The existence and uniqueness of the solution of the MLE equation have already been proved in Balakrishnan and Kateri (2008) and Farnum and Booth (1997) using Cauchy-Schwarz inequality. A different proof is presented here, leading to our proposed analytical solution for the \(\beta\) parameter.

Let us denote:

$$\begin{aligned} x_{k}& = {} \left( \frac{t_{k}}{t_{N}}\right) ^{\beta _{1}},\qquad k=\overline{1,N}, \\ \zeta& = {} \frac{\beta }{\beta _{1}}, \end{aligned}$$

(8)

where \(0<x_{k}\le 1\) and \(x_{N}=1\). Hence, it can be seen that

$$\begin{aligned} Z(\zeta )\equiv 1-\frac{1}{\zeta } +r(\zeta )=0, \end{aligned}$$

(9)

after multiplying Eq. (6) by \(\beta _{1}\). Here,

$$\begin{aligned} r(\zeta )\equiv \frac{ \sum \nolimits _{k=1}^{N} \left( \ln x_{k} \right) x_{k}^{\zeta }}{ \sum \nolimits _{k=1}^{N} x_{k}^{\zeta }}. \end{aligned}$$

(10)

In order to prove the existence and uniqueness of the solution of Eq. (9), the global monotonicity of \(Z(\zeta )\) and its asymptotic behavior, in the limits \(\zeta \rightarrow 0+\) and \(\zeta \rightarrow +\infty\), is developed. Let us consider that \(0<x_{1}\le x_{2}\le \cdots \le x_{N}=1\), where \(N\ge 2\), is a monotonous nondecreasing data sequence. Also, let us suppose that there must be at least two different statistical data sets, i.e.,

$$\begin{aligned} \exists \, m\in \{\overline{1,N-1}\}: x_{m}<x_{N}=1 \quad {\mathrm{and}} \quad x_{m+1}=x_{N}=1. \end{aligned}$$

(11)

It can be proved that \(r(\zeta )\) is continuous and monotonously increasing on \(\zeta >0\), since the derivative of \(r(\zeta )\) is positive for all \(\zeta >0\):

$$\begin{aligned} \frac{dr(\zeta )}{d\zeta }& = {} \frac{ \left( \sum \nolimits _{k=1}^{N}(\ln x_{k})^2 x_{k}^{\zeta } \right) \left( \sum \nolimits _{k=1}^{N} x_{k}^{\zeta } \right) - \left( \sum \nolimits _{k=1}^{N} (\ln x_{k}) x_{k}^{\zeta } \right) ^{2}}{ \left( \sum \nolimits _{k=1}^{N} x_{k}^{\zeta } \right) ^2 }\\& = {} \frac{ \sum \nolimits _{x_{i}>x_{j}} \left( \ln \frac{x_{i}}{x_{j}} \right) ^{2} x_{i}^{\zeta } x_{j}^{\zeta }}{ 2 \left( \sum \nolimits _{k=1}^{N} x_{k}^{\zeta } \right) ^2 } >0. \end{aligned}$$

This implies that \(Z(\zeta )\) is also continuous and monotonously increasing on \(\zeta >0\). On the other hand, \(r(\zeta )\) is bounded and its boundaries can be obtained from the asymptotic behavior of \(r(\zeta )\) in the limits \(\zeta \rightarrow 0+\) and \(\zeta \rightarrow +\infty\). In the limit \(\zeta \rightarrow 0+\), the asymptotic series expansions of \(\zeta\) is given by:

$$\begin{aligned} \sum \limits _{k=1}^{N} x_{k}^{\zeta }=\sum \limits _{k=1}^{N} e^{\zeta \ln x_{k}} = \sum \limits _{k=1}^{N} \left( 1+{\mathcal{O}}(\zeta ) \right) = N\left( 1+{\mathcal{O}}(\zeta ) \right) \end{aligned}$$

and

$$\begin{aligned} \sum \limits _{k=1}^{N} \left( \ln x_{k} \right) x_{k}^{\zeta } = \sum \limits _{k=1}^{N} \left( \ln x_{k} \right) \left( 1+{\mathcal{O}}(\zeta ) \right) \end{aligned}$$

where \({\mathcal{O}}(\zeta )\) is the big Landau O notation.

Substitution of these results into (10) yields the following asymptotic equation for \(r(\zeta )\) in the limit \(\zeta \rightarrow 0+\):

$$\begin{aligned} r(\zeta ) = \frac{1}{N}\sum \limits _{k=1}^{N} \ln x_{k}+{\mathcal{O}}(\zeta ) \end{aligned}$$

(12)

which states that \(r(\zeta )\rightarrow \frac{1}{N}\sum \nolimits _{k=1}^{N}\ln x_{k}\) as \(\zeta \rightarrow 0+\).

