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Спутники
РН
SpaceX
Роскосмос
NASA
СпутникиРНSpaceXРоскосмосNASAIn this paper, we propose a new efficient estimator for the weighted exponential family and its underlying components. These components constitute a flexible class of distributions within the standard exponential family, characterized by positive support and a generator function. For such models, maximum likelihood estimators (MLEs) are often unavailable in closed form and must be derived through numerical optimization. To address this limitation, an asymptotically efficient closed-form estimator was developed for these distributions. Monte Carlo simulations demonstrate that the proposed estimator achieves performance nearly identical to the numerically computed MLE while consistently outperforming previously proposed closed-form estimators.
As stated in Eq. (4), the score component \(S_i(x)\) can be expressed as follows:
$$\begin{aligned} S_i(x)&=: a_i(\varvec{\theta }) + b_{1i}(\varvec{\theta })\,T(x) + b_{2i}(\varvec{\theta })\,\log T(x), \end{aligned}$$
(10)
where \(a_i(\varvec{\theta })\) denotes the negative of the \(\theta _i\)-derivative of the log-normalizing constant \(Z(\varvec{\theta })\), while \(b_{1i}(\varvec{\theta })\) and \(b_{2i}(\varvec{\theta })\) denote the \(\theta _i\)-derivatives of the natural parameters \(\eta _1(\varvec{\theta })\) and \(\eta _2(\varvec{\theta })\), respectively. Finally, let \(m_1\) denote \(\mathbb {E}_\theta [T(X)]\) and \(m_2\) denote \(\mathbb {E}_\theta [\log T(X)]\). Differentiating the expectation of the score component with respect to \(\theta _j\) yields the following expression:
$$\begin{aligned} \frac{\partial }{\partial \theta _j}\,\mathbb {E}_{\varvec{\theta }}[S_i(X)]&= \frac{\partial }{\partial \theta _j}\,a_i(\varvec{\theta }) + \frac{\partial }{\partial \theta _j}\,b_{1i}(\varvec{\theta })\,m_1 + \frac{\partial }{\partial \theta _j}\,b_{2i}(\varvec{\theta })\,m_2 \nonumber \\&\quad + b_{1i}(\varvec{\theta })\,\frac{\partial }{\partial \theta _j}m_1 + b_{2i}(\varvec{\theta })\,\frac{\partial }{\partial \theta _j}m_2, \end{aligned}$$
(11)
where
$$\begin{aligned} \frac{\partial }{\partial \theta _j}m_1&= \int \frac{\partial }{\partial \theta _j}\,T(x)\,f(x;\varvec{\theta })\,dx = \int T(x)\,S_j(x)\,f(x;\varvec{\theta })\,dx \nonumber \\&= \mathbb {E}_{\varvec{\theta }}\!\big [T(X)\,S_j(X)\big ], \end{aligned}$$
(12)
$$\begin{aligned} \frac{\partial }{\partial \theta _j}m_2&= \int \frac{\partial }{\partial \theta _j}\,\log T(x)\,f(x;\varvec{\theta })\,dx = \int \log T(x)\,S_j(x)\,f(x;\varvec{\theta })\,dx \nonumber \\&= \mathbb {E}_{\varvec{\theta }} \!\big [\log T(X)\,S_j(X)\big ]. \end{aligned}$$
(13)
To justify the interchange of differentiation and integration in the above identities, dominating functions must be specified for \(\bigl |\frac{\partial }{\partial \theta _j}T(x)\,f(x;\varvec{\theta })\bigr |\) and \(\bigl |\frac{\partial }{\partial \theta _j}\log T(x)\,f(x;\varvec{\theta })\bigr |\) in accordance with Theorem 2.3.
