**4. Theory**

We now develop two propositions on both density and distribution functions for the quotient *Y* := *X*1 *X*2 and the ratio of one variable over the sum of two variables *Z* := *X*1 *X*1+*X*2 of two dependent random variables *X*1 and *X*2 by using copulas. We first develop the proposition on the density and distribution functions for the quotient *Y* = *X*1 *X*2as stated in the following:

**Proposition 1.** *Supposing that* (*<sup>X</sup>*1, *<sup>X</sup>*2) *is a vector of two absolutely continuous random variables X*1 *and X*2 *with the marginal distributions F*1 *and F*2*, respectively, let C be an absolutely continuous copula modeling dependence structure of the random vector* (*<sup>X</sup>*1, *<sup>X</sup>*2)*, and define Y as*

$$Y := \frac{X\_1}{X\_2}.\tag{7}$$

*Then, the density fY*(*y*) *and distribution functions FY*(*y*) *of Y are*

$$f\_Y(y) = \int\_0^1 |F\_2^{-1}(v)| c\left(F\_1\left(yF\_2^{-1}(v)\right), v\right) f\_1\left(yF\_2^{-1}(v)\right) dv,\tag{8}$$

$$F\_Y(y) \quad = \; F\_2(0) + \int\_0^1 \text{sgn}\left(F\_2^{-1}(v)\right) \frac{\partial}{\partial v} \mathcal{C}\left(F\_1\left(y F\_2^{-1}(v)\right), v\right) dv,\tag{9}$$

*respectively, where F*−<sup>1</sup> 2 *denotes the inverse function of F*2*, c is the density of copula C, and* sgn(·) *stands for a sign function such that*

$$\text{sgn}(\mathbf{x}) = \begin{cases} 1, & \text{if} \quad \mathbf{x} > 0, \\ -1, & \text{if} \quad \mathbf{x} < 0. \end{cases}$$

**Proof.** Letting

$$\begin{cases} \mathbf{y}\_1 := \frac{\mathbf{X}\_1}{\mathbf{X}\_2}, \\ \mathbf{y}\_2 := \mathbf{X}\_2. \end{cases}$$

We note that since *X*2 is absolutely continuous, *<sup>P</sup>*(*<sup>X</sup>*2 = 0) = 0; that is, *X*2 = 0 almost surely. Hence, the transformation *Y*1 = *X*1 *X*2 always exists with probability 1 and we can obtain its inverse transformation by using

$$\begin{cases} \mathcal{X}\_1 = \mathcal{Y}\_1 \mathcal{Y}\_{2\nu} \\ \mathcal{X}\_2 = \mathcal{Y}\_{2\nu} \end{cases}$$

and their corresponding Jacobian

$$J = \begin{vmatrix} \chi\_2 & \chi\_1 \\ 0 & 1 \end{vmatrix} = \chi\_2.$$

Then, we obtain the joint density of *Y*1 and *Y*2 such that

$$\begin{aligned} \left| h(y\_1, y\_2) \right| &= \left| f(y\_1 y\_2, y\_2) \, |y\_2| \right| \\ &= \left| y\_2 \right| c \left( F\_1(y\_1 y\_2), F\_2(y\_2) \right) f\_1(y\_1 y\_2) f\_2 \left( y\_2 \right) \, . \end{aligned}$$

This yields the density of *Y*1:

$$f\_{Y\_1}(y\_1) = \int\_{-\infty}^{\infty} |y\_2| \ c\left(F\_1(y\_1 y\_2), F\_2(y\_2)\right) f\_1(y\_1 y\_2) f\_2\left(y\_2\right) dy\_2 \tag{10}$$

$$=\int\_{0}^{1} \left| F\_{2}^{-1}(v) \right| c \left( F\_{1} \left( y\_{1} F\_{2}^{-1}(v) \right), v \right) f\_{1} \left( y\_{1} F\_{2}^{-1}(v) \right) dv. \tag{11}$$

As a result, the CDF of *Y*1 is determined by

$$F\_{Y\_1}(t) = \int\_0^1 \int\_{-\infty}^t \left| F\_2^{-1}(v) \right| c\left( F\_1\left(y\_1 F\_2^{-1}(v)\right), v \right) f\_1\left(y\_1 F\_2^{-1}(v)\right) dy\_1 dv. \tag{12}$$

By changing variable, *u* = *F*1 *<sup>y</sup>*1*F*−<sup>1</sup> 2 (*v*), we ge<sup>t</sup> *du* = *F*−<sup>1</sup> 2 (*v*) *f*1 *<sup>y</sup>*1*F*−<sup>1</sup> 2 (*v*) *dy*1 and we note that

$$F\_2^{-1}(v) \ge 0 \Longleftrightarrow v \in [0, F\_2(0)], \text{ and } F\_2^{-1}(v) \le 0 \Longleftrightarrow v \in [F\_2(0), 1].$$

This yields

$$\begin{split} F\_{Y\_1}(t) &= \ &-\int\_0^{F\_2(0)} \int\_1^{F\_1(tF\_2^{-1}(v))} \frac{\partial^2}{\partial u \partial v} \mathcal{C}\left(u,v\right) du dv + \int\_{F\_2(0)}^1 \int\_0^{F\_2(tF\_1^{-1}(v))} \frac{\partial^2}{\partial u \partial v} \mathcal{C}\left(u,v\right) dv du \\ &= &-\int\_0^{F\_2(0)} \left[\frac{\partial}{\partial v} \mathcal{C}\left(F\_1\left(tF\_2^{-1}(v)\right),v\right) - \frac{\partial}{\partial v} \mathcal{C}\left(1,v\right)\right] dv + \int\_{F\_2(0)}^1 \frac{\partial}{\partial v} \mathcal{C}\left(F\_1\left(tF\_2^{-1}(v)\right),v\right) dv \\ &= &-F\_2(0) + \int\_0^1 \text{sgn}\left(F\_2^{-1}(v)\right) \frac{\partial}{\partial v} \mathcal{C}\left(F\_1\left(tF\_2^{-1}(v)\right),v\right) dv. \end{split} \tag{13}$$

Thus, the assertions of Proposition 1 hold.

From Proposition 1 and applying Equation (8), we obtain the following corollary on both density and distribution functions for the quotient *Y* = *X*1 *X*2 of two independent random variables *X*1 and *X*2 by using copulas:

**Corollary 1.** *When X*1 *and X*2 *are independent, then its copula <sup>C</sup>*(*<sup>u</sup>*, *v*) = *uv has the density <sup>c</sup>*(*<sup>u</sup>*, *v*) = 1, ∀*<sup>u</sup>*, *v* ∈ I *and the density fY*(*y*) *of the ratio Y* := *X*1 *X*2*of two independent random variables becomes*

$$f\_Y(y) = \int\_{-\infty}^{\infty} |x| f\_1(xy) f\_2(x) dx.$$

This result is well known in the literature.

