Etor A.

In Section II.B of the paper, we claimed that Property 1 and Property 7 are incompatible: there cannot exist a comparison function with both of these properties at the same time. With an illustrative example, we will now show why.

We define the random variables \(X_A,X_{B1}\) and \(X_{B2}\) as

, where \(\mathcal{N}(\mu, \sigma)\) is the normal distribution of mean \(\mu\) and standard deviation \(\sigma\) and \(\tau \in [0,0.6]\) is a real valued parameters. Let \( X_B = \mathcal{M}_{[1-\tau,\tau]}(X_{B1},X_{B2})\) be the mixture distribution of \(X_{B1}\) and \(X_{B2}\), such that the density of the mixture is the weighted average of the densities of \(X_{B1}\) and \(X_{B2}\), with weights \(1-\tau,\tau\) respectively.

When \(\tau = 0\), we have that \(X_B = X_{B1} = \mathcal{N}(0.05025, 0.0015)\) and \(X_A = \mathcal{N}(0.05, 0.0015)\). This means that the cumulative distribution of \(X_B\) will be higher than the cumulative distribution of \(X_A\) in every point, hence, \(X_A \succ X_B\). According to Property 1, in this case, \(\mathcal{C}(X_A,X_B) = 1\). When \(\tau = 0.4\), the cumulative distribution of \(X_B\) is higher than the cumulative distribution of \(X_A\) in every point, and thus, \(X_B \succ X_A\). Now, because of Property 1, \(\mathcal{C}(X_A,X_B) = 0\). But this contradicts Property 7, as \(1 = \left| \mathcal{C}(\mathcal{M}_{[0.6,0.4]}(X_{B1},X_{B2}),X_A) - \mathcal{C}(X_{B1},X_A) \right| \nleq 0.4\)

We have just proven that no comparison function can satisfy Properties 1 and 7 simultaneously. The dominance rate satisfies Property 1, but not Property 7. The probability that \(X_A \lt X_B\) is exactly the opposite: it satisfies Poperty 7, but not Property 1.

In the interactive plot below, we show the dominance relationship between \(X_A\) and \(X_B\), their dominance rate and the probability that \(X_A \lt X_B\) for different \(\tau\) values.