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Aug 26, 2014 at 0:31 comment added willeM_ Van Onsem No offense ;), it was merely from a programming perspective. And well one could simply branch over the two cases with an if I guess?
Aug 26, 2014 at 0:21 comment added Douglas Zare @CommuSoft: I used the assumption that $p \gt q$. As far as I can see, this is a correct application of Hoeffding's inequality on random variables in $[-1,1]$. Perhaps what happens if you reverse that assumption indicates that Hoeffding's inequality can't be sharp.
Aug 25, 2014 at 23:13 comment added willeM_ Van Onsem Isn't there something wrong with your first formula: if you swap $p$ and $q$ (or $X$ and $Y$ if you want), you still have the same bound...
Feb 24, 2010 at 13:45 vote accept user4120
Feb 24, 2010 at 3:30 history edited Douglas Zare CC BY-SA 2.5
Added more explicit estimates and numerical examples.
Feb 21, 2010 at 23:12 comment added maks @Alekk: Isn't the bound you obtain essentially tight? (i.e loosing at most a factor of 1/sqrt(n))
Feb 21, 2010 at 7:28 comment added Alekk for this, even the simple Markov inequality is better than Hoeffding since one can carry out all the computations exactly. This gives something like: P[Y>X] < [ 2\sqrt{pq(1-p)(1-q)} + (1-p)(1-q) + pq ]^n
Feb 21, 2010 at 5:17 comment added user4120 The third point is not the source of the problem. Or, at least, it isn't the immediate problem-- Hoeffding's bound is still frustratingly loose for $p$ and $q$ near $\frac{1}{2}$. (Still processing the rest of your post.)
Feb 21, 2010 at 4:39 comment added Douglas Zare I forgot to add the following: Please clarify whether the third point is a source of problems with the Hoeffding bound in your case.
Feb 21, 2010 at 4:31 history answered Douglas Zare CC BY-SA 2.5