Sum of i.i.d. random variables for which Chebyshev inequalities are tight Chebyshev's inequalities: Let $X$ be a random variable with finite expected value $\mu$ and finite non-zero variance $\sigma^{2}$. Then for any real number $\delta > 0$,
$$ \Pr[|X - \mu| \geq \delta\sigma] \leq \frac {1}{\delta^{2}}$$
There is a tight example in wiki, $X_c$ is a random variable with $\sigma = 1/c$:
$$
\left\{
\begin{aligned}
&\Pr[X_c = -1] = \frac{1}{2c^{2}}  \\
&\Pr[X_c = 0] = 1 - \frac{1}{c^{2}}  \\
&\Pr[X_c = 1] = \frac{1}{2c^{2}}  \\
\end{aligned}
\right.
$$
If $\delta = c$, then we have
$$ \Pr[|X - \mu| \geq \delta\sigma] = \Pr[|X| \geq 1] = \frac {1}{\delta^{2}}$$
But for $\delta > c$, it is not tight.
Is there another example that is tight for infinite large $\delta$?
In addition, suppose $X_{1}, X_{2}, \ldots, X_{n}$ are i.i.d. random variables with finite expected value $\mu$ and finite non-zero variance $\sigma^{2}$.
According to Chebyshev's inequalities:
$$\Pr\left[\left|\sum_{i}^{n}X_{i} - n\mu\right| \geq \delta n\sigma\right] \leq \frac{1}{n\delta^{2}}$$
Is there also an (asymptotic) tight example for $\{ X_{i} \}_{i}$?
 A: For your first question, consider a r.v. $X$ with finite first and second moment, variance $\sigma^2>0$ and w.l.o.g. $\mu=E(X)=0$. Assume that the Chebyshev inequality $$P(|X|\geq\varepsilon)\leq\frac{\sigma^2}{\varepsilon^2}$$ is tight for all $\varepsilon\geq\varepsilon_0$. Let $f_X$ be the PDF of $X$. Then
$$P(|X|\geq\varepsilon)=\int_\varepsilon^\infty f_X(x)+f_X(-x)dx=\sigma^2\varepsilon^{-2}$$ which implies $f_X(x)+f_X(-x)=2\sigma^2x^{-3}$ for all $x>\varepsilon_0$. But now we have run into a contradiction, since $$E(X^2)\geq\int_{\varepsilon_0}^\infty x^2(f_X(x)+f_X(-x))dx=\int_{\varepsilon_0}^\infty 2\sigma^2x^{-1}dx=\infty$$ contrary to our assumption that the second moment is finite, which is what allowed us to use the inequality in the first place.
A: You can replace "infinite" by "2 or more."  That is, there is no example where the inequality is tight for two distinct values.

With no loss in generality (just change the units of measurement of $X$) simplify the notation by taking $\mu=0$ and $\sigma=1,$ so that
$$E\left[|X|^2\right] = \operatorname{Var}(X) = \sigma^2 = 1.$$
Let $Y = |X|^2,$ a positive variable, and write
$$S(y) = \Pr(Y\ge y) = \Pr(|X|^2 \ge y)$$
for its survival function.
Suppose there exist $\delta_2 \gt \delta_1 \gt 0$ for which the inequality is tight; that is,
$$S(\delta_1^2) = \Pr(Y \ge \delta_1^2) = \Pr(|X|\ge \delta_1) = \frac{1}{\delta_1^2},\quad S(\delta_2^2) =\Pr(Y \ge \delta_2^2) = \Pr(|X|\ge \delta_2) = \frac{1}{\delta_2^2}.$$
Then, because $S$ is non-increasing and non-negative,
$$\begin{aligned}
1 &= E\left[|X|^2\right]  = E[Y] = \int_0^\infty S(y)\,\mathrm{d}y \\
&\ge \int_0^{\delta_1^2} S(y)\,\mathrm{d}y\ +\ \int_{\delta_1^2}^{\delta_2^2} S(y)\,\mathrm{d}y\\
&\ge \int_0^{\delta_1^2} S(\delta_1^2)\,\mathrm{d}y\ +\  \int_{\delta_1^2}^{\delta_2^2}  S(\delta_2^2)\,\mathrm{d}y\\
&= \Pr(Y \ge \delta_1^2)\delta_1^2\ +\ \Pr(Y \ge \delta_2^2)\left(\delta_2^2 - \delta_1^2\right)\\
&= \frac{1}{\delta_1^2}\delta_1^2\ +\ \frac{1}{\delta_2^2}\left(\delta_2^2 - \delta_1^2\right)\\
&= 1 + \left(1 - \left(\frac{\delta_1}{\delta_2}\right)^2\right)\\
&\gt 1.
\end{aligned}$$
This contradiction shows the hypothetical existence of two tight equalities is false, QED.
