# Is the differential privacy definition with lower and upper bound equivalent to the definition with just an upper bound?

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?

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