Next, we apply Equation (9) to derive both density and distribution function for *Y* = *X*1 *X*2 in case *X*1 and *X*2 are normal random variables and their dependence structure is captured by Gaussian Copulas. We first obtain the following corollary:

**Corollary 2.** *Assume that X*1 ∼ *<sup>N</sup>*(*μ*1, *σ*21 )*, X*2 ∼ *<sup>N</sup>*(*μ*2, *σ*22 ) *and* (*<sup>X</sup>*1, *<sup>X</sup>*2) *follows Gaussian Copulas Cr*(*<sup>u</sup>*, *<sup>v</sup>*), |*r*| < 1, *given in* (46)*. Then, the density fY*(*y*) *and distribution function FY*(*y*) *of Y* = *X*1 *X*2 *have the forms*

$$f\_{\mathcal{Y}}(y) = \int\_0^1 \text{sgn}\left(\sigma\_2 \Phi^{-1}(v) + \mu\_2\right) \rho\left(\frac{(y\sigma\_2 - rv\_1)\Phi^{-1}(v) + y\mu\_2 - \mu\_1}{\sigma\_1\sqrt{1 - r^2}}\right) \frac{\sigma\_2 \Phi^{-1}(v) + \mu\_2}{\sigma\_1\sqrt{1 - r^2}} dv,\tag{14}$$

$$F\_Y(y) = -\Phi\left(-\frac{\mu\_2}{\sigma\_2}\right) + \int\_0^1 \text{sgn}\left(\sigma\_2 \Phi^{-1}(v) + \mu\_2\right) \Phi\left(\frac{(y\sigma\_2 - r\sigma\_1)\Phi^{-1}(v) + y\mu\_2 - \mu\_1}{\sigma\_1\sqrt{1 - r^2}}\right) dv,\tag{15}$$

*respectively, where ϕ*(*x*) *and* <sup>Φ</sup>(*x*) *are PDF and CDF of the standard normal distribution, respectively, and* <sup>Φ</sup>−<sup>1</sup>(*x*) *denotes for the inverse function of* <sup>Φ</sup>(*x*)*.*

**Proof.** Let *X*1 ∼ *<sup>N</sup>*(*μ*1, *σ*21 ), and *X*2 ∼ *<sup>N</sup>*(*μ*2, *σ*22 ); then, their CDFs and inverse functions can be expressed in the following form:

$$F\_i(x) = \Phi\left(\frac{x - \mu\_i}{\sigma\_i}\right), \quad F\_i^{-1}(v) = \sigma\_i \Phi^{-1}(v) + \mu\_{i\prime} \quad i = 1, 2.1$$

Given Gaussian Copulas *Cr*(*<sup>u</sup>*, *v*) with |*r*| < 1, one can obtain its derivative *∂Cr ∂v* (*<sup>u</sup>*, *<sup>v</sup>*), see Meyer (2013), as shown in the following:

$$\frac{\partial C\_r}{\partial \upsilon}(\mu, \upsilon) = \Phi \left( \frac{\Phi^{-1}(\mu) - r\Phi^{-1}(\upsilon)}{\sqrt{1 - r^2}} \right).$$

Now, applying Equation (9), we can simplify it to be

$$\begin{split} F\_{Y}(y) &=& \Phi\left(-\frac{\mu\_{2}}{\sigma\_{2}}\right) + \int\_{0}^{1} \operatorname{sgn}\left(\sigma\_{2}\Phi^{-1}(v) + \mu\_{2}\right) \Phi\left(\frac{\Phi^{-1}\left(\Phi\left(\frac{y\varphi\_{2}\Phi^{-1}(v) + y\mu\_{2} - \mu\_{1}}{\sigma\_{1}}\right)\right) - r\Phi^{-1}(v)}{\sqrt{1 - r^{2}}}\right) d\mu \\ &=& \Phi\left(-\frac{\mu\_{2}}{\sigma\_{2}}\right) + \int\_{0}^{1} \operatorname{sgn}\left(\sigma\_{2}\Phi^{-1}(v) + \mu\_{2}\right) \Phi\left(\frac{(y\varphi\_{2} - rv\_{1})\Phi^{-1}(v) + y\mu\_{2} - \mu\_{1}}{\sigma\_{1}\sqrt{1 - r^{2}}}\right) dv. \end{split}$$

Taking derivative of *Fy*(*y*) with respect to *y*, one gets the density *fY*(*y*) defined as in (14). The assertions of Corollary 2 hold.

We note that the probability exhibited in (15) can be easily computed by using the following Monte Carlo algorithm: For each *y* ∈ R, we generate *V* from the uniform distribution on the unit interval [0, 1] with sample size *N*, say *N* = 10,000, and then the estimated probability is given by

$$\hat{F}\_y(y) \approx \Phi\left(-\frac{\mu\_2}{\sigma\_2}\right) + \frac{1}{N} \sum\_{i=1}^N \text{sgn}\left(\sigma\_2 \Phi^{-1}(v\_i) + \mu\_2\right) \Phi\left(\frac{(y\sigma\_2 - r\sigma\_1)\Phi^{-1}(v\_i) + y\mu\_2 - \mu\_1}{\sigma\_1\sqrt{1 - r^2}}\right). \tag{16}$$

Using the result, we obtain the following corollary:

**Corollary 3.** *If X*1 ∼ *<sup>N</sup>*(*μ*1, *σ*21 )*, X*2 ∼ *N*(0, *σ*22 )*, and* (*<sup>X</sup>*1, *<sup>X</sup>*2) *follows Gaussian Copulas Cr*(*<sup>u</sup>*, *<sup>v</sup>*)*,* |*r*| < 1, *given in* (46)*, then the median of Y* = *X*1 *X*2*satisfies*

$$median(\boldsymbol{\chi}) = r \frac{\sigma\_1}{\sigma\_2}, \quad \text{for all } \mu\_1 \in \mathbb{R}. \tag{17}$$

**Proof.** Since *X*2 ∼ *N*(0, *σ*22 ), <sup>Φ</sup>(−*μ*<sup>2</sup> *σ*2 ) = Φ(0) = 0.5. Hence, it is sufficient to prove that the integral term given in (15) is equal to zero. In fact, we find that

$$\begin{split} F\_{Y} \Big( r \frac{\sigma\_{1}}{\sigma\_{2}} \Big) &=& 0.5 + \int\_{0}^{1} \operatorname{sgn}(\sigma\_{2} \Phi^{-1}(v)) \Phi \Big( \frac{-\mu\_{1}}{\sigma\_{1} \sqrt{1 - r^{2}}} \Big) dv \\ &=& 0.5 + \Phi \Big( \frac{-\mu\_{1}}{\sigma\_{1} \sqrt{1 - r^{2}}} \Big) \Big[ -\int\_{0}^{1/2} dv + \int\_{1/2}^{1} dv \Big] \\ &=& 0.5. \end{split}$$

Hence, the quantity *r σ*1*σ*2is the median of *Y*. The proof is complete.

We turn to develop the proposition on density and distribution functions for the ratio of one variable over the sum of two variables *Z* := *X*1 *X*1+*X*2 of two dependent random variables *X*1 and *X*2 by using copulas as stated in the following:

**Proposition 2.** *Suppose that* (*<sup>X</sup>*1, *<sup>X</sup>*2) *is a vector of two absolutely continuous random variables X*1 *and X*2 *with the marginal distributions F*1 *and F*2*, respectively, and let C be an absolutely continuous copula modeling dependence structure of the random vector* (*<sup>X</sup>*1, *<sup>X</sup>*2)*, and define Z as*

$$Z := \frac{X\_1}{X\_1 + X\_2}.\tag{18}$$

*Then, the density fZ*(*z*) *and distribution function FZ*(*z*) *of Z are*

$$f\_{\mathbb{Z}}(z) = \begin{cases} \int\_0^1 \frac{\left| F\_1^{-1}(u) \right|}{z^2} c \Big( u, F\_2 \left( \frac{1-z}{z} F\_1^{-1}(u) \right) \Big) f\_2 \left( \frac{1-z}{z} F\_1^{-1}(u) \right) du, & \text{if } z \neq 0, \\\\ f\_1(0) \int\_0^1 \left| F\_2^{-1}(v) \right| c \Big( F\_1(0), v \Big) dv, & \text{if } z = 0, \end{cases} \tag{19}$$

*JRFM* **2019**, *12*, 42

$$F\_{\mathcal{Z}}(\boldsymbol{z}) = \mathbf{1}\_{\{\boldsymbol{z} \geq 0\}} + \int\_0^1 \text{sgn}\left(F\_1^{-1}(\boldsymbol{u})\right) \left[\frac{\partial}{\partial \boldsymbol{u}} \mathcal{C}\left(\boldsymbol{u}, F\_2\left(-F\_1^{-1}(\boldsymbol{u})\right)\right) - \frac{\partial}{\partial \boldsymbol{u}} \mathcal{C}\left(\boldsymbol{u}, F\_2\left(\frac{1-\boldsymbol{z}}{\boldsymbol{z}} F\_1^{-1}(\boldsymbol{u})\right)\right)\right] d\boldsymbol{u},\tag{20}$$

