# Probability that random variable greater than its mean by some value, related to its variance

I'm attempting to show the following is true for all random variables $X$, where $\lambda > 0$ is a constant:

$$P(X - EX \geq \lambda) \leq \dfrac{\sigma^2(X)}{\sigma^2(X) + \lambda^2}$$

Here's what I've got so far:

$$P(|X-EX| \geq \lambda) = P((X-EX)^2 \geq \lambda^2) = P((X-EX)^2 + \sigma^2(X) \geq \lambda^2 + \sigma^2(X)) \leq \dfrac{E[(X-EX)^2 + \sigma^2(X)]}{\sigma^2(X)+\lambda^2} = 2\left(\dfrac{\sigma^2(X)}{\sigma^2(X)+\lambda^2}\right)$$

Now $P(|X-EX|\geq\lambda) = P(X-EX\geq \lambda) + P(EX-X \geq \lambda)$ $= P(X-EX \geq\lambda) + P((-X)-E(-X)\geq \lambda)$.

Therefore one of $P(X-EX\geq\lambda)$ and $P((-X)-E(-X)\geq\lambda)$ must be at most half of $P(|X-EX|\geq\lambda)$, and so given any $\lambda>0$ and random variable $X$, the desired statement is true for either $X$ or for $-X$.

Now my issue is in showing that this statement holds for any random variable $X$. My intuition is to assume that the statement holds for $-X$ and somehow show that it must also hold for $X$, but I've not been able to make any real progress on this. I basically keep working myself in circles at this point.

Is there perhaps an easier or better approach for me to take to this problem altogether? Is the statement even true for all random variables $X$ and constant $\lambda > 0$?

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See this answer on stats.SE for a more general version of this calculation. –  Dilip Sarwate Apr 30 '14 at 20:02
Wonderful! Thank you for your help, I like the clarity of that proof. –  mookystank Apr 30 '14 at 20:15