**Appendix A**

**Proof of Proposition 1 (cdf in the FBS model).** From Equation (7), *T* is a monotonically increasing function of *Z* ∼ *FSN*(*<sup>δ</sup>*, *<sup>λ</sup>*). Therefore the cdf of *T* is given by

$$F\_T(t) = F\_Z(a\_t) \tag{A1}$$

where *FZ*(·) denotes the cdf of *Z* and *at* was given in (6).

(i) First, we obtain the cdf of *Z* ∼ *FSN*(*<sup>δ</sup>*, *λ*)

It can be seen in Gómez et al. [2], Proposition 4, that the pdf of *Z* ∼ *FSN*(*<sup>δ</sup>*, *λ*) is

$$f\_Z(z) = \begin{cases} c\_\delta \Phi(z-\delta)\Phi(\lambda z), & \text{if } z < 0\\ c\_\delta \Phi(z+\delta)\Phi(\lambda z), & \text{if } z \ge 0. \end{cases}$$

Let us consider the case for *z* < 0

$$F\_Z(z) = \int\_{-\infty}^{z} f\_Z(t)dt = \int\_{-\infty}^{z} c\_\delta \Phi(t-\delta)\Phi(\lambda t)dt$$

By making the change of variable *v* = *t* − *δ*, and later, taking into account that <sup>Φ</sup>(·) is the cdf of a *N*(0, 1) distribution, we have that

$$F\_{\mathbb{Z}}(z) = c\_{\delta} \int\_{-\infty}^{z-\delta} \phi(v) \Phi(\lambda(v+\delta)) dv = c\_{\delta} \int\_{-\infty}^{z-\delta} \int\_{-\infty}^{\lambda(v+\delta)} \phi(v) \phi(s) ds dv \tag{A2}$$

The integrand in (A2) is the joint pdf of two independent *N*(0, 1) rv's, (*<sup>S</sup>*, *<sup>V</sup>*), i.e.,

$$
\left( \begin{array}{c} S \\ V \end{array} \right) \sim N\_2 \left( \left( \begin{array}{c} 0 \\ 0 \end{array} \right) , \left( \begin{array}{c} 1 & 0 \\ 0 & 1 \end{array} \right) \right)
$$

Note that (A2) can be rewritten as

$$\begin{split} F\_{\mathcal{Z}}(z) &= c\_{\delta} \Pr\left[S - \lambda V \le \lambda \delta, \ V \le z - \delta\right] \\ &= c\_{\delta} \Pr\left[\frac{S - \lambda V}{\sqrt{1 + \lambda^2}} \le \frac{\lambda \delta}{\sqrt{1 + \lambda^2}}, \ V \le z - \delta\right] \\ &= c\_{\delta} \Phi\_{\text{BN}\_{\lambda}}\left(\frac{\lambda \delta}{\sqrt{1 + \lambda^2}}, z - \delta\right) \end{split} \tag{A3}$$

where <sup>Φ</sup>*BNλ* (*<sup>x</sup>*, *y*) denotes the cdf of a bivariate normal distribution, with mean vector *μ* = (0, 0) and covariance matrix Ω*λ* given in (10).

For *z* > 0, we have that

$$F\_{\mathbf{Z}}(\mathbf{z}) = \int\_{-\infty}^{0} f\_{\mathbf{Z}}(\mathbf{t})d\mathbf{t} \, \, \, \int\_{0}^{z} f\_{\mathbf{Z}}(\mathbf{t})d\mathbf{t}$$

$$= F\_{\mathbf{Z}}(0) + \mathbf{c}\_{\delta} \int\_{0}^{z} \phi(\mathbf{t} + \delta)\Phi(\lambda \mathbf{t})d\mathbf{t} \,. \tag{A4}$$

*Symmetry* **2019**, *11*, 1305

> From (A3), it follows that

$$F\_{\mathbb{Z}}(0) = \lim\_{z \to 0^{-}} F\_{\mathbb{Z}}(z) \;= \; c\_{\delta} \Phi\_{\text{BN}\_{\lambda}} \left( \frac{\lambda \delta}{\sqrt{1 + \lambda^2}}, -\delta \right) \tag{A5}$$