*respectively, where* **<sup>1</sup>**{·} *denotes an indicator function, F*−<sup>1</sup> *i denotes the inverse function of Fi for i* = 1, 2*, c is the density of copula C, and* sgn(·) *is the sign function such that*

$$\mathbf{1}\_{\{z\geq 0\}} = \begin{cases} 1, & \text{if} \quad z \geq 0, \\ 0, & \text{if} \quad z < 0, \end{cases} \quad \text{and} \quad \text{sgn}(\mathbf{x}) = \begin{cases} 1, & \text{if} \quad \mathbf{x} > 0, \\ -1, & \text{if} \quad \mathbf{x} < 0. \end{cases}$$

**Proof.** By defining

$$\begin{cases} Z\_1 = \frac{X\_1}{X\_1 + X\_2},\\ Z\_2 = X\_1 + X\_2. \end{cases}$$

Here, we note that, since *X*1 and *X*2 are absolutely continuous, *<sup>P</sup>*(*<sup>X</sup>*1 + *X*2 = 0) = 0; that is, *X*1 + *X*2 = 0 almost surely. Hence, the transformation *Z*1 = *X*1 *X*1+*X*2 always exists with probability 1 and we obtain the following inverse transformation:

$$\begin{cases} \mathbf{X}\_1 = \mathbf{Z}\_1 \mathbf{Z}\_{2\nu} \\ \mathbf{X}\_2 = \mathbf{Z}\_2 - \mathbf{Z}\_1 \mathbf{Z}\_{2\nu} \end{cases}$$

and the Jacobian

$$J = \begin{vmatrix} Z\_2 & Z\_1 \\ -Z\_2 & 1 - Z\_1 \end{vmatrix} = Z\_2.$$

Thus, the joint density of *Z*1 and *Z*2 becomes

$$\begin{aligned} \left| h(z\_1, z\_2) \right| &= \left| f(z\_1 z\_2, z\_2 - z\_1 z\_2) \right| \left| z\_2 \right| \\ &= \left| z\_2 \right| c \left( F\_1(z\_1 z\_2), F\_2 \left( z\_2 - z\_1 z\_2 \right) \right) f\_1(z\_1 z\_2) f\_2 \left( z\_2 - z\_1 z\_2 \right) \,, \end{aligned}$$

which leads us to ge<sup>t</sup> the density of *Z*1 such that

$$f\_{\mathcal{Z}\_1}(z\_1) \quad = \int\_{-\infty}^{\infty} |z\_2| \, \mathcal{C}(F\_1(z\_1 z\_2), F\_2(z\_2 - z\_1 z\_2)) \, f\_1(z\_1 z\_2) f\_2(z\_2 - z\_1 z\_2) \, dz\_2. \tag{21}$$

If *z*1 = 0, then by taking *v* := *<sup>F</sup>*2(*<sup>z</sup>*2), we ge<sup>t</sup>

$$f\_{Z\_1}(0) = f\_1(0) \int\_0^1 \left| F\_2^{-1}(v) \right| c \left( F\_1(0), v \right) dv.$$

If *z*1 > 0, then by taking *u* := *<sup>F</sup>*1(*<sup>z</sup>*1*z*2), we obtain

$$f\_{Z\_1}(z\_1) = \int\_0^1 \frac{\left| F\_1^{-1}(u) \right|}{z\_1^2} c(u, F\_2\left(\frac{1 - z\_1}{z\_1} F\_1^{-1}(u)\right) ) f\_2\left(\frac{1 - z\_1}{z\_1} F\_1^{-1}(u)\right) du.$$

If *z*1 < 0, then also by taking *u* := *<sup>F</sup>*1(*<sup>z</sup>*1*z*2), we yield

$$f\_{Z\_1}(z\_1) = \int\_1^0 \frac{|F\_1^{-1}(u)|}{-z\_1^2} c\left(u, F\_2\left(\frac{1-z\_1}{z\_1} F\_1^{-1}(u)\right)\right) f\_2\left(\frac{1-z\_1}{z\_1} F\_1^{-1}(u)\right) du. \tag{22}$$

Hence, for *z*1 = 0, we obtain

$$f\_{Z\_1}(z\_1) = \int\_0^1 \frac{|F\_1^{-1}(u)|}{z\_1^2} c\left(u, F\_2\left(\frac{1 - z\_1}{z\_1} F\_1^{-1}(u)\right)\right) f\_2\left(\frac{1 - z\_1}{z\_1} F\_1^{-1}(u)\right) du \quad \text{for} \quad z\_1 \neq 0.$$

As a consequence, the distribution of *Z*1 becomes

$$F\_{Y\_1}(t) = \int\_0^1 \int\_{-\infty}^t \frac{\left| F\_1^{-1}(u) \right|}{z\_1^2} c\left(u, F\_2\left(\frac{1 - z\_1}{z\_1} F\_1^{-1}(u)\right)\right) f\_2\left(\frac{1 - z\_1}{z\_1} F\_1^{-1}(u)\right) dz\_1 du. \tag{23}$$

$$\text{Setting } v = F\_2\left(\frac{1 - z\_1}{z\_1} F\_1^{-1}(u)\right) \implies dv = -\frac{F\_1^{-1}(u)}{z\_1^2} f\_2\left(\frac{1 - z\_1}{z\_1} F\_1^{-1}(u)\right) dz\_1, \text{ and note that}$$

$$F\_1^{-1}(u) \ge 0 \Longleftrightarrow u \ge F\_1(0), \text{ and } F\_1^{-1}(u) \le 0 \Longleftrightarrow u \le F\_1(0),$$

we consider two cases as follows:

(i) **Case 1:** For *t* < 0, we have

*FZ*1 (*t*) = *<sup>F</sup>*1(0) 0 *<sup>F</sup>*2( 1−*t t F*−<sup>1</sup> 1 (*u*)) *<sup>F</sup>*2(−*F*−<sup>1</sup> 1 (*u*)) *∂*2 *<sup>∂</sup>u∂v<sup>C</sup>* (*<sup>u</sup>*, *v*) *dvdu* − 1*<sup>F</sup>*1(0) *<sup>F</sup>*2( 1−*t t F*−<sup>1</sup> 1 (*u*)) *<sup>F</sup>*2(−*F*−<sup>1</sup> 1 (*u*)) *∂*2 *<sup>∂</sup>u∂v<sup>C</sup>* (*<sup>u</sup>*, *v*) *dvdu* = *<sup>F</sup>*1(0) 0 *∂∂u<sup>C</sup> u*, *F*2 <sup>1</sup>−*<sup>t</sup> t F*−<sup>1</sup> 1 (*u*) − *∂∂u<sup>C</sup><sup>u</sup>*, *F*2 − *F*−<sup>1</sup> 1 (*u*) *du* − 1*<sup>F</sup>*1(0) *∂∂u<sup>C</sup> u*, *F*2 <sup>1</sup>−*<sup>t</sup> t F*−<sup>1</sup> 1 (*u*) − *∂∂u<sup>C</sup><sup>u</sup>*, *F*2 − *F*−<sup>1</sup> 1 (*u*) *du* = 10 sgn *<sup>F</sup>*−<sup>1</sup> 1 (*u*) *∂∂u<sup>C</sup><sup>u</sup>*, *F*2 − *F*−<sup>1</sup> 1 (*u*) − *∂∂u<sup>C</sup> u*, *F*2 <sup>1</sup>−*<sup>t</sup> t F*−<sup>1</sup> 1 (*u*) *du*. (24)

(ii) **Case 2:** For *t* ≥ 0, we first split the integrals

$$\begin{split} F\_{\rm Z1}(t) &= \quad \int\_{0}^{1} \int\_{-\infty}^{0} \frac{\left| F\_{1}^{-1}(u) \right|}{z\_{1}^{2}} c \Big( u, F\_{2} \left( \frac{1 - z\_{1}}{z\_{1}} F\_{1}^{-1}(u) \right) \Big) f\_{2} \left( \frac{1 - z\_{1}}{z\_{1}} F\_{1}^{-1}(u) \right) dz\_{1} du \\ &+ \int\_{0}^{1} \int\_{0}^{t} \frac{\left| F\_{1}^{-1}(u) \right|}{z\_{1}^{2}} c \Big( u, F\_{2} \left( \frac{1 - z\_{1}}{z\_{1}} F\_{1}^{-1}(u) \right) \Big) f\_{2} \left( \frac{1 - z\_{1}}{z\_{1}} F\_{1}^{-1}(u) \right) dz\_{1} du \end{split} \tag{25}$$

