Goodness-of-fit tests and applications for left-truncated Weibull distributions to non-life insurance

Abstract

In risk theory with application to insurance, the identification of the relevant distributions for both the counting and the claim size processes from given observations is of major importance. In some situations left-truncated distributions can be used to model, not only the single claim severity, but also the inter-arrival times between two consecutive claims. We show that left-truncated Weibull distributions are particularly relevant, especially for the claim severity distribution. For that, we first demonstrate how the parameters can be estimated consistently from the data, and then show how a Kolmogorov-Smirnov goodness-of-fit test can be set up using modified critical values. These critical values are universal to all left-truncated Weibull distributions, independent of the actual Weibull parameters. To illustrate our findings we analyse three applications using real insurance data, one from a Swiss excess of loss treaty over automobile insurance, another from an American private passenger automobile insurance and a third from earthquake inter-arrival times in California.

Appendices

Appendix 1: Proofs

1.1 Proof of Lemma 1

• 1. Starting from the Lemma 1 precondition $X_i>x_L>0$ for all $i=1,2,...,n$, we see from Eq. (7) with the new variables $\zeta _i \equiv X_i/x_L >1$ that the first derivative of $h(\cdot )$ with respect to $\beta$ is given by

  $$\begin{aligned}&h'(\beta ) \nonumber \\&= - \frac{ \left[ \sum _{i=1}^{n} ( \zeta _i^{\beta } - 1) \right] ^2 + \beta ^2 \sum _{i=1}^{n} (\zeta _i^{\beta } \log ^2{\zeta _i}) \cdot \sum _{i=1}^{n} ( \zeta _i^{\beta } - 1 ) - \beta ^2 \left( \sum _{i=1}^{n} \zeta _i^{\beta } \log {\zeta _i} \right) ^2 }{ \beta ^2 \left[ \sum _{i=1}^{n} ( \zeta _i^{\beta } - 1) \right] ^2 }\,. \end{aligned}$$

  (12)

  To prove that $h(\cdot )$ is monotonically decreasing, we need to show that $h'(\beta )<0$ for $\beta >0$. Thus we need to show that the numerator of Eq. (12) is positive. We prove this statement by mathematical induction over $n$

  $$\begin{aligned} A(n): \left[ \sum _{i=1}^{n} ( \zeta _i^{\beta } - 1) \right] ^2 + \beta ^2 \sum _{i=1}^{n} (\zeta _i^{\beta } \log ^2{\zeta _i}) \cdot \sum _{i=1}^{n} ( \zeta _i^{\beta } - 1 ) - \beta ^2 \left( \sum _{i=1}^{n} \zeta _i^{\beta } \log {\zeta _i} \right) ^2 > 0 \end{aligned}$$

  (13)

  For the base case $A(1)$ we have to show that for any $\zeta _1\,>\,1$ and $\beta >0$

  $$\begin{aligned} A(1): ( \zeta _1^{\beta } - 1)^2 + \beta ^2 ( \zeta _1^{\beta } \log ^2{\zeta _1} ) \cdot ( \zeta _1^{\beta } - 1 ) - \beta ^2 ( \zeta _1^{\beta } \log {\zeta _1} )^2 > 0 \end{aligned}$$

  (14)

  Simplifying Eq. (14) leads to

  $$\begin{aligned} e^{\beta \log {\zeta _1}} - 2 + e^{-\beta \log {\zeta _1}} &> (\beta \log {\zeta _1})^2 \nonumber \\ 2\,\left( \cosh \left[ y\right] -1 \right)&> y^2 \nonumber \\ \sum _{k=2}^{\infty } \frac{y^{2k}}{(2k)!}&> 0 \end{aligned}$$

  (15)

  where $y= \beta \log {\zeta _1}$ and the base case $A(1)$ is verified. Next we do the inductive step $A(n) \rightarrow A(n+1)$. $A(n+1)$ reads as

