TITLE: concentration inequality for entropy from sample
QUESTION [7 upvotes]: Consider a measure $\mu$ on a finite set, and let $x_1, \ldots, x_n$ be i.i.d samples from $\mu$.  Then the expression $S_n = -\frac{1}{n} \sum_{i=1}^n \log \mu(x_i)$ converges by a.s. to the entropy $H(\mu)$. 
What concentration inequalities exist for finite $n$? In other words, what upper bounds are known for the expression $P(|S_n - H(\mu)|>t)$? 
Of course, we can get some upper bounds using, for instance, Bernstein's inequality for bounded variables, with bound depending on the size of the smallest atom of $\mu$. However, using the sup norm bound seems very rough in this case, since small atoms inflate the norm but also have small contributions to the mean. On the other hand, using more general versions of Bernstein's inequality with moment bounds does not seem natural. ($\sum_{i \in X} \mu(i) \log^k \mu(i)$ does not seem like a natural quantity). 
Since this is a classical situation, there probably exist sharper inequalities than sup norm, using some natural quantities. What are they?

REPLY [5 votes]: Here's a step that seems nice enough to point out. It still leaves a parameter to pick, and I'm not sure it's ever better than applying Bernstein, but it does something different.
We can get a probability bound in terms of how much $S_n$ exceeds the Renyi entropy $H_{\alpha}$ of $\mu$ (equivalently, worded in terms of the $\ell_{\alpha}$ norm of $\mu$), for any $0 < \alpha < 1$.
The unresolved question is if we can to pick $\alpha$ to get a nice closed form of some kind. Maybe someone more clever than I can speak to that.
Claim. Let $X_1,\dots,X_n$ be i.i.d. according to $\mu$ and $Y_i = \log(1/\mu(X_i))$; let $S_n = \frac{1}{n} \sum_{i=1}^n Y_i$. Then for any $0 < \alpha < 1$,
\begin{align} \Pr[ S_n \geq t ] &\leq 2^{-n (1-\alpha) \left( t - H_{\alpha}(\mu) \right) }  \\
&= 2^{-n \left( (1-\alpha)t - \alpha \log \| \mu \|_{\alpha} \right) } .
\end{align}
Here I'm writing $\mu = (\mu_1,\dots,\mu_m)$ as a vector of probabilities. Note that $H_{\alpha}$ is decreasing in $\alpha$ and $H_1 = H$, Shannon entropy. So as $n \to \infty$, we can pick $\alpha \to 1$ and get tail bounds for $t \to H(\mu)$.
Proof. Using the general Chernoff method,
\begin{align}
 \Pr[S_n \geq t] &= \Pr\left[ 2^{\lambda S_n} \geq 2^{\lambda t}\right]  & (\forall \lambda \geq 0) \\
  &\leq \frac{\mathbb{E} 2^{\lambda S_n} }{2^{\lambda t}}  & (\text{Markov's}).
\end{align}
We have
\begin{align}
\mathbb{E} 2^{\lambda S_n} &= \left( \mathbb{E} 2^{\frac{\lambda}{n} Y_1} \right)^n  \\
 &= \left( \mathbb{E} \mu(X_1)^{-\lambda/n} \right)^n  \\
 &= \left( \sum_{j=1}^m \mu_j^{1-\lambda/n} \right)^n .
\end{align}
Hence
\begin{align}
 \Pr[S_n \geq t] \leq 2^{-n \left(\frac{\lambda}{n} t - \log \sum_j \mu_j^{1-\lambda/n} \right)} .
\end{align}
Pick $\lambda$ such that $1-\lambda/n = \alpha$, for a chosen $\alpha \in [0,1]$. In other words, $\frac{\lambda}{n} = 1-\alpha$, and factoring this out and substituting,
\begin{align}
 \Pr[S_n \geq t] \leq 2^{-n (1-\alpha) \left(t - \frac{1}{1-\alpha} \log \sum_j \mu_j^{\alpha} \right)} .
\end{align}