Appendix
Proof of Theorem 1:
The first expression follows from Eq. (1). For the second equation, we observe that
$$\begin{aligned} P(R_{s_i}^{A_1}=u,s_i=k)= & {} P(R_{s_i}^{A_1}=u|s_i=k)P(s_i=k)\nonumber \\= & {} \frac{\left( \begin{array}{c} {k-i} \\ {u-i} \end{array}\right) \left( \begin{array}{c} {N-n-k+i} \\ {M-n-u+i} \end{array}\right) }{ \left( \begin{array}{c} {N-n} \\ {M-n} \end{array}\right) } \frac{\left( \begin{array}{c} {k-1} \\ {i-1} \end{array}\right) \left( \begin{array}{c} {N-k} \\ {n-i} \end{array}\right) }{ \left( \begin{array}{c} {N} \\ {n} \end{array}\right) } \nonumber \\= & {} \frac{\left( \begin{array}{c} {N-k} \\ {M-u} \end{array}\right) \left( \begin{array}{c} {M-u} \\ {n-i} \end{array}\right) \left( \begin{array}{c} {k-1} \\ {u-1} \end{array}\right) \left( \begin{array}{c} {u-1} \\ {i-1} \end{array}\right) }{\left( \begin{array}{c} {N-n} \\ {M-n} \end{array}\right) \left( \begin{array}{c} {N} \\ {n} \end{array}\right) } \nonumber \\ P(R_{s_i}^{A_1}=u)= & {} \sum _{k=1}^N \frac{\left( \begin{array}{c} {N-k} \\ {M-u} \end{array}\right) \left( \begin{array}{c} {M-u} \\ {n-i} \end{array}\right) \left( \begin{array}{c} {k-1} \\ {u-1} \end{array}\right) \left( \begin{array}{c} {u-1} \\ {i-1} \end{array}\right) }{\left( \begin{array}{c} {N-n} \\ {M-n} \end{array}\right) \left( \begin{array}{c} {N} \\ {n} \end{array}\right) } = \frac{\left( \begin{array}{c} {u-1} \\ {i-1} \end{array}\right) \left( \begin{array}{c} {M-u} \\ {n-i} \end{array}\right) }{\left( \begin{array}{c} {M} \\ {n} \end{array}\right) }. \end{aligned}$$
(7)
We now consider the conditional distribution of \(X_{s_i}\) given \(R_{s_i}^{A_1}\). Using Eq. (7) and \(P(R^{A_q}_{s_i}=h)\), we write
$$\begin{aligned} P(X_{s_i}=x_k|R_{s_i}^{A_1}=u)= & {} \frac{P(X_{s_i}=x_k,R_{s_i}^{A_1}=u)}{P(R_{s_i}^{A_1}=u)}\\= & {} \frac{\frac{\left( \begin{array}{c} {N-k} \\ {M-u} \end{array}\right) \left( \begin{array}{c} {M-u} \\ {n-i} \end{array}\right) \left( \begin{array}{c} {k-1} \\ {u-1} \end{array}\right) \left( \begin{array}{c} {u-1} \\ {i-1} \end{array}\right) }{\left( \begin{array}{c} {N-n} \\ {M-n} \end{array} \right) \left( \begin{array}{c} N \\ n \end{array} \right) }}{ \frac{\left( \begin{array}{c} {u-1} \\ {i-1} \end{array}\right) \left( \begin{array}{c} {M-u} \\ {n-i} \end{array}\right) }{\left( \begin{array}{c} {M} \\ {n} \end{array}\right) }} \\= & {} \frac{\left( \begin{array}{c} {k-1} \\ {h-1} \end{array}\right) \left( \begin{array}{c} {N-k} \\ {M-u} \end{array}\right) }{\left( \begin{array}{c} {N} \\ {M} \end{array}\right) }. \end{aligned}$$
For the joint distribution of the sample ranks in Eq. (4) for design \(A_1\), we first observe that
$$\begin{aligned} P(s_i=k,s_j=t)= \frac{ \left( \begin{array}{c} {k-1} \\ {i-1} \end{array} \right) \left( \begin{array}{c} {t-k-1} \\ {j-i-1} \end{array} \right) \left( \begin{array}{c} {N-t} \\ {n-j} \end{array} \right) }{\left( \begin{array}{c} N \\ n \end{array}\right) }. \end{aligned}$$
The joint distribution of sample and population ranks are then given by