  $$\begin{aligned}&\left[ \sum _{i=1}^{n+1} ( \zeta _i^{\beta } - 1) \right] ^2 + \beta ^2 \left\{ \sum _{i=1}^{n+1} \zeta _i^{\beta } \log ^2{\zeta _i} \cdot \sum _{i=1}^{n+1} ( \zeta _i^{\beta } - 1 ) - \left( \sum _{i=1}^{n+1} \zeta _i^{\beta } \log {\zeta _i} \right) ^2 \right\} \nonumber \\&= \left[ \sum _{i=1}^{n} ( \zeta _i^{\beta } - 1) + ( \zeta _{n+1}^{\beta } - 1) \right] ^2 + \nonumber \\&+ \beta ^2 \left\{ \left[ \sum _{i=1}^{n} \zeta _i^{\beta } \log ^2{\zeta _i} + \zeta _{n+1}^{\beta } \log ^2{\zeta _{n+1}} \right] \cdot \left[ \sum _{i=1}^{n} ( \zeta _i^{\beta } - 1 ) + ( \zeta _{n+1}^{\beta } - 1 ) \right] \right. \nonumber \\& \left. - \left( \sum _{i=1}^{n} \zeta _i^{\beta } \log {\zeta _i} + \zeta _{n+1}^{\beta } \log {\zeta _{n+1} } \right) ^2 \right\} \nonumber \\&= \left[ \sum _{i=1}^{n} ( \zeta _i^{\beta } - 1) \right] ^2 + 2 \sum _{i=1}^{n} ( \zeta _i^{\beta } - 1) \cdot ( \zeta _{n+1}^{\beta } - 1) + ( \zeta _{n+1}^{\beta } - 1)^2 \nonumber \\&+ \beta ^2 \left\{ \sum _{i=1}^{n} ( \zeta _i^{\beta } \log ^2{\zeta _i} ) \cdot \sum _{i=1}^{n} ( \zeta _i^{\beta } - 1) + \sum _{i=1}^{n} ( \zeta _i^{\beta } \log ^2{\zeta _i} ) \cdot ( \zeta _{n+1}^{\beta } -1 ) \right. \nonumber \\& + \left. ( \zeta _{n+1}^{\beta } \log ^2{\zeta _{n+1}} ) \cdot \sum _{i=1}^{n} ( \zeta _i^{\beta } - 1) + ( \zeta _{n+1}^{\beta } \log ^2{\zeta _{n+1}} ) \cdot ( \zeta _{n+1}^{\beta } -1 ) \right. \nonumber \\& - \left. \left( \sum _{i=1}^{n} \zeta _i^{\beta } \log {\zeta _i} \right) ^2 - 2 \sum _{i=1}^{n} (\zeta _i^{\beta } \log {\zeta _i} ) \cdot ( \zeta _{n+1}^{\beta } \log {\zeta _{n+1}} ) - ( \zeta _{n+1}^{\beta } \log {\zeta _{n+1}} )^2 \right\} \nonumber \\&> 2 \sum _{i=1}^{n} ( \zeta _i^{\beta } - 1) \cdot ( \zeta _{n+1}^{\beta } - 1) \nonumber \\&+ \beta ^2 \left\{ \sum _{i=1}^{n} ( \zeta _i^{\beta } \log ^2{\zeta _i} ) \cdot ( \zeta _{n+1}^{\beta } -1 ) + ( \zeta _{n+1}^{\beta } \log ^2{\zeta _{n+1}} ) \sum _{i=1}^{n} ( \zeta _i^{\beta } - 1) \right. \nonumber \\&\quad - \left. 2 \sum _{i=1}^{n} (\zeta _i^{\beta } \log {\zeta _i} ) \cdot ( \zeta _{n+1}^{\beta } \log {\zeta _{n+1}} ) \right\} \ge 0 \end{aligned}$$

  (16)

  where we have used the Base Case $A(1)$, Eq. (14), and the induction assumption $A(n)$, Eq. (13), to arrive at Eq. (16). We shall prove this inequality Eq. (16) again by mathematical induction. Rewriting the inequality in terms of new variables $z_i\,\equiv \,\zeta _i^{\beta }>1$ the induction statment $B(n)$ reads as:

  $$\begin{aligned} B(n){:}\, &2 \sum _{i=1}^{n} (z_i - 1) \cdot ( z_{n+1} - 1) + \sum _{i=1}^{n} ( z_i \log ^2{z_i} ) \cdot ( z_{n+1} -1 ) \nonumber \\&+ \sum _{i=1}^{n} ( z_i - 1) \cdot ( z_{n+1} \log ^2{z_{n+1}} ) - 2 \sum _{i=1}^{n} (z_i \log {z_i} ) \cdot ( z_{n+1} \log {z_{n+1}} ) \ge 0\,. \end{aligned}$$

