Say an estimator converges with probability one and at the same time its variance goes to zero in the limit. How is it different than an estimator that converges with probability one but its variance does not go to zero? Does that achieve sure convergence? I am wondering what difference it makes.
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Convergence almost surely and convergence in distribution are not the same thing.
Take a simple example where you have i.i.d. mean zero random variables $Y_i$, $i= 1, 2, \cdots$ for which the law of iterated logarithm holds. Then
$$ \lim \sup_{n \rightarrow \infty} \frac{\frac{1}{\sqrt{n}}\sum_1^n Y_i}{\sqrt{\log\log n}} = \sqrt{2}, \; a.s. $$
In particular, $\frac{1}{\sqrt{n}}\sum_1^n Y_i$ diverges almost surely, but convergence in distribution. A sequence of random variables can diverge for a fixed sample path but converge in distribution across all sample paths.
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2$\begingroup$ I am not sure how my question is related to convergence in distribution, which you are talking about. By the way, whoever gave the -1, it would have been much more helpful if he told his concerns in the comment. $\endgroup$ Commented Oct 22, 2014 at 23:02
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$\begingroup$ Unless I misread your question, the variance of the asymptotic distribution is, well, part of the behavior of the limiting distribution. The point I was trying to make is that, in general, limiting behavior a.s. and in distribution are very different things. Here you have a sequence ( admittedly in the more general situation of r.v.'s rather than estimators) that diverges a.s. but has a finite limiting variance. $\endgroup$– MichaelCommented Oct 24, 2014 at 4:33