4
$\begingroup$

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!

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.