Goodness-of-fit tests for the Weibull distribution based on the Laplace transform and Stein’s method

Abstract

We propose novel goodness-of-fit tests for the Weibull distribution with unknown parameters. These tests are based on an alternative characterizing representation of the Laplace transform related to the density approach in the context of Stein’s method. Asymptotic theory of the tests is derived, including the limit null distribution, the behaviour under contiguous alternatives, the validity of the parametric bootstrap procedure, and consistency of the tests against a large class of alternatives. A Monte Carlo simulation study shows the competitiveness of the new procedure. Finally, the procedure is applied to real data examples taken from the materials science.

A Proofs

1.1 A.1 Proof of Theorem 1

We first assume that X has the Weibull distribution $W(\lambda ,k)$, and we write $F(\cdot ,\lambda ,k)$ and $f(\cdot ,\lambda ,k)$ for the distribution function and the density function of X, respectively. Furthermore, we put

$$\begin{aligned} \kappa _{f}(x)=\biggl \vert \frac{\frac{\textrm{d}}{\textrm{d}x}f(x,\lambda ,k) \min \big (F(x,\lambda ,k),1-F(x,\lambda ,k)\big )}{f^2(x,\lambda ,k)} \biggl \vert , \qquad 0< x < \infty . \end{aligned}$$

Letting $\tau =(-\lambda ^k\log (1/2))^{1/k}$, we have

$$\begin{aligned} \min \big (F(x,\lambda ,k),1-F(x,\lambda ,k)\big )=\left\{ \begin{array}{ll} F(x,\lambda ,k), &{} x \le \tau \\ 1-F(x,\lambda ,k), &{} x > \tau \end{array}\right. . \end{aligned}$$

Using L’Hôspital’s rule, we deduce $\lim _{x \rightarrow 0} \big (x^{-k} (1-\exp (-(x/\lambda )^k))\big )=\lambda ^{-k}$, and it follows that $\lim _{x \rightarrow 0} \kappa _{f}(x) [3] =\vert \frac{k-1}{k} \vert .$ It is easily seen that $\lim _{x \rightarrow \infty } \kappa _{f}(x)=1.$ The continuity of $\kappa _{f}(\cdot )$ then yields

$$\begin{aligned} \sup _{x\in (0,\infty )} \kappa _{f}(x) < \infty . \end{aligned}$$

(20)

A further application of L’Hôspital’s rule gives

$$\begin{aligned} \lim _{x \rightarrow 0} \frac{F(x,\lambda ,k)}{f(x,\lambda ,k)}= \lim _{x \rightarrow 0} \frac{f(x,\lambda ,k)}{\frac{\textrm{d}}{\textrm{d}x}f(x,\lambda ,k)} =0. \end{aligned}$$

(21)

In view of (20), (21) and

$$\begin{aligned} \int _0^{\infty } x \Big \vert \frac{\textrm{d}}{\textrm{d}x}f(x,\lambda ,k) \Big \vert \textrm{d}x \le k-1+\frac{k}{\lambda ^k}\mathbb {E}[X^k] < \infty , \end{aligned}$$

we can apply Corollary 2 of Betsch and Ebner (2021). Hence $X$ follows a $W(\lambda ,k)$-distribution if and only if its density is given by

$$\begin{aligned} f_{X}(s)=\mathbb {E}\biggl [ -\frac{\frac{\textrm{d}}{\textrm{d}x}f(x,\lambda ,k)\vert _{X}}{f(X,\lambda ,k)}1\{X>s\}\biggl ] =\mathbb {E}\biggl [ -\frac{k-1-\frac{kX^k}{\lambda ^k}}{X}1\{X>s\}\biggl ] \end{aligned}$$

for almost every $s > 0$. Next, we apply Tonelli’s theorem to conclude that

$$\begin{aligned} \int _0^{\infty } \textrm{e}^{-ts} \mathbb {E}\biggl [\biggl \vert -\frac{\frac{\textrm{d}}{\textrm{d}x}f(x,\lambda ,k)\vert _{X}}{f(X,\lambda ,k)}\biggl \vert 1\{X>s\}\biggl ]\textrm{d}s= & {} \int _0^{\infty } \textrm{e}^{-ts} \int _0^{\infty } \Big \vert \frac{\textrm{d}}{\textrm{d}x}f(x,\lambda ,k)\Big \vert 1\{x>s\}\textrm{d}x\textrm{d}s \\\le & {} \int _0^{\infty } x\Big \vert \frac{\textrm{d}}{\textrm{d}x}f(x,\lambda ,k)\Big \vert \textrm{d}x \le \vert k-1 \vert +k \end{aligned}$$

holds for $t > 0$. Using Fubini’s theorem, the Laplace transform of $X$ takes the form

$$\begin{aligned} \mathcal {L}_{X}(t)=\int _0^{\infty } \textrm{e}^{-ts} \mathbb {E}\biggl [ -\frac{k-1-\frac{kX^k}{\lambda ^k}}{X} 1\{X>s\}\biggl ]\textrm{d}s = \mathbb {E}\biggl [\frac{1}{X}\biggl (k \biggl (\frac{X}{\lambda }\biggl )^k -k+1 \biggl ) \biggl (\frac{1}{t}-\frac{1}{t}\textrm{e}^{-tX} \biggl ) \biggl ] \end{aligned}$$

for $t > 0$. The converse assertion follows since the Laplace transform determines the distribution.$\square$

