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

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

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.]