How does Chernoff-Hoeffding bound with limited independence reduce to the usual generic CH bound with complete independence

Question

As the title might suggest, I am referring to this paper https://www.cs.umd.edu/~srin/PDF/ch-bounds.pdf , titled : Chernoff-Hoeffding Bounds for Application with Limited Independence.

The theorem in question is Theorem 5 of the paper, specifically the following

If I wish to reduce back to the generic CH-bound, I would then set $k = n$, where $n$ is the number of $r.v$. But now base on the theorem, I could (potentially) get a better bound than the generic CH-bound. I feel like I am misunderstanding some nuances involving this theorem.

Any help or insight is deeply appreciated

Cheers

Answer 1

$\newcommand\de\delta$If $X$ is the sum of $n$ random variables each of them with values in $[0,1]$ and $\mu=EX$, then clearly $\mu\le n$. So, if $\de\in[0,1]$ and $k\le\lfloor\de^2\mu e^{-1/3}\rfloor$, then $k\le\mu e^{-1/3}\le e^{-1/3}n<n$, so that here you cannot "set $k=n$".