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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

2 Answers 2

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.

But assuming I'm right that you're interested in the probabilistic method for existence proofs (which is hard to tell since you didn't give any context), it's not really the right question to ask. The first and second moment methods have names at all because they're frequently useful and easy to carry out. I don't know offhand of applications where you can't use those but can use a higher moment method. Instead one typically has to turn to more sophisticated tools. Check out "The Probabilistic Method" by Alon and Spencer for a taste of lots of those.

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 $P(Y=-1)=P(Y=1)=1/2$.

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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 $p>1$).

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