Maximum Likelihood Estimation of Parameters of a Random Variable Using Monte Carlo Methods

Abstract

In a parametric estimation framework, this paper proposes different properties for the maximum likelihood estimators of unknown parameters of a given random variable having a known distribution, where different parameter estimation cases are studied. The Refined Descriptive Sampling (RDS) method is chosen to generate samples used for the estimation purpose. Then, we compare the RDS maximum likelihood estimators to their competitors provided by simple random samples with the same size and issued from the same distribution, through their Fisher information. Furthermore, the Maximum likelihood RDS mean is written as a function of its corresponding empirical estimator where the expression can be used to determine the estimator value when a refined descriptive sample is provided. All these results allow us to conclude that the proposed Maximum Likelihood Estimation (MLE) using refined descriptive samples is more efficient than that already obtained from simple random samples, which means that MLE using RDS has advantage in estimating parameters when the samples are not independent and identically distributed. Some Monte Carlo simulations are provided to validate the obtained theoretical results.

Appendices

Appendix 1

Proof of equation (5.2) given in Theorem 1 of the Section 5.1.1. Equation (5.2) has been subdivided into three parts denoted I_i,i = 1,2,3 such as

$$I_{RDS}(\mu )=I_{1}+I_{2}+I_{3}$$

where

$$\left\{ \begin{array}{l} I_{1}=\frac{1}{\sigma^{2}}{\sum}_{j=1}^{m}{\sum}_{i=1}^{p_{j}} E\left[ \frac{-f^{\prime \prime} (Z_{(i)}^{j})}{ f(Z_{(i)}^{j})}+\left( \frac{f^{\prime} (Z_{(i)}^{j})}{f(Z_{(i)}^{j})}\right)^{2} \right] \\ I_{2}=\frac{1}{\sigma^{2}}{\sum}_{j=1}^{m}{\sum}_{i=1}^{p_{j}} (i-1)E\left[ \frac{-f^{\prime }(Z_{(i)}^{j})}{ F(Z_{(i)}^{j})}+\left( \frac{f(Z_{(i)}^{j})}{F(Z_{(i)}^{j})}\right)^{2}\right] \\ I_{3}=\frac{1}{\sigma^{2}}{\sum}_{j=1}^{m}{\sum}_{i=1}^{p_{j}} (p_{j}-i)E\left[ \frac{f^{\prime} (Z_{(i)}^{j})}{ 1-F(Z_{(i)}^{j})}+\left( \frac{f(Z_{(i)}^{j})}{1-F(Z_{(i)}^{j})}\right)^{2}\right]. \end{array} \right.$$

• (1) Let us calculate the first term I₁.

Let

$$\ h_{1}(z)=\left( \frac{f^{\prime} (z)}{f(z)}\right)^{2}-\frac{f^{\prime \prime} (z)}{f(z)},$$

and according to the definition of expectation

$$ E\left[ h_{1}(Z_{(i)}^{j})\right]={\int}_{-\infty }^{+\infty }h_{1}(z)f_{(i)}^{j}(z)dz, $$

using the expression of $f_{(i)}^{j}(z)$ given in Section 5, we obtain

$$ E\left[ h_{1}(Z_{(i)}^{j})\right]={\int}_{-\infty }^{+\infty }h_{1}(z)\frac{p_{j}!}{(i-1)!(p_{j}-i)!}\left[ F(x)\right]^{i-1}f(x)\left[ 1-F(x)\right]^{p_{j}-i}dz.$$

It follows

$$ I_{1}=\frac{1}{\sigma^{2}}\sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} {\int}_{-\infty }^{+\infty }h_{1}(z)p_{j}C_{i-1}^{p_{j}-1}f(z)\left( F(z)\right)^{i-1}\left[ 1-F(z) \right]^{p_{j}-i}dz.$$

By making some rearrangements of the latter equation, we write

$$ \begin{array}{@{}rcl@{}} I_{1} &=&\frac{1}{\sigma^{2}}{\int}_{-\infty }^{+\infty }h_{1}(z)f(z)\sum\limits_{j=1}^{m} p_{j}\left[ \sum\limits_{i=0}^{p_{j}-1} C_{i}^{p_{j}-1}\left( F(z)\right)^{i}\left[ 1-F(z)\right]^{p_{j}-1-i} \right] dz \\ &&+\frac{1}{\sigma^{2}}{\int}_{-\infty }^{+\infty }h_{1}(z)f(z)\sum\limits_{j=1}^{m} p_{j}dz. \end{array} $$