  (17)

  The Base Case $B(1)$ is given by Proposition 1 below. The inductive step $B(n) \rightarrow B(n+1)$ is done by some simple algebraic mainipulations. We write $B(n+1)$ as:

  $$\begin{aligned}&2 \sum _{i=1}^{n+1} (z_i - 1) \cdot ( z_{n+2} - 1) + \sum _{i=1}^{n+1} ( z_i \log ^2{z_i} ) \cdot ( z_{n+2} -1 ) \nonumber \\&\qquad+ \sum _{i=1}^{n+1} ( z_i - 1) \cdot ( z_{n+2} \log ^2{z_{n+2}} ) - 2 \sum _{i=1}^{n+1} (z_i \log {z_i} ) \cdot ( z_{n+2} \log {z_{n+2}} ) \nonumber \\&\quad= \left\{ 2 \sum _{i=1}^{n} (z_i - 1) \cdot ( z_{n+2} - 1) + \sum _{i=1}^{n} ( z_i \log ^2{z_i} ) \cdot ( z_{n+2} -1 ) \right. \nonumber \\& \left. \qquad+ \sum _{i=1}^{n} ( z_i - 1) \cdot ( z_{n+2} \log ^2{z_{n+2}} ) - 2\sum _{i=1}^{n} (z_i \log {z_i} ) \cdot ( z_{n+2} \log {z_{n+2}} ) \right\} \nonumber \\&\qquad+ \Bigg \{ 2 (z_{n+1} - 1) \cdot ( z_{n+2} - 1) + ( z_{n+1} \log ^2{z_{n+1}} ) \cdot ( z_{n+2} -1 ) \nonumber \\& \qquad+ ( z_{n+1} - 1) \cdot ( z_{n+2} \log ^2{z_{n+2}} ) - 2 (z_{n+1} \log {z_{n+1}} ) \cdot ( z_{n+2} \log {z_{n+2}} ) \Bigg \} \nonumber \\&\quad \ge 0 \end{aligned}$$

  (18)

  Here, all terms in the first curly brackets are non-negative by the induction assumption $B(n)$, Eq. (17), (with some arbitrary number $z_{n+2}>1$ playing the role of $z_{n+1}>1$), and likewise the terms in the second curly brackets, constituting the Base Case $B(1)$, Eq. (19), (with two arbitrary numbers $z_{n+1},z_{n+2}>1$, playing the roles of $z_1,z_2>1$).

• 2. Note that the order statistic $X_{(n)} = \max {\{x_L, X_1,..., X_n \}}$. Rewrite Eq. (7) as

  $$\begin{aligned} \lim _{\beta \rightarrow +\infty } h(\beta )&= \lim _{\beta \rightarrow +\infty } \left( \frac{1}{\beta } - \frac{ \left( \frac{X_{(n)}}{x_L}\right) ^{\beta } \sum _{i=1}^{n} \left( \frac{X_i}{X_{(n)}}\right) ^{\beta } \log {\frac{X_i}{x_L}} }{ \left( \frac{X_{(n)}}{x_L}\right) ^{\beta } \sum _{i=1}^{n} \left[ \left( \frac{X_i}{X_{(n)}}\right) ^{\beta } - \left( \frac{x_L}{X_{(n)}}\right) ^{\beta } \right] } + \frac{1}{n} \sum _{i=1}^{n} \log {\frac{X_i}{x_L}} \right) \\&= 0 - \frac{ 1 \cdot \log { \frac{X_{(n)}}{x_L} }}{1-0} + \frac{1}{n} \sum _{i=1}^{n} \log {\frac{X_i}{x_L}} \\&= -\log { \frac{X_{(n)}}{x_L} } + \frac{1}{n} \sum _{i=1}^{n} \log {\frac{X_i}{x_L}} < 0 \end{aligned}$$

  because the geometric mean of $n$ different real numbers is smaller than their largest number.

