(Ir)relevant T-norm Joint Distributions in the Arithmetic of Fuzzy Quantities

Abstract

Irrelevance, a notion put forward in [5, 12], is a convenient tool to speed up computations in the arithmetic of interactive fuzzy numbers, which were first put forward in the seminal paper [3]. To make our point, below we deal with standard and less standard binary operations for fuzzy quantities whose interactivity is described by a t-norm joint distribution function.

Appendices

Appendix

A Montecatini Lemma

With caveat 2 below, the following lemma basically states that the arithmetic of fuzzy quantities, however unruly, is the very same as for crisp numbers. At an INdAM workshop held in Montecatini, Tuscany, the lemma gave rise to a fruitful discussion; we are tentatively using this odd “attribution” because we were unable to trace back a self-standing and explicit formulation of this result in the literature.

Lemma 4

Montecatini lemma:

$f(x_1, \dots , x_n)=g(x_1, \dots , x_n)$ is an identity for crisp numbers iff the two fuzzy quantities $Z_1 \dot{=} f (X_1, \dots , X_n)$ and $Z_2 \dot{=} g(X_1, \dots , X_n)$ are deterministically equal whatever the joint distribution of $X_1, \dots , X_n$.

E.g., since $x(y+z) = xy + xz$ for any crisp numbers x, y and z, one has $X(Y+Z) = XY + XZ$ for any fuzzy quantities X, Y and Z, whatever their joint distribution. Since $\log xy = \log x + \log y$ for positive x and y one has $\log XY = \log X + \log Y$ for any fuzzy quantities X and Y whatever their joint distribution with positive support. To prove the equidistribution of $Z_1$ and $Z_2$ just observe that one is taking the supremum of the same function over two sets, $\{ \underline{x} : \, f(\underline{x}) = z\}$ and $\{ \underline{x} : \, g(\underline{x}) = z\}$, which are however equal. As for deterministic equality $Z_1 = Z_2$, one cannot have $f(\underline{x}) \not = g(\underline{x})$, and so the joint distribution of the fuzzy couple $[Z_1 , Z_2]$ is zero outside the main diagonal $z_1 = z_2$.

Caveat 1: in more traditional literature on fuzzy arithmetic, where in practice only non-interactivity is used to “glue together” marginal distributions, one comes across statements as are $X \times X \not = X^2$ or $X(Y+Z) \not = XY + XZ$. What is meant is that equidistribution between $X_1$ and $X_2$ does not imply equidistribution between $X_1 \times X_2$ and $X_1^2$ or between $X_1(Y+Z)$ and $X_1Y + X_2Z$. We insist that in our approach a single symbol X is used for two fuzzy numbers only when they are not only equidistributed but also deterministically equal. Note that the analogue of Montecatini lemma holds also for random numbers, i.e. for random variables as used in probabilistic distribution calculus, where one takes much care to distinguish between equidistribution and deterministic equality, the latter being a very special form of joint probability. We recall that non-interactivity of fuzzy quantities is seen as an analogue of probabilistic independence between random variables.

Caveat 2: The lemma is stated in terms of arbitrary fuzzy quantities: if one insists on certain properties, e.g. upper continuity or unimodality, one should of course check whether the result $Z = f (X_1, \dots , X_n) = g(X_1, \dots , X_n)$ still verifies those properties, so as to ensure stability.

B Computations for Examples 5 and 6

We shortly cover the very familiar case of non-interactivity as in Example 5, when the t-norm is $x \wedge y$. We assume that X(x) and Y(y) are continuous and strictly increasing on [0,1] and $[0,\alpha ]$, respectively, $\alpha \ge 1$, cf. Lemma 3; so $Y(z-x)$ is strictly decreasing when seen as a functions of x. For fixed z in the interval $[0, 1 + \alpha ]$, one soon checks that the equation in x $X(x)= Y(z-x)$ has a single solution $\mu (z)$, $\, X\big ( \mu (z) \big ) = Y\big ( z - \mu (z) \big )$, and that $\mu (z)$ strictly increases in z from $\mu (0)=0$ to $\mu (1 + \alpha )=1$. One has three cases, cf. (3); in the first, $x \in [0,z]$, fix $z \in [0,1]$: on the border $x=0$ the increasing function X(x) and the decreasing function $Y(z-x)$ take the two values $0=X(0) < Y(z)$, while on the border $x=z$ they take the two values $X(z) > Y(0)=0$: the required maximum is found for $x=\mu (z)$. The remaining two cases are dealt in the same way and give the same solution $Z(z) = X\big (\mu (z)\big )= Y \big ( z - \mu (z) \big )$, which is found also when X(x) and Y(y) are both strictly decreasing. If the support-intervals are different just use Lemma 3. In Example 5 one has $\alpha =1$; since $X(x)=\sqrt{x}$, $Y(y)=y$, one gets $\mu (z) = z + {1 \over 2} \big (1 - \sqrt{4z + 1} \big )$, and so $Z(z) =Y \big ( z - \mu (z) \big ) = {1 \over 2} (\sqrt{4z + 1} -1)$.

As for Example 6, for fixed z, $z \le -1$, one has to maximize w.r. to x the minimum of two functions $\, x \wedge {-1 \over xz}$, the first increasing, the second decreasing, over the intersection $\{x: \, 0 \le x \le 1\} \cap \{x: \, 0 \le {-1 \over xz} \le 1\}$ i.e. over $\{x: \, -z^{-1} \le x \le 1 \}$. The two functions meet at $\mu (z) = \sqrt{-z^{-1}}$, $-z^{-1} \le \mu (z) \le 1 \}$.