By simplifying, we get

$$ \begin{array}{@{}rcl@{}} I_{1} &=&\frac{n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }h_{1}(z)f(z)dz \\ &=&\frac{n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\left( \frac{f\prime (z)}{ f(z)}\right)^{2}f(z)dz-\frac{n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }f^{\prime \prime} (z)dz, \end{array} $$

by integrating the second term, it follows

$$ I_{1}=\frac{n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\left( \frac{f^{\prime} (z) }{f(z)}\right)^{2}f(z)dz-\lim\limits_{x\rightarrow +\infty}\left[ \frac{n}{\sigma^{2}}f^{\prime} (z)\right]_{-x}^{x}. $$

Using the hypothesis of the Theorem 1, we have$\lim \limits _{x\rightarrow +\infty } f^{\prime } (x)=\lim \limits _{x\rightarrow -\infty } f^{\prime } (x)$, so,

$$ I_{1}=\frac{n}{\sigma^{2}}E\left[\left( \frac{f^{\prime} (Z_{k})}{f(Z_{k})} \right)^{2}\right]. $$

• (2) Let us calculate the second term I₂.

Let

$$ h_{2}(z)=\left( \frac{f(z)}{F(z)}\right)^{2}-\frac{f^{\prime} (z)}{F(z)}.$$

Using the same method of calculation as when developing I₁, we get

$$ \begin{array}{@{}rcl@{}} I_{2} &=&\frac{1}{\sigma^{2}}\sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} {\int}_{-\infty }^{+\infty }(i-1)h_{2}(z)p_{j}C_{i-1}^{p_{j}-1}f(z)\left( F(z)\right)^{i-1}\left[ 1-F(z)\right]^{p_{j}-i}dz \\ &=&\frac{1}{\sigma^{2}}{\int}_{-\infty }^{+\infty }h_{2}(z)f(z)\sum\limits_{j=1}^{m} p_{j}\left[ \sum\limits_{i=1}^{p_{j}} (i-1)C_{i-1}^{p_{j}-1}\left( F(z)\right)^{i-1}\left[ 1-F(z)\right] ^{p_{j}-i}\right] dz. \end{array} $$

As the term of the series is equal to zero for i = 1 so,

$$ I_{2}=\frac{1}{\sigma^{2}}{\int}_{-\infty }^{+\infty }h(z)f(z)\sum\limits_{j=1}^{m} p_{j}\left[ F(z)\sum\limits_{i=2}^{p_{j}} (i-1)C_{i-1}^{p_{j}-1}\left( F(z)\right)^{i-2}\left[ 1-F(z)\right]^{p_{j}-i}\right] dz. $$

By making some rearrangements of the latter equation, we obtain

$$ I_{2}=\frac{1}{\sigma^{2}}{\int}_{-\infty }^{+\infty }h_{2}(z)f(z)F(z) \sum\limits_{j=1}^{m}p_{j}\left[ \sum\limits_{i=1}^{p_{j}-1} \frac{(p_{j}-1)!}{(i-1)!(p_{j}-1-i)!}\left( F(z)\right)^{i-1}\left[ 1-F(z)\right]^{p_{j}-1-i}\right] dz $$

so,

$$ \begin{array}{@{}rcl@{}} I_{2} &=&\frac{1}{\sigma^{2}}{\int}_{-\infty }^{+\infty }h_{2}(z)f(z)F(z) \\ &&\sum\limits_{j=1}^{m}p_{j}\frac{(p_{j}-1)}{f(x)}\left[ \frac{1 }{(p_{j}-1)}\sum\limits_{i=1}^{p_{j}-1}\frac{(p_{j}-1)!}{ (i-1)!(p_{j}-1-i)!}f(x)\left( F(z)\right)^{i-1}\left[ 1-F(z)\right] ^{p_{j}-1-i}\right] dz. \end{array} $$

As

$$ \frac{1}{(p_{j}-1)}\sum\limits_{i=1}^{p_{j}-1}\frac{(p_{j}-1)!}{ (i-1)!(p_{j}-1-i)!}f(x)\left( F(z)\right)^{i-1}\left[ 1-F(z)\right]^{p_{j}-1-i}=f(x), $$

so,

$$ I_{2}=\frac{1}{\sigma^{2}}{\int}_{-\infty }^{+\infty }h_{2}(z)f(z)F(z) \sum\limits_{j=1}^{m}p_{j}(p_{j}-1)dz.$$

Taking ${\sum }_{j=1}^{m}(p_{j})^{2}=q$, it follows

$$ \begin{array}{@{}rcl@{}} I_{2} &=&\frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }f(z)h_{2}(z)F(z)dz \\ &=&\frac{q-n}{\sigma^{2}}E[\frac{\left( f(Z_{k})\right)^{2}}{ F(Z_{k})}-f^{\prime} (Z_{k})]. \end{array} $$

• (3) Let us calculate the third term I₃.

