slides_13_inferasymptotic

# Both converge in distribution to normal 01 under the

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Both converge in distribution to Normal 0,1 under the null, and they have exactly the sample local power function. 68

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An uncomfortable fact about large-sample inference is that there are many different statistics that are asymptotically equivalent in the sense that the have the same local power function. One could conceivably reject H 0 using one of the statistics but not the other. At a minimum, the asymptotic p -values will usually differ. Suppose with n 400 we obtain x ̄ .8. and so ̂ exp .8 .45. Rounding to two decimal places, 20 exp .8 .5 exp .8 .8 ≈ − 2.52 20 .8 log 2  .8 2.39 69
(Note that if the alternative is H 1 : .5 this is equivalent to H 1 : log 2 , which is why the signs are different.) So, the asymptotic p -values, gotten in both cases from the Normal 0,1 distribution, are different. (The one-sided p -values are about .0059 in the first case and about .0084 in the second case.) In practice, this has not been viewed as much of a problem, as there is often a “natural” way to compute the statistic. 70

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The key to testing a single hypothesis using asymptotic methods is to just remember the general formula for a t statistic. If is a parameter and n ̂ d Normal 0, 2  , so that 2 is the asymptotic variance of n ̂ (which depends on the parameter vector ), then the asymptotic standard error of ̂ is se ̂ 2 ̂ n ̂ n 71
The asymptotic t statistic for testing H 0 : 0 is ̂ 0 se ̂ n ̂ 0 ̂ , and this has a limiting standard normal distribution under H 0 . Local power analysis can also be carried out. 72

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Testing Multiple Restrictions Let be a p 1 vector in the open set Θ p , and let c : Θ q be a continuously differentiable function. Assume the q p restrictions to be tested are H 0 : c 0 For example, suppose p 3 and we want to test H 0 : 1 2 1, 3 0 so there are q 2 restrictions. Then c c 1 c 2 1 2 1 3 73
Given n ̂ n n d Normal 0 , V we use the delta method to obtain the asymptotic variance of n c ̂ n under the null hypothesis c 0 : Avar n c ̂ n  C VC and the Wald test statistic is n c ̂ n C ̂ n VC ̂ n 1 n c ̂ n c ̂ n C ̂ n VC ̂ n / n 1 c ̂ n 74
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