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

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. – Jonathan Christensen Nov 1 '12 at 1:44

I don't think that's a correct equation. Take the special case where $P(X = -1) = P(X = 1) = \frac{1}{2}$, so $EX = 0$ and $E(X - EX)^2 = 1$. The right hand side is $\frac{1}{a^2}$ and the left hand side is either $0$ or $1$. This phenomenon happens for all such $X$: the left hand side is a step function of $a$ and the right hand side is $var(X)/a^2$, a rational function of $a$.

The inequality (replace $=$ with $\le$) is of course true and indeed a special case of Chebyshev's inequality.

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Do you know Markov's inequality? Chebyshev's inequality is a corollary of Markov's inequality and says that for any $\alpha > 0$

$$P(|X - E(X)| \geq \alpha) \leq \frac{Var(X)}{\alpha^2}.$$

This is true for any random variable $X$ and in particular an $X$ such that $X(\omega) \in \{-1, 0, 1\}$. Since this is homework I will leave the rest to you.

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ah yes I have seen that before, it hadn't come to mind though, I will look into it thanks! – user38784 Nov 1 '12 at 15:08