Because of examples like this and others where we

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Because of examples like this – and others where we cannot compute the mean or the variance – we usually establish consistency using the law of large numbers and the algebra of plims. 14
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EXAMPLE : Consider the Poisson example where we want to estimate P X 0 exp . The natural estimator is ̂ n exp X ̄ n . Because 0 ̂ n 1, E ̂ n exists, but we cannot easily compute it. But it is easy to verify that ̂ n is consistent because the plim passes through continuous nonlinear functions (Slutsky’s Theorem): plim ̂ n plim exp X ̄ n  exp plim X ̄ n  exp . If we have a sequence of vectors ̂ n : n 1,2,. .. then consistency is defined element by element. 15
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THEOREM : Suppose ̂ n : n 1,2,. .. is consistent for Θ p . Define a parameter vector g where g : Θ s . Define ̂ n g ̂ n .If g is continuous on Θ then plim ̂ n ,all . This is simply a restatement of Slutsky’s Theorem: the plim passes through general nonlinear functions. 16
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Slutsky’s Theorem applied to sequences of estimators shows that consistency is much more broadly applicable than asymptotic unbiasedness: Even if E ̂ n is asymptotically unbiased, the expected value of g ̂ n can be very difficult to find or even approximate. 17
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EXAMPLE : Consider estimating the log-odds ration from a Bernoulli population with random sampling. Take Θ 0,1 (so 0 and 1 are ruled out). We are interested in estimating log 1 g . The function g : Θ→ is continuous. Therefore, a consistent estimator of is ̂ n log X ̄ n 1 X ̄ n . The expected value of ̂ n is not well defined because X ̄ n can take on the values zero and one with positive probability. 18
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Often we want to know more than just whether an estimator is consistent. We would like to have an idea of how fast the estimator converges to its plim as the sample size grows. The vast majority of estimators in statistics and econometrics converage at the rate n . 19
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DEFINITION : A sequence of estimators ̂ n : n 1,2,. .. is n - consistent for Θ if n ̂ n O p 1 ,all Θ If an estimator is n -consistent then it is consistent, but the condition implies much more. In particular, n a ̂ n p 0 for any a 1/2. 20
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To show this, use the simple algebra of o p and O p . Write n a ̂ n n a n 1/2 n 1/2 ̂ n n a 1/2 n ̂ n  o 1  O p 1 o p 1 , whereweuse n a 1/2 0 for all a 1/2, and so n a 1/2 o 1 .(A special case is a
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Because of examples like this and others where we cannot...

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