To derive dominating functions, first, the common factor \(\frac{\partial }{\partial \theta _j}\,f(x;\varvec{\theta })\) must be derived. Using Eq. (3), the \(\theta _j\)-derivatives of the WEF can be represented as follows:
$$\begin{aligned} \frac{\partial }{\partial \theta _j}\,f(x;\varvec{\theta }) = f(x;\varvec{\theta }) \left[ b_{1j}(\varvec{\theta })T(x) + b_{2j}(\varvec{\theta })\log (T(x)) -\frac{1}{Z(\varvec{\theta })}\frac{\partial }{\partial \theta _j}\,Z(\varvec{\theta }) \right] , \end{aligned}$$
and \(\frac{1}{Z(\varvec{\theta })}\frac{\partial }{\partial \theta _j}\,Z(\varvec{\theta })\) can be simply obtained as \(b_{1j}(\varvec{\theta }) m_1 + b_{2j}(\varvec{\theta }) m_2\) using the property of the score component in Eqs. (5) and (10). Finally, using triangle inequality and \(T(x) > 0\), the dominating functions can be derived.
$$\begin{aligned} \bigl |T(x)\bigr | \left| \frac{\partial }{\partial \theta _j}\,f(x;\varvec{\theta }) \right|&= f(x;\varvec{\theta })T(x)\, \Bigl |\, b_{1j}(\varvec{\theta })(T(x)-m_1) + b_{2j}(\varvec{\theta })(\log T(x)-m_2)\Bigr |\\&\le f(x;{\varvec{\theta }})\Bigl ( \bigl |b_{1j}(\varvec{\theta })\bigr |\,T(x)^2 + \bigl |b_{2j}(\varvec{\theta })\bigr |\,T(x)\,|\log T(x)|\\&\quad + \bigl |b_{1j}(\varvec{\theta })\bigr |\,\bigl |m_1\bigr |\,T(x) + \bigl |b_{2j}(\varvec{\theta })\bigr |\,\bigl |m_2\bigr |\,T(x) \Bigr )\\&=: g_1(x), \end{aligned}$$
$$\begin{aligned} \bigl |\log T(x)\bigr | \left| \frac{\partial }{\partial \theta _j}\,f(x;\varvec{\theta }) \right|&=f(x;\varvec{\theta })\, |\log T(x)|\,|\, b_{1j}(\varvec{\theta })(T(x)-m_1) \\ &\quad + b_{2j}(\varvec{\theta })(\log T(x) - m_2)\bigr |\\&\le f(x;\varvec{\theta })\Bigl ( \bigl |b_{1j}(\varvec{\theta })\bigr |\,T(x)\,\bigl |\log T(x)\bigr | + \bigl |b_{2j}(\varvec{\theta })\bigr |\,\bigl (\log T(x)\bigr )^2 \\&\quad + \bigl |b_{1j}(\varvec{\theta })\bigr |\,\bigl |m_1\bigr |\,\bigl |\log T(x)\bigr | + \bigl |b_{2j}(\varvec{\theta })\bigr |\,\bigl |m_2\bigr |\,\bigl |\log T(x)\bigr | \Bigr )\\&=: g_2(x), \end{aligned}$$
Under the WEF assumptions, \(\mathbb {E}_{\varvec{\theta }} [T(X)^2]\), \(\mathbb {E}_{\varvec{\theta }} [T(X) \left| \log T(X) \right| ]\), \(\mathbb {E}_{\varvec{\theta }} [(\log T(X))^2] < \infty \), we have \(g_1(x)\), \(g_2(x) \in L^1\). Hence, the interchange of differentiation and integration is justified by Theorem 2.3. Now, substituting Eqs. (12) and (13) into Eq. (11) yields:
$$\begin{aligned} \frac{\partial }{\partial \theta _j}\,\mathbb {E}_{\varvec{\theta }}[S_i (X)]&= \frac{\partial }{\partial \theta _j}a_i(\varvec{\theta }) + \frac{\partial }{\partial \theta _j}\,b_{1i}(\varvec{\theta })\,m_1 + \frac{\partial }{\partial \theta _j}\,b_{2i}(\varvec{\theta })\,m_2 \nonumber \\&\quad + b_{1i}(\varvec{\theta })\,\mathbb {E}_{\varvec{\theta }}[T(X)\,S_j(X)] + b_{2i}(\varvec{\theta })\,\mathbb {E}_{\varvec{\theta }} [\log T(X)\,S_j(X)]. \end{aligned}$$
(14)
To conclude the proof, differentiating the score component \(S_i(x)\) with respect to another parameter \(\theta _j\) yields:
$$\begin{aligned} \frac{\partial }{\partial \theta _j}\,S_i(x) = \frac{\partial }{\partial \theta _j}\,a_i(\varvec{\theta )} + \frac{\partial }{\partial \theta _j}\,b_{1i}(\varvec{\theta )}T(x) + \frac{\partial }{\partial \theta _j}\,b_{2i}(\varvec{\theta )}\log T(x). \end{aligned}$$
(15)
Taking the expectation for Eq. (15) and substituting it into Eq. (14) yields:
$$\begin{aligned} \frac{\partial }{\partial \theta _j}\,\mathbb {E}_{\varvec{\theta }} [S_i (X)]&= \mathbb {E}_{\varvec{\theta }} \!\left[ \frac{\partial }{\partial \theta _j}\,S_i(X)\right] + b_{1i}\,\mathbb {E}_{\varvec{\theta }} [T(X)\,S_j(X)]\\ &\quad + b_{2i}\,\mathbb {E}_{\varvec{\theta }} [\log T(X)\,S_j(X)] \nonumber \\&= \mathbb {E}_{\varvec{\theta }} \!\left[ \frac{\partial ^2}{\partial \theta _j\,\partial \theta _i}\log f(X; {\varvec{\theta }})\right] \\ &\quad + \mathbb {E}_{\varvec{\theta }} \!\big [(b_{1i}(\varvec{\theta })T(X)+b_{2i}(\varvec{\theta })\log T(X))\,S_j(X)\big ]. \end{aligned}$$
From Eq. (10), \(b_{1i}(\varvec{\theta })T(x)+b_{2i}(\varvec{\theta })\log T(x)=S_i(x)-a_i(\varvec{\theta })\); hence:
$$\begin{aligned} 0 = \frac{\partial }{\partial \theta _j}\,\mathbb {E}_{\varvec{\theta }} [S_i (X)] = \mathbb {E}_{\varvec{\theta }} \!\left[ \frac{\partial ^2}{\partial \theta _j\,\partial \theta _i}\log f(X; {\varvec{\theta }})\right] + \mathbb {E}_{\varvec{\theta }}\!\big [(S_i(X)-a_i({\varvec{\theta }}))\,S_j(X)\big ]. \end{aligned}$$
Because we already verified that the expectation of the score component is zero using Theorem 2.2, the Fisher information matrix satisfies the following relationship:
$$ \;\mathbb {E}_{\varvec{\theta }} [S_i (X) S_j (X)] = -\,\mathbb {E}_{\varvec{\theta }} \!\left[ \frac{\partial ^2}{\partial \theta _j\,\partial \theta _i}\log f(X; {\varvec{\theta }}) \right] . $$
Let \(K_{\delta _{ab}}:=\frac{\sigma (\mu \sigma )^\mu }{(\sigma +\delta _{ab})\Gamma (\mu )},\) and \(f_W(w)=K_{\delta _{ab}}(1+\delta _{ab}w)w^{\mu -1}e^{-\mu \sigma w}\). Then we can derive the bounds of \(f_W(w)\) as follows:
$$ f_W(w) \le {\left\{ \begin{array}{ll} 2K_{\delta _{ab}}w^{\mu -1}, & 0< w < 1, \\ 2K_{\delta _{ab}}w^{\mu -1+\delta _{ab}}e^{-\mu \sigma w}, & w \ge 1. \end{array}\right. } $$
First, suppose that \(0<w<1\), we have
$$\begin{aligned} 1+\delta _{ab}w \le 1+w < 2, \qquad e^{-\mu \sigma w} \le 1, \end{aligned}$$
