# Refinement Orders for Quantitative Information Flow and Differential Privacy +
## 6 Application: Comparing DP Mechanisms
### Refinement Order within Families of Mechanisms  
We first ask which refinement orders hold within a family of mechanisms. That is, when does reducing \(\varepsilon\) for a particular mechanism produce a refinement? Since we have the convenient order \(\sqsubseteq^{\text{avg}}\subset\sqsubseteq^{\text{max}}\subset\sqsubseteq^{ \text{prv}}\), it is useful to first check if \(\sqsubseteq^{\text{avg}}\) holds as we get the other refinements 'for free'. For the (infinite) geometric mechanism, we have the following result. **Theorem 9**. : _Let \(G^{\varepsilon},G^{\varepsilon^{\prime}}\) be geometric mechanisms. Then, \(G^{\varepsilon}\sqsubseteq^{\text{avg}}G^{\varepsilon^{\prime}}\) iff \(\varepsilon\geq\varepsilon^{\prime}\). That is, decreasing \(\varepsilon\) produces a refinement of the mechanism._  
This means that reducing \(\varepsilon\) in an infinite geometric mechanism is safe against _any_ adversary that can be modelled using, for example, max-case or average-case vulnerabilities. For the truncated geometric mechanism, we get the same result. **Theorem 10**. : _Let \(TG^{\varepsilon},TG^{\varepsilon^{\prime}}\) be truncated geometric mechanisms. Then, \(TG^{\varepsilon}\sqsubseteq^{\text{avg}}TG^{\varepsilon^{\prime}}\) iff \(\varepsilon\geq\varepsilon^{\prime}\). That is, decreasing \(\varepsilon\) produces a refinement of the mechanism._  
However, the over-truncated geometric mechanism does not behave so well. **Theorem 11**. : _Let \(OTG^{\varepsilon},OTG^{\varepsilon^{\prime}}\) be over-truncated geometric mechanisms. Then, \(OTG^{\varepsilon}\sqsubseteq^{\text{avg}}OTG^{\varepsilon^{\prime}}\) for any \(\varepsilon\
eq\varepsilon^{\prime}\). That is, decreasing \(\varepsilon\) does **not** produce a refinement._  
We can think of this last class of geometrics as'skinny' mechanisms, that is, corresponding to a channel with a smaller output space than input space. Intuitively, this theorem means that we can _always_ find some (average-case) adversary who prefers the over-truncated geometric mechanism with the smaller \(\varepsilon\). We remark that the gain function we found can be easily calculated by treating the columns of channel \(A\) as vectors, and finding a vector orthogonal to both of these. This follows from the results in Section 5.1. Since the columns of \(A\) cannot span the space \(\mathbb{R}^{3}\), it is always possible to find such a vector, and, when this vector is not orthogonal to the 'column space' of \(B\), it can be used to construct a gain function preferring \(B\) to \(A\). Even though the \(\sqsubseteq^{\text{avg}}\) refinement does not hold, we can check whether the other refinements are satisfied. **Theorem 12**. : _Let \(OTG^{\varepsilon}\) be an over-truncated geometric mechanism. Then, reducing \(\varepsilon\) does **not** produce a \(\sqsubseteq^{\max}\) refinement; however, it **does** produce a \(\sqsubseteq^{\mathrm{prv}}\) refinement._  
This means that, although a smaller \(\varepsilon\) does not provide safety against all max-case adversaries, it _does_ produce a safer mechanism wrt \(d\)-privacy for _any_ choice of metric we like. Intuitively, the \(\sqsubseteq^{\mathrm{prv}}\) order relates mechanisms based on how they distinguish _inputs_. Specifically, if \(A\sqsubseteq^{\mathrm{prv}}B\), then, for any pair of inputs \(x,x^{\prime}\), the corresponding output distributions are 'further apart' in channel \(A\) than in channel \(B\), and thus the inputs are more distinguishable using channel \(A\). When \(\sqsubseteq^{\mathrm{prv}}\) fails to hold, it means that there are some inputs in \(A\) which are more distinguishable than in \(B\), and vice versa. This means an adversary who is interested in distinguishing some particular pair of inputs would prefer one mechanism to the other. We now consider the exponential mechanism. In this case, we do not have a theoretical result, but, experimentally, it appears that the exponential mechanism respects refinement, so we present the following conjecture. **Conjecture 1**.