Let

$$ h_{3}(z)=\left( \frac{f(z)}{1-F(z)}\right)^{2}+\frac{f^{\prime }(z)}{ 1-F(z)}. $$

Using the same method of calculation as when developing I₁, we get

$$ \begin{array}{@{}rcl@{}} I_{3} &=&\frac{1}{\sigma^{2}}\sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} {\int}_{-\infty }^{+\infty }(p_{j}-i)h_{3}(z)p_{j}C_{i-1}^{p_{j}-1}f(z)\left( F(z)\right)^{i-1}\left[ 1-F(z)\right]^{p_{j}-i}dz \\ &=&\frac{1}{\sigma^{2}}{\int}_{-\infty }^{+\infty }h_{3}(z)f(z)\sum\limits_{j=1}^{m} p_{j}\left[ \sum\limits_{i=1}^{p_{j}} (p_{j}-i)C_{i-1}^{p_{j}-1}\left( F(z)\right)^{i-1}\left[ 1-F(z)\right] ^{p_{j}-i}\right] dz. \end{array} $$

As the term of the series is equal to zero for i = p_j so,

$$ I_{3}=\frac{1}{\sigma^{2}}{\int}_{-\infty }^{+\infty }h_{3}(z)f(z)\sum\limits_{j=1}^{m} p_{j}\left[ (1-F(z))\sum\limits_{i=1}^{p_{j}-1} (p_{j}-i)C_{i-1}^{p_{j}-1}\left( F(z)\right)^{i-1}\left[ 1-F(z) \right]^{p_{j}-1-i}\right] dz, $$

it follows

$$ \begin{array}{@{}rcl@{}} I_{3} &=&\frac{1}{\sigma^{2}}{\int}_{-\infty }^{+\infty }h_{3}(z)f(z)(1-F(z)) \\ &&\sum\limits_{j=1}^{m}\frac{p_{j}(p_{j}-1)}{f(z)}\left[ \frac{1 }{p_{j}-1}\sum\limits_{i=1}^{p_{j}-1}\frac{(p_{j}-1)!}{ (i-1)!(p_{j}-1-i)}f(z)\left( F(z)\right)^{i-1}\left[ 1-F(z)\right] ^{p_{j}-1-i}\right] dz. \end{array} $$

As

$$\frac{1}{p_{j}-1}\sum\limits_{i=1}^{p_{j}-1}\frac{(p_{j}-1)!}{ (i-1)!(p_{j}-1-i)}f(z)\left( F(z)\right)^{i-1}\left[ 1-F(z)\right]^{p_{j}-1-i}=f(z),$$

so,

$$ \begin{array}{@{}rcl@{}} I_{3} &=&\frac{1}{\sigma^{2}}{\int}_{-\infty }^{+\infty }h_{3}(z)f(z)(1-F(z)) \sum\limits_{j=1}^{m}p_{j}(p_{j}-1)dz \\ &=&\frac{q-n}{\sigma^{2}}E\left[\frac{\left( f(Z_{k})\right)^{2}}{ 1-F(Z_{k})}+f^{\prime} (Z_{k})\right]. \end{array} $$

By summing the three terms I₁,I₂ and I₃, we obtain

$$ \begin{array}{@{}rcl@{}} I_{RDS}(\mu ) &=&\frac{n}{\sigma^{2}}E\left[\left( \frac{f^{\prime} (Z_{k})}{ f(Z_{k})}\right)^{2}\right] +\frac{q-n}{\sigma^{2}}E\left[\frac{\left( f(Z_{k})\right)^{2}}{F(Z_{k})}-f^{\prime} (Z_{k})\right] \\ &&+\frac{q-n}{\sigma^{2}}E\left[\frac{\left( f(Z_{k})\right)^{2}}{ 1-F(Z_{k})}+f^{\prime} (Z_{k})\right] \end{array} $$

which yields to the result.

