# What is the implication for differential privacy if $\epsilon = 0$?

In pure differential privacy, the parameter $$\epsilon$$ represents the desired privacy loss. The smaller the $$\epsilon$$ is, the more privacy we can obtain. What happens when we want the privacy loss $$\epsilon = 0$$. Does that mean we need to keep the data permanently private and we do not answer any queries from outside ?

Thanks

Yes.

$$\varepsilon$$-differential privacy with $$\varepsilon=0$$ implies that the output of the mechanism must be independent of the input -- i.e., it provides no information whatsoever.

When we look at the definition of Differencial Privacy in the Dwork papers, e.g. The Algorithmic Foundations of Differential Privacy, we have that a $$\epsilon$$-Differential Privacy for algorithm $$M$$ as:

$$Pr[M(x_1) \in S] \leq e^\epsilon Pr[M(x_2) \in S]$$

with $$S \subseteq range(M)$$, for any two neighboring data sets $$x_1,x_2$$, differing by the addition or removal of a single row.

If your concern is about an $$\epsilon = 0$$, and so $$e^\epsilon=1$$, which means that the probability distribution of $$M$$ is not influenced by the addition or removal of any single row; in other words, no single row can impact the output of $$M$$.

Maybe we have several examples, but let us imagine that the size of the data sets $$x_1, x_2$$ are big enough to make $$\epsilon$$ negligible, such that the influence of a single row is almost null in the output of $$M$$.

So, I think that the answer is no: whatever the mechanism $$M$$ answers, it will leak no information about any individual row.