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How to prove the one-sided Chebyshev's inequality which states that if X has mean 0 and variance $\sigma^2$, then for any $a > 0$

$P(X \geq a) \leq \frac{\sigma^2}{\sigma^2+a^2}$

Solution. I know the Chebyshev's inequality which States that

$P(|X-\mu| \geq a) \leq \frac{Var(X)}{a^2}$ If I first argue that for any $b > 0$

$P(X \geq a) \leq P{[(X+b)^2 \geq (a+b)^2]}$

$P(X\geq a) \leq \frac{E(X+b)^2}{(a+b)^2}$

$P(X \geq a) \leq \frac{[E(X^2)+2E(X)b+b^2]}{(a+b)^2}$

$P(X \geq a) \leq \frac{\sigma^2+ b^2}{(a+b)^2}$

I got the correct answer.

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    $\begingroup$ Hint: Expand $(X+b)^2$, take the expected value, and write in terms of $\sigma^2$. Substitute what you get into $\frac{E(X+b)^2}{(a+b)^2}$ and then minimize that. $\endgroup$ – grndl Oct 1 '16 at 15:03
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    $\begingroup$ There is a full proof on the Wikipedia page en.wikipedia.org/wiki/Cantelli%27s_inequality $\endgroup$ – Jack D'Aurizio Oct 1 '16 at 15:09
  • $\begingroup$ I got the correct answer. $\endgroup$ – Dhamnekar Winod Oct 1 '16 at 15:25
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I have a couple of proofs at http://www.se16.info/hgb/cheb.htm#OTProof and http://www.se16.info/hgb/cheb2.htm

One of these, loosely based on Probability and Random Processes by Grimmett and Stirzaker, would give a proof like this:

With $a>0$, for any $b\ge 0$ $$P(X\ge a) = P(X+b \ge a+b) \le E\left[\dfrac{(X+b)^2}{(a+b)^2}\right] = \dfrac{\sigma^2+b^2}{(a+b)^2}$$

But treating $\dfrac{\sigma^2+b^2}{(a+b)^2}$ as a function of $b$, the minimum occurs at $b = \sigma^2 / a$, so $$P(X\ge a) \le \dfrac{\sigma^2+(\sigma^2/a)^2}{(a+\sigma^2/a)^2} =\dfrac{\sigma^2(a^2+\sigma^2)}{(a^2+\sigma^2)^2} = \dfrac{\sigma^2}{ \sigma^2+a^2}.$$

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