1.2 A.2 Proof of Theorem 5

Recall (15) and the definition of $V_n$ given in (14). The proof consists of two steps. We first write $V_n$ as a sum of i.i.d. random elements of $\mathscr {L}_w^2$ plus a term that is $o_{\mathbb {P}}(1)$. Then, a Hilbert space central limit theorem completes the proof. As for step 1, we apply two Taylor expansions in order to approximate the estimator ${\widehat{k}}_n$ in the exponent by $k_n$ and ${\widehat{\lambda }}_n$ in the denominator by $\lambda _n$. Starting with ${\widehat{k}}_n$, a second-order Taylor expansion yields

$$\begin{aligned} \left( \frac{X_{n,j}}{{\widehat{\lambda }}_{n}}\right) ^{{\widehat{k}}_{n}}&=\left( \frac{X_{n,j}}{{\widehat{\lambda }}_{n}}\right) ^{k_n}+\log \left( \frac{X_{n,j}}{{\widehat{\lambda }}_{n}}\right) \left( \frac{X_{n,j}}{{\widehat{\lambda }}_{n}}\right) ^{k_n}({\widehat{k}}_{n}-k_n)+R_{n,j}({\widehat{k}}_{n}-k_n)^2, \end{aligned}$$

where

$$\begin{aligned} R_{n,j}= \frac{1}{2} \biggl (\log \biggl (\frac{X_{n,j}}{{\widehat{\lambda }}_{n}}\biggl )\biggl )^2\biggl (\frac{X_{n,j}}{{\widehat{\lambda }}_{n}}\biggl )^{k_n^*} \end{aligned}$$

and $|k_n^* - k_n| \le |{\widehat{k}}_{n} -k_n|$. We now define

$$\begin{aligned} V_{n}^{(1)}(t)=&\frac{1}{\sqrt{n}} \sum _{j=1}^{n} \bigg {[} \frac{1}{X_{n,j}}\bigg (\left( \frac{X_{n,j}}{{\widehat{\lambda }}_{n}}\right) ^{k_n}\left( {\widehat{k}}_{n}+\log \left( \frac{X_{n,j}}{{\widehat{\lambda }}_{n}}\right) {\widehat{k}}_{n}({\widehat{k}}_{n}-k_n)\right) -{\widehat{k}}_{n}+1\bigg ) \\ {}&\times \big (1-\textrm{e}^{-tX_{n,j}}\big )-t \textrm{e}^{-tX_{n,j}}\bigg {]} \end{aligned}$$

and show that

$$\begin{aligned} \Vert V_{n}-V_{n}^{(1)}\Vert ^2 = o_{\mathbb {P}}(1). \end{aligned}$$

(22)

To this end, notice that

$$\begin{aligned} \Vert V_{n}-V_{n}^{(1)}\Vert ^2&= \int _0^{\infty } \big {|}V_{n}(t)-V_{n}^{(1)}(t)\big {|}^2w(t) \textrm{d}t \nonumber \\&=\int _{0}^{\infty }\bigg | \frac{{\widehat{k}}_n}{\sqrt{n}} \sum _{j=1}^{n} \frac{1-\exp (-tX_{n,j})}{X_{n,j}} ({\widehat{k}}_{n}-k_n)^2R_{n,j} \bigg |^2w(t)\textrm{d}t\nonumber \\&\le {\widehat{k}}_n^2\big (\sqrt{n}({\widehat{k}}_{n}-k_n)({\widehat{k}}_{n}-k_n)\big )^2 \cdot \bigg (\frac{1}{n} \sum _{j=1}^n R_{n,j}\bigg )^2 \cdot \int _{0}^{\infty } t^2w(t)\textrm{d}t, \end{aligned}$$

(23)

since $1-\textrm{e}^{-t} \le t$ for $t \ge 0$. The first factor of (23) converges to zero in probability in view of the tightness of $\sqrt{n}({\widehat{k}}_{n}-k_n)$, and assumption (5) ensures the existence of the integral. It thus remains to show that $n^{-1}\sum _{j=1}^n R_{n,j}$ is a tight sequence. Since $(a-b)^2 \le 2a^2+2b^2$ $(a,b \in \mathbb {R}$), the definition of $R_{n,j}$ yields

$$\begin{aligned} 0 \le \frac{1}{n}\sum _{j=1}^n R_{n,j} \le \frac{1}{{\widehat{\lambda }}_n^{k_n^*}} \cdot \frac{1}{n}\sum _{j=1}^n \big (\log X_{n,j}\big )^2 X_{n,j}^{k_n^*} + \frac{\big (\log {\widehat{\lambda }}_n\big )^2}{{\widehat{\lambda }}_n^{k_n^*}} \cdot \frac{1}{n}\sum _{j=1}^n X_{n,j}^{k_n^*}. \end{aligned}$$

