依r阶矩收敛:Convergence in the r-th mean
矩的收敛:Convergence of moment
依分布收敛:Convergence in distribution
Xn是随机变量序列中的第n个随机变量。X是一个随机变量。
一方面,依r阶矩收敛蕴含(implies)依分布收敛。
另一方面,依r阶矩收敛也蕴含矩的收敛。
但是在Greene (2012)第1115页,在定义完矩的收敛以后,作者认为即使随机变量Xn与随机变量X的各阶矩都存在且Xn的各阶矩收敛于X的各阶矩,也不能保证Xn的分布和X的分布是一样的,因为一个随机变量的分布不由其各阶矩唯一决定。截图如下:
https://drive.google.com/open?id=1w6_B_ml1WyfNNQyUkS791FApbscrMPXp
(如果看不到图,这里是相关文本): Theorem D.15 raises an interesting question. Suppose we let r grow, and suppose that Xn converges in the r-th mean to X, in addition, all moments are finite. If this holds for any r, do we conclude that these random variables have the same distribution? The answer to this longstanding problem in probability theory - the problem of the sequence of moments - is no. The sequence of moments does not uniquely determine the distribution. Although convergence in r-th mean and almost surely still both imply convergence in probability, it remains true, even with convergence to a random variable instead of a constant, that these are different forms of convergence.
单独看这一段推导感觉没问题。但是我想问的是,依r阶矩收敛蕴含依分布收敛,那不就是说两个随机变量的分布最终一致了吗?这个想法好像就跟截图的内容相左了:
依r阶矩收敛-->依分布收敛-->Xn与X分布最终一致
依r阶矩收敛-->矩的收敛-->不保证Xn与X分布最终一致
请问这两者之间的“冲突”应该怎么理解?是作者错了?还是我对依分布收敛的理解(概率分布趋于相同)错了?
谢谢。
附 Greene (2012): William H. Greene, Econometric Analysis, 7th Edition
https://www.researchgate.net/file.PostFileLoader.html?id=568181165cd9e37af18b458f&assetKey=AS%3A311705391828994%401451327765378
(我不具有任何权利)