LFD Book Forum - View Single Post - Chapter 1

henry2015 · #5 11-05-2017, 06:17 AM

Quote:

Originally Posted by magdon

If $\mu=0.9$ and $\nu\le0.1$ , it implies that $|\nu-\mu|>0.8^-$ (any number slightly less than 0.8). By the implication bound,

$P[\nu\le0.1] \le P[|\nu-\mu|>0.8^-]$

By looking at the RHS, one can identify $\epsilon$ for applying the Hoeffding bound.

Hi, I am a bit confused.

I thought when it says $\nu\le0.1$ , it means $0\le\nu\le0.1$ because smallest probably is 0; hence, $0.8\le|\nu-\mu|\le0.9$ ....I am sure I miss something.

Any pointer?

Thanks!