The factors that precede the arithmetic means converge almost surely and are thus tight sequences. Hence, it remains to show that $Z_{n,1}= n^{-1}\sum _{j=1}^n X_{n,j}^{k_n^*}$ and $Z_{n,2}= n^{-1}\sum _{j=1}^n \big (\log X_{n,j}\big )^2 X_{n,j}^{k_n^*}$ are tight sequences. We tackle $Z_{n,1}$ since the reasoning for $Z_{n,2}$ is the same. Given $\varepsilon >0$, we have to find $K >0$ such that $\mathbb {P}(Z_{n,1} > K) \le \varepsilon$ for each n. Since $k_n^*$ converges almost surely, there is some positive $k^+$ such that $\mathbb {P}(k_n^* \le k^+) \ge 1- \varepsilon /2$, $n \ge 1$, whence $\mathbb {P}\big (Z_{n,1} \le 1 + n^{-1}\sum _{j=1}^n X_{n,j}^{k^+}\big ) \ge 1-\varepsilon /2$ for each n. In view of the almost sure convergence of $n^{-1}\sum _{j=1}^n X_{n,j}^{k^+}$, there is some $L>0$ such that $\mathbb {P}\big (n^{-1}\sum _{j=1}^n X_{n,j}^{k^+} \le L\big ) \ge 1-\varepsilon /2$ for each n. Taking $K= 1+L$, it follows that $\mathbb {P}(Z_{n,1} \le K) \ge 1-\varepsilon$ for each n, as was to be shown.

In a similar way, a Taylor expansion yields

$$\begin{aligned} \left( \frac{X_{n,j}}{{\widehat{\lambda }}_{n}}\right) ^{k_n}&=\left( \frac{X_{n,j}}{\lambda _n}\right) ^{k_n}-k_n\frac{X_{n,j}^{k_n}}{\lambda _n^{k_n+1}}({\widehat{\lambda }}_{n}-\lambda _n)+{\widetilde{R}}_{n,j}({\widehat{\lambda }}_{n}-\lambda _n)^2, \end{aligned}$$

where

$$\begin{aligned} {\widetilde{R}}_{n,j}=\frac{1}{2}\, (k_n+1)k_n\frac{X_{n,j}^{k_n}}{(\lambda _n^*)^{k_n+2}} \end{aligned}$$

and $|\lambda _n^*-\lambda _n| \le |{\widehat{\lambda }}_n-\lambda _n|$. Putting

$$\begin{aligned} V_{n}^{(2)}(t)=&\frac{1}{\sqrt{n}} \sum _{j=1}^{n} \Bigg [ \frac{1}{X_{n,j}}\left( \left( \frac{X_{n,j}}{\lambda _n}\right) ^{k_n}-k_n\frac{X_{n,j}^{k_n}}{\lambda _n^{k_n+1}}({\widehat{\lambda }}_{n}-\lambda _n)\right) \\&\times \left( {\widehat{k}}_{n}+\log \left( \frac{X_{n,j}}{{\widehat{\lambda }}_{n}}\right) {\widehat{k}}_{n}({\widehat{k}}_{n}-k_n)\right) -{\widehat{k}}_{n}+1\Biggl )\big (1-\textrm{e}^{-tX_{n,j}}\big )-t \textrm{e}^{-tX_{n,j}}\Bigg ], \end{aligned}$$

and

$$\begin{aligned} A_{n,j} \!&\!:= \!&\! {\widehat{k}}_{n} \! +\! \log \left( \frac{X_{n,j}}{{\widehat{\lambda }}_{n}}\right) {\widehat{k}}_{n}({\widehat{k}}_{n}-k_n),\nonumber \\ B_{n,j} \!&\!:= \!&\! \left( 1\! +\! \log \left( \frac{X_{n,j}}{{\widehat{\lambda }}_{n}}\right) ({\widehat{k}}_{n}\! -\! k_n)\right) , \end{aligned}$$

(24)

it follows by complete analogy with the first expansion that

$$\begin{aligned} \Vert V_{n}^{(1)}-V_{n}^{(2)}\Vert ^2 =&\int _0^{\infty } |V_{n}^{(1)}(t)\! -\! V_{n}^{(2)}(t)|^2w(t)\textrm{d}t \\ =&\int _{0}^{\infty }\biggl | \frac{1}{\sqrt{n}} \sum _{j=1}^{n} \frac{1\! -\! \exp (-tX_{n,j})}{X_{n,j}} A_{n,j} {\widetilde{R}}_{n,j}({\widehat{\lambda }}_{n}\! -\! \lambda _n)^2 \biggl |^2w(t)\textrm{d}t \\ \le&\big (\sqrt{n}({\widehat{\lambda }}_{n}\! -\! \lambda _n)^2{\widehat{k}}_{n}\big )^2 \bigg (\frac{1}{n} \sum _{j=1}^n B_{n,j} {\widetilde{R}}_{n,j}\bigg )^2 \! \int _{0}^{\infty } t^2w(t)\textrm{d}t =o_{\mathbb {P}}(1). \end{aligned}$$

To finish the first step, we show

$$\begin{aligned} \Biggl \Vert V_{n}^{(2)}(\cdot ) - \frac{1}{\sqrt{n}}\sum _{j=1}^n W_{n,j}(\cdot ) \Biggl \Vert ^{2}=o_{\mathbb {P}}(1), \end{aligned}$$