We then apply (24) to obtain the following expression for the integral *I*1:

$$\begin{array}{rcl} I\_1 = F\_{\mathbb{Z}\_1}(0) &=& \int\_0^1 \text{sgn}\left( F\_1^{-1}(u) \right) \left[ \frac{\partial}{\partial u} \mathbb{C} \left( u, F\_2 \left( -F\_1^{-1}(u) \right) \right) - \frac{\partial}{\partial u} \mathbb{C} \left( u, \lim\_{t \to 0^-} F\_2 \left( \frac{1-t}{T} F\_1^{-1}(u) \right) \right) \right] du \\ &=& \int\_0^1 \text{sgn}\left( F\_1^{-1}(u) \right) \frac{\partial}{\partial u} \mathbb{C} \left( u, F\_2 \left( -F\_1^{-1}(u) \right) \right) du \\ &+& \int\_0^{F\_1(0)} \frac{\partial}{\partial u} \mathbb{C} \left( u, 1 \right) du - \int\_{f\_1(0)}^1 \frac{\partial}{\partial u} \mathbb{C} \left( u, 0 \right) du \\ &=& \int\_0^1 \text{sgn}\left( F\_1^{-1}(u) \right) \frac{\partial}{\partial u} \mathbb{C} \left( u, F\_2 \left( -F\_1^{-1}(u) \right) \right) du + F\_1(0), \end{array} \tag{26}$$

and obtain the following expression for the integral *I*2:

$$\begin{split} I\_{2} &= \quad \int\_{0}^{\overline{r}\_{1}(0)} \int\_{0}^{\overline{r}\_{2}\left(\frac{1-t}{t}\overline{r}\_{1}^{-1}(u)\right)} \frac{\partial^{2}}{\partial u \partial v} \mathbb{C}\left(u,v\right) dv du - \int\_{\overline{r}\_{1}(0)}^{1} \int\_{1}^{\overline{r}\_{2}\left(\frac{1-t}{t}\overline{r}\_{1}^{-1}(u)\right)} \frac{\partial^{2}}{\partial u \partial v} \mathbb{C}\left(u,v\right) dv du \\ &= \quad \int\_{0}^{\overline{r}\_{1}(0)} \left[ \frac{\partial}{\partial u} \mathbb{C}\left(u, \overline{r}\_{2}\left(\frac{1-t}{t}\overline{r}\_{1}^{-1}(u)\right)\right) - \frac{\partial}{\partial u} \mathbb{C}\left(u,0\right) \right] du \\ & \quad \quad - \int\_{\overline{r}\_{1}(0)}^{1} \left[ \frac{\partial}{\partial u} \mathbb{C}\left(u, \overline{r}\_{2}\left(\frac{1-t}{t}\overline{r}\_{1}^{-1}(u)\right)\right) - \frac{\partial}{\partial u} \mathbb{C}\left(u,1\right) \right] du \\ & \quad = \quad 1 - F\_{1}(0) - \int\_{0}^{1} \text{sgn}\left(\overline{r}\_{1}^{-1}(u)\right) \frac{\partial}{\partial u} \mathbb{C}\left(u, \overline{r}\_{2}\left(-\overline{r}\_{1}^{-1}(u)\right)\right) du. \tag{27} \end{split} \tag{27}$$

From (26) and (27), we ge<sup>t</sup> the following for *t* ≥ 0,

$$F\_{\mathcal{Z}\_1}(t) = 1 + \int\_0^1 \text{sgn}\left(F\_1^{-1}(u)\right) \left[\frac{\partial}{\partial u}\mathcal{C}\left(u, F\_2\left(-F\_1^{-1}(u)\right)\right) - \frac{\partial}{\partial u}\mathcal{C}\left(u, F\_2\left(\frac{1-t}{t}F\_1^{-1}(u)\right)\right)\right] du. \tag{28}$$

Combining (24) and (28) imply (20), we complete the proof.

In the situation *X*1 and *X*2 are independent, applying Proposition 2, we obtain the following corollary:

**Corollary 4.** *When X*1 *and X*2 *are independent, then its copula <sup>C</sup>*(*<sup>u</sup>*, *v*) = *uv has the density <sup>c</sup>*(*<sup>u</sup>*, *v*) = 1, ∀*<sup>u</sup>*, *v* ∈ I *and the density fZ*(*z*) *and distribution function FZ*(*z*) *for the ratio of one variable over the sum of two variables Z* := *X*1 *X*1+*X*2*of two independent random variables X*1 *and X*2 *become*

$$f\_Z(z) = \int\_{-\infty}^{\infty} |x| \, f\_1(xz) f\_2\left((1-x)z\right) dx$$

*and*

$$\begin{aligned} F\_{\mathbb{Z}}(z) &= \mathbf{1}\_{\{z \ge 0\}} + \int\_0^1 \text{sgn}\left(F\_1^{-1}(u)\right) \left[F\_2\left(-F\_1^{-1}(u)\right) - F\_2\left(\frac{1-z}{z} F\_1^{-1}(u)\right)\right] du, \\ &= \mathbf{1}\_{\{z \ge 0\}} + \int\_{-\infty}^\infty \text{sgn}(\mathbf{x}) \left[F\_2(-\mathbf{x}) - F\_2\left(\frac{1-z}{z} \mathbf{x}\right)\right] f\_1(\mathbf{x}) d\mathbf{x}, \end{aligned}$$

*respectively.*

Next, we apply Equation (20) to derive the distribution function of *Z* := *X*1 *X*1+*X*2 in the situation that both *X*1 and *X*2 are normal distributed such that their dependence structure can be captured by Gaussian Copulas as shown in the following corollary:

**Corollary 5.** *Assume that X*1 ∼ *<sup>N</sup>*(*μ*1, *σ*21 )*, X*2 ∼ *<sup>N</sup>*(*μ*2, *σ*22 )*, and* (*<sup>X</sup>*1, *<sup>X</sup>*2) *follows Gaussian Copulas Cr*(*<sup>u</sup>*, *<sup>v</sup>*), |*r*| < 1, *given in* (46)*. Then, distribution function FZ*(*z*) *of Z* := *X*1 *X*1+*X*2*has the form*

$$\begin{split} \text{Fz}(\mathbf{z}) &= \quad \mathbf{1}\_{\{z \ge 0\}} + 2\Phi\left(\frac{\mu\_1}{\sigma\_1}\right) - 1 - \int\_0^1 \text{sgn}\left(\sigma\_1 \Phi^{-1}(u) + \mu\_1\right) \Phi\left(\frac{(\sigma\_1 + r\varphi\_2)\Phi^{-1}(u) + \mu\_1 - \mu\_2}{\sigma\_2\sqrt{1 - r^2}}\right) du \\ &- \int\_0^1 \text{sgn}\left(\sigma\_1 \Phi^{-1}(u) + \mu\_1\right) \Phi\left(\frac{[(1 - z)\sigma\_1 - zr\varphi\_2]\Phi^{-1}(u) - z(\mu\_1 + \mu\_2) + \mu\_1}{z\sigma\_2\sqrt{1 - r^2}}\right) du \end{split} \tag{29}$$

*where* <sup>Φ</sup>(*x*) *and* <sup>Φ</sup>−<sup>1</sup>(*x*) *are CDF and its inverse of the standard normal random variable, respectively.*

**Proof.** Let *X*1 ∼ *<sup>N</sup>*(*μ*1, *σ*21 ) and *X*2 ∼ *<sup>N</sup>*(*μ*2, *σ*22 ), the CDFs and their inverse functions can be written in the form 

$$F\_i(x) = \Phi\left(\frac{\chi - \mu\_i}{\sigma\_i}\right), \quad F\_i^{-1}(v) = \sigma\_i \Phi^{-1}(v) + \mu\_i, \quad i = 1, 2.1$$