Appendix 2

Proof of equation (5.1) given in Theorem 3 of the Section 5.2.1. Equation 5.1 has been subdivided into three parts denoted I_i,i = 4,5,6 such as

$$ I_{RDS}(\sigma)=I_{4}+I_{5}+I_{6} $$

where

$$ \left\{ \begin{array}{l} I_{4}=E\left[ \frac{\partial }{\partial \sigma }\left[ \frac{-1}{\sigma } \left( {\sum}_{j=1}^{m}{\sum}_{i=1}^{p_{j}} (i-1)\frac{Z_{(i)}^{j}f(Z_{(i)}^{j})}{F(Z_{(i)}^{j})}\right) \right] \right] \\ I_{5}=E\left[ \frac{\partial }{\partial \sigma }\left[ \frac{-1}{\sigma } \left( {\sum}_{j=1}^{m}{\sum}_{i=1}^{p_{j}} \frac{Z_{(i)}^{j}f^{\prime} (Z_{(i)}^{j})}{f(Z_{(i)}^{j})}\right) \right] \right] \\ I_{6}=E\left[ \frac{\partial }{\partial \sigma }\left[ \frac{-1}{\sigma } \left( {\sum}_{j=1}^{m}{\sum}_{i=1}^{p_{j}} (p_{j}-i)\frac{Z_{(i)}^{j}f(Z_{(i)}^{j})}{1-F(Z_{(i)}^{j})}\right) \right] \right]. \end{array} \right.$$

• (1) Let us calculate the first term I₄.

We have

$$ I_{4}=E\left[ \frac{\partial }{\partial \sigma }\left[ \frac{-1}{\sigma } \left( \sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} (i-1)\frac{Z_{(i)}^{j}f(Z_{(i)}^{j})}{F(Z_{(i)}^{j})}\right) \right] \right], $$

using the product derivation rule, we write

$$I_{4}=E\left[ \frac{1}{\sigma^{2}}\left( \sum\limits_{j=1}^{m} \sum\limits_{i=1}^{p_{j}}(i-1)\frac{Z_{(i)}^{j}f(Z_{(i)}^{j})}{ F(Z_{(i)}^{j})}\right) -\frac{1}{\sigma }\times \frac{\partial }{\partial \sigma }\left( \sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} (i-1)\frac{Z_{(i)}^{j}f(Z_{(i)}^{j})}{F(Z_{(i)}^{j})}\right) \right]. $$

By the derivation of the second term of I₄ according to σ and by simplifying, we get

$$ I_{4}=\frac{1}{\sigma^{2}}E\left[ \sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} (i-1)\left( \frac{2Z_{(i)}^{j}f(Z_{(i)}^{j})}{ F(Z_{(i)}^{j})}+\frac{\left( Z_{(i)}^{j}\right)^{2}f^{\prime} (Z_{(i)}^{j})}{ F(Z_{(i)}^{j})}-\left( \frac{Z_{(i)}^{j}f(Z_{(i)}^{j})}{F(Z_{(i)}^{j})}\right)^{2}\right) \right]. $$

Let

$$ h_{4}(z)=\frac{2zf(z)}{F(z)}+\frac{\left( z\right)^{2}f^{\prime} (z)}{F(z)} -\left( \frac{Z_{(i)}^{j}f(z)}{F(z)}\right)^{2}, $$

(A.1)

the latter equality becomes

$$ \begin{array}{@{}rcl@{}} I_{4}&=&\frac{1}{\sigma^{2}}\sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} (i-1){\int}_{-\infty }^{+\infty }h_{4}(z)f_{(i)}^{j}(z)dz \\ &=&\frac{1}{\sigma^{2}}{\int}_{-\infty }^{+\infty }h_{4}(z)\sum\limits_{j=1}^{m} \sum\limits_{i=1}^{p_{j}}(i-1)\frac{p_{j}! }{(i-1)!(p_{j}-i)!}\left[ F(x)\right]^{i-1}f(x)\left[ 1-F(x)\right] ^{p_{j}-i}dz. \end{array} $$

Using the same method of calculation as when developing I₂, we get

$$ \begin{array}{@{}rcl@{}} I_{4} &=&\frac{1}{\sigma^{2}}{\int}_{-\infty }^{+\infty }h_{4}(z)f(z)F(z) \sum\limits_{j=1}^{m}p_{j}(p_{j}-1)dz \\ &=&\sum\limits_{j=1}^{m}p_{j}(p_{j}-1)\frac{1}{\sigma^{2}} {\int}_{-\infty }^{+\infty }h_{4}(z)f(z)F(z)dz, \end{array} $$

substituting the Eq. A.1 in the latter equality, we obtain

$$ I_{4}=\frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\left[ 2z\left( f(z)\right)^{2}+\left( z\right)^{2}f^{\prime} (z)f(z)-\frac{\left( zf(z)\right)^{2}f(z)}{F(z)}\right] dz. $$

(A.2)

• (2) Let us calculate the second term I₅.