(25)

where $W_{n,j}(\cdot )$ is defined by

$$\begin{aligned} W_{n,j}(t)=&\frac{1}{X_{n,j}}\biggl (\left( \frac{X_{n,j}}{\lambda _n}\right) ^{k_n} k_n-k_n+1\biggl ) \big (1- \textrm{e}^{-tX_{n,j}}\big ))-t \textrm{e}^{-tX_{n,j}} \\&-\psi _{1}(X_{n,j},\lambda _n,k_n)\frac{k_n^2}{\lambda _n^{k_n+1}} \mathbb {E}\left[ X^{k_n-1} \big (1- \textrm{e}^{-tX}\big )\right] \\&+\psi _{2}(X_{n,j},\lambda _n,k_n)\biggl (\frac{k_n}{\lambda _n^{k_n}} \mathbb {E}\left[ X^{k_n-1}\log (X/\lambda _n) \big (1-\textrm{e}^{-tX}\big )\right] \\&\quad -\mathbb {E}\left[ X^{-1}\big (1- \textrm{e}^{-tX}\big )\right] +\frac{1}{\lambda _n^{k_n}} \mathbb {E}\left[ X^{k_n-1}\big (1-\textrm{e}^{-tX}\big )\right] \biggl ). \end{aligned}$$

Here, X has the Weibull distribution $W(\lambda _0,k_0)$, and $\psi _1,\psi _2$ satisfy (7) – (11). To verify (25) we successively eliminate the remaining estimators in $V_n^{(2)}$. Note that – with $A_{n,j}$ given in (24) –

$$\begin{aligned} V_{n}^{(2)}(t)=&\frac{1}{\sqrt{n}} \sum _{j=1}^{n} \Bigg \{ \frac{1}{X_{n,j}}\Biggl (\left( \frac{X_{n,j}}{\lambda _n}\right) ^{k_n} A_{n,j} -{\widehat{k}}_{n}+1\Biggl ) \big (1-\textrm{e}^{-tX_{n,j}}\big )-t\textrm{e}^{-tX_{n,j}} \Bigg \} \\&-\sqrt{n} ({\widehat{\lambda }}_{n}-\lambda _n) \bigg ( \frac{k_n^2}{\lambda _n^{k_n+1}} \mathbb {E}\left[ X^{k_n-1} \big (1-\textrm{e}^{-tX}\big )\right] +K_{n}^{(1)}(t) \bigg ), \end{aligned}$$

where

$$\begin{aligned} K_{n}^{(1)}(t)=&\frac{1}{n}\sum _{j=1}^n \frac{1}{X_{n,j}}k_n\frac{X_{n,j}^{k_n}}{\lambda _n^{k_n+1}}\left( {\widehat{k}}_{n}+\log \left( \frac{X_{n,j}}{{\widehat{\lambda }}_{n}}\right) {\widehat{k}}_{n}({\widehat{k}}_{n}-k_n)\right) \big (1-\textrm{e}^{-tX_{n,j}}\big )\\ {}&-\frac{k_n^2}{\lambda _n^{k_n+1}}\mathbb {E}\left[ X^{k_n-1}\big (1-\textrm{e}^{-tX}\big )\right] . \end{aligned}$$

We have

$$\begin{aligned} \Vert K_{n}^{(1)}\Vert ^2 =&\int _{0}^{\infty } \! \biggl ( \frac{1}{n}\sum _{j=1}^n\frac{k_n X_{n,j}^{k_n-1}}{\lambda _n^{k_n+1}}{\widehat{k}}_{n} \big (1\! -\! \textrm{e}^{-tX_{n,j}}\big ) -\frac{k_n^2}{\lambda _n^{k_n+1}}\mathbb {E} \! \left[ \! X^{k_n-1}\big (1\! -\! \textrm{e}^{-tX}\big )\right] \! \biggl )^2 \! w(t)\textrm{d}t \\&+2\int _{0}^{\infty } \biggl ( \frac{1}{n}\sum _{j=1}^nk_n\frac{X_{n,j}^{k_n-1}}{\lambda _n^{k_n+1}}{\widehat{k}}_{n}\big (1-\textrm{e}^{-tX_{n,j}}\big ) -\frac{k_n^2}{\lambda _n^{k_n+1}}\mathbb {E}\left[ X^{k_n-1}\big (1-\textrm{e}^{-tX}\big )\right] \biggl ) \\&\quad \times \biggl (\frac{1}{n}\sum _{j=1}^nk_n\frac{X_{n,j}^{k_n-1}}{\lambda _n^{k_n+1}}\log \left( \frac{X_{n,j}}{{\widehat{\lambda }}_{n}}\right) {\widehat{k}}_{n}({\widehat{k}}_{n}-k_n)\big (1-\textrm{e}^{-tX_{n,j}}\big )\biggl )w(t)\textrm{d}t \\&+\int _{0}^{\infty } \biggl (\frac{1}{n}\sum _{j=1}^nk_n\frac{X_{n,j}^{k_n-1}}{\lambda _n^{k_n+1}}\log \left( \frac{X_{n,j}}{{\widehat{\lambda }}_{n}}\right) {\widehat{k}}_{n}({\widehat{k}}_{n}-k_n)\big (1-\textrm{e}^{-tX_{n,j}}\big )\biggl )^2w(t)\textrm{d}t\\ =&: I_{n,1} + 2I_{n,2} + I_{n,3}, \end{aligned}$$

