According to Wikipedia, given a randomized algorithm $\mathcal{A}$, two neighboring datasets $D_1, D_2$, a real number $\epsilon > 0$, $\mathcal{A}$ provides $\epsilon$-differential privacy, if $$ \frac{\mathbb{P}[\mathcal{A}(D_1) \in S]}{\mathbb{P}[\mathcal{A}(D_2) \in S]} \leq e^\epsilon $$ for all subsets $S$ of $\mathcal{A}$’s support.

However, I've stumbled across a different definition of $\epsilon$-DP, where also a lower bound is required: $$ e^{-\epsilon} \leq \frac{\mathbb{P}[\mathcal{A}(D_1) \in S]}{\mathbb{P}[\mathcal{A}(D_2) \in S]} \leq e^\epsilon. $$ Are these two definitions equivalent?


1 Answer 1


Yes, the definitions are equivalent. That the second implies the first is obvious. To see that the first implies the second, note that the datasets $D_1,D_2$ can be interchanged. So, if

$$ \frac{\mathbb{P}[\mathcal{A}(D_1) \in S]}{\mathbb{P}[\mathcal{A}(D_2) \in S]} \leq e^\epsilon $$

for all $D_1,D_2$, then in particular

$$ \frac{\mathbb{P}[\mathcal{A}(D'_2) \in S]}{\mathbb{P}[\mathcal{A}(D'_1) \in S]} \leq e^\epsilon, $$ hence $$ \frac{\mathbb{P}[\mathcal{A}(D'_1) \in S]}{\mathbb{P}[\mathcal{A}(D'_2) \in S]} \geq e^{-\epsilon}, $$ because all values are positive.

  • 1
    $\begingroup$ Thanks for pointing that $D_1$ and $D_2$ are interchangeable. I missed that. $\endgroup$ Aug 16, 2021 at 11:43

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