# Prove $\mu(\{|f-\mu f|>K\}) \le \frac{1}{K^2}(\mu f^2 -(\mu f)^2)$

Let $f$ be a random variable on a probability space $(\Omega, \Sigma,\mu)$ where $\mu f^2 < \infty$.

How would I prove (or disprove) that $$\mu(\{|f-\mu f|>K\}) \le \frac{1}{K^2}(\mu f^2 -(\mu f)^2),$$ for any $K>0$?

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Any relation to Chebychev's inequality? – marty cohen Jul 3 '12 at 15:50
To use symbols I am more comfortable with, note that $E(X^2)-(E(X))^2$ turns out to be the variance of $X$. – André Nicolas Jul 3 '12 at 15:57
Thanks. Didnt realise the connection/relation to Chebychev's inequality. – Steven Jul 3 '12 at 16:18

For $K>0$, write \begin{align} K^2\mu(|X-EX|>K)&=K^2\mu((X-EX)^2>K^2)\\ &=\int_{\Omega} K^2\chi_{\{(X-EX)^2>K^2\}}d\mu\\ &\leq \int_{\Omega}(X-EX)^2d\mu\\ &=\int_{\Omega}\left(X^2-2XEX-(EX)^2\right)d\mu\\ &=EX^2-2(EX)^2+(EX)^2\\ &=EX^2-(EX)^2. \end{align}

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Nice +1 apply Chebyshev inside the integal. – Michael Chernick Jul 3 '12 at 18:31

The Chebyshev inequality says that μ({|f−μf|>Kσ})≤1/K$^2$ for any K>0. Here σ$^2$=(μf$^2$−(μf)$^2$). This is very similar to your expression but I do not see how removing the sigma from the left hand side of the inequality would bring a factor of sigma square to the right hand side.

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Thanks! I was asked to prove/disprove the expression, so I guess it's false. – Steven Jul 3 '12 at 16:18
@Steven: you don't seem to make a right conclusion. – Ilya Jul 3 '12 at 16:22
@Steven: Please guess again. – Did Jul 3 '12 at 16:36
@Steven I just said that I didn't see how to complete the connection. I did not say that it was false. – Michael Chernick Jul 3 '12 at 16:41
Thanks, sorry I misunderstood it. – Steven Jul 3 '12 at 17:42