tag:blogger.com,1999:blog-2198942534740642384.post5434930583494923192..comments2023-10-24T03:16:41.009-07:00Comments on Econometrics Beat: Dave Giles' Blog: On the Asymptotic Properties of Sample MeansDave Gileshttp://www.blogger.com/profile/05389606956062019445noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-2198942534740642384.post-53406554202337469732012-04-21T20:08:05.831-07:002012-04-21T20:08:05.831-07:00You are right. Thanks a lot.You are right. Thanks a lot.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-55179199161616757502012-04-17T21:28:19.442-07:002012-04-17T21:28:19.442-07:00Sorry, but what you read is simply wrong. As n goe...Sorry, but what you read is simply wrong. As n goes to infinity, it is NOT the case that AM/GM converges to one, or that GM/HM converges to one. You can verify this really quickly with a couple of lines of code.Dave Gileshttps://www.blogger.com/profile/05389606956062019445noreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-7084859751257845712012-04-13T10:09:36.442-07:002012-04-13T10:09:36.442-07:00I have one question about this convergence. If HM ...I have one question about this convergence. If HM and the GM cannot be consistent estimators of E(x) and AM is, then AM,GM and HM cannot converge to the same finite number, yes? I read from somewhere that AM/GM converges to 1, and GM/HM converges to 1. Do the claims conflict with each other?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-71355286215644306952012-02-09T08:07:57.896-08:002012-02-09T08:07:57.896-08:00Jack: you're absolutely right, of course. Than...Jack: you're absolutely right, of course. Thanks for clarifying this!Dave Gileshttps://www.blogger.com/profile/05389606956062019445noreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-12411379275787477282012-02-09T01:24:14.130-08:002012-02-09T01:24:14.130-08:00Not really. Lack of asymptotic bias is not the sam...Not really. Lack of asymptotic bias is not the same thing as consistency. Let me be more explicit.<br /><br />As per one of the many versions of the WLLN, with iid observations the existence of E(x)=\mu implies that the sample average of X converges in probability to \mu.<br /><br />Of course, the continuous mapping theorem implies that, if E(g(x)) also exists, then the sample average of g(x) will converge in probability to it. However, this limit will in general be different from g(\mu) unless g() is linear. As a consequence, <br /><br />plim \frac{1}{n} \sum g(x_i) = E(g(x)) \ne g(E(x))<br /><br />so <br /><br />plim g^{-1}[\frac{1}{n} \sum g(x_i)] = \ne E(x)<br /><br />which proves that g^{-1}(1/n \sum g(x_i)) is not consistent for E(x) from the definition of consistency (ie convergence in probability to the parameter of interest).jack lucchettinoreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-7941868863074216452012-02-08T08:33:41.708-08:002012-02-08T08:33:41.708-08:00Nick: Thanks for the comment. However, what you...Nick: Thanks for the comment. However, what you've established relates to the finite-sample bias. This doesn't say anything about the mean of the asymptotic distribution, and hence the asymptotic bias.Dave Gileshttps://www.blogger.com/profile/05389606956062019445noreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-55184245146434298052012-02-08T08:30:09.834-08:002012-02-08T08:30:09.834-08:00Nick: Thanks for the comment. We know already that...Nick: Thanks for the comment. We know already that the answer to those questions is "yes". The question that was raised in the earlier comment was explicitly about estimators of E[X] - the arithmetic mena of the population.Dave Gileshttps://www.blogger.com/profile/05389606956062019445noreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-30210259161162547002012-02-08T03:25:08.957-08:002012-02-08T03:25:08.957-08:00Dave: Wouldn't it make more sense to ask:
1. ...Dave: Wouldn't it make more sense to ask:<br /><br />1. Is the sample GM a consistent estimator of the population GM?<br /><br />2. Is the sample HM a consistent estimator of the population HM?Nick Rowehttps://www.blogger.com/profile/04982579343160429422noreply@blogger.comtag:blogger.com,1999:blog-2198942534740642384.post-27317169176709035142012-02-07T23:23:56.532-08:002012-02-07T23:23:56.532-08:00There's a very simple way to prove analyticall...There's a very simple way to prove analytically that the HM and the GM cannot be consistent estimators of E(x). Both can be written as<br /><br />g^{-1}(1/n \sum g(x_i))<br /><br />where g() is a non-linear function (eg the natural log for the GM).<br /><br />Even if E[g(X)] exists (which may not), then it's different from g[E(x)], unless g() is linear.jack lucchettinoreply@blogger.com