27-10-2014: Unfortunately (for me that is), no-one has as yet contributed an answer here -perhaps because it looks like a weird, "pathological" theoretical issue and nothing more?
Well to quote a comment for user Cardinal (which I will subsequently explore)
"Here is an admittedly absurd, but simple example. The idea is to
illustrate exactly what can go wrong and why. It does have practical
applications (my emphasis). Example: Consider the typical i.i.d. model with finite
second moment. Let $\hat θ_n=\bar X_n+Z_n$ where $Z_n$ is independent of
$\bar X_n$ and $Z_n=\pm an$ each with probability $1/n^2$ and is zero
otherwise, with $a>0$ arbitrary. Then $\hat θ_n$ is unbiased, has
variance bounded below by $a^2$, and $\hat θ_n→\mu$ almost surely
(it's strongly consistent). I leave as an exercise the case regarding
the bias".
The maverick random variable here is $Z_n$, so let's see what we can say about it.
The variable has support $\{-an,0,an\}$ with corresponding probabilities $\{1/n^2,1-2/n^2,1/n^2\}$. It is symmetric around zero, so we have
$$E(Z_n) = 0,\;\; \text{Var}(Z_n) = \frac {(-an)^2}{n^2} + 0 + \frac {(an)^2}{n^2} = 2a^2$$
These moments do not depend on $n$ so I guess we are allowed to trivially write
$$\lim_{n\rightarrow \infty} E(Z_n) = 0,\;\;\lim_{n\rightarrow \infty}\text{Var}(Z_n) = 2a^2$$
In Poor Man's Asymptotics, we know of a condition for the limits of moments to equal the moments of the limiting distribution. If the $r$-th moment of the finite case distribution converges to a constant (as is our case), then, if moreover,
$$\exists \delta >0 :\lim \sup E(|Z_n|^{r+\delta}) < \infty $$
the limit of the $r$-th moment will be the $r$-th moment of the limiting distribution. In our case
$$E(|Z_n|^{r+\delta}) = \frac {|-an|^{r+\delta}}{n^2} + 0 + \frac {|an|^{r+\delta}}{n^2} = 2a^{r+\delta}\cdot n^{r+\delta-2}$$
For $r\geq2$ this diverges for any $\delta >0$, so this sufficient condition does not hold for the variance (it does hold for the mean).
Take the other way: What is the asymptotic distribution of $Z_n$? Does the CDF of $Z_n$ converge to a non-degenerate CDF at the limit?
It doesn't look like it does: the limiting support will be $\{-\infty, 0, \infty\}$ (if we are permitted to write this), and the corresponding probabilities $\{0,1,0\}$. Looks like a constant to me.
But if we don't have a limiting distribution in the first place, how can we talk about its moments?
Then, going back to the estimator $\hat \theta_n$, since $\bar X_n$ also converges to a constant, it appears that
$\hat \theta_n$ does not have a (non-trivial) limiting distribution, but it does
have a variance at the limit. Or, maybe this variance is infinite? But an infinite variance with a constant distribution?
How can we understand this? What does it tell us about the estimator? What is the essential difference, at the limit, between $\hat \theta_n = \bar X_n + Z_n$ and $\tilde \theta_n = \bar X_n$?
