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The equation below shows the proof of Laplace mechanism for differential privacy. I am not understanding the product operator, is this a common rule?

$$ \frac{p_x(z)}{p_y(z)} = \prod_{i=1}^{k}\left(\frac{exp(-\frac{\varepsilon |f(x)_i - z_i|}{\Delta})}{exp(-\frac{\varepsilon |f(y)_i - z_i|}{\Delta})}\right) $$

While $p_x$ and $p_y$ denote the pdf of mechanisms $\mathcal{M}_L(x,f,\varepsilon)$ and $\mathcal{M}_L(y,f,\varepsilon)$ respectively. Important to notice that $x$ and $y$ are neighbor datasets. $x, y, z \in \mathbb{R}^k$

Thank you in advance.

Iago Chaves
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1 Answers1

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The definition of Laplace Mechanism: $$ \mathcal{M}_L(x, f(\cdot), \varepsilon) = f(x) + (Y_1,\ldots,Y_k) $$ Once each $Y_i \sim Lap(\Delta f/\varepsilon)$ they are indepedent of each other. So we can calcule the joint probability of them using the product of marginals.

As we know $p_x = \mathcal{M}_L(x,f,\varepsilon)$ and $z \in \mathbb{R}^k$, thus is possible to do:

$$ p_x(z) = \frac{\varepsilon}{2\Delta f} exp\left(-\frac{\varepsilon |f(x) - z|}{\Delta f}\right) = \prod_{i=1}^k \frac{\varepsilon}{2\Delta f} exp\left(-\frac{\varepsilon |f(x)_i - z_i|}{\Delta f}\right) $$

I think that is the answer.

Iago Chaves
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  • 6