Therefore,
$$\begin{aligned} K_{\delta _{ab}}(1+\delta _{ab}w)w^{\mu -1}e^{-\mu \sigma w} \le 2K_{\delta _{ab}}w^{\mu -1}. \end{aligned}$$
Next, suppose that \(w \ge 1\), then
$$\begin{aligned} 1+\delta _{ab}w \le 2w^{\delta _{ab}}, \end{aligned}$$
It follows that
$$\begin{aligned} f_W(w) = K_{\delta _{ab}}(1+\delta _{ab}w)w^{\mu -1}e^{-\mu \sigma w}\le 2K_{\delta _{ab}}w^{\mu -1+\delta _{ab}}e^{-\mu \sigma w}. \end{aligned}$$
This proves the desired bounds. \(\square \)
Proposition B.2Assume that there exist constants \(0<w_0<1<w_1\) and nonnegative functions
$$ a_0,\; b_0:(0,w_0)\rightarrow [0,\infty ), \qquad a_\infty ,\; b_\infty :[w_1,\infty )\rightarrow [0,\infty ), $$
such that, for \(0<w<w_0\):
$$ |q(w)|\le w\,a_0(w), \qquad \left| \frac{q(w)}{w}\right| +|q'(w)|+|\eta _{\delta _{ab}}(w)| \le b_0(w), $$
and for \(w \ge w_1\):
$$ |q(w)|\le w\,a_\infty (w), \qquad \left| \frac{q(w)}{w}\right| +|q'(w)|+|\eta _{\delta _{ab}}(w)| \le b_\infty (w). $$
Assume further that
$$ \lim _{w\downarrow 0} w^\mu a_0(w)=0, \qquad \int _0^{w_0} w^{\mu -1}\bigl \{a_0(w)^2+b_0(w)^2+w^2\bigr \}\,dw<\infty , $$
and
$$ \lim _{w\rightarrow \infty } w^{\mu +\delta _{ab}}e^{-\mu \sigma w}a_\infty (w)=0, $$
$$ \int _{w_1}^{\infty } w^{\mu -1+\delta _{ab}}e^{-\mu \sigma w} \bigl \{a_\infty (w)^2+b_\infty (w)^2+w^2\bigr \}\,dw <\infty . $$
Then conditions (i) and (ii) of Theorem 2.7 hold.
ProofFirst, by Lemma B.1, for \(0<w<w_0\),
$$\begin{aligned} |q(w)f_W(w)| \le 2K_{\delta _{ab}}\,w^\mu a_0(w), \quad \text {so that} \quad \lim _{w\downarrow 0}q(w)f_W(w)=0, \end{aligned}$$
Similarly, for \(w\ge w_1\),
$$\begin{aligned} |q(w)f_W(w)| \le 2K_{\delta _{ab}}\,w^{\mu +\delta _{ab}}e^{-\mu \sigma w}a_\infty (w), \quad \text {so that} \quad \lim _{w\rightarrow \infty }q(w)f_W(w)=0. \end{aligned}$$
Therefore, condition (ii) of Theorem 2.7 holds. Next, by the definition of \(\varvec{U}_1\), it is enough to show that quantities below are finite.
$$ \mathbb {E}\!\left[ \eta _{\delta _{ab}}(W)^2\right] , \qquad \mathbb {E}\!\left[ \left( \frac{q(W)}{W}\right) ^2\right] , \qquad \mathbb {E}\!\left[ q(W)^2\right] , \qquad \mathbb {E}(W^2). $$
For \(0<w<w_0\), Lemma B.1 gives
$$\begin{aligned} \eta _{\delta _{ab}}(w)^2f_W(w)&\le 2K_{\delta _{ab}}\,b_0(w)^2w^{\mu -1}, \qquad \left( \frac{q(w)}{w}\right) ^2f_W(w) \le 2K_{\delta _{ab}}\,b_0(w)^2w^{\mu -1},\\ q(w)^2f_W(w)&\le 2K_{\delta _{ab}}\,a_0(w)^2w^{\mu +1}, \qquad w^2f_W(w)\le 2K_{\delta _{ab}}w^{\mu +1}, \end{aligned}$$
These are integrable on \((0,w_0)\) by assumption. For \(w\ge w_1\), Lemma B.1 yields
$$ \eta _{\delta _{ab}}(w)^2f_W(w) \le 2K_{\delta _{ab}}\,b_\infty (w)^2w^{\mu -1+\delta _{ab}}e^{-\mu \sigma w}, $$
