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 Ii,i = 1,2,3 such as
$$I_{RDS}(\mu )=I_{1}+I_{2}+I_{3}$$
where
$$\left\{ \begin{array}{l} I_{1}=\frac{1}{\sigma^{2}}{\sum}_{j=1}^{m}{\sum}_{i=1}^{p_{j}} E\left[ \frac{-f^{\prime \prime} (Z_{(i)}^{j})}{ f(Z_{(i)}^{j})}+\left( \frac{f^{\prime} (Z_{(i)}^{j})}{f(Z_{(i)}^{j})}\right)^{2} \right] \\ I_{2}=\frac{1}{\sigma^{2}}{\sum}_{j=1}^{m}{\sum}_{i=1}^{p_{j}} (i-1)E\left[ \frac{-f^{\prime }(Z_{(i)}^{j})}{ F(Z_{(i)}^{j})}+\left( \frac{f(Z_{(i)}^{j})}{F(Z_{(i)}^{j})}\right)^{2}\right] \\ I_{3}=\frac{1}{\sigma^{2}}{\sum}_{j=1}^{m}{\sum}_{i=1}^{p_{j}} (p_{j}-i)E\left[ \frac{f^{\prime} (Z_{(i)}^{j})}{ 1-F(Z_{(i)}^{j})}+\left( \frac{f(Z_{(i)}^{j})}{1-F(Z_{(i)}^{j})}\right)^{2}\right]. \end{array} \right.$$
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,
$$ I_{1}=\frac{n}{\sigma^{2}}E\left[\left( \frac{f^{\prime} (Z_{k})}{f(Z_{k})} \right)^{2}\right]. $$
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
$$ \begin{array}{@{}rcl@{}} I_{2} &=&\frac{q-n}{\sigma^{2}}{\int}_{-\infty }^{+\infty }f(z)h_{2}(z)F(z)dz \\ &=&\frac{q-n}{\sigma^{2}}E[\frac{\left( f(Z_{k})\right)^{2}}{ F(Z_{k})}-f^{\prime} (Z_{k})]. \end{array} $$
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 = pj 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 Ii,i = 4,5,6 such as
$$ I_{RDS}(\sigma)=I_{4}+I_{5}+I_{6} $$
where
$$ \left\{ \begin{array}{l} I_{4}=E\left[ \frac{\partial }{\partial \sigma }\left[ \frac{-1}{\sigma } \left( {\sum}_{j=1}^{m}{\sum}_{i=1}^{p_{j}} (i-1)\frac{Z_{(i)}^{j}f(Z_{(i)}^{j})}{F(Z_{(i)}^{j})}\right) \right] \right] \\ I_{5}=E\left[ \frac{\partial }{\partial \sigma }\left[ \frac{-1}{\sigma } \left( {\sum}_{j=1}^{m}{\sum}_{i=1}^{p_{j}} \frac{Z_{(i)}^{j}f^{\prime} (Z_{(i)}^{j})}{f(Z_{(i)}^{j})}\right) \right] \right] \\ I_{6}=E\left[ \frac{\partial }{\partial \sigma }\left[ \frac{-1}{\sigma } \left( {\sum}_{j=1}^{m}{\sum}_{i=1}^{p_{j}} (p_{j}-i)\frac{Z_{(i)}^{j}f(Z_{(i)}^{j})}{1-F(Z_{(i)}^{j})}\right) \right] \right]. \end{array} \right.$$
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)
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.