say. Regarding $I_{n,1}$, we have

$$\begin{aligned} I_{n,1}=&\int _{0}^{\infty } \biggl ( \frac{1}{n}\sum _{j=1}^nk_n\frac{X_{n,j}^{k_n-1}}{\lambda _n^{k_n+1}}k_{n} \big (1-\textrm{e}^{-tX_{n,j}}\big ) -\frac{k_n^2}{\lambda _n^{k_n+1}}\mathbb {E}\left[ X^{k_n-1}\big (1-\textrm{e}^{-tX}\big )\right] \biggl )^2w(t)\textrm{d}t \\&+\int _{0}^{\infty } \biggl ( \frac{1}{n}\sum _{j=1}^nk_n\frac{X_{n,j}^{k_n-1}}{\lambda _n^{k_n+1}}({\widehat{k}}_{n}-k_{n}) \big (1-\textrm{e}^{-tX_{n,j}}\big ) \biggl )^2w(t)\textrm{d}t \\&+2\int _{0}^{\infty } \biggl ( \frac{1}{n}\sum _{j=1}^nk_n\frac{X_{n,j}^{k_n-1}}{\lambda _n^{k_n+1}}({\widehat{k}}_{n}-k_{n}) \big (1-\textrm{e}^{-tX_{n,j}}\big ) \biggl ) \\&\quad \times \biggl ( \frac{1}{n}\sum _{j=1}^nk_n\frac{X_{n,j}^{k_n-1}}{\lambda _n^{k_n+1}}k_{n} \big (1-\textrm{e}^{-tX_{n,j}}\big ) -\frac{k_n^2}{\lambda _n^{k_n+1}}\mathbb {E}\left[ X^{k_n-1}\big (1-\textrm{e}^{-tX}\big )\right] \biggl ) w(t)\textrm{d}t \\ =&: I_{n,1}^{(1)} + I_{n,1}^{(2)} + 2I_{n,1}^{(3)}, \end{aligned}$$

say. To tackle $I_{n,1}^{(2)}$, we use Fubini’s theorem and the convergence in distribution of $(X_{n,i},X_{n,j},k_n)$ to $(X^{(1)},X^{(2)},k_0)$ as $n \rightarrow \infty$ for $i\ne j$, where $X^{(1)},X^{(2)}$ are i.i.d. random variables having the Weibull distribution $W(\lambda _0,k_0)$. Invoking the continuous mapping theorem, the inequality $1-\textrm{e}^{-t}\le t$ for $t \ge 0$ and assumption (5), it follows that

$$\begin{aligned}&\sup _{n \in \mathbb {N}} \mathbb {E} \bigg [ \int _{0}^{\infty } \biggl ( \frac{1}{n}\sum _{j=1}^nk_n\frac{X_{n,j}^{k_n-1}}{\lambda _n^{k_n+1}} \big (1-\textrm{e}^{-tX_{n,j}}\big ) \biggl )^2w(t)\textrm{d}t \bigg ] \nonumber \\ \le&\sup _{n \in \mathbb {N}} \frac{1}{n^2} \sum _{i,j=1}^n \mathbb {E} \bigg [ \bigg (k_n\frac{X_{n,i}^{k_n}}{\lambda _n^{k_n+1}} \biggl )\bigg (k_n\frac{X_{n,j}^{k_n}}{\lambda _n^{k_n+1}} \biggl ) \bigg ] \int _{0}^{\infty } t^2 w(t)\textrm{d}t \nonumber \\ =&\sup _{n \in \mathbb {N}} \frac{k_n^2}{\lambda _n^{2(k_n-1)}} \frac{1}{n^2} \Big (n(n-1) \mathbb {E} \big [ X_{n,1}^{k_n} X_{n,2}^{k_n} \big ] +n \mathbb {E} \big [ X_{n,1}^{2k_n} \big ] \Big ) \int _{0}^{\infty } t^2 w(t)\textrm{d}t < \infty . \end{aligned}$$

(26)

By Markov’s inequality, the expression inside the expectation in (26) is a tight sequence. Since ${\widehat{k}}_n - k_n \rightarrow 0$ almost surely as $n \rightarrow \infty$, we have $I_{n,1}^{(2)} = o_{\mathbb {P}}(1)$. We now show that $I_{n,1}^{(1)}$ converges to $0$ in $\mathscr {L}^1(\Omega ,\mathscr {A},\mathbb {P})$. With the same arguments as above, it follows that

$$\begin{aligned} \mathbb {E}\big [ I_{n,1}^{(1)} \big ]=&\int _{0}^{\infty } \Biggl \{ \mathbb {E}\Biggl [ \frac{1}{n^2}\sum _{i,j=1}^n \left( \frac{k_n^2}{\lambda _n^{k_n+1}}\right) ^2 X_{n,i}^{k_n-1}X_{n,j}^{k_n-1}\big (1- \textrm{e}^{-tX_{n,i}}\big )\big (1-\textrm{e}^{-tX_{n,j}}\big ) \Biggl ] \\&-2\frac{k_n^2}{\lambda _n^{k_n+1}}\mathbb {E}\left[ X^{k_n-1}\big (1-\textrm{e}^{-tX}\big )\right] \mathbb {E}\Biggl [ \frac{1}{n}\sum _{j=1}^n\frac{k_n^2}{\lambda _n^{k_n+1}}X_{n,j}^{k_n-1}\big (1-\textrm{e}^{-tX_{n,j}}\big ) \Biggl ] \\&+\left( \frac{k_n^2}{\lambda _n^{k_n+1}}\right) ^2\mathbb {E}\left[ X^{k_n-1}\big (1-\textrm{e}^{-tX}\big )\right] ^2 \Biggl \}w(t)\textrm{d}t \\ =&\left( \frac{k_n^2}{\lambda _n^{k_n+1}}\right) ^2 \Biggl \{\frac{1}{n}\int _{0}^{\infty }\text{ Var }\left[ X_{n,1}^{k_n-1}\big (1-\textrm{e}^{-tX_{n,1}}\big )\right] w(t)\textrm{d}t \\&+\int _{0}^{\infty } \biggl ( \mathbb {E}\left[ X^{k_n-1}\big (1-\textrm{e}^{-tX}\big )\right] -\mathbb {E}\left[ X_{n,1}^{k_n-1}\big (1-\textrm{e}^{-tX_{n,1}}\big )\right] \biggl )^2 w(t)\textrm{d}t \Biggl \}. \end{aligned}$$

