# x-th moment method

For a real-valued random variable, $X$, the first moment method, is simply

$P(X\ge\mathbb{E}[X])>0$

This can be extended to the second moment quite easily:

$P(X\ge\mathbb{E}[X]+\sqrt{Var[X]})>0$

$P(|X-\mathbb{E}[X]|\ge\sqrt{Var[X]})>0$

The question must be asked: How does one generalize this to higher (probably centralized) moments?

Edit: Good catch Mark! Let me rephrase the question in another way

Let $X$ be a real-valued random variable. Given only the first $n$ moments of $X$: $\mathbb{E}(X), \ldots, \mathbb{E}(X^n)$, what is the largest value for $|X-\mathbb{E}[X]|$ that can be guaranteed to have positive probability?

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Wikipedia link for those like me who'd forgotten what the terminology means: en.wikipedia.org/wiki/Method_of_moments_%28statistics%29 –  Yemon Choi Sep 20 '10 at 20:19
@Yemon: I don't think the statistical method of moments is what the OP was asking about. This "first/second moment method" terminology is used in basic treatments of the probabilistic method of existence proofs. –  Mark Meckes Sep 20 '10 at 23:56
@Mark: Thanks for the correction and retagging –  Yemon Choi Sep 21 '10 at 0:11

Edit: The "second moment method" you've stated is false, as shown for example by $P(X=1)=p$ and $P(X=0)=1-p$ with $p>1/2$. See this Wikipedia article for a discussion of the more complicated inequalities sometimes called the first or second moment methods.
Added: Okay, here's the answer to your revised question: $$P(\vert X - \mathbb{E}X \vert \ge (\mathbb{E}\vert X- \mathbb{E} X \vert^n)^{1/n} ) > 0,$$ and when $n$ is even you cannot replace $(\mathbb{E}\vert X- \mathbb{E} X \vert^n)^{1/n}$ with any larger quantity depending only on the first $n$ moments of $X$. To see the latter claim, let $Y=X-\mathbb{E}X$ and observe that knowledge of the first $n$ moments of $X$ is equivalent to knowledge of $\mathbb{E}X$ and the first $n$ moments of $Y$. My claim is that there is a random variable $Y$ with $\mathbb{E}Y=0$ such that $$P(Y^n > \mathbb{E} Y^n) = 0,$$ and indeed this is true if $P(Y=-1)=P(Y=1)=1/2$.
Not that I have ever seen this used anywhere (so this might be the wrong direction), but since the second moment method is essentially using Cauchy Schwartz on the variable $X=X 1_{X>0}$, can't we use Holder's inequality on this expression to obtain higher order inequalities? edit: and obtain something like $P(X>0)\geq (\frac{E[X]^p}{E[X^p]})^{1/(p-1)}$ (assuming $X\geq 0$ and $p>1$).