On the other hand, proceeding similarly to the previous case (change of variable *v* = *t* + *δ*), it can be proved that

$$\begin{split} \int\_{0}^{z} \Phi(t+\delta)\Phi(\lambda t)dt &= Pr\left[\frac{S-\lambda V}{\sqrt{1+\lambda^{2}}} \le -\frac{\lambda \delta}{\sqrt{1+\lambda^{2}}}, \ \delta < V \le z+\delta \right] \\ &= \Phi\_{\text{BN}\_{\lambda}}\left(-\frac{\lambda \delta}{\sqrt{1+\lambda^{2}}}, z+\delta\right) - \Phi\_{\text{BN}\_{\lambda}}\left(-\frac{\lambda \delta}{\sqrt{1+\lambda^{2}}}, \delta\right) \end{split} \tag{A6}$$

Therefore, from (A3)–(A6), we have just proved that the cdf of *Z* is

$$F\_{\mathcal{Z}}(\mathbf{z}) = \begin{cases} \ c\_{\delta} \boldsymbol{\Phi}\_{\mathrm{BN}\_{\mathrm{l}}} \left( \frac{\lambda \delta}{\sqrt{1+\lambda^{2}}}, \boldsymbol{z} - \delta \right), & \text{if } \boldsymbol{z} < \mathbf{0} \\\\ c\_{\delta} \left[ \boldsymbol{\Phi}\_{\mathrm{BN}\_{\mathrm{l}}} \left( \frac{\lambda \delta}{\sqrt{1+\lambda^{2}}}, -\delta \right) + \boldsymbol{\Phi}\_{\mathrm{BN}\_{\mathrm{l}}} \left( -\frac{\lambda \delta}{\sqrt{1+\lambda^{2}}}, \boldsymbol{z} + \delta \right) - \boldsymbol{\Phi}\_{\mathrm{BN}\_{\mathrm{l}}} \left( -\frac{\lambda \delta}{\sqrt{1+\lambda^{2}}}, \delta \right) \right], & \text{if } \boldsymbol{z} \ge \mathbf{0} \end{cases} \tag{A7}$$

(ii) Finally, the expression for the cdf of *T* ∼ *FBS*(*<sup>α</sup>*, *β*, *δ*, *λ*) given in (9) follows from (A1) and (A7).

**Proof of Proposition 3 (Modes in the FBS model).** Recall that, from (A1), the pdf of *T* is given by

$$f\_T(t) = f\_Z(a\_t)a\_t' = c\_\delta a\_t' \Phi(|a\_t| + \delta)\Phi(\lambda a\_t)$$

where *fZ*(·) denotes the pdf of *Z* ∼ *FSN*(*<sup>δ</sup>*, *<sup>λ</sup>*), *at* was given in (6) and

$$a\_t' = \frac{\partial}{\partial t} a\_t = \frac{t^{-3/2} \beta^{-1/2}}{2a} (t + \beta) \,. \tag{A8}$$

For *at* < 0, or equivalently 0 < *t* < *β*, consider the first derivative with respect to *t* of *fT*(·) and equating to zero, we have

$$\{f\_T^{\prime}(t) = c\_\delta \frac{\partial}{\partial t} \left\{a\_t^{\prime} \phi(-a\_t + \delta) \Phi(\lambda a\_t)\right\} = 0\tag{A9}$$

By using that *φ* (*z*) = <sup>−</sup>*<sup>z</sup>φ*(*z*), it can be proved that (A9) is equivalent to

$$\left\{ \left\{ a\_t' \right\}^2 \left[ (\delta - a\_t) \Phi(\lambda a\_t) + \lambda \phi(\lambda a\_t) \right] + a\_t'' \Phi(\lambda a\_t) = 0 \tag{A10}$$

Since *a t* > 0, ∀*t* > 0 (*β* > 0), we have that (A10) is equivalent to (12). Similarly, for *at* > 0, i.e., *t* > *β*, from *f T*(*t*) = *cδ ∂∂t* {*<sup>a</sup> tφ*(*at* + *<sup>δ</sup>*)Φ(*<sup>λ</sup>at*)} = 0,(13) is obtained.

