# Two-sided Chebyshev inequality for event not symmetric around the mean?

Let $X$ be a random variable with finite expected value $μ$ and non-zero variance $σ^2$. Then for any real number $k > 0$, two-sided Chebyshev inequality states $$\Pr(|X-\mu|\geq k\sigma) \leq \frac{1}{k^2}.$$

1. I saw a paper applies two-sided Chebyshev inequality to $\Pr(|X|\geq b)$ and got an upper bound $Var(X)/b^2$, when $X$ does not necessarily have mean zero. I think it is not correct. Is there any way to apply two-sided Chebyshev inequality to this case?
2. Instead, for $\Pr(|X-a|\geq b)$, I think it is only possible to apply one-sided Chebyshev inequality to $\Pr(X-a\geq b)$ and $\Pr(X-a \leq -b)$ respectively. Am I right?

Thanks and regards!

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## 2 Answers

Edited in response to OP's comment and request

It depends on what you mean by applying the two-sided Chebyshev Inequality. The same method that is used to prove the Chebyshev Inequality (viz. bound $E[\mathbf 1_{(-\infty, \mu-b]} + \mathbf 1_{[\mu+b,\infty)}]$ from above by $E[(X-\mu)^2/b^2] = \sigma^2/b^2$ can be used to show that $$P\{|X-a| \geq b\} \leq \frac{E[(X-a)^2]}{b^2} = \frac{\sigma^2 + (\mu-a)^2}{b^2}.$$ In more detail, $\displaystyle \mathbf 1_{(-\infty, a-b]} + \mathbf 1_{[a+b,\infty)} \leq \left(\frac{x-a}{b}\right)^2$ for all $x \in \mathbb R$ since the parabola $\displaystyle \left(\frac{x-a}{b}\right)^2$ passes through $(a-b,1), (a,0)$, and $(a+b,1)$. Consequently, $$\begin{align*} P\{|X-a| \geq b\} &= P\{X \leq a-b\} + P\{X \geq a + b\}\\ &= \int_{-\infty}^{a-b} f_X(x)\,\mathrm dx + \int_{a+b}^{\infty} f_X(x)\,\mathrm dx\\ &= \int_{-\infty}^{\infty}(\mathbf 1_{(-\infty, a-b]} + \mathbf 1_{[a+b,\infty)})f_X(x)\,\mathrm dx\\ &\leq \int_{-\infty}^{\infty}\left(\frac{x-a}{b}\right)^2f_X(x)\,\mathrm dx\\ &= \frac{1}{b^2}\int_{-\infty}^{\infty}(x-a)^2f_X(x)\,\mathrm dx\\ &= \frac{E[(X-a)^2]}{b^2} \end{align*}$$ If $a$ equals the mean $\mu$, that expectation on the right is the variance $\sigma^2$ and we get Chebyshev's Inequality. More generally, it is a standard result in probability theory that $$\begin{align*} E[(X-a)^2] &= E[((X-\mu) + (\mu - a))^2\\ &= E[(X-\mu)^2] + (\mu-a)^2 + 2(\mu-a)E[X-\mu]\\ &= \sigma^2 + (\mu-a)^2 \end{align*}$$ and we get the inequality I stated initially.

With regard to the second question in the OP's comment,

I saw a paper applies two-sided Chebyshev inequality to $\Pr(|X|\geq b)$ and got an upper bound $Var(X)/b^2$, when $X$ does not necessarily have mean zero. I think it is not correct.

Yes, that is not right when the mean is not zero; the upper bound from the variation on the Chebyshev Inequality described above is $(\sigma^2+\mu^2)/b^2$

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Thanks! (1) I wonder how to prove the inequality in your reply using the same method for proving Chebyshev inequality? (2) Are you saying the paper is right, and I am wrong about it? – Tim May 13 '12 at 19:02

There can exist no such non-trivial upper bound since, for every fixed $b$, $$\sup\limits_{a\in\mathbb R}\mathrm P(|X-a|\geqslant b)=1.$$

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Thanks! Is the example easy to construct? – Tim May 13 '12 at 18:42
Every random variable $X$ and real number $b$ do the job. – Did May 13 '12 at 18:45
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