Suppose $X_1, X_2, \ldots, X_n$ are independent and each has the same distribution with a sub-exponential random variable $X$ (for example, $X$ is the square of a standard normal Gaussian variable). Can I obtain a concentration inequality for the square of sub-exponential $X_i$, say,
$$\mathbb{P}\left( \frac{1}{n} \left( X_1^2+\cdots+X_n^2 \right) \ge \mathbb{E}\left[X^2\right] + t \right) \le C \exp\left( – n \cdot \min\left( C_1 t^2, C_2 t, C_3 \sqrt{t} \right) \right),$$
where $C, C_1, C_2, C_3$ are constants?
This problem arises in my research.
Remark:
Actually, for i.i.d. sub-Gaussian random variables $Y_i\ (i=1,\ldots,n)$, I knew that
$$\mathbb{P}\left( \frac{1}{n} \left(Y_1+\cdots+Y_n\right) \ge \mathbb{E}\left[Y\right] + t \right) \le \exp\left( – n\cdot C_1 t^2 \right).$$
Besides, since $Y_i^2\ (i=1,2,\ldots,n)$ are sub-exponential, I also knew that
$$\mathbb{P}\left( \frac{1}{n} \left(Y_1^2+\cdots+Y_n^2\right)\ge \mathbb{E}\left[ Y^2 \right] + t \right) \le \exp\left( – n\cdot \min(C_1 t^2, C_2 t) \right).$$
These two inequalities can be proved by a Chernoff bound, since the moment generating functions of $Y$ (sub-Gaussian) and $Y^2$ (sub-exponential) both exist.
However, I want to know whether there is an inequality like
$$\mathbb{P}\left( \frac{1}{n} \left( Y_1^4+\cdots+Y_n^4 \right) \ge \mathbb{E}\left[Y^4\right] + t \right) \le C \exp\left( – n \cdot \min\left( C_1 t^2, C_2 t, C_3 \sqrt{t} \right) \right),$$
even though the moment generating function of $Y^4$ (square of sub-exponential) does not exist.
Best Answer
No. For non-negative i.i.d. $Y_i$, $$P(Y_1\ge (\mu+t)n)\le P\Bigg(\sum_{i=1}^n Y_i\ge (\mu+t)n\Bigg)\le\exp(-nf(t))$$ implies that $Y_1$ is sub-exponential and a square of a sub-exponential is not guaranteed to be sub-exponential.
However you can obtain $$P(X_1^2+\dots+X_n^2>nt)\sim nP(X_1^2>nt)=n\exp(-\lambda\sqrt {nt})$$ for $X_i\sim \exp(\lambda)$ which you can extend to all subexponential $X_i$ that have exponential tails.