We have

$$ I_{5}=E\left[ \frac{\partial }{\partial \sigma }\left[ \frac{-1}{\sigma } \left( \sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} \frac{Z_{(i)}^{j}f^{\prime} (Z_{(i)}^{j})}{f(Z_{(i)}^{j})}\right) \right] \right]. $$

By the derivation according to σ, we obtain

$$ \begin{array}{@{}rcl@{}} I_{5}&=&E\left[ \frac{1}{\sigma^{2}}\sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} \frac{Z_{(i)}^{j}f^{\prime} (Z_{(i)}^{j})}{f(Z_{(i)}^{j}) }\right] \\ &&-E\left[ \frac{1}{\sigma }\left[\sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} \frac{\frac{-Z_{(i)}^{j}f^{\prime} (Z_{(i)}^{j})f(Z_{(i)}^{j})}{\sigma }+\frac{-\left( Z_{(i)}^{j}\right)^{2}f^{\prime \prime} (Z_{(i)}^{j})f(Z_{(i)}^{j})}{\sigma }+\frac{\left( Z_{(i)}^{j}\right)^{2}\left( f^{\prime} (Z_{(i)}^{j})\right)^{2}}{\sigma }}{\left( f(Z_{(i)}^{j})\right)^{2}}\right] \right]. \end{array} $$

By simplifying the latter equality, we get

$$ I_{5}=E\left[ \frac{1}{\sigma^{2}}\sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} \frac{Z_{(i)}^{j}f^{\prime} (Z_{(i)}^{j})}{ f(Z_{(i)}^{j})}+\frac{Z_{(i)}^{j}f^{\prime} (Z_{(i)}^{j})}{f(Z_{(i)}^{j})}+\frac{\left( Z_{(i)}^{j}\right)^{2}f^{\prime \prime} (Z_{(i)}^{j})}{f(Z_{(i)}^{j})}-\left( \frac{ Z_{(i)}^{j}f^{\prime} (Z_{(i)}^{j})}{f(Z_{(i)}^{j})}\right)^{2}\right]. $$

Let

$$ h_{5}(z)=\frac{2zf^{\prime} (z)}{f(z)}+\frac{\left( z\right)^{2}f^{\prime \prime} (z)}{f(z)}-\left( \frac{zf^{\prime} (z)}{f(z)}\right)^{2}, $$

(A.3)

and using the same method of calculation as when developing I₁, we obtain

$$ I_{5}=\frac{n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\left[ 2zf^{\prime} (z)+\left( z\right)^{2}f^{\prime \prime} (z)-\left( \frac{zf^{\prime} (z)}{f(z)} \right)^{2}f(z)\right] dz. $$

(A.4)

(3) Let us calculate the third term I₆.

We have

$$ I_{6}=E\left[ \frac{\partial }{\partial \sigma }\left[ \frac{-1}{\sigma } \left( \sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} (p_{j}-i)\frac{Z_{(i)}^{j}f(Z_{(i)}^{j})}{1-F(Z_{(i)}^{j})}\right) \right] \right].$$

By the derivation according to σ, we obtain

$$ \begin{array}{@{}rcl@{}} I_{6}&=&E\left[ \frac{1}{\sigma^{2}}\sum\limits_{j=1}^{m} \sum\limits_{i=1}^{p_{j}}(p_{j}-i)\frac{Z_{(i)}^{j}f(Z_{(i)}^{j}) }{1-F(Z_{(i)}^{j})}\right] \\ &&+E\left[ -\frac{1}{\sigma }\left[ \sum\limits_{j=1}^{m} \sum\limits_{i=1}^{p_{j}}(p_{j}-i)\frac{\frac{ -Z_{(i)}^{j}f(Z_{(i)}^{j})\left[ 1-F(Z_{(i)}^{j})\right] }{\sigma }+\frac{ -\left( Z_{(i)}^{j}\right)^{2}f^{\prime} (Z_{(i)}^{j})\left[ 1-F(Z_{(i)}^{j}) \right] }{\sigma }+\frac{-\left( Z_{(i)}^{j}\right)^{2}\left( f(Z_{(i)}^{j})\right)^{2}}{\sigma }}{\left[ 1-F(Z_{(i)}^{j})\right]^{2}} \right] \right], \end{array} $$