Given Gaussian Copulas *Cr*(*<sup>u</sup>*, *<sup>v</sup>*), |*r*| < 1, we apply the results from Meyer (2013) to obtain its derivative *∂Cr ∂u*(*<sup>u</sup>*, *v*) as shown in the following:

$$\frac{\partial C\_r}{\partial \mu}(\mu, v) = \Phi \left( \frac{\Phi^{-1}(v) - r\Phi^{-1}(\mu)}{\sqrt{1 - r^2}} \right).$$

Applying Equation (20), one can simplify it to be

*FZ*(*z*) = **<sup>1</sup>**{*z*≥<sup>0</sup>} + 10 *sgn<sup>σ</sup>*1Φ−<sup>1</sup>(*u*) + *<sup>μ</sup>*1Φ<sup>Φ</sup>−<sup>1</sup>Φ − *<sup>σ</sup>*1Φ−<sup>1</sup>(*u*)+*μ*1−*μ*<sup>2</sup> *σ*2 − *r*Φ−<sup>1</sup>(*u*) √1 − *r*2 *du* − 10 *sgn<sup>σ</sup>*1Φ−<sup>1</sup>(*u*) + *<sup>μ</sup>*1Φ<sup>Φ</sup>−<sup>1</sup>Φ 1−*<sup>z</sup> z* !*<sup>σ</sup>*1Φ−<sup>1</sup>(*u*)+*μ*1"−*μ*<sup>2</sup> *σ*2 − *r*Φ−<sup>1</sup>(*u*) √1 − *r*2 *du* = **<sup>1</sup>**{*z*≥<sup>0</sup>} + 10 *sgn<sup>σ</sup>*1Φ−<sup>1</sup>(*u*) + *<sup>μ</sup>*1Φ − (*<sup>σ</sup>*1 + *<sup>r</sup>σ*2)<sup>Φ</sup>−<sup>1</sup>(*u*) + *μ*1 − *μ*2 *<sup>σ</sup>*2√<sup>1</sup> − *r*2 *du* − 10 *sgn<sup>σ</sup>*1Φ−<sup>1</sup>(*u*) + *<sup>μ</sup>*1Φ[(<sup>1</sup> − *<sup>z</sup>*)*<sup>σ</sup>*1 − *zrσ*2]<sup>Φ</sup>−<sup>1</sup>(*u*) − *<sup>z</sup>*(*μ*1 + *μ*2) + *μ*1 *<sup>z</sup>σ*2√<sup>1</sup> − *r*2 *du* = **<sup>1</sup>**{*z*≥<sup>0</sup>} + 2Φ*<sup>μ</sup>*<sup>1</sup> *σ*1 − 1 − 10 *sgn<sup>σ</sup>*1Φ−<sup>1</sup>(*u*) + *<sup>μ</sup>*1Φ(*<sup>σ</sup>*<sup>1</sup> + *<sup>r</sup>σ*2)<sup>Φ</sup>−<sup>1</sup>(*u*) + *μ*1 − *μ*2 *<sup>σ</sup>*2√<sup>1</sup> − *r*2 *du* − 10 *sgn<sup>σ</sup>*1Φ−<sup>1</sup>(*u*) + *<sup>μ</sup>*1Φ[(<sup>1</sup> − *<sup>z</sup>*)*<sup>σ</sup>*1 − *zrσ*2]<sup>Φ</sup>−<sup>1</sup>(*u*) − *<sup>z</sup>*(*μ*1 + *μ*2) + *μ*1 *<sup>z</sup>σ*2√<sup>1</sup> − *r*2 *du*.

In the last step of the above, we use the property <sup>Φ</sup>(−*<sup>x</sup>*) = 1 − <sup>Φ</sup>(*x*) and

$$\int\_0^1 \operatorname{sgn} \left( \sigma\_1 \Phi^{-1} (u) + \mu\_1 \right) du = -\int\_0^{\Phi \left( -\frac{\mu\_1}{\sigma\_1} \right)} du + \int\_{\Phi \left( -\frac{\mu\_1}{\sigma\_1} \right)}^1 du = 1 - 2\Phi \left( -\frac{\mu\_1}{\sigma\_1} \right) = 2\Phi \left( \frac{\mu\_1}{\sigma\_1} \right) - 1.$$

The proof is complete.

We note that the probability given in (29) can also be easily computed by using the following Monte Carlo algorithm: For each *z* ∈ R, we first generate *U* from the uniform distribution on the unit interval [0, 1] with sample size *N*, say *N* = 10, 000. Then, we obtain the following estimated probability:

$$\hat{F}\_{\mathcal{Z}}(z) \approx \ \mathbf{1}\_{\{z \ge 0\}} + 2\Phi\left(\frac{\mu\_1}{\sigma\_1}\right) - 1 - \frac{1}{N} \sum\_{i=1}^{N} \text{sgn}\left(\sigma\_1 \Phi^{-1}(u\_i) + \mu\_1\right) \Phi\left(\frac{(\sigma\_1 + r\sigma\_2)\Phi^{-1}(u\_i) + \mu\_1 - \mu\_2}{\sigma\_2\sqrt{1 - r^2}}\right)$$

$$- \frac{1}{N} \sum\_{i=1}^{N} \text{sgn}\left(\sigma\_1 \Phi^{-1}(u\_i) + \mu\_1\right) \Phi\left(\frac{[(1 - z)\sigma\_1 - zr\sigma\_2]\Phi^{-1}(u\_i) - z(\mu\_1 + \mu\_2) + \mu\_1}{z\sigma\_2\sqrt{1 - r^2}}\right). \tag{30}$$

Using the above results, we obtain the following corollary:

**Corollary 6.** *Assume that X*1 ∼ *N*(0, *<sup>σ</sup>*<sup>2</sup>)*, X*2 ∼ *N*(0, *σ*<sup>2</sup>) *and* (*<sup>X</sup>*1, *<sup>X</sup>*2) *follows Gaussian Copulas Cr*(*<sup>u</sup>*, *<sup>v</sup>*)*,* |*r*| < 1, *given in* (46)*. Then, the median of Z* := *X*1 *X*1+*X*2 *is equal to* 12 *.*

**Proof.** Because *X*1 ∼ *N*(0, *<sup>σ</sup>*<sup>2</sup>), *X*2 ∼ *N*(0, *σ*<sup>2</sup>) and *sgnσ*Φ−<sup>1</sup>(*u*) = *sgn*<sup>Φ</sup>−<sup>1</sup>(*u*), we obtain CDF of the ratio *Z* := *X*1 *X*1+*X*2 from (29) as shown in the following:

$$F\_{\mathbb{Z}}\left(z\right) = \mathbf{1}\_{\{z \ge 0\}} - \int\_0^1 \operatorname{sgn}\left(\Phi^{-1}(u)\right) \left[\Phi\left(\frac{(1+r)\Phi^{-1}(u)}{\sqrt{1-r^2}}\right) + \Phi\left(\frac{[1-z-zr]\Phi^{-1}(u)}{z\sqrt{1-r^2}}\right)\right] du. \tag{31}$$

We ge<sup>t</sup>

$$F\_{\mathbb{Z}}\left(\frac{1}{2}\right) = 1 - \int\_0^1 \text{sgn}\left(\Phi^{-1}(u)\right) \left[\Phi\left(\frac{(1+r)\Phi^{-1}(u)}{\sqrt{1-r^2}}\right) + \Phi\left(\frac{[1-r]\Phi^{-1}(u)}{\sqrt{1-r^2}}\right)\right] du.\tag{32}$$

