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George Li avatar Crypto

Proof of Basic Composition in Differential Privacy

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George Li

I'm currently reading the proof of basic composition from the paper https://link.springer.com/content/pdf/10.1007/11761679_29.pdf. In particular, Theorem 1 in Section 2.2.

The proof starts as follows:

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My question is why we can assume the set $S$ is in the form $S_1\times S_2\times\ldots\times S_T$. In general, I remember that for differential privacy, we need to prove the inequality for all sets $S$.

I don't believe the statement holds without the assumption, and I don't think we proving it for this case implies the general case (there are counter examples even in the discrete case).

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Mark
Are they assuming that $S$ takes that form? I would assume they meant implicitly that $S_i$ may depend on both $i$, and $x_1,\dots, x_{i-1}$, i.e. $S_i$ could more precisely be written as $S_{i, x_1, \dots, x_{i-1}}$. I think their proof still goes through in this setting, and makes no assumption that $S = S_1\times \dots \times S_T$.
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George Li avatar
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George Li
You are completely right. Since they wrote it as a product, it felt like assuming independence and hence being in the form $S_1\times\cdots\times S_T$, but that is indeed not the case. Thanks so much!
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