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