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The Chebyshev's inequality is $$P(|X-E(X)|>\varepsilon)\leq \frac{\operatorname{Var(X)}}{\varepsilon^2}.$$

I saw a proof which goes like this:

$$ \begin{align} \operatorname{Var(X)}(X) &= E((X-E(X))^2) \\ &= \sum_{x\in S}(x-E(X))^2\cdot P(X=x) \\ &\geq \sum_{|x-E(X)|>\varepsilon}(x-E(X))^2\cdot P(X=x) \\ &> \sum_{|x-E(X)|>\varepsilon}\varepsilon^2\cdot P(X=x) \\ &= \varepsilon^2 P(|X-E(X)|>\varepsilon) \\ \end{align} $$

from which the equation should follow by dividing by $\varepsilon^2$.

What I don't understand here is the 4th step:

$$\sum_{|x-E(X)|>\varepsilon}(x-E(X))^2\cdot P(X=x) > \sum_{|x-E(X)|>\varepsilon}\varepsilon^2\cdot P(X=x) $$

Doen't this imply $$P(|X-E(X)|>\varepsilon)< \frac{\operatorname{Var(X)}}{\varepsilon^2}$$ rather then $$P(|X-E(X)|>\varepsilon)\leq \frac{\operatorname{Var(X)}}{\varepsilon^2}.$$

Why is this correct?

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@In a technical sense, it is not incorrect. If I say $5\le 7$, I am not lying. – André Nicolas Jan 14 '12 at 6:33
@AndréNicolas: I understand, thank you! – Aufwind Jan 14 '12 at 6:34
up vote 2 down vote accepted

Note that $$P(|X-E(X)|>\varepsilon)< \frac{Var(X)}{\varepsilon^2}$$ is (sometimes) false. For let $X=a$ with probability $1$. The variance of $X$ is $0$, but no probability can be $<0$. But if we assume non-zero variance, your reasoning is correct.

The usual version of the Chebyshev Inequality is
$$P(|X-E(X)|\ge \varepsilon) \le \frac{Var(X)}{\varepsilon^2}.$$

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Thank you for your answer. So this Wikibooks entry is in general wrong, then? I'm sorry that this is in german. But the proof is in formulas, and they are international (at least I hope so). :-) – Aufwind Jan 14 '12 at 5:46
@Aufwind: The Wikibooks entry divides by $\sigma_X$. This makes sense only when the variance is non-zero. Perhaps that is explicitly mentioned somewhere, though I only see the requirement that the variance is $<\infty$. But $0$ variance is not interesting! – André Nicolas Jan 14 '12 at 5:50
Actually the German Wikibooks entry says $$P(|X-EX| > \varepsilon) \le \dfrac{\mathrm{Var}(X)}{\varepsilon^2}$$ with a $\le$ rather than $\lt$, so it is true. – Henry Jan 14 '12 at 11:10
@Henry: Yes, and the OP quoted it correctly. I was just noting that in the entry there is division by the standard deviation, so implicitly at least non-zero is assumed. The "$\dots$ is (sometimes) false" in my answer refers to the version the OP was thinking about. – André Nicolas Jan 14 '12 at 15:58
Fine - I was really responding to "So this Wikibooks entry is in general wrong, then?", to which the answer is no. – Henry Jan 14 '12 at 16:07

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