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 every $\delta$ satisfying $|\delta| > \delta_{0}$? ($\delta_{0}$ is a constant only depending on the distribution.)
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 $\{ X_{i} \}_{i = 1}^{n}$?

No to both.

*

*Suppose $X$ attains the bound for $\delta=\delta_1$.   Construct $\tilde X$ by moving any probability mass of $X$ further than $\delta_1$ from $\mu$ to distance $\delta_1$ from $\mu$. If there is any such mass then
$$P(|\tilde X-\mu|\geq\delta_1)=P(|X-\mu|\geq\delta_1),$$
but
$$\mathrm{var}[\tilde X]<\mathrm{var}[X].$$
Since Chebyshev's inequality is tight for $X$ is it violated for $\tilde X$: a contradiction.  So there must be no probability mass with $X-\mu$ greater than $\delta_1$. But in that case Chebyshev's inequality cannot be sharp for any $\delta>\delta_1$.


*Write $\sigma^2_n$ for $\mathrm{var}[X_n]=\mathrm{var}[X]/n$. The Central Limit Theorem says that as $n\to\infty$
$$P[|\bar X_n-\mu|\geq\delta\sigma_n]\to 2\Phi(-\delta)$$
so it is not possible to have
$$P[|\bar X_n-\mu|\geq\delta\sigma_n]=1/\delta^2$$
for any fixed $X$ and arbitrarily large $n$.
[2. Alternative proof sketch: the inequality is only sharp when all the probability mass of $X$ is at $\mu$ or $\mu\pm\delta$, a distribution supported on more than one and no more than three points. A mean of $n$ iid observations supported on more than one point must be supported on at least $n$ points.]