Using again $1-\textrm{e}^{-t} \le t$ for $t \ge 0$, the variance is bounded from above by $t^2 \mathbb {E}[X_{n,1}^{2k_n}]$, and the last integral converges to zero as $n \rightarrow \infty$ by dominated convergence. Hence, $\mathbb {E}\big [ I_{n,1}^{(1)} \big ] \rightarrow 0$ and thus $I_{n,1}^{(1)} = o_{\mathbb {P}}(1)$. Likewise, the Cauchy-Schwarz inequality implies $I_{n,1}^{(3)} = o_{\mathbb {P}}(1)$. Moreover, with a similar reasoning, one obtains $I_{n,2} = o_{\mathbb {P}}(1)$ and $I_{n,3} = o_{\mathbb {P}}(1)$ and thus $\Vert K_{n}^{(1)}\Vert ^2 = o_{\mathbb {P}}(1)$. Using the tightness of the sequence $\sqrt{n} ({\widehat{\lambda }}_{n}-\lambda _n)$ and display (7) we conclude $\Vert V_{n}^{(2)}(\cdot )-V_{n}^{(3)}(\cdot )\Vert ^2=o_{\mathbb {P}}(1)$, where

$$\begin{aligned} V_{n}^{(3)}(t)=&\frac{1}{\sqrt{n}} \sum _{j=1}^{n} \Bigg \{\frac{1}{X_{n,j}}\Biggl (\left( \frac{X_{n,j}}{\lambda _n}\right) ^{k_n} \left( {\widehat{k}}_{n}+\log \left( \frac{X_{n,j}}{{\widehat{\lambda }}_{n}}\right) {\widehat{k}}_{n}({\widehat{k}}_{n}-k_n)\right) -{\widehat{k}}_{n}+1\Biggl )\\&\times \big (1-\textrm{e}^{-tX_{n,j}}\big )-t \textrm{e}^{-tX_{n,j}} -\psi _{1}(X_{n,j},\lambda _n,k_n)\frac{k_n^2}{\lambda _n^{k_n+1}} \mathbb {E}\left[ X^{k_n-1}\big (1-\textrm{e}^{-tX}\big )\right] \Bigg \}. \end{aligned}$$

We can write

$$\begin{aligned} V_{n}^{(3)}(t)=&\frac{1}{\sqrt{n}} \sum _{j=1}^{n} \Bigg \{ \frac{1}{X_{n,j}}\Biggl (\left( \frac{X_{n,j}}{\lambda _n}\right) ^{k_n} {\widehat{k}}_{n}-{\widehat{k}}_{n}+1\Biggl ) \big (1-\textrm{e}^{-tX_{n,j}}\big )-t\textrm{e}^{-tX_{n,j}} \\&-\psi _{1}(X_{n,j},\lambda _n,k_n)\frac{k_n^2}{\lambda _n^{k_n+1}} \mathbb {E}\left[ X^{k_n-1}\big (1-\textrm{e}^{-tX}\big )\right] \Bigg \} \\&+\sqrt{n} ({\widehat{k}}_{n}-k_n) \Bigg (\frac{k_n}{\lambda _n^{k_n}}\mathbb {E}\left[ \log \left( \frac{X}{\lambda _n}\right) X^{k_n-1}\big (1-\textrm{e}^{-tX}\big )\right] + K_{n}^{(2)}(t) \Bigg ), \end{aligned}$$

where

$$\begin{aligned} K_{n}^{(2)}(t) =&\frac{1}{n}\sum _{j=1}^n \log \! \left( \frac{X_{n,j}}{{\widehat{\lambda }}_{n}}\right) \frac{{\widehat{k}}_{n} X_{n,j}^{k_n-1}}{\lambda _n^{k_n}} \big (1\! -\! \textrm{e}^{-t X_{n,j}}\big ) - \frac{k_n}{\lambda _n^{k_n}}\mathbb {E} \! \left[ \log \! \left( \frac{X}{\lambda _n}\right) X^{k_n-1}\big (1\! - \! \textrm{e}^{-t X}\big )\right] . \end{aligned}$$

