I am trying to understand the main result (Theorem 1.1) in this paper by Shao, which gives a large deviation bound for the self-normalized sum of iid variables $$ \frac{\sum X_i}{\sqrt{n}\sqrt{\sum X_i^2}} $$ without any conditions on the moments of $X$. One of the key steps involves applying Cramer's theorem to a sum of the variables $$ bX_i -x\left(\frac{X_i^2+b^2}{2}\right) $$ for $b>0$, $x>\mathbb{E}X/\sqrt{\mathbb{E}X^2}$, and $\mathbb{E}X\geq 0$. And, to this end, the author claims without proof that $$ \mathbb{E}e^{t\left[bX_i -x\left((X_i^2+b^2)/2\right)\right]}<\infty $$ for all such $b$, $x$, and $t\geq 0$. Perhaps I am missing something obvious, but I don't see how to establish this. I tried using a Taylor expansion as well as the AM-GM inequality to get a bound $$ \mathbb{E}e^{t\left[bX_i -x\left((X_i^2+b^2)/2\right)\right]}\leq\min\left\lbrace \mathbb{E}e^{tb(1-x)X},c\mathbb{E}e^{t(1-x)X^2/2} \right\rbrace $$ but I couldn't see how to get the result from either of these methods. Any help with this would be greatly appreciated!

## 1 Answer

$\begingroup$
$\endgroup$

The conditions $x>EX/\sqrt{EX^2}$ and $EX\ge0$ imply $x>0$. So, $t[bX_i -x(X_i^2+b^2)/2]$ is a quadratic polynomial in $X_i$ whose leading coefficient $-tx/2$ is less than $0$ (except for the trivial case $t=0$). Therefore this polynomial is bounded from above, and hence the random variable $e^{t[bX_i -x(X_i^2+b^2)/2]}$ is bounded. So, its expectation is finite.