Can an estimator of the mean of a distribution with no variance have a variance? Suppose you have a sample from a distribution with a mean but no defined variance, like the Pareto with tail parameter between 1 and 2, or Student’s t with 2 degrees of freedom.  
Can an unbiased estimator of the mean of such a distribution have a (defined, finite) variance?  
 A: Sure. For example, the sample median will have a variance and for a continuous symmetric distribution with a mean, it should be a consistent estimator of it (as long as there's some density at and around the population median).
http://en.wikipedia.org/wiki/Median#Variance
(Note also the subsection there "Estimation of variance from sample data" about some of the difficulties and peculiarities of it.)
The median isn't the only such choice, of course - various robust estimators should work - at least in the sense that they should produce consistent estimators of the location and have finite variance.
Similarly, consider for simplicity a Pareto with known lower bound (which we can take to be 1, wlog). Estimators of the index ($\alpha$) parameter in the Pareto (which imply an estimator for the mean) exist which have finite variance even with $1<\alpha<2$.
Now take a sample of $n$ independent observations from the that Pareto (assume for the moment that $n$ is large enough that the discussion carries through; it won't work at $n=1$ for example). 
The log of such a Pareto r.v. is exponential with mean $1/\alpha\,$. The ML of $\alpha$ in that exponential has finite variance (it has an inverse Gamma distribution whose mean and variance is a function of $n$ as well as $\alpha$). It is also ML for the same parameter in the Pareto. I believe the resulting ML estimator for the mean of the Pareto has finite variance if $n$ is not too small.
(I haven't checked all the details carefully here, though - it's more an outline of an argument --- don't rely on it unless you do work it through.)
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To address the bias issue:
The ML estimator itself in general won't be unbiased (that's not even true for the ML of the variance, even at something nice like the normal), but with finite variance, you can sometimes multiply by a simple function of $n$ that will make it unbiased (as with variance), or in other cases (ones with behaviour more akin to estimating $\sigma$ at the normal, for example), a more complicated one.
In most cases you just find a nice finite-variance estimator and then see how to unbias it if you care about bias (though generally, I'm not sure why you'd pick bias over say a smaller MSE).
[Even with the complicated ones, you can sometimes derive a simple correction factor that makes the bias term in $1/n$ disappear, leaving the bias as $O(1/n^2)$ - which means even at quite small $n$ it may be effectively unbiased.]