In a similar way as for $K_{n}^{(1)}$, one can show that $\Vert K_{n}^{(2)}\Vert ^2 = o_{\mathbb {P}}(1)$. Using the tightness of the sequence $\sqrt{n} ({\widehat{k}}_{n}-k_n)$ and display (8) we conclude $\Vert V_{n}^{(3)}(\cdot )-V_{n}^{(4)}(\cdot )\Vert ^2=o_{\mathbb {P}}(1)$, where

$$\begin{aligned} V_{n}^{(4)}(t)=&\frac{1}{\sqrt{n}} \sum _{j=1}^{n} \Bigg \{ \frac{1}{X_{n,j}}\Biggl (\left( \frac{X_{n,j}}{\lambda _n}\right) ^{k_n} {\widehat{k}}_{n}-{\widehat{k}}_{n}+1\Biggl ) \big (1-\textrm{e}^{-tX_{n,j}}\big )-t\textrm{e}^{-tX_{n,j}} \\&-\psi _{1}(X_{n,j},\lambda _n,k_n)\frac{k_n^2}{\lambda _n^{k_n+1}} \mathbb {E}\left[ X^{k_n-1}\big (1-\textrm{e}^{-tX}\big )\right] \\&+\psi _{2}(X_{n,j},\lambda _n,k_n)\frac{k_n}{\lambda _n^{k_n}} \mathbb {E}\left[ X^{k_n-1}\log \left( \frac{X}{\lambda _n}\right) \big (1-\textrm{e}^{-tX}\big )\right] \Bigg \}. \end{aligned}$$

Next, we rewrite

$$\begin{aligned} V_{n}^{(4)}(t)=&\frac{1}{\sqrt{n}} \sum _{j=1}^{n} \Bigg \{ \frac{1}{X_{n,j}}\Biggl ( \Biggl (\left( \frac{X_{n,j}}{\lambda _n}\right) ^{k_n}-1 \Biggl )k_n+1\Biggl ) \big (1- \textrm{e}^{-tX_{n,j}}\big )-t\textrm{e}^{-tX_{n,j}} \\&-\psi _{1}(X_{n,j},\lambda _n,k_n)\frac{k_n^2}{\lambda _n^{k_n+1}} \mathbb {E}\left[ X^{k_n-1}\big (1-\textrm{e}^{-tX}\big )\right] \\&+\psi _{2}(X_{n,j},\lambda _n,k_n)\frac{k_n}{\lambda _n^{k_n}} \mathbb {E}\left[ X^{k_n-1}\log \left( \frac{X}{\lambda _n}\right) \big (1-\textrm{e}^{-tX}\big )\right] \Bigg \} \\&+\sqrt{n} ({\widehat{k}}_{n}\! - \! k_n) \bigg (\! -\mathbb {E} \! \left[ X^{-1}\big (1\! -\! \textrm{e}^{-tX}\big )\right] +\frac{1}{\lambda _n^{k_n}} \mathbb {E} \! \left[ X^{k_n-1} \big (1\! -\! \textrm{e}^{-tX}\big )\right] + K_{n}^{(3)}(t) \! \bigg ), \end{aligned}$$

where

$$\begin{aligned} K_{n}^{(3)}(t) =&\frac{1}{n}\sum _{j=1}^n \frac{1}{X_{n,j}} \Biggl (\left( \frac{X_{n,j}}{\lambda _n}\right) ^{k_n}-1 \Biggl ) \big (1-\textrm{e}^{-tX_{n,j}}\big ) +\mathbb {E}\left[ X^{-1}\big (1-\textrm{e}^{-tX}\big )\right] \\ {}&-\frac{1}{\lambda _n^{k_n}} \mathbb {E}\left[ X^{k_n-1}\big (1-\textrm{e}^{-tX}\big )\right] . \end{aligned}$$

It is an easy task to show that $\Vert K_{n}^{(3)}\Vert ^2 = o_{\mathbb {P}}(1)$. Due to the tightness of $\sqrt{n} ({\widehat{k}}_{n}-k_n)$ and (8) we obtain (25).

Note that $W_{n,j}, j=1,\ldots ,n$, are centered and row-wise i.i.d. random elements of $\mathscr {L}_w^2$ with finite second moments, i.e., we have $\mathbb {E} \Vert W_{n,1} \Vert ^2 < \infty$ for all $n$. Furthermore, by dominated convergence we conclude that $\lim _{n \rightarrow \infty } \mathbb {E}[W_{n,1}(s)W_{n,1}(t)] = \mathbb {E}[W(s)W(t)]$, where $W$ is defined in the claim of the theorem.

Step 2: By assumptions (9) and (10), there is a function ${\widetilde{c}}$ such that $\vert \mathbb {E}[W_{n,1}(s)W_{n,1}(t)] \vert \le {\widetilde{c}}(s,t)$ for each n and for each $s,t \in [0, \infty ) \times [0, \infty )$. Moreover, by assumption (5),

$$\begin{aligned} \int _0^{\infty } \int _0^{\infty } {\widetilde{c}}(s,t)^i w(s)w(t)\, \textrm{d}s\, \textrm{d}t < \infty , \qquad i=1,2. \end{aligned}$$

(27)