Hence, it is sufficient to prove that the integral term given in (32) is equal to 12. We let

$$I\_1 := \int\_0^1 \operatorname{sgn} \left( \Phi^{-1}(u) \right) \Phi \left( \frac{(1+r)\Phi^{-1}(u)}{\sqrt{1-r^2}} \right) du, \quad I\_2 := \int\_0^1 \operatorname{sgn} \left( \Phi^{-1}(u) \right) \Phi \left( \frac{(1-r)\Phi^{-1}(u)}{\sqrt{1-r^2}} \right) du, \quad I\_3 := \int\_0^1 \operatorname{sgn} \left( \Phi^{-1}(u) \right) \Phi \left( \frac{(1+r)\Phi^{-1}(u)}{\sqrt{1-r^2}} \right) du$$

and denote *∂iCr*(*<sup>u</sup>*, *<sup>v</sup>*), *i* = 1, 2 to be the partial derivative of *Cr*(*<sup>u</sup>*, *v*) with respect to the *i*th variable, That is, *∂*1*Cr*(*<sup>u</sup>*, *v*) = *∂∂uCr*(*<sup>u</sup>*, *v*) and *∂*2*Cr*(*<sup>u</sup>*, *v*) = *∂∂vCr*(*<sup>u</sup>*, *<sup>v</sup>*). One can observe that

$$
\partial\_1 \mathbb{C}\_r(u, u) = \Phi \left( \frac{(1 - r)\Phi^{-1}(u)}{\sqrt{1 - r^2}} \right).
$$

Since Gaussian Copulas is symmetric; that is, *Cr*(*<sup>u</sup>*, *v*) = *Cr*(*<sup>v</sup>*, *<sup>u</sup>*), we have *∂*1*Cr*(*<sup>u</sup>*, *v*) = *∂*2*Cr*(*<sup>v</sup>*, *<sup>u</sup>*), and, thus, for *u* = *v*, we can derive *∂*1*Cr*(*<sup>u</sup>*, *u*) = *∂*2*Cr*(*<sup>u</sup>*, *<sup>u</sup>*). Thereafter, the differentiation of *Cr*(*<sup>u</sup>*, *u*) can be obtained:

$$d\mathbb{C}\_r(\boldsymbol{\mu}, \boldsymbol{\mu}) = \left[\partial\_1 \mathbb{C}\_r(\boldsymbol{\mu}, \boldsymbol{\mu}) + \partial\_2 \mathbb{C}\_r(\boldsymbol{\mu}, \boldsymbol{\mu})\right] d\boldsymbol{\mu} = 2\partial\_1 \mathbb{C}\_r(\boldsymbol{\mu}, \boldsymbol{\mu})\_r$$

and we ge<sup>t</sup>

$$\begin{split} l\_{2} = \frac{1}{2} \int\_{0}^{1} \text{sgn} \left( \Phi^{-1}(u) \right) d\mathbb{C}\_{r}(u,u) &= \ -\frac{1}{2} \int\_{0}^{1/2} d\mathbb{C}\_{r}(u,u) + \frac{1}{2} \int\_{1/2}^{1} d\mathbb{C}\_{r}(u,u) \\ &= \ -\frac{1}{2} \mathbb{C}\_{r} \left( \frac{1}{2}, \frac{1}{2} \right) + \frac{1}{2} \left[ 1 - \mathbb{C}\_{r} \left( \frac{1}{2}, \frac{1}{2} \right) \right] \\ &= \ \frac{1}{2} - \mathbb{C}\_{r} (\frac{1}{2}, \frac{1}{2}) . \end{split} \tag{33}$$

Similarly, for *I*1, since <sup>Φ</sup>−<sup>1</sup>(<sup>1</sup> − *u*) = <sup>−</sup>Φ−<sup>1</sup>(*u*), we obtain

$$\partial\_1 \mathbb{C}\_r(u, 1 - u) = \Phi\left(\frac{\Phi^{-1}(1 - u) - r\Phi^{-1}(u)}{\sqrt{1 - r^2}}\right) = \Phi\left(-\frac{(1 + r)\Phi^{-1}(u)}{\sqrt{1 - r^2}}\right) = 1 - \Phi\left(\frac{(1 + r)\Phi^{-1}(u)}{\sqrt{1 - r^2}}\right),$$

and ge<sup>t</sup>

$$I\_1 = \int\_0^1 \text{sgn}\left(\Phi^{-1}(u)\right) du - \int\_0^1 \text{sgn}\left(\Phi^{-1}(u)\right) \partial\_1 \mathbb{C}\_r(u, 1-u) du = -\int\_0^1 \text{sgn}\left(\Phi^{-1}(u)\right) \partial\_1 \mathbb{C}\_r(u, 1-u) du \ge 0$$

in which we apply 10 *sgn*<sup>Φ</sup>−<sup>1</sup>(*u*)*du* = 0. From symmetry of the Gaussian Copulas, we also have *∂*1*Cr*(<sup>1</sup> − *u*, *u*) = *∂*2*Cr*(*<sup>u</sup>*, 1 − *<sup>u</sup>*), obtain the differentiation of *Cr*(*<sup>u</sup>*, 1 − *u*) given by

$$\frac{d}{dt}\mathcal{C}\_{\tau}(\boldsymbol{u},\boldsymbol{1}-\boldsymbol{u}) = \left[\partial\_{1}\mathcal{C}\_{\tau}(\boldsymbol{u},\boldsymbol{1}-\boldsymbol{u}) - \partial\_{2}\mathcal{C}\_{\tau}(\boldsymbol{u},\boldsymbol{1}-\boldsymbol{u})\right]d\boldsymbol{u} = \left[\partial\_{1}\mathcal{C}\_{\tau}(\boldsymbol{u},\boldsymbol{1}-\boldsymbol{u}) - \partial\_{1}\mathcal{C}\_{\tau}(\boldsymbol{1}-\boldsymbol{u},\boldsymbol{u})\right]d\boldsymbol{u}\_{\tau}$$

and ge<sup>t</sup>

$$\begin{split} I\_{1} &= \quad \int\_{0}^{1/2} \partial\_{1} \mathbb{C}\_{r}(u, 1-u) du - \int\_{1/2}^{1} \partial\_{1} \mathbb{C}\_{r}(u, 1-u) du, \\ &= \quad \int\_{0}^{1/2} \partial\_{1} \mathbb{C}\_{r}(u, 1-u) du + \int\_{1/2}^{0} \partial\_{1} \mathbb{C}\_{r}(1-u, u) du \\ &= \quad \int\_{0}^{1/2} \partial\_{1} \mathbb{C}\_{r}(u, 1-u) du - \int\_{0}^{1/2} \partial\_{1} \mathbb{C}\_{r}(1-u, u) du \\ &= \quad \int\_{0}^{1/2} d \mathbb{C}\_{r}(u, 1-u) \\ &= \quad \mathbb{C}\_{r}\left(\frac{1}{2}, \frac{1}{2}\right). \end{split} \tag{34}$$

Combining (32)–(34), we find that *FZ*12= 12 . Hence, the quantity 12 is the median of *Z*. The proof is complete.

