Units Recovery Methods in Compositional Data Analysis

Abstract

Compositional data carry relative information. Hence, their statistical analysis has to be performed on coordinates with respect to a log-ratio basis. Frequently, the modeler is required to back-transform the estimates obtained with the modeling to have them in the original units such as euros, kg or mg/liter. Approaches for recovering original units need to be formally introduced and its properties explored. Here, we formulate and analyze the properties of two procedures: a simple approach consisting of adding a residual part to the composition and an approach based on the use of an auxiliary variable. Both procedures are illustrated using a geochemical data set where the original units are recovered when spatial models are applied.

Appendices

Appendix 1

Let f(T) be the function

$$\begin{aligned} f(T)=\frac{T}{\sum _{j=1}^D Gx_j +Gr}, \end{aligned}$$

then it holds

• $f({T})>1$, for any total T. To prove this property, one can use the well-known inequality between the geometric and arithmetic means

  $$\begin{aligned} \sum _{j=1}^D Gx_j +Gr \le \sum _{j=1}^D \left( \frac{1}{n}\sum _{i=1}^n x_{ij}\right) + \frac{1}{n}\sum _{i=1}^n {\text {Res}}_{i} , \end{aligned}$$

  where the equality holds only for a constant series, which is not the case in our context. Therefore,

  $$\begin{aligned} \sum _{j=1}^D Gx_j +Gr < \frac{1}{n} \sum _{i=1}^n \left( \sum _{j=1}^D x_{ij} + {\text {Res}}_{i}\right) = {T}. \end{aligned}$$

• $\lim _{{T} -> +\infty } f({T}) = 1$. For any ${T}>0$, the expression

  $$\begin{aligned} f({T})=\frac{{T}}{\sum _{j=1}^D Gx_j +Gr}=\frac{{T}}{\sum _{j=1}^D Gx_j + \left( \prod _{i=1}^n \left( {T} - \sum _{j=1}^D x_{ij}\right) \right) ^{1/n} }, \end{aligned}$$

  is equal to

  $$\begin{aligned} f({T})=\frac{1}{\frac{\sum _{j=1}^D Gx_j}{{T}} + \left( \prod _{i=1}^n \left( 1 - \frac{\sum _{j=1}^D x_{ij}}{{T}}\right) \right) ^{1/n} }, \end{aligned}$$

  where $\lim _{{T} -> +\infty } \frac{\sum _{j=1}^D Gx_j}{{T}} = 0$ and $\lim _{{T} -> +\infty } \frac{\sum _{j=1}^D x_{ij}}{{T}} = 0$.

• f(T) is a monotonically decreasing function. To prove this behavior, one can prove that the function $g({T})=1/f({T})$ is a monotonically increasing function. The derivative function $g'({T})$ is equal to

  $$\begin{aligned} g'({T})= & {} \frac{\frac{1}{n}\left( \prod _{i=1}^n \left( {T} - \sum _{j=1}^D x_{ij}\right) \right) ^{1/n}\left( \sum _{i=1}^n \frac{1}{{T} - \sum _{j=1}^D x_{ij}} \right) {T}}{\mathrm {T}^2}\\&-\frac{\sum _{j=1}^D Gx_j + \left( \prod _{i=1}^n \left( {T} - \sum _{j=1}^D x_{ij}\right) \right) ^{1/n}}{{T}^2}, \end{aligned}$$

  where using the inequality between the geometric and arithmetic means it holds

  $$\begin{aligned} g'({T})> & {} \frac{\frac{1}{n}\left( \prod _{i=1}^n \left( {T} - \sum _{j=1}^D x_{ij}\right) \right) ^{1/n}\left( \sum _{i=1}^n \frac{1}{{T} - \sum _{j=1}^D x_{ij}} \right) {T}-\mathrm {T}}{{T}^2} \\= & {} \frac{\frac{1}{n}\left( \prod _{i=1}^n \left( {T} - \sum _{j=1}^D x_{ij}\right) \right) ^{1/n}\left( \sum _{i=1}^n \frac{1}{{T} - \sum _{j=1}^D x_{ij}} \right) -1}{{T}} \end{aligned}$$

  Because the term $\prod _{i=1}^n \left( {T} - \sum _{j=1}^D x_{ij}\right) ^{1/n}$ is the geometric mean of the residuals and the term $\frac{1}{n}\left( \sum _{i=1}^n \frac{1}{{T} - \sum _{j=1}^D x_{ij}} \right)$ is the inverse of the harmonic mean of the residuals, the sign of the numerator is positive. Therefore, $g'({T})>0$.

Appendix 2

Let T be a total fixed but as large as we need, like a “big T.” In consequence, for $i=1,2,\ldots ,n$, the residual ${\text {Res}}_i$ is as large as we need, that is ${T}>> \sum _{j=1}^D x_{ij}$. For $i=1,2,\ldots ,n$, it holds that

$$\begin{aligned} \sqrt{\frac{D}{D+1}}\cdot \ln \frac{{{\text {Res}}_i}}{m_i}= & {} \sqrt{\frac{D}{D+1}}\cdot \ln {{{\text {Res}}_i}}-\sqrt{\frac{D}{D+1}}\cdot \ln {m_i} \\= & {} \sqrt{\frac{D}{D+1}}\cdot \ln \left( {{T}- \sum _{j=1}^D x_{j}}\right) -\sqrt{\frac{D}{D+1}}\cdot \ln {m_i} \\= & {} \sqrt{\frac{D}{D+1}}\cdot \ln \left( {T}\cdot \left( 1-\frac{ \sum _{j=1}^D x_{ij}}{{T}}\right) \right) -\sqrt{\frac{D}{D+1}} \cdot \ln {m_i} \\= & {} \sqrt{\frac{D}{D+1}}\cdot \ln {T}+\sqrt{\frac{D}{D+1}}\cdot \ln \left( 1-\frac{ \sum _{j=1}^D x_{ij}}{{T}}\right) \\&-\sqrt{\frac{D}{D+1}}\cdot \ln {m_i}. \end{aligned}$$

In consequence, because ${T}>> \sum _{j=1}^D x_{ij}$, it holds that

$$\begin{aligned} \sqrt{\frac{D}{D+1}}\cdot \ln \frac{{{\text {Res}}_i}}{m_i}\approx \sqrt{\frac{D}{D+1}}\cdot \ln {T}-\sqrt{\frac{D}{D+1}}\cdot \ln {m_i}. \end{aligned}$$