# One sided Chebyshev's inequality

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 $b > 0$

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

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

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

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

• 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. – grndl Oct 1 '16 at 15:03
• There is a full proof on the Wikipedia page en.wikipedia.org/wiki/Cantelli%27s_inequality – Jack D'Aurizio Oct 1 '16 at 15:09
• I got the correct answer. – Dhamnekar Winod Oct 1 '16 at 15:25

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 $$P(X\ge a) \le \dfrac{\sigma^2+(\sigma^2/a)^2}{(a+\sigma^2/a)^2} =\dfrac{\sigma^2(a^2+\sigma^2)}{(a^2+\sigma^2)^2} = \dfrac{\sigma^2}{ \sigma^2+a^2}.$$