• 3. Rewrite Eq. (7) as

  $$\begin{aligned} h(\beta )&= \frac{ \sum _{i=1}^{n} \left[ \left( \frac{X_i}{x_L}\right) ^{\beta } - 1 \right] - \beta \cdot \sum _{i=1}^{n} \left( \frac{X_i}{x_L}\right) ^{\beta } \log {\frac{X_i}{x_L}} }{ \beta \cdot \sum _{i=1}^{n} \left[ \left( \frac{X_i}{x_L}\right) ^{\beta } - 1 \right] } + \frac{1}{n} \sum _{i=1}^{n} \log {\frac{X_i}{x_L}} \end{aligned}$$

  and apply L’Hospital’s rule for $\beta \rightarrow 0+$ twice to obtain

  $$\begin{aligned} \lim _{\beta \rightarrow 0+} h(\beta ) = \frac{2\left( \frac{1}{n}\sum _{i=1}^{n} \log {\frac{X_i}{x_L}} \right) ^{2} - \frac{1}{n} \sum _{i=1}^{n} \log ^{2}{\frac{X_i}{x_L} } }{\frac{2}{n} \sum _{i=1}^{n} \log {\frac{X_i}{x_L} }} \end{aligned}$$

  and the proof is finished.

$\square$

Proposition 1

For two real numbers $z_1,z_2\ge 1$ we have

$$\begin{aligned} F(z_1,z_2)&= (z_1-1) \cdot \left[ z_2 \cdot \left( 1+ \log ^2{z_2} \right) -1 \right] +(z_2-1) \cdot \left[ z_1 \cdot \left( 1+ \log ^2{z_1} \right) -1 \right] \nonumber \\&\quad- 2 (z_1 \log {z_1}) (z_2 \log {z_2}) \ge 0\,. \end{aligned}$$

(19)

Proof

We start from the following inequality

$$\begin{aligned} (\log x- \log y)^2 \,=\, \log ^2 x-2\log x \log y + \log ^2 y&\ge 0 \end{aligned}$$

By the monotonicity of the Riemann integral we have

$$\begin{aligned} \int _1^{z1}\, \int _1^{z2} &\, dx dy\, (\log ^2 x-2\log x \log y + \log ^2 y)\ge 0 \\ &\Leftrightarrow 2 (z_1 - 1) \cdot ( z_{2} - 1) + ( z_1 \log ^2{z_1} ) \cdot ( z_{2} -1 )\\&\quad+ ( z_1 - 1) \cdot ( z_{2} \log ^2{z_{2}} )-2 (z_1 \log {z_1} ) \cdot ( z_{2} \log {z_{2}} )\ge 0 \end{aligned}$$

which is the desired inequality, Eq. (19). $\square$

1.2 Proof of Theorem 1: checking the assumptions of [20]

For the proof we apply [20], Theorem 5.1 of section 6.5 (p. 463). Thus, we only need to check conditions (A0)–(A2) of section 6.3 and assumptions (A)–(D) from Sect. 6.5. Let us define as in this reference the general parameter vector $\varvec{\theta }\equiv (\alpha ,\beta )$ and the true parameter vector of the distribution as $\varvec{\theta }^0 \equiv (\alpha ^0,\beta ^0)$. Then the calculations for this are as follows:

1.3 Conditions

• (A0): requires that the distributions of the observations are distinct, i.e. for different sets of parameters $\varvec{\theta } \ne \varvec{\theta }'$ the corresponding pdf’s are different. This is readily checked because for $\varvec{\theta }=(\alpha ,\beta )$ and $\varvec{\theta }'=(\alpha ',\beta ')$ we see that $f(X|\alpha ,\beta ,x_L) \ne f(X|\alpha ',\beta ',x_L)$ almost everywhere in $X$.

• (A1): requires that all distributions have a common support, which is true, since $X\in (x_L,\infty )$.

• (A2): requires the observations $X_1,...,X_n$ are i.i.d. with a probability density $f(X|\varvec{\theta }^0,x_L)$, which follows from our assumption.

1.4 Assumptions

• (A): There exists an open subset $\omega$ of $\Omega$ containing the true parameter point $\varvec{\theta }^0$ such that for almost all $X$ the pdf $f(X|\varvec{\theta },x_L)$ admits all third derivatives $(\partial ^3 / \partial \theta _j \partial \theta _k \partial \theta _l )f(X|\varvec{\theta },x_L)$ for all $\varvec{\theta }$ in $\omega$. This condition is also readily checked because the left-truncated Weibull distribution is contineously differentiable with respect to its parameters $0<\theta _j<\infty$, with $j=1,2$ as mentioned in the first section.

