New answers tagged

1

OK, I did this quickly. Hope it’s correct. When $r*\geq 0,$ the relationship holds as you observed. And when $r^*\leq -1,$ the same expression for both probabilities you want to compare enables a direct proof. Let $r^*\in(-1,0),$ so that $1+r^* \in (0,1).$ Then what you want to show is $$\frac{1}{2} e^{-\epsilon(1+r^*)}\geq e^{-\epsilon}\left(1-\frac{1}{2} e^...


1

By allowing a slack in $\tilde \delta$, one can get a higher privacy of $\tilde \epsilon = O(k \epsilon^2 + \sqrt{k}\epsilon)$, compared with the basic composition $\tilde \epsilon = O(k\epsilon)$. That's the advantage of advanced composition theorem. Of course when $\tilde \delta$ close to 0 it is very likely that advanced composition guarantee will be ...


Top 50 recent answers are included