by simplifying the latter equation, we have

$$ \begin{array}{@{}rcl@{}} I_{6} &=&E\left[ \frac{1}{\sigma^{2}}\sum\limits_{j=1}^{m} \sum\limits_{i=1}^{p_{j}}(p_{j}-i)\frac{Z_{(i)}^{j}f(Z_{(i)}^{j}) }{1-F(Z_{(i)}^{j})}\right] \\ &&+E\left[ \frac{1}{\sigma^{2}}\sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} (p_{j}-i)\left[ \frac{Z_{(i)}^{j}f(Z_{(i)}^{j})}{ 1-F(Z_{(i)}^{j})}+\frac{\left( Z_{(i)}^{j}\right)^{2}f^{\prime} (Z_{(i)}^{j})}{ 1-F(Z_{(i)}^{j})}+\left( \frac{Z_{(i)}^{j}f(Z_{(i)}^{j})}{1-F(Z_{(i)}^{j})} \right)^{2}\right] \right], \end{array} $$

and by making some rearrangements of the latter equation’s terms, we obtain

$$ I_{6}=\frac{1}{\sigma^{2}}E\left[ \sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} (p_{j}-i)\left[ \frac{2Z_{(i)}^{j}f(Z_{(i)}^{j}) }{1-F(Z_{(i)}^{j})}+\frac{\left( Z_{(i)}^{j}\right)^{2}f^{\prime} (Z_{(i)}^{j})}{ 1-F(Z_{(i)}^{j})}+\left( \frac{Z_{(i)}^{j}f(Z_{(i)}^{j})}{1-F(Z_{(i)}^{j})} \right)^{2}\right] \right]. $$

Let

$$ h_{6}(z)=\frac{2zf(z)}{1-F(z)}+\frac{\left( z\right)^{2}f^{\prime} (z)}{1-F(z)} +\left( \frac{zf(z)}{1-F(z)}\right)^{2}. $$

(A.5)

Using the same method of calculation as when developing I₃, it follows

$$ \begin{array}{@{}rcl@{}} I_{6} &=&\frac{1}{\sigma^{2}}{\int}_{-\infty }^{+\infty }h_{6}(z)f(z)(1-F(z)) \sum\limits_{j=1}^{m}p_{j}(p_{j}-1)dz \\ &=&\frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }h_{6}(z)f(z)(1-F(z))dz, \end{array} $$

so,

$$ I_{6}=\frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\left[ 2z\left( f(z)\right)^{2}+\left( z\right)^{2}f^{\prime} (z)f(z)+\frac{\left( zf(z)\right)^{2}}{1-F(z)}f(z)\right] dz. $$

(A.6)

By summing the three terms I₄,I₅ and I₆, we obtain

$$ \begin{array}{@{}rcl@{}} I_{RDS}(\sigma )&=&\frac{-n}{\sigma^{2}}-\left[ \frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\left[ 2z\left( f(z)\right)^{2}+\left( z\right)^{2}f^{\prime} (z)f(z)-\frac{\left( zf(z)\right)^{2}f(z)}{F(z)}\right] dz\right] \\ &&-\left[ \frac{n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\left[ 2zf^{\prime} (z)+\left( z\right)^{2}f^{\prime \prime} (z)-\left( \frac{zf^{\prime} (z)}{f(z)} \right) f(z)\right] dz\right] \\ &&+\frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\left[ 2z\left( f(z)\right)^{2}+\left( z\right)^{2}f^{\prime} (z)f(z)+\frac{\left( zf(z)\right)^{2}}{1-F(z)}f(z)\right] dz. \end{array} $$

By simplifying the latter equality, it follows

$$ \begin{array}{@{}rcl@{}} I_{RDS}(\sigma )&=&\frac{-n}{\sigma^{2}}+\frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty } \frac{\left( zf(z)\right)^{2}f(z)}{F(z)}dz \\ &&-\frac{n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }2zf^{\prime} (z)dz-\frac{n}{ \sigma^{2}}{\int}_{-\infty }^{+\infty }\left( z\right)^{2}f^{\prime \prime} (z)dz \\ &&+\frac{n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\left( \frac{zf^{\prime} (z)}{f(z)}\right)^{2}f(z)dz+\frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\frac{\left( zf(z)\right)^{2}}{1-F(z)}f(z)dz, \end{array} $$

using the integration part of the fourth term of the right hand side of the latter equation, we get

