# Refinement Orders for Quantitative Information Flow and Differential Privacy +
## 8 Conclusions
of \[\tilde{c}_{y}^{z}\ \ :=\ \ c_{y}^{z}\,\frac{(\hat{\pi}A)_{y}\,(\pi B)_{z}}{(\pi A )_{y}\,(\hat{\pi}B)_{z}}\]5 \[\tilde{c}_{y}^{z}\ \ :=\ \ c_{y}^{z}\,\frac{(\hat{\pi}A)_{y}\,(\pi B)_{z}}{(\pi A )_{y}\,(\hat{\pi}B)_{z}}\]6 Theorem A1 \[\tilde{c}_{y}^{z}\ \ :=\ \ c_{y}^{z}\,\frac{(\hat{\pi}A)_{y}\,(\pi B)_{z}}{(\pi A )_{y}\,(\hat{\pi}B)_{z}}\]7 quasi-convexity of \[\tilde{c}_{y}^{z}\ \ :=\ \ c_{y}^{z}\,\frac{(\hat{\pi}A)_{y}\,(\pi B)_{z}}{(\pi A )_{y}\,(\hat{\pi}B)_{z}}\]8 \[\tilde{c}_{y}^{z}\ \ :=\ \ c_{y}^{z}\,\frac{(\hat{\pi}A)_{y}\,(\pi B)_{z}}{(\pi A )_{y}\,(\hat{\pi}B)_{z}}\]9 \[\sum_{y}c_{y}^{z}\frac{(\hat{\pi}A)_{y}}{(\pi A)_{y}}\]0  
from which \(A\sqsubseteq^{\max}_{\mathbb{Q}}B\) follows. Now, assume that \(A\
ot\sqsubseteq^{\max}B\), let \(S=ch\,supp\,[\pi,A]\subseteq\mathbb{D}\mathcal{X}\) and define a vulnerability function \(V\colon\mathbb{Q}\mathcal{X}\) that maps every prior \(\sigma\colon\mathbb{D}\mathcal{X}\) to its Euclidean distance from \(S\), that is,  
\[\sum_{y}c_{y}^{z}\frac{(\hat{\pi}A)_{y}}{(\pi A)_{y}}\]1 (A4)  
Since \(S\) is a convex set, it is well known that \(V(\sigma)\) is convex on \(\sigma\) (hence also quasi-convex). Note that \(V(\sigma)=0\) for all \(\sigma\in S\) and strictly positive anywhere else. By definition of \(S\), we have that \(\alpha^{y}\in S\) and hence \(V(\alpha^{y})=0\) for all posteriors of \(A\), as a consequence \(V^{\max}[\pi,A]=0\). On the other hand, since \(A\
ot\sqsubseteq^{\max}B\), from Theorem A1, we get that there exists some posterior of \(B\) such that \(\delta^{z}\
ot\in S\). As a consequence, \(V^{\max}[\pi,B]\geq V(\delta^{z})>0=V^{\max}[\pi,A]\), which implies that \(A\
ot\sqsubseteq^{\max}_{\mathbb{Q}}B\). **Theorem A3**. : \(\sqsubseteq^{\max}_{\mathbb{Q}}\) _is strictly stronger than \(\sqsubseteq^{\max}\)._  
\[ \begin{array}{|c|c|c|}\hline A&y_{1}&y_{2}\\ \hline x_{1}&\
icefrac{{2}}{{3}}&\
icefrac{{1}}{{4}}\\ x_{2}&\
icefrac{{1}}{{3}}&\
icefrac{{3}}{{4}}\\ \hline\end{array}\begin{array}{|c|c|c|}\hline B&y_{1}&y_{2}\\ \hline x_{1}&\
icefrac{{1}}{{3}}&\
icefrac{{3}}{{4}}\\ x_{2}&\
icefrac{{2}}{{3}}&\
icefrac{{1}}{{4}}\\ \hline\end{array}  \tag{104}\]  
Since \((\
icefrac{{3}}{{4}},\
icefrac{{1}}{{4}})\) cannot be written as a convex combination of \((\
icefrac{{2}}{{3}},\
icefrac{{1}}{{3}})\) and \((\
icefrac{{1}}{{4}},\
icefrac{{3}}{{4}})\), and similarly \((\
icefrac{{1}}{{4}},\
icefrac{{3}}{{4}})\) cannot be written as a convex combination of \((\
icefrac{{1}}{{3}},\
icefrac{{2}}{{3}})\) and \((\
icefrac{{3}}{{4}},\
icefrac{{1}}{{4}})\), from Theorem 1, we conclude that \(A\
ot\sqsubseteq^{\max}B\
ot\sqsubseteq^{\max}A\). \(\Box\)  
**Theorem 1.5**. : _Let \(f:\mathcal{X}\rightarrow\mathcal{Y}\) be any query and \(A,B\) be two mechanisms on \(\mathcal{Y}\). If \(A\sqsubseteq^{\mathrm{prv}}B\), then \(A\circ f\sqsubseteq^{\mathrm{prv}}B\circ f\)._  
Proof. : Define \(d_{A\circ f}(x_{1},x_{2})=d_{A}(f(x_{1}),f(x_{2}))\), and similarly for \(d_{B\circ f}\). Then, we have:  
\[d_{A\circ f}(x_{1},x_{2})=d_{A}(f(x_{1}),f(x_{2}))\geq d_{B}(f(x_{1}),f(x_{2}) )=d_{B\circ f} \tag{105}\]  
\(\Box\)  
**Theorem 1.6**. : \(\mathcal{ML}_{d}^{+,\max}\) _is always achieved on a uniform prior \(\pi^{u}\). Namely_  
\[\max_{\pi}\mathcal{L}_{d}^{+,\max}(\pi,C) = \mathcal{L}_{d}^{+,\max}(\pi^{u},C) = V_{d}^{\max}[\pi^{u},C] \tag{106}\]  
Proof. : Fix \(\pi,C\), and let \((a,\delta^{y})\) and \((b,\rho^{y})\) be the outer and inners of \([\pi,V_{d}]\) and \([\pi^{u},V_{d}]\), respectively.