*2.2. Log–Logistic Distributions*

The CDF of a uniparametric log–logistic distributed random variable *X* is given by

$$F\_{LL}\left(\mathbf{x};\theta\right) = \left(1 + \mathbf{x}^{-\theta}\right)^{-1},\tag{7}$$

for *x* > 0, with *θ* > 0. The mode of these distributions depends on *θ*. If 0 < *θ* ≤ 1, then *M* = 0, and

$$\nu\_{LL}\left(z;\theta\right) = \frac{1}{1+z^{\theta}},$$

and *S*<sup>+</sup> (*FLL* (*θ*)) = 1. The functionals *νLL* (*z*; *θ*) for different values of *θ* inside the rank cross each other at *z* = 1, and these distributions are ordered neither by skewness function nor by skewness indexes. Nevertheless, for *θ* > 1, the mode is

$$0 < M = \left(\frac{\theta - 1}{\theta + 1}\right)^{1/\theta} < 1. \tag{8}$$

Notice that *M* is an increasing function of *θ* when *θ* > 1, because

$$\frac{dM}{d\theta} = M \cdot \left[ \frac{2}{\left(\theta^2 - 1\right)\theta} - \frac{1}{\theta^2} \ln \frac{\theta - 1}{\theta + 1} \right] > 0. \tag{9}$$

When *θ* > 1, it is also known from Arnold and Groeneveld (1995) that

$$\nu\_{LL}\left(0; \theta\right) = \frac{1}{\theta}.$$

As *νL*Ł (*z*; *θ*) is a decreasing function, it is then stated that 1 < *θ*1 < *θ*2 implies *FLL* (*<sup>θ</sup>*1) ≥+ *FLL* (*<sup>θ</sup>*2). Furthermore, the skewness functions are ordered, as we prove below.

**Proposition 3.** *Let be FLL* (*<sup>θ</sup>*1) *and FLL* (*<sup>θ</sup>*2) *log-logistic distributions with CDF as in (7), where* 1 < *θ*1 < *θ*2*. Then,*

$$F\_{LL}\left(\theta\_1\right) \ge\_{\nu} F\_{LL}\left(\theta\_2\right). \tag{10}$$

**Proof.** Let *θ* > 1. Then,

$$\nu\_{LL}\left(z;\theta\right) = \begin{cases} \frac{1-\left(M^2-z^2\right)^{\theta}}{1+\left(M+z\right)^{\theta}+\left(M-z\right)^{\theta}+\left(M^2-z^2\right)^{\theta}}, & 0 \le z \le M\_{\theta} \\\\ \frac{1}{1+\left(M+z\right)^{\theta}}, & z > M\_{\theta} \end{cases}$$

If we consider 1 < *θ*1 < *θ*2, such that the respective modes verify 0 < *M*1 < *M*2 < 1, we can then denote

$$a = (M\_1 - z)^{\theta\_1} < b = \left(M\_1 + z\right)^{\theta\_1}.$$

$$c = \left(M\_2 - z\right)^{\theta\_2} < d = \left(M\_2 + z\right)^{\theta\_2}.$$

.

and consider the function *h* given by

$$h\left(\theta\right) = \left(M \pm z\right)^{\theta}, \quad 0 \le z \le M\_r$$

with *M* as in (8). Then,

$$\frac{dh}{d\theta} \quad = \frac{\left(M \pm z\right)^{\theta - 1}}{\theta \left(\theta + 1\right)^{2}} \left(\frac{2\theta}{M^{\theta - 1}} + \theta \left(1 + \theta\right)^{2} \left(M \pm z\right) \ln\left(M \pm z\right) + \left(1 + \theta\right)^{2} M \ln\frac{\theta + 1}{\theta - 1}\right),$$

For *z* < *M*, this implies that *a* < *c*, *b* < *d*. With this notation, we can write *νLL* (*z*; *<sup>θ</sup>*1) − *νLL* (*z*; *<sup>θ</sup>*2) as follows.

Firstly, for 0 ≤ *z* ≤ *M*1,

$$\begin{aligned} \left(\nu\_{LL}\left(z;\theta\_{1}\right)-\nu\_{LL}\left(z;\theta\_{2}\right)\right) &= \frac{1-ab}{\left(1+a\right)\left(1+b\right)} - \frac{1-cd}{\left(1+c\right)\left(1+d\right)} \\ &= \frac{\left(c-a\right)+\left(d-b\right)+ac\left(d-b\right)+bd\left(c-a\right)+2\left(cd-ab\right)}{\left(a+1\right)\left(b+1\right)\left(c+1\right)\left(d+1\right)} > 0. \end{aligned}$$

Secondly, for *M*1 < *z* ≤ *M*2, we only need to compare *d* − *b*, because

$$\begin{aligned} \nu\_{LL} \left( z; \theta\_1 \right) - \nu\_{LL} \left( z; \theta\_2 \right) &=& \frac{1}{1+b} - \frac{1-cd}{\left( 1+c \right) \left( 1+d \right)} \\ &=& \frac{c + (d-b) + 2cd + bcd}{\left( 1+b \right) \left( 1+c \right) \left( 1+d \right)} > 0. \end{aligned}$$

Finally, when *z* > *M*2,

$$\nu\_{LL} \left( z; \theta\_1 \right) - \nu\_{LL} \left( z; \theta\_2 \right) = \frac{1}{1+b} - \frac{1}{1+d} = \frac{d-b}{\left( 1+b \right) \left( 1+d \right)} > 0.1$$

Hence, the proof is completed.

*2.3. Lognormal Variance Distributions*

$$LN\left(x;\sigma\right) = \Phi\left(\frac{\ln x}{\sigma}\right),\tag{11}$$

for *x*, *σ* > 0, where Φ (·) is the standard normal distribution function. The mode is given by *Mσ* = exp −*σ*<sup>2</sup> and

$$\nu\_{LN}\left(z;\sigma\right) = 1 - \Phi\left(\frac{\ln\left[z + \exp\left(-\sigma^2\right)\right]}{\sigma}\right) - \Phi\left(\frac{\ln\left[\exp\left(-\sigma^2\right) - z\right]}{\sigma}\right).$$

**Proposition 4.** *Let LN* (*<sup>σ</sup>*1) *and LN* (*<sup>σ</sup>*2) *be lognormal distributions with CDF as in (11), where* 0 < *σ*1 < *σ*2*. Then,*

$$LN\left(\sigma\_2\right) \ge\_{\nu} LN\left(\sigma\_1\right).$$

**Proof.** For 0 < *σ*1 < *σ*2, the corresponding modes are *M*1 > *M*2, and

$$
\Phi\left(\frac{\ln\left(z+M\_1\right)}{\sigma\_1}\right) > \Phi\left(\frac{\ln\left(z+M\_2\right)}{\sigma\_2}\right),
$$

$$
\Phi\left(\frac{\ln\left(-z+M\_1\right)}{\sigma\_1}\right) > \Phi\left(\frac{\ln\left(-z+M\_2\right)}{\sigma\_2}\right).
$$

because Φ is a strictly increasing function. Thus, we obtain that *νLN* (*z*; *<sup>σ</sup>*1) > *νLN* (*z*; *<sup>σ</sup>*2) for all *z* > 0 and the proof is completed.
