Introduction

Let's assume we have N options A₁, A₂, ..., A_N and look for a linear combination of these options. Every combination is desired to be equally likely. Hence we look for weights for these options. The weights need to sum up to one.

• if we have two options, it's easy: $f = w_1 \cdot A_1 + (c-w_1) \cdot A_2$ where we randomly sample $w_1 \sim \mathcal{U}[0,c]$ and $w_2$ is the remainder ($c-w_1$)
• with more options more complicated
• example: we want to sample soil textures and have the three options ($A_1$) sand content, ($A_2$) silt, and ($A_3$) clay content. The percentage of each option are $w_1$, $w_2$, and $w_3$. They need to sum up to 1. The order of weights should not matter, i.e. a high sand content should be as likely as a high clay or silt content. hence we want to uniformly sample from the sand-silt-clay classification diagram
• one common approach strategy (references) is to sample the first weight $w_1 \sim \mathcal{U}[0,1]$. then sample the next weight in the remaining range $w_2 \sim \mathcal{U}[0,1-w_1]$. The last weight is then set as the remainder $w_3 = 1-w_1-w_2$ to obtain a sum of 1.
• This approach can be extended to any target constant $c$:
  $w_1 \sim \mathcal{U}[0,c]$, $w_2 \sim \mathcal{U}[0,c-w_1]$, $w_3 = c-w_1-w_2$
• And can also be extended to $N$ alternatives and weights:
  $w_1 \sim \mathcal{U}[0,c]$,
  $w_2 \sim \mathcal{U}[0,c-w_1]$,
  $w_3 \sim \mathcal{U}[0,c-w_1-w_2]$,
  $w_4 \sim \mathcal{U}[0,c-w_1-w_2-w_3]$,
  $\ldots$
  $w_N = c-w_1-w_2-\ldots-w_{N-1}$
• The problem is that the distribution functions of the weights are not identical, leading to the fact that the order of the alternatives matters (large values of the first alternative/weight are always more likely than the large weights of the last alternative). See Figure 1 Panels A-C.
• This in turn means for the soil texture example that the sand-silt-clay classification diagram is not sampled uniformly. See Figure 1 Panels D.
• We propose a sampling strategy to obtain N weights for N alternatives that are all identically distributed (see Figure 1 Panels E-G) leading to a uniform sampling of the full domain (See Figure 1 Panels H)

Figure 1: The sampling of three weights w₁, w₂, and w₃ is shown for a naive approach (row 1) and the proposed strategy (row 2). The distribution of the three weights is not identical using the naive method (A-C) but yields the same distribution for the three weights when the proposed method is applied (E-G). The ternary diagrams (D and H) show that the domain is uniformly sampled only when the proposed method is applied. The naive method favoring small weights for weights w₂ and w₃ which leads to a more dense sampling close to the upper edge of the ternary diagram.