# Quantifying Privacy Risks for Continuous Trait Data
## 6 Attack Validation
### Mathematical Analysis
#### 6.3.1 Binary Classification Analysis
**  
\[(r^{2}-1)x^{2}+2(r+1)zx-[(1+r)^{2}z^{2}+2r^{2}\ln\frac{r}{p}]=0 \tag{12}\]  
\(b^{2}-4 ac\geqslant 0\) is the condition under which Equation (12) has solutions, which can be represented as  
\[b^{2}-4 ac=4r^{2}(r+1)^{2}(z^{2}+2\times\frac{r-1}{r+1}\times\ln\frac{r}{p}) \tag{13}\]  
Obviously \(r>0\), so \(r^{2}>0\), \((r+1)^{2}>0\), then the case where \(p_{1}f_{1}(x)\) and \(p_{2}f_{2}(x)\) have intersections can be expressed as  
\[z^{2}+2\times\frac{r-1}{r+1}\times\ln\frac{r}{p}\geqslant 0 \tag{14}\]  
It needs to be divided into two cases to discuss: \(0<r<1\) and \(r>1\). For \(0<r<1\), we have \(\frac{r-1}{r+1}<0\). If \(p>r\times\exp(\frac{z^{2}(r+1)}{2(r-1)})\), then \(\ln\frac{r}{p}<0\), \(\frac{r-1}{r+1}\times\ln\frac{r}{p}>0\), \(p_{1}f_{1}(x)\) and \(p_{2}f_{2}(x)\) always have intersections; similarly, for \(r>1\), we have \(\frac{r-1}{r+1}>0\). If \(p<r\times\exp(\frac{z^{2}(r+1)}{2(r-1)})\), then \(\ln\frac{r}{p}>0\), \(\frac{r-1}{r+1}\times\ln\frac{r}{p}>0\), we still have the conclusion that \(p_{1}f_{1}(x)\) and \(p_{2}f_{2}(x)\) always have intersections. Then we should have  
\[p \begin{cases}\geqslant r\times\exp(\frac{z^{2}(r+1)}{2(r-1)})&0<r<1\\ \leqslant r\times\exp(\frac{z^{2}(r+1)}{2(r-1)})&r>1\end{cases}  \tag{15}\]  
to make \(b^{2}-4 ac\geqslant 0\). Figure 5 is a typical example for \(f_{1}(x)\), \(f_{2}(x)\), \(p_{1}f_{1}(x)\), and \(p_{2}f_{2}(x)\) with given \(r,z,p\). In this example the \(r,z,p\) fix Equation (15), so we have \(p_{2}f_{2}(x)>p_{1}f_{1}(x)\) when \(x_{1}<x<x_{2}\), where \(x_{1},x_{2}\) is the solution of Equation (12) and \(x_{1}<x_{2}\). Let \(f_{1},f_{2}\) be \(f_{f_{1}^{*}|f_{1}}f_{f_{1}^{*}|g_{2}}\), respectively, the privacy breaching method will match \(\mathbb{T}^{*}(i,j^{\prime})\) to a \(\mathbb{S}^{*}(i,l^{\prime})\) with value \(s_{2}\) when \(\mathbb{T}^{*}(i,j^{\prime})\) is in the interval \((x_{1},x_{2})\). If \(r,z,p\) do not fix Equation (15), for example \(p=0.1\), we have \(p_{2}f_{2}(x)<p_{1}f_{1}(x)\) for any \(x\), then the privacy breaching method will only match \(\mathbb{T}^{*}(i,\cdot)\) to the same value \(s_{1}\) in \(\mathbb{S}^{*}(i,\cdot)\), which makes \(s_{2}\) indistinguishable. Next, we discuss how distinguishable \(s_{1}\) and \(s_{2}\) are. Let the area of the blue part be \(A_{b}\), and the area of the orange part be \(A_{o}\). It is obvious that \(A_{b}=p_{1},A_{o}=p_{2}\). The area  
Figure 5: **Schematic diagram of the probability density distribution of \(x\) when \(r=0.9,z=1.0\), \(p=0.5\). The horizontal coordinate axis is \(x\), while the vertical coordinate axis is the probability density. Blue and orange dashed lines are curves of function \(f_{1}(x)\) and \(f_{2}(x)\), respectively, while the blue and orange solid lines are curves of function \(p_{1}f_{1}(x)\) and \(p_{2}f_{2}(x)\), respectively. The areas of the blue and orange regions are integrals of \(f_{1}(x)\) and \(f_{2}(x)\), respectively. **  
where the blue part does not overlap with the orange part is \(A_{b-o}\); similarly, the area where the orange part does not overlap with the blue part is \(A_{o-b}\). As shown in Figure 5, the theoretical minimum error probability is given by the area under the minimum value of the two curves, which is the area of the overlapping area of blue and orange \(A_{b\cap o}\).