4.1.2. Score Test

The statistic for the ST (say SR) to test *H*(*j*) 0 , *j* = 1, 2, 3, is defined as

$$\mathbf{S}\mathbf{R}\_{\circ} = \left[\mathbf{S}\left(\widehat{\theta}\_{0\circ}\right)\right]^{\top}\left[IF\left(\widehat{\theta}\_{0\circ}\right)\right]^{-1}\mathbf{S}\left(\widehat{\theta}\_{0\circ}\right)\mathbf{J}$$

where *θ j* is the ML estimator under *H*(*j*) 0 . Under *H*(*j*) 0 , *j* = 1, 2, SR*j* ∼ *<sup>χ</sup>*<sup>2</sup>(1) and under *H*(3) 0 , SR3 ∼ *<sup>χ</sup>*<sup>2</sup>(2).

4.1.3. Gradient Test

> The statistic for the GT (say ST) to tests *H*(*j*) 0 , *j* = 1, 2, 3, is defined as

$$\mathrm{ST}\_{\dot{\jmath}} = \mathcal{S}(\widehat{\theta}\_{\dot{\jmath}}) \left( \widehat{\theta} - \widehat{\theta}\_{\dot{\jmath}} \right) \dots$$

Again, under *H*(*j*) 0 , *j* = 1, 2, ST*j* ∼ *<sup>χ</sup>*<sup>2</sup>(1) and under *H*(3) 0 , ST3 ∼ *<sup>χ</sup>*<sup>2</sup>(2). After some algebraic manipulations, we obtain that

$$\begin{split} \mathrm{ST}\_{1} &= (\widehat{\boldsymbol{n}} - 1) \left\{ \boldsymbol{n} \left[ 1 - \log \left( \widehat{\boldsymbol{v}}\_{01} \right) \right] + \sum\_{i=1}^{n} \left[ \log z\_{i} - \left( \left( \frac{z\_{i}}{\widehat{\boldsymbol{v}}\_{01}} \right) - \widehat{\boldsymbol{\lambda}}\_{01} \right) \left( \frac{z\_{i}}{\widehat{\boldsymbol{v}}\_{01}} \right) \log \left( \frac{z\_{i}}{\widehat{\boldsymbol{v}}\_{01}} \right) \right] \right\}. \\ \mathrm{ST}\_{2} &= \widehat{\boldsymbol{\lambda}} \left[ -n \sqrt{\frac{2}{\pi}} + \sum\_{i=1}^{n} \left( \frac{z\_{i}}{\widehat{\boldsymbol{v}}\_{02}} \right)^{\widehat{\boldsymbol{v}}\_{02}} \right]. \\ \mathrm{ST}\_{3} &= (\widehat{\boldsymbol{n}} - 1) \left\{ n \left[ 1 - \log \left( \widehat{\boldsymbol{v}}\_{03} \right) \right] + \sum\_{i=1}^{n} \left[ \log z\_{i} - \left( \frac{z\_{i}}{\widehat{\boldsymbol{v}}\_{03}} \right)^{2} \log \left( \frac{z\_{i}}{\widehat{\boldsymbol{v}}\_{03}} \right) \right] \right\} + \widehat{\boldsymbol{\lambda}} \left[ -n \sqrt{\frac{2}{\pi}} + \sum\_{i=1}^{n} \left( \frac{z\_{i}}{\widehat{\boldsymbol{v}}\_{03}} \right) \right]. \end{split}$$
