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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 $\{ X_{i} \}_{i = 1}^{n}$?

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    $\begingroup$ Could you explain what an "infinite large $\delta$" means? $\endgroup$
    – whuber
    Sep 29, 2022 at 16:21
  • $\begingroup$ @whuber I have modified the expression. $\endgroup$
    – Blanco
    Sep 30, 2022 at 3:41
  • $\begingroup$ When you contemplate the idea behind Chebyshev's Inequality, the answer is immediate. $\endgroup$
    – whuber
    Sep 30, 2022 at 12:38
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    $\begingroup$ Also cross-posted to math.SE where it has also been answered. $\endgroup$ Sep 30, 2022 at 13:33

1 Answer 1

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No to both.

  1. 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 $\delta>\delta_1$.

  2. 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.]

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