• (B): This condition requires the first and second logarithmic derivatives of $f$ satisfy

  $$\begin{aligned} \mathbb {E}_{\varvec{\theta }} \left[ \frac{\partial }{\partial \theta _j} \log {f(X|\varvec{\theta },x_L)} \right] = 0\, {\rm for}\, j=1,2 \end{aligned}$$

  and

  $$\begin{aligned} I_{jk}(\varvec{\theta })&= \mathbb {E}_{\varvec{\theta }} \left[ \frac{\partial }{\partial \theta _j} \log {f(X|\varvec{\theta },x_L)} \cdot \frac{\partial }{\partial \theta _k} \log {f(X|\varvec{\theta },x_L)} \right] \\&= - \mathbb {E}_{\theta } \left[ \frac{\partial ^2}{\partial \theta _j \partial \theta _k} \log {f(X|\varvec{\theta },x_L)} \right] {\rm for}\, j,k=1,2 \end{aligned}$$

  Here, the expectation operator $\mathbb {E}_{\theta } \left[ \cdot \right]$ denotes the expectation over the absolute continuous probability measure $f(X|\varvec{\theta },x_L) dX$. These two conditions are readily verified as the left-truncated Weibull distribution functions are continously differentiable and in $C^{\infty }((x_L,\infty )\times (0,\infty ) \times (0,\infty ))$. Thus integration and differentiation can be interchanged and integration by parts leads to the desired result since the integral of $f$ over the integration domain $(x_L,\infty )$ is 1 by normalisation and thus any of the derivatives vanishes.

• (C): All $I_{jk}(\varvec{\theta })$ defined in Assumption (B) are finite and the $2\times 2$ matrix $I(\varvec{\theta })$ is positive definite for all $\varvec{\theta }$ in $\omega$: the relevant integrals can be computed explicitly and are finite, due to the asymptotic condition for $x\rightarrow \infty$ we have $f(x|\alpha , \beta , x_L) =O( \exp { [-(x/\alpha )^{\beta '}]})$ for $\alpha >0$ and any $\beta ' \in (0,\beta )$. Hence also the $I_{jk}(\varvec{\theta })$ are finite. Thus the matrix $I_{jk}$ is well-defined and as a covariance matrix by construction positive definite. That the functions and are affinely linear independent with probability 1 can be seen immediately by explicit computation.

• (D): The absolute values of all third derivatives $\left| (\partial ^3 / \partial \theta _j \partial \theta _k \partial \theta _l )\log {f(X|\varvec{\theta },x_L)}\right|$ again can be bounded by integrable functions and their expectations $\mathbb {E}_{\theta } \left[ \cdot \right]$ can be computed and are finite. This is a consequence of the properties of the left-truncated Weibull distributions, namely as $x\rightarrow \infty$ we have $f(x|\alpha , \beta , x_L) =O( \exp { [-(x/\alpha )^{\beta '}]})$ for $\alpha >0$ and any $\beta ' \in (0,\beta )$.

Appendix 2: Truncated Weibull random variates and their representation by exponential variates

Let $u_i \in (0,1)$ denote the standard uniform random variable. Then from the cdf in Eq. (2) we obtain a left truncated Weibull distributed random variable $X_i$

$$\begin{aligned} X_i = \alpha \cdot \left[ \left( \frac{x_L}{\alpha } \right) ^{\beta } + \log {\frac{1}{u_i}} \right] ^{1/\beta } = \alpha \cdot \left[ \left( \frac{x_L}{\alpha } \right) ^{\beta } + y_i \right] ^{1/\beta } = \alpha \cdot \left[ \eta + y_i \right] ^{1/\beta } \end{aligned}$$

(20)

where $y_i$ is a standard exponential random variate and $\eta \equiv \left( {x_L}/{\alpha } \right) ^{\beta }$.

Appendix 3: Universality of pivotal functions

As in [32] we wish to demonstrate that for a given sample size $n$ and the truncation parameter $\eta >0$ the pivotal functions $(\alpha ^0 / \hat{\alpha })^{\hat{\beta }}$ and $\hat{\beta }/\beta ^0$ are distributed independently with the choice of $(\alpha ^0,\beta ^0)$ and have the same distribution as $(1/\hat{\alpha }_{(1,1)})^{\hat{\beta }_{(1,1)}}$ and $\hat{\beta }_{(1,1)}$ respectively for the same $n$ and $\eta >0$, see Eq. (11). Our preference of $(\alpha ^0,\beta ^0) = (1,1)$ is the simplest choice for the Weibull distribution, i.e. a standard exponential distribution. Demonstrating universality of the KS distance can be achieved by confirming the equality in distribution of the pivotal functions, namely Eq. (11). For this purpose we show that the MLE equations Eqs. (5) and (6) can be written in terms of the pivotal functions, $\hat{\beta }/\beta ^0$, $(\hat{\alpha }/\alpha ^0)^{\hat{\beta }}$, the truncation parameter $\eta =(x_L/\alpha ^0)^{\beta ^0}$ and $n$. Here $(\hat{\alpha },\hat{\beta })$ denote the solutions of the MLE equations and $(\alpha ^0,\beta ^0)$ are the true values.