$$ \begin{array}{@{}rcl@{}} I_{RDS}(\sigma )&=&\frac{-n}{\sigma^{2}}+\frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty } \frac{\left( zf(z)\right)^{2}f(z)}{F(z)}dz-\frac{n}{\sigma^{2}} {\int}_{-\infty }^{+\infty }2zf^{\prime} (z)dz \\ &&-\frac{n}{\sigma^{2}}\left( \left[ \left( z\right)^{2}f^{\prime} (z)\right]_{-\infty }^{+\infty }-2{\int}_{-\infty }^{+\infty }zf^{\prime} (z)dz\right) \\ &&+\frac{n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\left( \frac{zf^{\prime} (z)}{ f(z)}\right)^{2}f(z)dz+\frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\frac{\left( zf(z)\right)^{2}}{1-F(z)}f(z)dz, \end{array} $$

after simplifying, we obtain

$$ \begin{array}{@{}rcl@{}} I_{RDS}(\sigma )&=&\frac{-n}{\sigma^{2}}+\frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty } \frac{\left( zf(z)\right)^{2}f(z)}{F(z)}dz \\ &&-\frac{n}{\sigma^{2}}\left[ \left( z\right)^{2}f^{\prime} (z)\right]_{-\infty }^{+\infty }+\frac{n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\left( \frac{zf^{\prime} (z)}{f(z)}\right)^{2}f(z)dz \\ &&+\frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\frac{\left( zf(z)\right)^{2}}{1-F(z)}f(z)dz. \end{array} $$

According to the hypothesis of the Theorem 3, we have $\left [ \left (z\right )^{2}f^{\prime } (z)\right ]_{-\infty }^{+\infty }=0$, so,

$$ \begin{array}{@{}rcl@{}} I_{RDS}(\sigma )&=&\left[ \frac{n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\left( \frac{ zf^{\prime} (z)}{f(z)}\right)^{2}f(z)dz-\frac{n}{\sigma^{2}}\right] \\ &&+\frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\left( \frac{\left( zf(z)\right)^{2} }{F(z)}+\frac{\left( zf(z)\right)^{2}}{1-F(z)}\right) f(z)dz \\ &=&\left[ \frac{n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\left( \frac{ zf^{\prime} (z)}{f(z)}\right)^{2}f(z)dz-\frac{n}{\sigma^{2}}\right] \\ &&+\frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\left( \frac{\left( zf(z)\right)^{2} }{F(z)\left[ 1-F(z)\right] }\right) f(z)dz \end{array} $$

which yields to the result.

Appendix 3

Proof of equation (5.3) given in Theorem 5 of the Section 5.3. By the derivation of each term of the right hand side of the Eq. 5.8, we get

$$ \begin{array}{@{}rcl@{}} I_{RDS}(\mu ,\sigma )&=&-E\left[ \frac{1}{\sigma^{2}}\left( \sum\limits_{j=1}^{m} \sum\limits_{i=1}^{p_{j}}(i-1)\frac{ f(Z_{(i)}^{j})}{F(Z_{(i)}^{j})}+\sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} \frac{f^{\prime} (Z_{(i)}^{j})}{f(Z_{(i)}^{j})}\right) \right] \\ &&-E\left[ \frac{-1}{\sigma }\frac{\partial }{\partial \sigma }\left( \sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}}(i-1) \frac{f(Z_{(i)}^{j})}{F(Z_{(i)}^{j})}+\sum\limits_{j=1}^{m} \sum\limits_{i=1}^{p_{j}}\frac{f^{\prime} (Z_{(i)}^{j})}{f(Z_{(i)}^{j}) }\right) \right] \\ &&+E\left[ \frac{1}{\sigma^{2}}\left( \sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} (p_{j}-i)\frac{f(Z_{(i)}^{j})}{1-F(Z_{(i)}^{j})}\right) \right] \\ &&-E\left[ \frac{1}{\sigma }\frac{\partial }{\partial \sigma }\left( \sum\limits_{j=1}^{m} \sum\limits_{i=1}^{p_{j}} (p_{j}-i)\frac{f(Z_{(i)}^{j})}{1-F(Z_{(i)}^{j})}\right) \right]. \end{array} $$

(A.7)