**Remark A1** (Comments to Proposition 3)**.** *In order to illustrate the use of Equations (12) and (13) next cases are considered.*


**Proof of Theorem 1 (***pth* **quantile, change of scale and reciprocity).** (i) (14) follows from the fact that (7) is one-to-one function preserving the order from R to R+.

(ii) Note that the pdf of *T* can be rewritten as

$$f\_T(t) = c\_\delta a\_t'(\mathfrak{a}, \beta) \phi(|a\_l(\mathfrak{a}, \beta)| + \delta) \Phi(\lambda a\_l(\mathfrak{a}, \beta))\tag{A11}$$

with *cδ* = (1 − <sup>Φ</sup>(*δ*))−1, *at* = *at*(*<sup>α</sup>*, *β*) given in (6) and

$$a\_t' = a\_t'(a, \beta) = \frac{\partial}{\partial t} a\_t(a, \beta) = \frac{t^{-3/2} \beta^{-1/2}}{2a} (t + \beta) \,. \tag{A12}$$

Let *Y* = *kT* with *k* > 0. By applying the Jacobian technique *fY*(*y*) = |*J*| *fT*( *yk* ; *α*, *β*, *δ*, *λ*) with |*J*| = 1*k* . From (6), *ay*/*<sup>k</sup>*(*<sup>α</sup>*, *β*) = *ay*(*<sup>α</sup>*, *kβ*), and from (A12)

$$|J|a\_{y/k}'(\mathfrak{a}, \mathfrak{z} \mathfrak{z}) = \frac{y^{-3/2} (k \mathfrak{z} \mathfrak{z})^{-1/2}}{2a} (y + k \mathfrak{z} \mathfrak{z}) = a\_y'(\mathfrak{a}, k \mathfrak{z}) \text{ .}$$

Therefore

$$f\_Y(y) = c\_\delta a'\_y(\alpha, k\beta) \phi(|a\_y(\alpha, k\beta)| + \delta) \Phi(\lambda a\_y(\alpha, k\beta)),$$

i.e., *Y* ∼ *FBS*(*<sup>α</sup>*, *kβ*, *δ*, *<sup>λ</sup>*).

(iii) Let be *Y* = *T*−1. In this case |*J*| = *<sup>Y</sup>*−2, *ay*−<sup>1</sup> (*<sup>α</sup>*, *β*) = <sup>−</sup>*ay*(*<sup>α</sup>*, *β*−<sup>1</sup>), and |*J*|*<sup>a</sup> <sup>y</sup>*−<sup>1</sup>(*<sup>α</sup>*, *β*) = *<sup>a</sup> y*(*<sup>α</sup>*, *β*−<sup>1</sup>). Therefore

$$\begin{aligned} f\_Y(y) &= |\boldsymbol{\beta}| f\_T(y^{-1}; a, \boldsymbol{\beta}, \boldsymbol{\delta}, \lambda) = c\_{\boldsymbol{\delta}} a'\_{\boldsymbol{\beta}}(a, \boldsymbol{\beta}^{-1}) \boldsymbol{\phi} \left( |a\_{\boldsymbol{\beta}}(a, \boldsymbol{\beta}^{-1})| + \boldsymbol{\delta} \right) \boldsymbol{\Phi} \left( -\lambda a\_{\boldsymbol{\beta}}(a, \boldsymbol{\beta}^{-1}) \right), \\ \boldsymbol{\Upsilon} &= T^{-1} \sim F \rm{BS}(a, \boldsymbol{\beta}^{-1}, \boldsymbol{\delta}, -\lambda). \end{aligned}$$
