# Deduce upper bound of variance from Chernoff-type tail bound

For a random variable $X$, I have a large deviation inequality of the form $$P(|X-\mathbb EX|\geq r)\leq ce^{-\alpha r}\,.$$

Consider a sample mean $S_n=\frac{1}{n}(X_1+X_2+\dots+X_n)$. I would to obtain a central limit theorem of the form $\sqrt n(S_n-\mu)\equiv N(0,\sigma^2)$. Is this possible?

It seems that I need to know the variance of $X$ to apply the Lindeberg–Lévy CLT. I don't see how I can get the variance of $X$ from its tail bound.

• Typed a more relevant title. – Did Aug 28 '18 at 19:09

You have, for a non-negative random variable $Z$, $$\mathbb{E}[Z] = \int_{0}^\infty \mathbb{P}\{Z \geq z\}dz \tag{1}$$ from which you can bound the variance of $X$: $$\operatorname{Var} X = \mathbb{E}[(X-\mathbb{E}[X])^2] = \int_{0}^\infty \mathbb{P}\{(X-\mathbb{E}[X])^2 \geq z\}dz = \int_{0}^\infty \mathbb{P}\{\lvert X-\mathbb{E}[X]\rvert\geq \sqrt z\} dz\tag{2}$$ and using your concentration bound you thus obtain $$\operatorname{Var} X \leq \int_{0}^\infty c e^{-\alpha \sqrt{z}}dz = \frac{c}{\alpha^2} \tag{3}$$ Does that suffice for your purposes?