*2.1. Uniparametric Gamma Distributions*

Let *X* be a uniparametric gamma distributed random variable, *<sup>G</sup>*(*α*). That is, its CDF *G* (*x*; *α*) is given by

$$G\left(x;a\right) = \frac{1}{\Gamma\left(1+a\right)} \int\_0^x t^a e^{-t} dt,\tag{6}$$

for *x* > 0, where −1 < *α* < <sup>∞</sup>, and the mode is given by *M* = max {*<sup>α</sup>*, <sup>0</sup>}. Then, for −1 < *α* ≤ 0, the density function decreases on *x* along the positive real line and we obtain that

$$\nu\_G\left(z;\alpha\right) = 1 - G\left(z;\alpha\right) = \frac{1}{\Gamma\left(1+\alpha\right)} \int\_z^\infty t^\alpha e^{-t} dt.$$

In these cases, *νG* (*z*; *α*) is a decreasing function on *z*, and *S*<sup>+</sup> (*G* (*<sup>α</sup>*1)) = *νG* (0; *<sup>α</sup>*1) = 1.

**Proposition 2.** *Let G* (*<sup>α</sup>*1) *and G* (*<sup>α</sup>*2) *be gamma distributions with CDF as in (6). Then:*

*1. If* −1 < *α*1 < *α*2 < 0*, then G* (*<sup>α</sup>*2) ≥*ν G* (*<sup>α</sup>*1). *2.If*0<*α*1<*α*2*,then G* (*<sup>α</sup>*1)≥+*G*(*<sup>α</sup>*2).

 **Proof.**Part1. Wewrite

 can

$$
\nu\_G\left(z; \mathfrak{a}\_1\right) - \nu\_G\left(z; \mathfrak{a}\_2\right) = G\left(z; \mathfrak{a}\_2\right) - G\left(z; \mathfrak{a}\_1\right).
$$

By denoting *α*2 = *α*1 + *ε*, *ε* > 0, and then considering *u* (*z*) = *ddz* [*<sup>ν</sup>G* (*z*; *<sup>α</sup>*1) − *νG* (*z*; *<sup>α</sup>*2)] , we obtain

$$u\left(z\right) = \frac{z^{a\_2}e^{-z}}{\Gamma\left(1+a\_2\right)} - \frac{z^{a\_1}e^{-z}}{\Gamma\left(1+a\_1\right)} = z^{a\_1}e^{-z} \left[\frac{z^{\varepsilon}\Gamma\left(1+a\_1\right) - \Gamma\left(1+a\_2\right)}{\Gamma\left(1+a\_2\right)\Gamma\left(1+a\_1\right)}\right]$$

.

Therefore, *u* (*z*) = 0 when

$$z = z\_0 = \left[\frac{\Gamma\left(1 + \alpha\_2\right)}{\Gamma\left(1 + \alpha\_1\right)}\right]^{1/\varepsilon},$$

*u* (*z*) is negative for 0 < *z* < *z*0, and positive for *z* > *z*0. Also, *νG* (0+; *<sup>α</sup>*1) − *νG* (0+; *<sup>α</sup>*2) = 0, *νG* (∞; *<sup>α</sup>*1) − *νG* (∞; *<sup>α</sup>*2) = 0. Then,

$$\nu\_G\left(z\_0; a\_1\right) - \nu\_G\left(z\_0; a\_2\right) = \frac{1}{\Gamma\left(1 + a\_1\right)\Gamma\left(1 + a\_2\right)} \int\_0^{x\_0} t^{a\_1} e^{-t} \left[\Gamma\left(1 + a\_1\right)t^{\varepsilon} - \Gamma\left(1 + a\_2\right)\right] dt\_{\prime\prime}$$

is the integral of a negative function, so it is negative, and the proof is complete.

Part 2. For 0 < *α* < <sup>∞</sup>, we have that

$$\nu\_G(z;a) = \begin{cases} \frac{1}{\Gamma(1+a)} \left( \int\_{z+a}^{\infty} t^a e^{-t} dt - \int\_0^{a-z} t^a e^{-t} dt \right), & 0 \le z < a, \\\\ \frac{1}{\Gamma(1+a)} \int\_{z+a}^{\infty} t^a e^{-t} dt, & z \ge a. \end{cases}$$

and,

$$\frac{d\nu\_G}{dz} = \begin{cases} \frac{1}{\Gamma(1+\alpha)} \left[ (\alpha - z)^a e^{-(\alpha - z)} - (\alpha + z)^a e^{-(\alpha + z)} \right], & 0 \le z < \alpha, \\\\ \frac{-1}{\Gamma(1+\alpha)} z^a e^{-z}, & z \ge \alpha. \end{cases}$$

Then, clearly we have that *d<sup>ν</sup>G*/*dz* < 0 for all *z* ≥ *α*. For 0 ≤ *z* < *α*, if we denote

$$w\left(z\right) = \left(\mathfrak{a} - z\right)^{\mathfrak{a}} e^z - \left(\mathfrak{a} + z\right)^{\mathfrak{a}} e^{-z},$$

then the sign of *d<sup>ν</sup>G*/*dz* is the sign of *w* (*x*). As *w* (0) = 0, *w* (*α*) = − (<sup>2</sup>*α*)*<sup>α</sup> e*<sup>−</sup>*<sup>α</sup>* < 0, and

$$\frac{dw}{dz} = z \left( z + \alpha \right)^{\alpha - 1} e^{-z} - z \left( \alpha - z \right)^{\alpha - 1} e^z \le 0\_{\prime}$$

we conclude that *νG* is a decreasing function on *z* ≥ 0 and *S*<sup>+</sup> (*G* (*α*)) = *νG* (0; *<sup>α</sup>*).

$$\mathcal{S}^+\left(G\left(\mathfrak{a}\right)\right) = \frac{\Gamma\left(1+\mathfrak{a},\mathfrak{a}\right)}{\Gamma\left(1+\mathfrak{a}\right)},$$

where Γ (1 + *α*, *α*) is the incomplete Gamma function, and then *S*<sup>+</sup> (*G* (*α*)) is a decreasing function on *α*, when *α* → ∞. Nevertheless, a simple plotting of the functionals *νG* (*z*; *<sup>α</sup>i*) for any 0 < *α*1 < *α*2 shows that both functionals cross each other and that they are not ordered by "≥*ν*". Thus, the proof is completed.