Making use of Eq. (20) to re-write the MLE equation for $\hat{\alpha }$, Eq. (5), becomes

$$\begin{aligned} \left( \frac{\hat{\alpha }}{\alpha ^0} \right) ^{\hat{\beta }} = \frac{1}{n} \sum _{i=1}^{n} \left[ (\eta +y_i)^{\hat{\beta }/\beta ^0}- \eta ^{\hat{\beta }/\beta ^0} \right] \,, \end{aligned}$$

(21)

where the $y_i$ are standard exponential random variates. Note that Eq. (21) is just the inverse of the first pivotal function $(\alpha ^0 / \hat{\alpha })^{\hat{\beta }}$. If we can demonstrate that $\hat{\beta }/\beta ^0$ is distributed as $\hat{\beta }_{(1,1)}$ (for fixed $n$ and $\eta$), then we can conclude that $(\alpha ^0 / \hat{\alpha })^{\hat{\beta }}$ is distributed as $1/(\hat{\alpha }_{(1,1)})^{\hat{\beta }_{(1,1)}}$ because the right-hand side of Eq. (21) only depends on $n$, $\eta$, the sample of standard exponential random variables $y_1,...,y_n$ and $\hat{\beta }_{(1,1)}$, and hence the first line in Eq. (11) is shown.

Let us now show that $\hat{\beta }/\beta ^0$ is distributed as $\hat{\beta }_{(1,1)}$. Using Eq. (20) and performing some algebraic manipulations we can rewrite the MLE equation for $\hat{\beta }$, Eq. (6), in terms of a new random variable $\xi \equiv \hat{\beta }/\beta ^0$

$$\begin{aligned} \frac{1}{n} \sum _{i=1}^{n} \log {(\eta + y_i)^{\xi }}&= \frac{1}{n} \sum _{i=1}^{n} \frac{(\eta +y_i)^{\xi }}{\frac{1}{n} \sum _{i=1}^{n} (\eta +y_i)^{\xi }- \eta ^{\xi } } \cdot \log { \frac{(\eta +y_i)^{\xi }}{\frac{1}{n} \sum _{i=1}^{n} (\eta +y_i)^{\xi }- \eta ^{\xi } } } \nonumber \\&\quad+ \log { \frac{1}{n} \sum _{i=1}^{n} \left[ (\eta + y_i)^{\xi } - \eta ^{\xi } \right] } -1 \nonumber \\&\quad+ \frac{\eta ^{\xi }}{\frac{1}{n} \sum _{i=1}^{n} (\eta +y_i)^{\xi }- \eta ^{\xi } } \cdot \log { \frac{\eta ^{\xi }}{\frac{1}{n} \sum _{i=1}^{n} (\eta +y_i)^{\xi }- \eta ^{\xi } }} \end{aligned}$$

(22)

Solving Eq. (22)⁶ for the unknown $\xi$, which is actually the second pivotal function, we see that $\xi =\xi (y_1,...y_n|\eta ,n)$ is a uniquely defined random variable depending on the sample of standard exponential random variates $y_1,...,y_n$, sample size $n$ and parameter $\eta$ only - but not on the original parameters $(\alpha ^0,\beta ^0)$. Next we consider the special case of Weibull distributions with $(\alpha ^0,\beta ^0)=(1,1)$ and conclude that the unique solution of Eq. (22), $\xi =\hat{\beta }_{(1,1)}=\xi (y_1,...y_n|\eta ,n)$ depends only on sample size $n$, parameter $\eta$ and the same sample of standard exponential random variables $y_1,...,y_n$. Thus our claim and the second line in Eq. (11) is shown.