Applying the proof of Corollary 6, we extend the result to obtain the following corollary for a larger family of symmetric distribution and symmetric copulas:

**Corollary 7.** *Assume that X*1 *and X*2 *are identically and symmetrically distributed with distribution F that has zero median and the dependence structure of* (*<sup>X</sup>*1, *<sup>X</sup>*2) *is modelled by a family of symmetric copulas <sup>C</sup>*(*<sup>u</sup>*, *<sup>v</sup>*)*, i.e., <sup>C</sup>*(*<sup>u</sup>*, *v*) = *<sup>C</sup>*(*<sup>v</sup>*, *u*) *for all u*, *v* ∈ I*. Then, the median of Z* := *X*1 *X*1+*X*2 *is equal to* 12 *.*

**Proof.** Since *X*1 and *X*2 are identically and symmetrically distributed with distribution *F* and zero median, we have *<sup>F</sup>*(−*<sup>x</sup>*) = 1 − *<sup>F</sup>*(*x*), for all *x* ∈ R. By applying Equation (20), we obtain the CDF of the ratio *Z*, which is defined by

$$\begin{split} F\_{\mathcal{Z}}(z) &= \mathbf{1}\_{\{z \ge 0\}} + \int\_{0}^{1} \text{sgn}\left(F^{-1}(u)\right) \left[ \frac{\partial}{\partial u} \mathbb{C}\left(u, F\left(-F^{-1}(u)\right)\right) - \frac{\partial}{\partial u} \mathbb{C}\left(u, F\left(\frac{1-z}{z} F^{-1}(u)\right)\right) \right] du, \\ &= \mathbf{1}\_{\{z \ge 0\}} + \int\_{0}^{1} \text{sgn}\left(F^{-1}(u)\right) \left[ \frac{\partial}{\partial u} \mathbb{C}\left(u, 1-u\right) \right) - \frac{\partial}{\partial u} \mathbb{C}\left(u, F\left(\frac{1-z}{z} F^{-1}(u)\right)\right) \right] du. \end{split} \tag{35}$$

Since copulas *<sup>C</sup>*(*<sup>u</sup>*, *v*) are symmetric, i.e., *<sup>C</sup>*(*<sup>v</sup>*, *u*) = *<sup>C</sup>*(*<sup>u</sup>*, *<sup>v</sup>*), we have *∂*1*<sup>C</sup>*(*<sup>v</sup>*, *u*) = *∂*2*<sup>C</sup>*(*<sup>u</sup>*, *<sup>v</sup>*), and, thus, for *v* = 1 − *u*, one can easily obtain *∂*1*<sup>C</sup>*(<sup>1</sup> − *u*, *u*) = *∂*2*<sup>C</sup>*(*<sup>u</sup>*, 1 − *u*) and find that the differentiation of *<sup>C</sup>*(*<sup>u</sup>*, 1 − *u*) with respect to *u* satisfies

$$d\mathbb{C}(u, 1-u) = \left[\partial\_1 \mathbb{C}(u, 1-u) - \partial\_2 \mathbb{C}(u, 1-u)\right] du = \left[\partial\_1 \mathbb{C}(u, 1-u) - \partial\_1 \mathbb{C}(1-u, u)\right] du \dots$$

In addition, because the distribution *F* has zero median; that is, *F*(0) = 0.5, we have

$$\begin{split} \int\_{0}^{1} \text{sgn}\left(F^{-1}(u)\right) \frac{\partial}{\partial u} \mathbb{C}\left(u, 1-u\right) &=& -\int\_{0}^{1/2} \partial\_{1} \mathbb{C}(u, 1-u) du + \int\_{1/2}^{1} \partial\_{1} \mathbb{C}(u, 1-u) du \\ &=& -\int\_{0}^{1/2} \partial\_{1} \mathbb{C}(u, 1-u) du - \int\_{1/2}^{0} \partial\_{1} \mathbb{C}(1-u, u) du \\ &=& -\int\_{0}^{1/2} \partial\_{1} \mathbb{C}(u, 1-u) du + \int\_{0}^{1/2} \partial\_{1} \mathbb{C}(1-u, u) du \\ &=& -\int\_{0}^{1/2} d\mathbb{C}(u, 1-u) = -\mathbb{C}\left(\frac{1}{2}, \frac{1}{2}\right). \end{split}$$

Therefore, we ge<sup>t</sup>

$$F\_{\mathbb{Z}}(z) \quad = \mathbf{1}\_{\{z \ge 0\}} - \mathbb{C}\left(\frac{1}{2}, \frac{1}{2}\right) - \int\_0^1 \text{sgn}\left(F^{-1}(u)\right) \frac{\partial}{\partial u} \mathbb{C}\left(u, F\left(\frac{1-z}{z} F^{-1}(u)\right)\right) du,\tag{36}$$

and thus,

$$F\_{\mathcal{Z}}(0.5) \quad = \; 1 - \mathcal{C}\left(\frac{1}{2}, \frac{1}{2}\right) - \int\_0^1 \text{sgn}\left(F^{-1}(u)\right) \frac{\partial}{\partial u} \mathcal{C}\left(u, u\right) du. \tag{37}$$

Similarly, since copulas *<sup>C</sup>*(*<sup>u</sup>*, *v*) are symmetric, i.e., *<sup>C</sup>*(*<sup>v</sup>*, *u*) = *<sup>C</sup>*(*<sup>u</sup>*, *<sup>v</sup>*), we have *∂*1*<sup>C</sup>*(*<sup>v</sup>*, *u*) = *∂*2*<sup>C</sup>*(*<sup>u</sup>*, *<sup>v</sup>*), and, thus, for *v* = *u*, we ge<sup>t</sup> *∂*1*<sup>C</sup>*(*<sup>u</sup>*, *u*) = *∂*2*<sup>C</sup>*(*<sup>u</sup>*, *<sup>u</sup>*). Thus, the differentiation of *<sup>C</sup>*(*<sup>u</sup>*, *u*) with respect to *u* satisfies

$$d\mathbb{C}(\mathfrak{u},\mathfrak{u}) = \left[\partial\_1\mathbb{C}(\mathfrak{u},\mathfrak{u}) + \partial\_2\mathbb{C}(\mathfrak{u},\mathfrak{u})\right]d\mathfrak{u} = 2\partial\_1\mathbb{C}(\mathfrak{u},\mathfrak{u})d\mathfrak{u}.$$

Applying this relation, we find

$$\int\_0^1 \text{sgn}\left(F^{-1}(u)\right) \frac{\partial}{\partial u} \mathbb{C}\left(u, u\right) du = \frac{1}{2} \int\_0^1 \text{sgn}\left(F^{-1}(u)\right) d\mathbb{C}\left(u, u\right) = \frac{1}{2} - \mathbb{C}\left(\frac{1}{2}, \frac{1}{2}\right).$$

Hence, *FZ*(0.5) = 0.5, i.e., 1 2 is median of *Z*. The proof is complete.

Remark: In the literature, they are many symmetric distributions with zero median, for example, normal *N*(0, *<sup>σ</sup>*), Student-*t t<sup>ν</sup>*, Cauchy distribution with location parameter *α* = 0, uniform *<sup>U</sup>*(−*a*, *<sup>a</sup>*), *a* ∈ R+, and logistic distribution with zero location. In addition, Elliptical copulas (Gaussian, Student-*t* copulas) and Archimedian copulas (Clayton, Gumbel, Frank, Joe,...) are classes of symmetric copulas. Thus, if we apply (*<sup>X</sup>*1, *<sup>X</sup>*2) with these distributions and these copulas, Proposition 7 tells us that the random variable *Z* := *X*1 *X*1+*X*2 always gets the median one-half. This theoretical result is consistent with our simulation results displayed in the next section.

We note that the formulas given in (8), (9), (19) and (20) may not have closed forms. However, they are easily computed by using the Monte Carlo (MC) simulation method or any techniques of numerical integration. We provide simulation studies in the next section.

### **5. A Simulation Study**

Since the density and the CDF formula of the ratio *Y* = *X*1 *X*2 [*Z* = *X*1 *X*1+*X*2 ] expressed in (8) and (9) ((19) and (20)) are in terms of integrals and are very complicated, we cannot obtain the exact forms of the density and the CDF. To circumvent the difficulty, in this paper, we propose to use the Monte Carlo algorithm, numerical analysis and graphical approach to study behavior of density and distribution and the changes of their shapes when parameters are changing.

Suppose that *X*1 and *X*2 are normally distributed and denoted by *Xi* ∼ *<sup>N</sup>*(*μi*, *σ*<sup>2</sup> *i* ) for *i* = 1, 2 with PDF given by

$$f\_{\mathcal{X}\_i}(\mathbf{x}) = \frac{1}{\sqrt{2\pi}\sigma\_i} \exp\left(-\frac{(\mathbf{x} - \mu\_i)^2}{2\sigma\_i^2}\right).$$

Without loss of generality, we consider *μ*1 = *μ*2 = 0 and *σ*1 = *σ*2 = 1. We note that, if *X*1 and *X*2 are independent and standard normal distributed, then it is well known that *Y* = *X*1 *X*2 follows standard Cauchy distribution. The Cauchy distribution is a type of distribution that has no mean and also does not exist any higher moments. To circumvent this problem, one may use median to measure the central tendency and use range or interquartile range to measure the spread of the distribution. The general Cauchy distribution has the following PDF:

$$f(\mathbf{x}) := \frac{\beta}{\pi(\beta^2 + (\mathbf{x} - \mathbf{a})^2)}, \quad \alpha \in \mathbb{R}, \beta > 0.$$

Thus, location and scale parameter of *Y* = *X*1 *X*2 are *α* = 0 and *β* = 1, respectively, if *X*1 and *X*2 are identically independent standard normal distributed.