$$ \left( \frac{q(w)}{w}\right) ^2f_W(w) \le 2K_{\delta _{ab}}\,b_\infty (w)^2w^{\mu -1+\delta _{ab}}e^{-\mu \sigma w}, $$
$$ q(w)^2f_W(w) \le 2K_{\delta _{ab}}\,a_\infty (w)^2w^{\mu +1+\delta _{ab}}e^{-\mu \sigma w}, $$
and
$$ w^2f_W(w)\le 2K_{\delta _{ab}}w^{\mu +1+\delta _{ab}}e^{-\mu \sigma w}. $$
These are integrable on \([w_1,\infty )\) by assumption. Therefore, condition (i) of Theorem 2.7 also holds. \(\square \)
Proposition B.3(Logarithmic class) Let \(T(x)=\bigl [\log (1+x^{\pm c})\bigr ]^r\) for \(c,r>0\). This class includes the new weighted exponentiated Lindley, new weighted exponentiated Nakagami, new exponentiated generalized gamma, new exponentiated generalized inverse gamma, Burr type XII, and Dagum distributions. Then, for both \(\delta _{ab}=0\) and \(\delta _{ab}=1\), conditions (i) and (ii) of Theorem 2.7 hold.
ProofBy Proposition B.2, it suffices to construct bounding functions \(a_0, b_0, a_\infty , b_\infty \) satisfying the conditions therein. Set \(u:=w^{1/r}\), then a direct calculation gives:
$$\begin{aligned} q(w)&=r\,u^{r-1}(1-e^{-u})\log (e^u-1),\\ q'(w)&= (r-1)u^{-1}(1-e^{-u})\log (e^u-1) + e^{-u}\log (e^u-1) + 1. \end{aligned}$$
For \(0<u<1\), the inequalities \(0<1-e^{-u}\le u\) and \(u\le e^u-1\le (e-1)u\) imply \(|\log (e^u-1)|\le C_1(1+|\log u|)\) for some constant \(C_1>0\). Hence, for \(0<w<1\),
$$\begin{aligned} |q(w)|&\le C_2w(1+|\log w|), \qquad \left| \frac{q(w)}{w}\right| \le C_2(1+|\log w|), \\ |q'(w)|&\le C_3(1+|\log w|), \qquad |\eta _{\delta _{ab}}(w)|\le C_4(1+|\log w|). \end{aligned}$$
For \(u\ge 1\), we can derive \(0<1-e^{-u}\le 1\) and \(0<\log (e^u-1)\le u\). Therefore, for \(w\ge 1\),
$$\begin{aligned} |q(w)|&\le C_5w, \qquad \left| \frac{q(w)}{w}\right| \le C_5,\\ |q'(w)|&\le C_6, \qquad |\eta _{\delta _{ab}}(w)|\le C_7. \end{aligned}$$
Thus Proposition B.2 applies with \(a_0(w)=C(1+|\log w|)\), \(b_0(w)=C(1+|\log w|)\) for \(0<w<1\) and \(a_\infty (w)=C\), \(b_\infty (w)=C\) for \(w\ge 1\). The required integrability and limit conditions are immediate, since
$$ \int _0^1 w^{\alpha -1}(1+|\log w|)^m\,dw<\infty , $$
for every \(\alpha >0\) and finite m, while the exponential factor \(e^{-\mu \sigma w}\) dominates polynomial growth on \([1,\infty )\). \(\square \)
Proposition B.4(Exponential class) Let \(T(x)=\bigl [e^{a x^{\pm c}}-1\bigr ]^r\) for \(a,c,r>0\). This class includes the new weighted log-Lindley, new weighted log-Nakagami, new log-generalized gamma, new log-generalized inverse gamma, Gompertz, and modified Weibull extension distributions. Then, for both \(\delta _{ab}=0\) and \(\delta _{ab}=1\), conditions (i) and (ii) of Theorem 2.7 hold.