Appendix 4: Computing the fisher information matrix

The covariance matrix in Theorem 1, $Z(\alpha ,\beta )^{-1}$ is derived from the logarithm of the pdf $f(\cdot )$ from Eq. (3). One readily obtains the second derivatives necessary for the calculation of the Fisher information matrix $Z(\alpha ,\beta )$, writing as usual $\eta =(x_L/\alpha )^{\beta }$,

$$\begin{aligned} \frac{\partial ^2}{\partial \alpha ^2} \log { f(X|\alpha ,\beta ,x_L)}&= -\frac{\beta }{\alpha ^2} \left\{ 1+ (\beta +1) \left[ \eta - \left( \frac{X}{\alpha } \right) ^{\beta } \right] \right\} \\ \frac{\partial ^2}{\partial \alpha \partial \beta } \log { f(X|\alpha ,\beta ,x_L)}&= -\frac{1}{\alpha } -\frac{1}{\alpha } \left[ \eta - \left( \frac{X}{\alpha } \right) ^{\beta } \right] \nonumber \\&\quad-\frac{1}{\alpha } \left[ \eta \log {\eta } - \left( \frac{X}{\alpha } \right) ^{\beta } \log {\left( \frac{X}{\alpha } \right) ^{\beta } } \right] \\ \frac{\partial ^2}{\partial \beta ^2} \log { f(X|\alpha ,\beta ,x_L)}&= -\frac{1}{\beta ^2} +\frac{1}{\beta ^2} \left\{ \eta [\log {\eta }]^2 - \left( \frac{X}{\alpha } \right) ^{\beta } \left[ \log {\left( \frac{X}{\alpha } \right) ^{\beta } } \right] ^2 \right\} \end{aligned}$$

To evaluate all relevant expectations $\mathbb {E} \left( \cdot \right)$ in the definition of the Fisher information matrix, the following integrals are needed

$$\begin{aligned} \mathbb {E} \left[ \left( \frac{X}{\alpha } \right) ^{\beta } \right]&= \int _{x_L}^{\infty } dx \frac{\beta }{\alpha } \left( \frac{x}{\alpha } \right) ^{\beta -1} e^{ \left( \frac{x_L}{\alpha } \right) ^{\beta } - \left( \frac{x}{\alpha } \right) ^{\beta } } \cdot \left( \frac{x}{\alpha } \right) ^{\beta }\\&= e^{\eta } \int _{\eta }^{\infty } dz e^{ -z } \cdot z \\&= 1 + \eta \\ \mathbb {E} \left[ \left( \frac{X}{\alpha } \right) ^{\beta } \log { \left( \frac{X}{\alpha } \right) ^{\beta } } \right]&= 1+ \eta \log {\eta } + \left[ \log {\eta }+e^{\eta }E_1(\eta ) \right] \\ \mathbb {E} \left[ \left( \frac{X}{\alpha } \right) ^{\beta } \left\{ \log { \left( \frac{X}{\alpha } \right) ^{\beta } } \right\} ^2 \right]&= \eta \left( \log {\eta } \right) ^2 + 2 \left[ \log {\eta }+e^{\eta }E_1(\eta ) \right] \\&+ \left[ \left( \log {\eta } \right) ^2 + 2 e^{\eta } E_2(\eta ) \right] \end{aligned}$$

where we have used the functions

$$\begin{aligned} E_1(s)&= \int _{s}^{\infty } dy \frac{e^{-y}}{y} \\ E_2(s)&= \int _{s}^{\infty } dy e^{-y}\frac{\log {y}}{y} \end{aligned}$$

In the limit $\eta \longrightarrow 0+$ we recover the covariance matrix for the untruncated system as given by [27] (Eq. 11.17)

$$\begin{aligned} Z(\alpha ,\beta )^{-1} = \left( \begin{array}{ccc} 1.1087 \cdot \frac{\alpha ^2}{\beta ^2} &{} &{} 0.2570 \cdot \alpha \\ 0.2570 \cdot \alpha &{} &{} 0.6079 \cdot \beta ^2 \\ & \end{array} \right) \end{aligned}$$

Note that in our representation of $Z(\alpha ,\beta )^{-1}$ the factor $1/n$ is displayed separately. For an estimate of the error in the MLE estimation of $\hat{\alpha }$ and $\hat{\beta }$, one may apply this expression with the hat-ed parameters rather than the unknown “true” parameters in conjunction with Theorem 1, statements 2 and 3.