We now investigate different dependence structures of *X*1 and *X*2 through several families of copulas and observe the shapes of the corresponding distributions for both *Y* and *Z* as well as estimate their percentiles at different levels including 0.05, 0.25, 0.5, 0.75, 0.95 and denote the corresponding percentiles to be *Q*0.05, *Q*0.25, *Q*0.50, *Q*0.75, and *Q*0.95, respectively. In risk analysis, the percentile *Q*0.05 is often used as the Value-at-Risk (VaR) 5% while *Q*0.25 and *Q*0.75 are, respectively, called the first and third quartile of the random variable and the interquartile range (IQR) is defined by their difference; that is, *IQR* = *Q*0.75 − *Q*0.25. The median measures the center of the distribution, which is equal to *Q*0.5.

For each copula *Cθ* (*<sup>u</sup>*, *<sup>v</sup>*), the PDF and CDF of *Y* and *Z* can be plotted on the interval [−4, 4] by using the following steps:

*JRFM* **2019**, *12*, 42


$$\hat{f}\gamma(y) \quad \approx \quad \frac{1}{N} \sum\_{i=1}^{N} |F\_2^{-1}(v\_i)| c\left(F\_1\left(yF\_2^{-1}(v\_i)\right), v\_i\right) f\_1\left(yF\_2^{-1}(v\_i)\right), \tag{38}$$

$$\hat{F}\_Y(y) \quad \approx \quad F\_2(0) + \frac{1}{N} \sum\_{i=1}^N \text{sgn}\left(F\_2^{-1}(v\_i)\right) \frac{\partial}{\partial v} \mathbb{C}\left(F\_1\left(y F\_2^{-1}(v\_i)\right), v\_i\right), \tag{39}$$

$$\hat{f}\_{\mathcal{Z}}(z) \approx \begin{cases} \frac{1}{N} \sum\_{i=1}^{N} \frac{|F\_1^{-1}(u\_i)|}{z^2} c\left(u\_i, F\_2\left(\frac{1-z}{z} F\_1^{-1}(u\_i)\right)\right) f\_2\left(\frac{1-z}{z} F\_1^{-1}(u\_i)\right), & \text{if } z \neq 0, \\\\ \frac{f\_1(0)}{N} \sum\_{i=1}^{N} \left|F\_2^{-1}(v\_i)\right| c\left(F\_1(0), v\_i\right), & \text{if } z = 0, \end{cases} \tag{40}$$

$$\hat{F}\_{\mathcal{Z}}(z) \approx \mathbf{1}\_{\{z \ge 0\}} + \frac{1}{N} \sum\_{i=1}^{N} \text{sgn}\left(F\_1^{-1}(u\_i)\right) \left[\frac{\partial \mathbb{C}}{\partial u}\left(\mu\_i, F\_2\left(-F\_1^{-1}(u\_i)\right)\right) - \frac{\partial \mathbb{C}}{\partial t}\left(\mu\_i, F\_2\left(\frac{1-z}{z} F\_1^{-1}(u\_i)\right)\right)\right],\tag{41}$$

in which the density copula *cθ* (*ui*, *vi*), the derivatives *∂∂uC<sup>θ</sup>* (*ui*, *vi*) and *∂∂vC<sup>θ</sup>* (*ui*, *vi*) can be obtained by using the packages of *VineCopula* in *R* language; and

### (iii) plot *f* . *<sup>Y</sup>*(*y*), *F* . *<sup>Y</sup>*(*y*), *f* . *<sup>Z</sup>*(*z*), and *F* . *Z*(*z*) with *y*, *z* ∈ {−4, −3.9, −3.8, ··· , 3.9, <sup>4</sup>}.

To estimate percentiles *Qα*'s of *Y* and *Z*, we first construct the joint distribution of (*<sup>X</sup>*1, *<sup>X</sup>*2) by using Sklar's Theorem as shown in the following: For each copula *Cθ* (*<sup>u</sup>*, *<sup>v</sup>*), we first obtain the joint CDF of (*<sup>X</sup>*1, *<sup>X</sup>*2) such that

$$H\_{\theta}(\mathfrak{x}\_1, \mathfrak{x}\_2) = \mathbb{C}\_{\theta} \left( F\_1(\mathfrak{x}\_1), F\_2(\mathfrak{x}\_2) \right).$$

We then repeat 5000 times, *k* = 1, 2, . . . , 5000 for the following steps in the computation:

	- (i) generate (*<sup>X</sup>*1, *<sup>X</sup>*2) from *Hθ* (*<sup>x</sup>*1, *<sup>x</sup>*2) of sample size 10<sup>4</sup> by using the package *copula* in *R* language and define

$$y\_i^{(k)} = \begin{array}{c} x\_{1i}^{(k)} \\ x\_{2i}^{(k)'} \end{array} \quad i = 1, 2, \cdots, 10^4; \tag{42}$$

$$z\_i^{(k)} = \begin{array}{c} \frac{x\_{1i}^{(k)}}{x\_{1i}^{(k)} + x\_{2i}^{(k)'}} \quad i = 1, 2, \cdots, 10^4; \tag{43}$$

 (ii) estimate the percentiles *Qα* with *α* = 0.05, 0.25, 0.5, 0.75, 0.95 for both *Y* and *Z* by using the following formula

$$
\begin{aligned}
\widehat{Q}\_{\mathfrak{k}}(Y^{(k)}) &= \quad \mathcal{Y}^{(k)}\_{(\lfloor h \rfloor)} + (h - \lfloor h \rfloor) \left( \mathcal{Y}^{(k)}\_{(\lfloor h \rfloor + 1)} - \mathcal{Y}^{(k)}\_{(\lfloor h \rfloor)} \right), \quad \text{with } h = \left( 10^4 - 1 \right) a + 1, \text{ (44)} \\
\widehat{Q}\_{\mathfrak{k}}(Z^{(k)}) &= \quad \mathcal{Z}^{(k)}\_{(\lfloor h \rfloor)} + (h - \lfloor h \rfloor) \left( \mathcal{Z}^{(k)}\_{(\lfloor h \rfloor + 1)} - \mathcal{Z}^{(k)}\_{(\lfloor h \rfloor)} \right), \end{aligned}
\tag{45}
$$

where *y*(*k*) (1) ≤ *y*(*k*) (2) ≤ ... ≤ *y*(*k*) (*N*) and *z*(*k*) (1) ≤ *z*(*k*) (2) ≤ ... ≤ *z*(*k*) (*N*) denote the order statistics of both *y*(*k*) and *<sup>z</sup>*(*k*), respectively, and *h* denotes integer part of *h*.

(2) Finally, we take average for each of the above quantities by using the following formula:

$$\begin{array}{rcl}\widehat{Q}\_{\mathfrak{a}}(\mathcal{Y})&=&\frac{1}{5000}\sum\_{k=1}^{5000}\widehat{Q}\_{\mathfrak{a}}(\mathcal{Y}^{(k)}),\\\widehat{Q}\_{\mathfrak{a}}(Z)&=&\frac{1}{5000}\sum\_{k=1}^{5000}\widehat{Q}\_{\mathfrak{a}}(Z^{(k)}),\end{array}$$

to obtain the estimates of the percentiles for *Y* and *Z*.

We note that the algorithms discussed in the above can be applied to any non-symmetric marginal distribution and skewed copulas family with absolutely continuous random variable that could contain negative range, for example, generalized skewed-*t* distribution and skewed-*t* Copulas. 2