ProofBy Proposition B.2, it suffices to construct bounding functions \(a_0, b_0, a_\infty , b_\infty \) satisfying the conditions therein. Set \(u:=w^{1/r}\), then a direct calculation gives:
$$\begin{aligned} q(w)&= r\,u^{r-1}(1+u)\log (1+u)\log \!\left( \frac{1}{a}\log (1+u)\right) ,\\ q'(w)&= \bigl (\log (1+u)+1\bigr )\log \!\left( \frac{1}{a}\log (1+u)\right) +1 \\ &\quad +(r-1)u^{-1}(1+u)\log (1+u)\log \!\left( \frac{1}{a}\log (1+u)\right) . \end{aligned}$$
For \(0<u<1\), the inequalities \(\frac{u}{2}\le \log (1+u)\le u\) imply \(\left| \log \!\left( \frac{1}{a}\log (1+u)\right) \right| \le C_1(1+|\log u|).\) Hence, for \(0<w<1\),
$$\begin{aligned} |q(w)|&\le C_2w(1+|\log w|), \qquad \left| \frac{q(w)}{w}\right| \le C_2(1+|\log w|),\\ |q'(w)|&\le C_3(1+|\log w|), \qquad |\eta _{\delta _{ab}}(w)|\le C_4(1+|\log w|). \end{aligned}$$
For \(u\ge 1\), we can derive \(1+u\le 2u\), \(\log (1+u)\le 1+\log u\) and for \(u\ge e\), \(\left| \log \!\left( \frac{1}{a}\log (1+u)\right) \right| \le C_5(1+\log \log u)\). Therefore, for \(w\ge e^r\),
$$ |q(w)|\le C_6w(1+\log w)(1+\log \log w), $$
$$ \left| \frac{q(w)}{w}\right| \le C_6(1+\log w)(1+\log \log w), $$
and
$$ |q'(w)|\le C_7(1+\log w)(1+\log \log w), \qquad |\eta _{\delta _{ab}}(w)|\le C_8(1+\log w)(1+\log \log w). $$
Thus Proposition B.2 applies with \(a_0(w)=C(1+|\log w|)\), \( b_0(w)=C(1+|\log w|)\) for \(0<w<1\) and \(a_\infty (w)=C(1+\log w)(1+\log \log w)\), \(b_\infty (w)=C(1+\log w)(1+\log \log w)\) for \(w\ge e^r\). The required limits and integrability conditions are immediate. \(\square \)
Proposition B.5(First hybrid class) Let \(T(x)=x^b[e^{ax^d}-1]^r\) for \(a,b,d,r>0\). This class includes the new extended log-generalized gamma and traditional Weibull distributions. Then, for both \(\delta _{ab}=0\) and \(\delta _{ab}=1\), conditions (i) and (ii) of Theorem 2.7 hold.
ProofBy Proposition B.2, it suffices to construct bounding functions \(a_0, b_0, a_\infty , b_\infty \) satisfying the conditions therein. Set \(k:=b+rd>0\) and define \(h(x):=xT'(x)/T(x)\), so that:
$$ h(x)=b+r\frac{adx^de^{ax^d}}{e^{ax^d}-1}. $$
For the local bounds, since \(e^{ax^d}-1=ax^d+O(x^{2d})\) as \(x\downarrow 0\), there exist \(x_0\in (0,1)\) and \(C_1>1\) such that \(C_1^{-1}x^k\le T(x)\le C_1x^k\) for \(0<x<x_0\). Hence, for \(0<w<w_0:=T(x_0)\),