Now, it can be shown that \(r(\zeta )\rightarrow 0\) as \(\zeta \rightarrow +\infty\). Hence,

$$\begin{aligned} \sum _{k=1}^{N} x_k^{\zeta }=x_1^{\zeta }+x_2^{\zeta }+\cdots +x_m^{\zeta }+(N-m)= (N-m) \Bigl \{ 1+{\mathcal{O}} \left( x_{m}^{\zeta } \right) \Bigr \} \end{aligned}$$

due to (11) and noticing that \(x_{k}=x_{N}=1\) for all \(m<k\le N\) and \(0<x_{m}<1\), \(x_m^{\zeta }\rightarrow 0\) as \(\zeta \rightarrow +\infty\). Therefore, Eq. (10) and this last result, along with the fact that \(\ln x_{k}=0\) for \(k>m\), show that the asymptotic expansion for \(r(\zeta )\) in the limit \(\zeta \rightarrow +\infty\) is given by:

$$\begin{aligned} r(\zeta ) = \frac{ \sum \nolimits _{k=1}^m \left( \ln x_{k} \right) x_k^{\zeta } }{ (N-m) \Bigl \{ 1+{\mathcal{O}} \left( x_{m}^{\zeta } \right) \Bigr \} } = \frac{1}{N-m}\sum \limits _{k=1}^m \left( \ln x_{k} \right) x_{k}^{\zeta } \Bigl \{ 1+{\mathcal{O}} \left( x_{m}^{\zeta } \right) \Bigr \} = {\mathcal{O}} \left( x_{m}^{\zeta } \right) , \end{aligned}$$

(13)

i.e., \(r(\zeta )\rightarrow 0\) as \(\zeta \rightarrow +\infty\).

It follows from the asymptotic equations (12) and (13) that \(Z(\zeta )\) is a continuous and monotonously increasing function for \(\zeta >0\) with the following asymptotic behavior:

$$\begin{aligned} Z(\zeta ) = 1-\frac{1}{\zeta }+\frac{1}{N}\sum \limits _{k=1}^{N} \ln x_{k}+{\mathcal{O}}(\zeta )\rightarrow -\infty \quad {\mathrm{as}}\quad \zeta \rightarrow 0+, \end{aligned}$$

(14)

$$\begin{aligned} Z(\zeta )=1-\frac{1}{\zeta }+{\mathcal{O}} \left( x_{m}^{\zeta } \right) \rightarrow 1 \quad {\mathrm{as}}\quad \zeta \rightarrow +\infty . \end{aligned}$$

(15)

As a result, there exists a unique value \(\zeta ^*\) such that \(Z(\zeta ^*)=0\) since (14) and (15).

Let us determine the solution interval of \(\zeta ^{*}\). It follows from Eq. (9) that \(0=Z(\zeta ^*)<1-{\zeta ^*}^{-1}\) since \(r(\zeta )<0\). As a result \(\zeta ^*>1\). On the other hand, it can be seen that \(r(1)<r(\zeta ^*)=-1+{\zeta ^*}^{-1}\) for \(\zeta ^*>1\), i.e., \({\zeta ^*}^{-1}>1+r(1)\) since \(r(\zeta )\) is monotonously increasing. At the same time, \({\zeta ^*}^{-1}>0\). Therefore, \({\zeta ^*}^{-1}>\max \left\{ 1+r(1),0\right\}\), where

$$\begin{aligned} r(1)=\frac{ \sum \nolimits _{k=1}^{N} \left( \ln x_{k} \right) x_{k}}{ \sum \nolimits _{k=1}^{N} x_{k}}. \end{aligned}$$

(16)

Summarizing, it can be concluded that

$$\begin{aligned} 1+r(1)<\frac{1}{\zeta ^{*}}<1. \end{aligned}$$

(17)

The asymptotical behavior of \(Z(\zeta )\) (in the limits \(\zeta \rightarrow 0+\) and \(\zeta \rightarrow +\infty\)) and \(\zeta ^{*}\) are illustrated in Fig. 1. However, the expression \(1+r(1)\) can be negative for some statistical data sets. In this case, the boundary estimation for \(\zeta ^{*}\) becomes:

$$\begin{aligned} 1<\zeta ^{*}<\infty . \end{aligned}$$

(18)

It can be observed that this interval is too large. Then, a better estimation of the right boundary is required. For this, let us denote:

$$\begin{aligned} \zeta =1+z, \end{aligned}$$

(19)

where \(z>0\) in virtue of (17). Then, Eq. (9) can be written as:

$$\begin{aligned} (1+z)\sum \limits _{k=1}^{N} (\ln x_{k})x_{k}^{1+z} + z \sum \limits _{k=1}^{N} x_{k}^{1+z}=0. \end{aligned}$$

(20)

The solution \(\beta ^{*}\) is found from Eq. (20), which can be substituted into \(\beta ^{*}=\beta _{1}\,\zeta ^{*}=\beta _{1} (1+z)\) according to Eqs. (8) and (19). This way, \(\alpha ^{*}\) can be calculated by substituting the estimated \(\beta ^{*}\) value into Eq. (5).