∙ suppose that in the earlier poisson example the

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Unformatted text preview: ∙ Suppose that, in the earlier Poisson example, the null is H : 2. The three statistics are T n 1 n X ̄ n − 2 2 , T n 2 n X ̄ n − 2 X ̄ n , T n 3 n X ̄ n − 2 S n ∙ Not only do these all have limiting Normal 0,1 distributions under H , they all have limiting Normal / 2 ,1 distributions under H n 1 . 46 ∙ We can show this by using more general results. Let n be the mean of the random sample X ni : i 1,2,..., n and n 2 Var X ni . We take as given that the CLT holds, that is, n X ̄ n − n n d → Normal 0,1 . ∙ Now, suppose that n → 0. Then n X ̄ n − n n X ̄ n − n n n d → Normal 0,1 . ∙ If we assume n o / n then n X ̄ n − o n X ̄ n − n d → Normal / ,1 47 ∙ We can apply this result immediately to the statistic T n 1 in the Poisson example. For the other two statistics, we use another general result. 48 ∙ If W n p → – for example, if we can show W n − n p → 0, along with n → – then n X ̄ n − n W n n X ̄ n − n 1 W n − 1 n X ̄ n − n n X ̄ n − n o p 1 O p 1 n X ̄ n − n o p 1 . 49 ∙ We cannot use local power analysis to choose among these three statistics when the Poisson distribution holds. Recall that T n 3 is valid more generally (because it is just the usual asymptotic t statistic that can be used for any population distribution). ∙ In the vast majority of econometric applications, how one estimates the asymptotic variance has no effect on asymptotic size or local power if the variance estimator is consistent under the null (and hence under local alternatives). 50 5 . A General Framework for Testing Hypotheses ∙ We now cover a general framework for testing multiple linear restrictions on a set parameters. We use local power analysis. ∙ Let ̂ n : n 1,2,... be a sequence of estimators p 1 such that n ̂ n − n d → Normal , V where n / n and is a fixed p 1 vector. 51 ∙ The null hypothesis is H : or ∙ The sequence of local alternatives is given above: H n 1 : n / n ∙ Let V ̂ n be a consistent estimator of V (under the null and local alternatives). The test statistic is T n n ̂ n − ′ V ̂ n − 1 n ̂ n − n ̂ n − ′ V − 1 n ̂ n − o p 1 52 ∙ Under H , n ̂ n − d → Normal , V , and so, under H , T n d → p 2 (Recall that if the p 1 vector W Normal , then W ′ − 1 W p 2 .) ∙ To find the asymptotic distribution of T n under H n 1 : n 1,2,... , we need to introduce a new distribution: the noncentral chi- square distribution....
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∙ Suppose that in the earlier Poisson example the null is...

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