$$ C_2^{-1}w^{1/k}\le x\le C_2w^{1/k}, \qquad |\log x|\le C_3(1+|\log w|). $$
Since \(adx^de^{ax^d}/(e^{ax^d}-1)\rightarrow d\) as \(x\downarrow 0\), we have \(h(x)\rightarrow b+rd=k\), so \(|h(x)|\le C_4\) for \(0<x<x_1\) with some \(x_1\in (0,x_0)\). Since \(q(w)=h(x)\,w\log x\) and \(q'(w)=\{xT''(x)/T'(x)\}\log x+\log x+1\) with \(xT''(x)/T'(x)=h(x)-1+xh'(x)/h(x)\), the boundedness of both h(x) and \(xh'(x)\) near the origin yields, for \(0<w<w_1:=T(x_1)\),
$$\begin{aligned} |q(w)|&\le C_5w(1+|\log w|), \qquad \left| \frac{q(w)}{w}\right| \le C_5(1+|\log w|),\\ |q'(w)|&\le C_6(1+|\log w|), \qquad |\eta _{\delta _{ab}}(w)|\le C_7(1+|\log w|). \end{aligned}$$
For the tail bounds, since \(e^{ax^d}-1\sim e^{ax^d}\) as \(x\rightarrow \infty \), there exist \(x_3>1\) and \(C_8>1\) such that \(C_8^{-1}x^be^{rax^d}\le T(x)\le C_8x^be^{rax^d}\) for \(x\ge x_3\). Hence, for \(w\ge w_3:=T(x_3)\),
$$ x^d\le C_9(1+\log w), \qquad \log x\le C_9(1+\log (1+\log w)). $$
Since \(h(x)=b+radx^de^{ax^d}/(e^{ax^d}-1)\asymp x^d\) for large x, it follows that \(|h(x)|\le C_{10}(1+\log w)\), and therefore, for \(w\ge w_3\),
$$\begin{aligned} |q(w)|&\le C_{11}w(1+\log w)(1+\log (1+\log w)),\\ \left| \frac{q(w)}{w}\right|&\le C_{11}(1+\log w)(1+\log (1+\log w)),\\ |q'(w)|&\le C_{12}(1+\log w)(1+\log (1+\log w)),\\ |\eta _{\delta _{ab}}(w)|&\le C_{13}(1+\log w)(1+\log (1+\log w)). \end{aligned}$$
Thus Proposition B.2 applies with \(a_0(w)=b_0(w)=C(1+|\log w|)\) for \(0<w<w_1\) and \(a_\infty (w)=b_\infty (w)=C(1+\log w)(1+\log (1+\log w))\) for \(w\ge w_3\). The required limits and integrability conditions are immediate. \(\square \)
Proposition B.6(Second hybrid class) Let \(T(x)=e^{ax^d-bx^{-e}}\) for \(a,b,d,e>0\). This class includes the new modified log-generalized gamma and flexible Weibull distributions. Then, for both \(\delta _{ab}=0\) and \(\delta _{ab}=1\), conditions (i) and (ii) of Theorem 2.7 hold.
ProofBy Proposition B.2, it suffices to construct bounding functions \(a_0, b_0, a_\infty , b_\infty \) satisfying the conditions therein. Let \(g(x):=ax^d-bx^{-e}\), so that \(T(x)=e^{g(x)}\) and \(\log w = g(x)\), with
$$ g'(x)=adx^{d-1}+bex^{-e-1}, \qquad xg'(x)=adx^d+bex^{-e}, $$
and \(q(w)=xg'(x)\,w\log x\).
