# Understanding a proof of Chebyshev's inequality

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

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.
$$P(|X-E(X)|\ge \varepsilon) \le \frac{Var(X)}{\varepsilon^2}.$$
@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