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Show that $P\{|X-E(X)| \ge a\} = \frac{E(X - E(X))^2}{a^2}$, $a \gt 0 $ and $X$ is a random variable taking the values $\{-1,0,1\}$.

I recognize that this is a special case of Chebyshev's inequality, this one in particular: $P\{|X| \ge a\} \le \frac{E(|X|^p)}{a^p} = \frac{||X||^p_p}{a^p}$. Where $p=2$ and $X=X - E(X)$. I am unsure how to proceed with this, perhaps I have to use indicator variables to indicate what portions of the sample space take on the values?

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    You need only consider the cases $a < 1$ and $a \geq 1$. All the expressions in terms of $X$ are easy enough to solve in terms of $P(X = -1), P(X=0), P(X=1)$. Work from there.2012-11-01

2 Answers 2