## 2 Materials and Methods
### Derivation of the pdf Assigned to True Analyte Concentration in the Aliquot  
The analyte concentration in an aliquot is the ratio \(\mathbf{R}\) = \(\mathbf{Y}\)/\(\mathbf{X}\). The cumulative distribution of \(\mathbf{R}\) is [31]:  
\[\mathbf{F}_{\mathbf{R}}(\mathbf{r}_{\mathbf{o}})=\iint_{(\mathbf{x},\mathbf{y}|\mathbf{y}/\mathbf{x}\leq\mathbf{r }_{\mathbf{o}})}\mathbf{f}_{\mathbf{X}}(\mathbf{x})\cdot\mathbf{f}_{\mathbf{Y}}(\mathbf{y})\cdot\mathbf{dx} \cdot\mathbf{dy} \tag{3}\]  
where \(\mathbf{f}_{\mathbf{X}}(\mathbf{x})\) and \(\mathbf{f}_{\mathbf{Y}}(\mathbf{y})\) are the **pdfs** assigned to the variables \(\mathbf{X}\) and \(\mathbf{Y}\), respectively. Based on Equation (3), the \(\mathbf{pdf}\) of \(R,f_{\mathbf{R}}(\mathbf{r})\), is derived as follows [31]:  
\[\mathbf{f}_{\mathbf{R}}(\mathbf{r})=\int_{(\mathbf{u})}\mathbf{f}_{\mathbf{X}}(\mathbf{u})\cdot\mathbf{f}_\mathbf{{Y}}(\mathbf{u})\cdot\mathbf{u}\cdot\mathbf{du} \tag{4}\]  
The upper limit of integration in Equation (4) depends on the \(\mathbf{r}\) value, as shown in Figure 1; i.e., for \(\mathbf{r<b/(m+a)}\), \(\mathbf{u}\) belongs to the interval \([m-\mathbf{a};\mathbf{m}+\mathbf{a}]\). In cases where \(\mathbf{r}\) ranges from \(b/(m+a)\) to \(b/(m-a)\), (green line in Figure 1), the upper limit of integration is \(\mathbf{b/r}\). Therefore, \(\mathbf{f}_{\mathbf{R}}(\mathbf{r})\) is a branched function defined on the intervals \([0;\,b/(m+a)]\), \([b/(m+a);\,b/m]\), and \([b/m;\,b/(m-a)]\). The value of the \(\mathbf{R}\) variable assigned to the analyte concentration in a specimen ranges from \(0\) to \(\mathrm{b/(m-a)}\). The range \([0,\,\mathrm{b/(m-a)}]\) is divided into 3 intervals--\([0;\,b/(m+a)]\), \([b/(m+a);\,b/m]\), and \([b/m;\,b/(m-a)]\)--because of the different ways of superposing between uniform distribution and triangular distribution. In the range \([0;\,b/(m+a)]\), \(\mathbf{R}\) intersects the entire distribution of \(\mathbf{X}\), while in the intervals \([b/(m+a);\,b/m]\) and \([b/m;\,b/(m-a)]\), \(\mathbf{R}\) only intersects a part of the triangular distribution of \(\mathbf{X}\), as shown in Equations (5)-(7). In cases where \(\mathbf{r}\in[0;b/(m+a)]\), the \(f_{R}(\mathbf{r})\) is calculated as follows:  
\[f_{\mathbf{R1}}(\mathbf{r})=\int_{m-a}^{m+a}f_{X}(\mathbf{u})\cdot\frac{\mathbf{1}}{b}\cdot\mathbf{u} \cdot d\mathbf{u}=\frac{\mathbf{1}}{b}\mathbf{m} \tag{5}\]  
Equation (5) shows that \(\mathbf{R}\) is uniformly distributed for \(r\in[0;b/(m+a)]\). In cases where \(\mathbf{r}\in[b/(m+a);b/m]\), the second branch of the \(\mathbf{pdf}\) of \(\mathbf{R}\), denoted as \(f_{\mathbf{R2}}(\mathbf{r})\), is calculated as follows:  
\[f_{\mathbf{R2}}(\mathbf{r})=\int_{m-a}^{m}\frac{\mathbf{u}-\mathbf{m}+a}{a^{2}}\cdot\frac\mathbf{{1}}{b}\cdot\mathbf{u}\cdot d\mathbf{u}+\int_{m}^{b/r}\frac{\mathbf{m}+a-\mathbf{u}}{a^{2}} \cdot\frac{\mathbf{1}}{b}\cdot\mathbf{u}\cdot d\mathbf{u} \tag{6}\]  
Respectively,  
\[f_{\mathbf{R2}}(\mathbf{r})=\frac{\mathbf{b}(\mathbf{m}+\mathbf{a})}{2\cdot a^{2}}\cdot\frac\mathbf{{1}}{\mathbf{r}^{2}}-\frac{\mathbf{b}^{2}}{3\cdot a^{2}}\cdot\frac{\mathbf{1}}{\mathbf{r}^{3} }+\frac{6a^{2}\cdot\mathbf{m}-(\mathbf{m}+\mathbf{a})^{3}}{6a^{2}b} \tag{7}\]  
In cases where \(\mathbf{r}\in[b/m;b/(m-a)]\), the third branch of the pdf of \(R\), denoted as \(f_{\mathbf{R3}}(\mathbf{r})\), is calculated as follows:  
\[f_{\mathbf{R3}}(\mathbf{r})=\int_{m-a}^{b/r}\frac{\mathbf{u}-\mathbf{m}+\mathbf{a}}{a^{2}}\cdot \frac{\mathbf{1}}{b}\cdot\mathbf{u}\cdot d\mathbf{u}=\frac{\mathbf{b}^{2}}{3\cdot a^{2}}\frac{ \mathbf{1}}{\mathbf{r}^{3}}-\frac{\mathbf{b}(\mathbf{m}-\mathbf{a})}{2\cdot a^{2}}\cdot\frac{\mathbf{ 1}}{\mathbf{r}^{2}}+\frac{(\mathbf{m}-\mathbf{a})^{3}}{6a^{2}b} \tag{8}\]  
The true analyte concentration is calculated as the mean of R, denoted by \(\overline{\mathbf{r}}\), i.e.,  
\[