TITLE: Analog of Chebyshev's inequality for higher moments
QUESTION [12 upvotes]: I have a positive random variable $X$ with $E[X] = 1$ and a small number $k$ more moments bounded by constants:
$$E[(X-1)^i] = O(1) \forall i \in \{2, ..., k\}.$$
I'd like to bound the average of $n$ independent samples of $X$.  Markov's inequality only uses the first moment to get:
$$\Pr[\frac{1}{n}\sum_{i=1}^n x_i \geq c] \leq 1/c$$
Chebyshev's inequality also uses the second moment to get:
$$\Pr[\frac{1}{n}\sum x_i \geq 1 + c] \leq \frac{O(1)}{nc^2}$$
which is better asymptotic behavior in $n$.  If infinite moments converged, I could use Azuma/Chernoff/Hoeffding/McDiarmid to get
$$\Pr[\frac{1}{n}\sum x_i \geq 1 + c] \leq e^{-f(c)n}$$
for some function $f(c)$.  But what if I have somewhere between 2 and infinite moments, say 10 moments.  Is there a theorem with better asymptotic behavior for intermediate moments?

REPLY [12 votes]: Tom is right: the proof of Chebyshev's inequality can be easily adapted to every nondecreasing nonnegative function. The proof of this generalization that I prefer has a principle worth remembering:


First find an inequality between random variables, then integrate it.


To apply the principle, let $g$ denote a nondecreasing nonnegative function defined on $[0,+\infty)$ and $Z$ a nonnegative random variable. Let $z\ge0$ such that $g(z)>0$. Then,
$$
g(z)\mathbf{1}_A\le g(Z)\ \mbox{with}\ A=[Z\ge z].
$$
Proof: if $\omega\notin A$, the assertion reduces to $0\le g(Z(\omega))$, which holds because $g$ is nonnegative everywhere; if $\omega\in A$, the assertion reduces to $g(z)\le g(Z(\omega))$, which holds because $Z(\omega)\ge z$ and $g$ is nondecreasing. 
Integrating the inequality yields
$$
g(z)P(A)\le E(g(Z)),
$$
and, dividing both sides by $g(z)$, we are done. 
The usual case is when $Z=|X-E(X)|$ and $g(z)=z$. The case mentioned by Thomas is when $Z=|X-E(X)|$ and $g(z)=z^p$, for every positive $p$ (and not only for $p\ge2$). Another case, mentioned by Tom and at the basis of the whole field of large deviations principles, is when $Z=\mathrm{e}^{rX}$ for a nonnegative $r$ and $g(z)=z$. But to deal with the $u\log(u)$ case mentioned by Tom requires to be more careful because the function $u\mapsto u\log(u)$ is non monotonous on $[0,1]$ and not of a constant sign on $[0,+\infty)$ (but everything works fine for $g(z)=z\log(z)$ on $[1,+\infty)$, that is, if $Z\ge1$ almost surely).