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
$$
    \Pr(|X-\mu|\geq k\sigma) \leq \frac{1}{k^2}. 
$$


*

*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?

*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!
 A: 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 $(\sigma^2+\mu^2)/b^2$
A: There can exist no such non-trivial upper bound since, for every fixed $b$, $$\sup\limits_{a\in\mathbb R}\mathrm P(|X-a|\geqslant b)=1.$$