Therefore, the Lindeberg–Feller central limit theorem and Slutzky’s lemma imply

$$\begin{aligned} \frac{1}{\sqrt{n}} \sum _{j=1}^n \langle W_{n,j},g\rangle {\mathop {\longrightarrow }\limits ^{D}} N(0,\sigma _{(\lambda _0,k_0)}^2(g)), \qquad g \in \mathscr {L}_w^2 \setminus \{0\}, \end{aligned}$$

where $\sigma _{(\lambda _0,k_0)}^2(g) = \lim _{n\rightarrow \infty } \mathbb {E} \big [ \langle W_{n,1},g\rangle ^2 \big ] = \mathbb {E} \big [ \langle W,g\rangle ^2 \big ].$ The last equality follows from (27). Note that Lindeberg’s condition is easily verified since $W_{n,j}$ are i.i.d. for $j=1,\ldots ,n$. Thus, an application of Lemma 3.1 of Chen and White (1998) yields $V_n {\mathop {\longrightarrow }\limits ^{D}} \mathcal {W}$ for some centered Gaussian random element $\mathcal {W}$ of $\mathscr {L}_w^2$ with covariance operator ${\widetilde{\Sigma }}_{(\lambda _0,k_0)}$ satisfying $\sigma _{(\lambda _0,k_0)}^2(g)=\langle {\widetilde{\Sigma }}_{(\lambda _0,k_0)}g,g \rangle$ for each $g \in \mathscr {L}_w^2 {\setminus } \{0\}$. By Fubini’s theorem and dominated convergence, we obtain

$$\begin{aligned} \sigma _{(\lambda _0,k_0)}^2(g)&= \lim _{n\rightarrow \infty } \int _0^{\infty } \int _0^{\infty } \mathbb {E} \big [ W_{n,1}(t)W_{n,1}(s) \big ] g(t)g(s)w(t)w(s)\textrm{d}t\textrm{d}s\\ {}&= \int _0^{\infty } (\Sigma _{(\lambda _0,k_0)}g)(s)g(s)w(s)\textrm{d}s, \end{aligned}$$

where $\Sigma _{(\lambda _0,k_0)}$ is given by (16). Thus ${\widetilde{\Sigma }}_{(\lambda _0,k_0)}= \Sigma _{(\lambda _0,k_0)}$ and the assertion follows. $\square$

1.3 A.3 Proof of Theorem 7

Let $\mu _n$ and $\nu _n$ denote the probability measures of $(X_{n,1},\ldots X_{n,n})$ under $H_0$ and in the situation of the assertion, respectively. As in the proof of Theorem 5, we have

$$\begin{aligned} \biggl \Vert V_n - \frac{1}{\sqrt{n}} \sum _{j=1}^n W_{n,j}^* \biggl \Vert ^2= o_{\mu _n}(1), \end{aligned}$$

where

$$\begin{aligned} W_{n,j}^*(t)=&\frac{1}{X_{n,j}}\biggl (\left( \frac{X_{n,j}}{\lambda }\right) ^{k} k-k+1\biggl )\big (1-\textrm{e}^{-tX_{n,j}}\big )-t\textrm{e}^{-tX_{n,j}} \\ {}&-\psi _{1}(X_{n,j},\lambda ,k)\frac{k^2}{\lambda ^{k+1}} \mathbb {E}\left[ X^{k-1}\big (1-\textrm{e}^{-tX}\big )\right] \\&+\psi _{2}(X_{n,j},\lambda ,k)\biggl (\frac{k}{\lambda ^{k}} \mathbb {E}\left[ X^{k-1}\log (X/\lambda )\big (1-\textrm{e}^{-tX}\big )\right] -\mathbb {E}\left[ X^{-1}\big (1-\textrm{e}^{-tX}\big )\right] \\ {}&+\frac{1}{\lambda ^{k}} \mathbb {E}\left[ X^{k-1}\big (1-\textrm{e}^{-tX}\big )\right] \biggl ). \end{aligned}$$

By contiguity, it follows that

$$\begin{aligned} \biggl \Vert V_n - \frac{1}{\sqrt{n}} \sum _{j=1}^n W_{n,j}^* \biggl \Vert ^2= o_{\nu _n}(1). \end{aligned}$$

(28)

Putting

$$\begin{aligned} \delta (g)= \lim _{n \rightarrow \infty } \text{ Cov } \biggl [ \langle W_{n,1}^*,g \rangle ,c(X_{n,1})-\frac{1}{2\sqrt{n}}c(X_{n,1})^2 \biggl ] \end{aligned}$$

for $g \in \mathscr {L}_w^2$, a combination of Slutzky’s lemma and the multivariate Lindeberg–Feller central limit theorem give

$$\begin{aligned} \left( \begin{array}{c} \frac{1}{\sqrt{n}} \sum _{j=1}^n \langle W_{n,j}^*,g \rangle \\ \log L_n(X_{n,1},\ldots ,X_{n,n}) \end{array}\right) {\mathop {\longrightarrow }\limits ^{D_{\mu _n}}} N \left( \left( \begin{array}{c} 0 \\ -\frac{\tau ^2}{2} \end{array} \right) , \left( \begin{array}{cc} \sigma ^2(g) &{} \delta (g) \\ \delta (g) &{} \tau ^2 \end{array} \right) \right) \end{aligned}$$

for some $\sigma ^2(g)>0$. Now LeCam’s third lemma yields the convergence in distribution of $\frac{1}{\sqrt{n}} \sum _{j=1}^n \langle W_{n,j}^*,g \rangle$ to the $N(\delta (g),\sigma ^2(g))$-law under $\nu _n$ for every $g \ne 0$, i.e. the convergence of the finite-dimensional distributions. The tightness under $\nu _n$ follows by contiguity. Therefore, $n^{-1/2} \sum _{j=1}^n W_{n,j}^{*} {\mathop {\longrightarrow }\limits ^{D_{\nu _n}}} \mathcal {W} + \zeta$, where $\zeta$ is defined in the assertion. $\square$