By developing the latter equality, we write the following expressions

$$ \frac{\partial }{\partial \sigma }\left( \sum\limits_{j=1}^{m} {\sum}_{i=1}^{p_{j}}(i-1)\frac{f(Z_{(i)}^{j})}{ F(Z_{(i)}^{j})}\right) =\sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} (i-1)\left[ \frac{Z_{(i)}^{j}f^{\prime} (Z_{(i)}^{j})}{ F(Z_{(i)}^{j})}-Z_{(i)}^{j}\left( \frac{f(Z_{(i)}^{j})}{F(Z_{(i)}^{j})}\right)^{2}\right], $$

$$ \frac{\partial }{\partial \sigma }\left( \sum\limits_{j=1}^{m} \sum\limits_{i=1}^{p_{j}}\frac{f^{\prime} (Z_{(i)}^{j})}{f(Z_{(i)}^{j}) }\right) =\sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} \left[ \frac{Z_{(i)}^{j}f^{\prime \prime} (Z_{(i)}^{j})}{f(Z_{(i)}^{j})} -Z_{(i)}^{j}\left( \frac{f^{\prime} (Z_{(i)}^{j})}{f(Z_{(i)}^{j})}\right)^{2} \right], $$

$$\frac{\partial }{\partial \sigma }\left( \sum\limits_{j=1}^{m} \sum\limits_{i=1}^{p_{j}}(p_{j}-i)\frac{f(Z_{(i)}^{j})}{ 1-F(Z_{(i)}^{j})}\right) =\sum\limits_{j=1}^{m}\sum\limits_{i=1}^{p_{j}} (p_{j}-i)\left[ \frac{Z_{(i)}^{j}f^{\prime} (Z_{(i)}^{j})}{ 1-F(Z_{(i)}^{j})}+Z_{(i)}^{j}\left( \frac{f(Z_{(i)}^{j})}{1-F(Z_{(i)}^{j})} \right)^{2}\right], $$

so, by replacing these expressions in Eq. A.7, we obtain

$$ \begin{array}{@{}rcl@{}} I_{RDS}(\mu ,\sigma )& = &-E\left[ \frac{1}{\sigma^{2}}\left( \sum\limits_{j=1}^{m} \sum\limits_{i=1}^{p_{j}}(i-1)\left[ \frac{f(Z_{(i)}^{j})}{F(Z_{(i)}^{j})}+\frac{Z_{(i)}^{j}f^{\prime} (Z_{(i)}^{j})}{ F(Z_{(i)}^{j})}-Z_{(i)}^{j}\left( \frac{f(Z_{(i)}^{j})}{F(Z_{(i)}^{j})}\right)^{2}\right] \right) \right] \\ &&-E\left[ \frac{1}{\sigma^{2}}\left( \sum\limits_{j=1}^{m} \sum\limits_{i=1}^{p_{j}}\frac{f^{\prime} (Z_{(i)}^{j})}{f(Z_{(i)}^{j}) }+\frac{Z_{(i)}^{j}f^{\prime \prime} (Z_{(i)}^{j})}{f(Z_{(i)}^{j})} -Z_{(i)}^{j}\left( \frac{f^{\prime} (Z_{(i)}^{j})}{f(Z_{(i)}^{j})}\right)^{2}\right) \right] \\ &&+E\left[ \frac{1}{\sigma^{2}}\left( \sum\limits_{j=1}^{m} \sum\limits_{i=1}^{p_{j}}(p_{j}-i)\left[ \frac{f(Z_{(i)}^{j})}{ 1-F(Z_{(i)}^{j})}+\frac{Z_{(i)}^{j}f^{\prime} (Z_{(i)}^{j})}{1-F(Z_{(i)}^{j})} +Z_{(i)}^{j}\left( \frac{f(Z_{(i)}^{j})}{1-F(Z_{(i)}^{j})}\right)^{2}\right] \right) \right]. \end{array} $$

Using the same method of calculation as in appendices 1 and 2, we get

$$ \begin{array}{@{}rcl@{}} I_{RDS}(\mu ,\sigma ) &=&-\frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty } \left[ f(z)+zf^{\prime} (z)-\frac{z(f(z))}{F(z)}^{2}\right] f(z)fdz \\ &&-\frac{n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\left[ f^{(1)}(z)+zf^{\prime \prime} (z)-z\left( \frac{f^{\prime} (z)}{f(z)}\right)^{2}f(z)\right] dz \\ &&+\frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }\left[ f(z)+zf^{\prime} (z)+ \frac{z(f(z))^{2}}{1-F(z)}\right] f(z)dz, \end{array} $$

by simplifying the latter equation, we get

$$ \begin{array}{@{}rcl@{}} I_{RDS}(\mu ,\sigma ) &=&\frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty } \left[ \frac{z(f(z))^{2}}{F(z)\left( 1-F(z)\right) }\right] f(z)dz \\ &&+\frac{n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }z\left( \frac{f^{\prime} (z)}{ f(z)}\right)^{2}f(z)dz \end{array} $$

which yields to the result.