*3.5. Order Statistics*

Assume that *X*1, ... , *Xn* is a random sample (RS) from the OGE2-G . Furthermore, assume that *Xi*:*n* denote the *i*th order statistic (OS). Consequently, pdf of *Xi*:*n* may be interpreted as

$$\begin{aligned} f\_{i:n}(\mathbf{x}) &= \frac{1}{\beta(i, n-i+1)} f(\mathbf{x}) F(\mathbf{x})^{i-1} \left\{ 1 - F(\mathbf{x}) \right\}^{n-i} . \\ &= \frac{1}{\beta(i, n-i+1)} \sum\_{j=0}^{n-i} (-1)^j \binom{n-i}{j} f(\mathbf{x}) F(\mathbf{x})^{j+i-1} . \end{aligned}$$

Inserting Equations (1) and (2) in the last equation, and expanding it as in Section 3.1, we ge<sup>t</sup>

$$f\_{\rm i:n}(\mathbf{x}) = \sum\_{j=0}^{n-i} \eta\_j \, h\_{\rm m}(\mathbf{x}) \, . \tag{14}$$

where

$$\eta\_{\vec{\jmath}} = \frac{(-1)^{\vec{\jmath}}}{\beta(\vec{\imath}, n-\vec{\imath}+1)} \binom{n-\vec{\imath}}{\vec{\jmath}} \sum\_{m=0}^{\infty} \xi\_m^\* \eta\_m$$

and

$$\xi\_m^\* = (-1)^m \sum\_{i,k=0}^{\infty} \sum\_{l=0}^i \frac{(-1)^{i+k+l} (k+1)^i \Gamma\{\beta(i+j)\}}{i! k! \Gamma\{\beta(i+j) - k\}} \binom{i}{l} \binom{-a(i+l+1)-1}{m}.$$

*3.6. Stress-Strength Reliability*

Supp *X*1 ∼ OGE2-G(*<sup>α</sup>*, *β*1; *ψ*) and *X*2 ∼ OGE2-G(*<sup>α</sup>*, *β*2; *ψ*) are two continuous rvs with pdfs *f*1(*x*) and *f*2(*x*) and cdfs *<sup>F</sup>*1(*x*) and *<sup>F</sup>*2(*x*), therefore the reliability *R* is supplied via

$$R = \mathbb{P}(X\_1 > X\_2) = \int\_0^\infty f\_1(\mathbf{x}) \, F\_2(\mathbf{x}) \, d\mathbf{x}.\tag{15}$$

**Theorem 2.** *Assume that X*1 *and X*2 *are two independent rvs established previously with constant parameters β*1 *and β*<sup>2</sup>*. Eventually,*

$$\mathcal{R}\_{-}=\beta\_{1}\sum\_{i=0}^{\infty}{\binom{\beta\_{1}-1}{i}}(-1)^{i}\Gamma(i+1)-\beta\_{1}\sum\_{i=0}^{\infty}{\binom{\beta\_{1}-1}{i}}\binom{\beta\_{2}}{j}(-1)^{i+j}\Gamma(i+j+1). \text{(16)}$$

**Proof.** Using Equations (1) and (2) in Equation (15), we have

$$\begin{split} \int\_{0}^{\infty} f\_{1}(\mathbf{x}) F\_{2}(\mathbf{x}) \, d\mathbf{x} &= \quad \int\_{0}^{\infty} a \beta\_{1} k(\mathbf{x}; \boldsymbol{\psi}) \overline{\mathcal{K}}(\mathbf{x}; \boldsymbol{\psi})^{-(a+1)} \mathbf{e}^{-\left\{\frac{1-\overline{\mathcal{K}}(\mathbf{x}; \boldsymbol{\psi})^{a}}{\overline{\mathcal{K}}(\mathbf{x}; \boldsymbol{\psi})^{a}}\right\}} \\ &\times \sum\_{i=0}^{\infty} \binom{\beta\_{1}-1}{i} (-1)^{i} \left\{\frac{1-\overline{\mathcal{K}}(\mathbf{x}; \boldsymbol{\psi})^{a}}{\overline{\mathcal{K}}(\mathbf{x}; \boldsymbol{\psi})^{a}}\right\}^{i} \\ &\times \left[1-\sum\_{j=0}^{\infty} \binom{\beta\_{2}}{j} (-1)^{j} \left\{\frac{1-\overline{\mathcal{K}}(\mathbf{x}; \boldsymbol{\psi})^{a}}{\overline{\mathcal{K}}(\mathbf{x}; \boldsymbol{\psi})^{a}}\right\}^{j}\right] d\mathbf{x}. \end{split}$$

Equation (16) follows immediately after solving the integral with any mathematical software.
