TITLE: Concentration of Measure for Power Law
QUESTION [5 upvotes]: I have a power law distribution $X$ with exponent $c$: $$p(X=t) = \left\\{\begin{array}{cl}(c-1)/t^{c} & t \geq 1 \\\\ 0 & t < 1\end{array}\right.$$
From $X$ I take $n$ independent samples $X_1, \ldots, X_n$ and would like to give a tail bound for the mean; i.e. a good bound on $\Pr[\frac{1}{n}\sum X_i > t]$.
Unfortunately, my $X$ is unbounded and much of the work on concentration of measure assumes bounded variables.  However, I do know that $X$ has bounded moments, so I can bound the variance and get a weak concentration.  But if $c$ is larger, say $10$, I ought to be able to use the higher moments to get a stronger result.  In particular it seems like the moment method outlined in http://terrytao.wordpress.com/2010/01/03/254a-notes-1-concentration-of-measure/ should be applicable, but the treatment there assumes bounded variables.
I can work through the modifications to that treatment myself and get a reasonable result, but it's kind of messy.  So my question is: is there any existing published theorem that does what I want, so I don't have to add an ugly proof to my paper's appendix?

REPLY [3 votes]: This topic is examined by Olivier Catoni in his recent paper: High confidence estimates of the mean of heavy-tailed real random variables. Denote the mean, variance and kurtosis of your distribution by $m$, $v$ and $\kappa$, respectively. Then, by Catoni's proposition 7.1, the following holds with probability at least $1-2\epsilon$:
$$\frac{|(\frac{1}{n}\sum_{i=1}^n X_i) - m|}{\sqrt{v}} \leq \frac{2\log (\frac{3}{2}\epsilon^-1)\sqrt{\kappa}}{5n} + \sqrt{\frac{2\log (\frac{3}{2}\epsilon^-1)}{n}} + \left(\frac{3\kappa}{2\epsilon n^3}\right)^{1/4}\left(1 + \frac{3^5(n-1)\log (\frac{3}{2}\epsilon^-1)^2\kappa}{2500n^2} + \frac{12\sqrt{2}\log(\frac{3}{2}\epsilon^-1)^{3/2}\sqrt{\kappa}}{25n^{3/2}}\right)^{1/4}$$
There is no mention of boundedness but of course you will want to check his proof.
Are you using the empirical average as an estimate of the mean? If so, you may prefer Catoni's estimator, which is conceptually similar to the truncation trick described in Tao's notes. The raw empirical average performs poorly because a single $X_i$ can throw it wildly off course. Instead, blunt the incoming $X_i$'s and use them to update a guess of $m$. Of course, an initial guess is needed, so the scheme is naturally iterative.