For the local bounds, since \(g(x)\sim -bx^{-e}\) as \(x\downarrow 0\), there exist \(x_0\in (0,1)\) and \(C_1>0\) such that \(C_1^{-1}x^{-e}\le |\log w|\le C_1x^{-e}\) for \(0<x<x_0\). Hence, for \(0<w<w_0:=T(x_0)\),
$$ x^{-e}\le C_2(1+|\log w|), \qquad |\log x|\le C_3(1+\log (1+|\log w|)). $$
Since \(xg'(x)=adx^d+bex^{-e}\) and \(x^{-e}\le C_2(1+|\log w|)\), we have \(|xg'(x)|\le C_4(1+|\log w|)\). Since \(q'(w)=(1+xg''(x)/g'(x))\log x + xg'(x)\log x+1\) and \(xg''(x)/g'(x)\) remains bounded for small x, it follows that, for \(0<w<w_0\),
$$\begin{aligned} |q(w)|&\le C_5w(1+|\log w|)(1+\log (1+|\log w|)),\\ \left| \frac{q(w)}{w}\right|&\le C_5(1+|\log w|)(1+\log (1+|\log w|)),\\ |q'(w)|&\le C_6(1+|\log w|)(1+\log (1+|\log w|)),\\ |\eta _{\delta _{ab}}(w)|&\le C_7(1+|\log w|)(1+\log (1+|\log w|)). \end{aligned}$$
For the tail bounds, since \(g(x)\sim ax^d\) as \(x\rightarrow \infty \), there exist \(x_1>1\) and \(C_8>0\) such that \(x^d\le C_8(1+\log w)\) and \(|\log x|\le C_8(1+\log (1+\log w))\) for \(w\ge w_1:=T(x_1)\). Since \(xg'(x)=adx^d+bex^{-e}\asymp x^d\) for large x, we have \(|xg'(x)|\le C_9(1+\log w)\). Since \(xg''(x)/g'(x)\) remains bounded for large x, it follows that, for \(w\ge w_1\),
$$\begin{aligned} |q(w)|&\le C_{10}w(1+\log w)(1+\log (1+\log w)),\\ \left| \frac{q(w)}{w}\right|&\le C_{10}(1+\log w)(1+\log (1+\log w)),\\ |q'(w)|&\le C_{11}(1+\log w)(1+\log (1+\log w)),\\ |\eta _{\delta _{ab}}(w)|&\le C_{12}(1+\log w)(1+\log (1+\log w)). \end{aligned}$$
Thus Proposition B.2 applies with \(a_0(w)=b_0(w)=C(1+|\log w|)(1+\log (1+|\log w|))\) for \(0<w<w_0\) and \(a_\infty (w)=b_\infty (w)=C(1+\log w)(1+\log (1+\log w))\) for \(w\ge w_1\). The required limits and integrability conditions are immediate. \(\square \)
Proof for condition 6From the lower bound on the trigamma function in Lemma 2.4, the first diagonal element of the Fisher information matrix is strictly positive for all \(\varvec{\theta }\); that is:
$$\begin{aligned} (I(\varvec{\theta }))_{11}&= \psi ^{(1)}(\mu +1)+\frac{1}{\mu ^{2}}-\frac{1}{\mu }> \frac{1}{\mu +1} + \frac{1}{\mu ^2} - \frac{1}{\mu } = \frac{1}{\mu ^2(\mu +1)} > 0, \end{aligned}$$
because \(\mu > 0\). Then, we need to compute the determinant when \(\delta _{ab} = 1\). Hence,
$$\begin{aligned} \det (I)&= \left( \psi ^{(1)}(\mu +1)+\frac{1}{\mu ^{2}}-\frac{1}{\mu }\right) \left( \frac{\mu +1}{\sigma ^{2}}-\frac{1}{(\sigma +1)^{2}}\right) - \frac{1}{\mu ^{2}\sigma ^{2}(\sigma +1)^{2}} \\&= \frac{1}{\sigma ^{2}(\sigma +1)^{2}} \left[ \left( \psi ^{(1)}(\mu +1)+\frac{1}{\mu ^{2}}-\frac{1}{\mu }\right) N_{1} - \frac{1}{\mu ^{2}} \right] , \end{aligned}$$
where \(N_1 = (\mu +1)(\sigma +1)^2 - \sigma ^2 = \mu \sigma ^2+2(\mu +1)\sigma +(\mu +1)\). For \(\mu ,\sigma > 0\), it is clear that \(N_1 > \mu +1\). Using the established inequalities, \((I(\varvec{\theta }))_{11} > 1/(\mu ^2(\mu +1)),\) and \(N_1 > \mu +1\):
$$\begin{aligned} (I(\varvec{\theta }))_{11} \cdot N_1> \frac{1}{\mu ^2(\mu +1)} \cdot N_1 > \frac{\mu +1}{\mu ^2(\mu +1)} = \frac{1}{\mu ^2}. \end{aligned}$$
This shows that \((I(\varvec{\theta }))_{11}N_1 - \frac{1}{\mu ^2} > 0\). Therefore, the bracketed term in the determinant expression is strictly positive, implying that \(\det (I(\varvec{\theta })) > 0\).