(([itex]\sum[/itex]Xi)/n

    = 1/n2[itex]\sum[/itex]V(Xi)

    Similarly for V( [itex]\sigma[/itex]2(estimate):

    1/n2[itex]\sum[/itex]V(Xi-X(bar))2

    How do I proceed for these two estimators from here?
     
  2. jcsd
  3. Jul 22, 2013 #2

    Ray Vickson

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    If you are estimating ##\sigma^2## using
    [tex] \text{est}(\sigma^2) = \frac{1}{n} \sum_{i=1}^n (X_i - \bar{X})^2[/tex]
    then your estimate is biased.
     
  4. Jul 22, 2013 #3

    Then I established that result wrong. I did this:

    Assuming [itex]\sigma[/itex]2 estimate:

    = 1/n ([itex]\sum[/itex]E(Xi-X(bar))2

    = 1/n ([itex]\sum[/itex][itex]\sigma[/itex]2

    = 1/n (n[itex]\sigma[/itex]2)

    = [itex]\sigma[/itex]2
     
  5. Jul 23, 2013 #4

    statdad

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    Your error is stating that
    [tex]
    E\left(X_i - \overline{X}\right)^2 = \sigma^2
    [/tex]
     
  6. Jul 23, 2013 #5

    Ray Vickson

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    As 'statdad' has explained to you, ##E(X_i - \bar{X})^2 \neq \sigma^2##.

    Take ##i = 1##, for example. We have [tex]X_1 - \bar{X} = \left(1-\frac{1}{n}\right) X_1 - \frac{1}{n} X_2 - \cdots - \frac{1}{n} X_n[/tex]
    You need to square this, expand it out, then take the expectation----and yes, I am perfectly serious! Try it yourself; it is not as bad as you might think at first. The basic properties you need are ##E X_j^2 = \sigma^2 + \mu^2## and the fact that the different ##X_j## are independent, so that ##E X_i X_j## is easy to get for ##i \neq j##.
     
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