$$\begin{aligned} P(R_{s_i}^{A_1}=u, R_{s_j}^{A_1}=u',s_i=k,s_j=t)= & {} \frac{ \left( \begin{array}{c} {k-1} \\ {i-1} \end{array} \right) \left( \begin{array}{c} {t-k-1} \\ {j-i-1} \end{array} \right) \left( \begin{array}{c} {N-t} \\ {n-j} \end{array} \right) }{\left( \begin{array}{c} N \\ n \end{array}\right) } \frac{ \left( \begin{array}{c} {k-i} \\ {u-i} \end{array} \right) \left( \begin{array}{c} {t-k-j+i} \\ {u'-u-j+i} \end{array} \right) \left( \begin{array}{c} {N-t-n+j} \\ {M-n+j-u'} \end{array} \right) }{\left( \begin{array}{c} {N-n} \\ {m- n} \end{array}\right) }\\= & {} \frac{ \left( \begin{array}{c} {k-1} \\ {u-1} \end{array} \right) \left( \begin{array}{c} {u-1} \\ {i-1} \end{array} \right) \left( \begin{array}{c} {t-k-1} \\ {u'-u-1} \end{array} \right) \left( \begin{array}{c} {u'-u-1} \\ {j-i-1} \end{array} \right) \left( \begin{array}{c} {N-t} \\ {M-u'} \end{array} \right) \left( \begin{array}{c} {M-u'} \\ {n-j} \end{array} \right) }{\left( \begin{array}{c} N \\ n \end{array}\right) \left( \begin{array}{c} {N-n} \\ {M- n} \end{array}\right) }. \end{aligned}$$
The marginal distribution of sample ranks then follows from summation over k and t
$$\begin{aligned} P(R_{s_i}^{A_1}=u, R_{s_j}^{A_1}=u')= & {} \sum _{t=1}^N \sum _{k=1}^{t-1}\frac{ \left( \begin{array}{c} {k-1} \\ {u-1} \end{array} \right) \left( \begin{array}{c} {u-1} \\ {i-1} \end{array} \right) \left( \begin{array}{c} {t-k-1} \\ {u'-u-1} \end{array} \right) \left( \begin{array}{c} {u'-u-1} \\ {j-i-1} \end{array} \right) \left( \begin{array}{c} {N-t} \\ {M-u'} \end{array} \right) \left( \begin{array}{c} {M-u'} \\ {n-j} \end{array} \right) }{\left( \begin{array}{c} N \\ n \end{array}\right) \left( \begin{array}{c} {N-n} \\ {M- n} \end{array}\right) } \\= & {} \frac{\left( \begin{array}{c} N \\ M \end{array}\right) \left( \begin{array}{c} {u-1} \\ {i-1} \end{array} \right) \left( \begin{array}{c} {u'-u-1} \\ {j-i-1} \end{array} \right) \left( \begin{array}{c} {M-u'} \\ {n-j} \end{array} \right) }{\left( \begin{array}{c} N \\ n \end{array}\right) \left( \begin{array}{c} {N-n} \\ {M- n} \end{array}\right) }. \end{aligned}$$
The conditional distribution of \(X_{s_i}\) and \(X_{s_j}\) given their sample ranks is obtained as follows
$$\begin{aligned} P(X_{s_i}=x_k,X_{s_j}=x_t|R_{s_i}^{A_1}=u, R_{s_j}^{A_1}=u')= & {} \frac{P(R_{s_i}^{A_1}=u, R_{s_j}^{A_1}=u',s_i=k,s_j=t)}{P(R_{s_i}^{A_1}=u, R_{s_j}^{A_1}=u')}\\= & {} \frac{ \left( \begin{array}{c} {k-1} \\ {u-1} \end{array} \right) \left( \begin{array}{c} {t-k-1} \\ {u'-u-1} \end{array} \right) \left( \begin{array}{c} {N-t} \\ {M-u'} \end{array} \right) }{\left( \begin{array}{c} N \\ M \end{array}\right) }. \end{aligned}$$
\(\square \)
Proof of Lemma 1:
For the proof of (i) and (ii) when \(q=1\), we first observe that \(\frac{I_h^{A_1}I_{hu}^{A_1}}{n_h^{A_1}d_n^{A_1}}; u \in D_h\), are identically distributed. Let \(A= E\left( \frac{I_h^{A_1}I_{hu}^{A_1}}{n_h^{A_1}d_n^{A_1}}\right) \) for \( u \in D_h\). Note that the following equalities hold
$$\begin{aligned} \sum _{u \in D_h} \frac{I_h^{A_1}I_{hu}^{A_1}}{n_h^{A_1}d_n^{A_1}}= \frac{I_h^{A_1}}{d_n^{A_1}}. \end{aligned}$$
We take the expected value in both sides of the above equation and write
$$\begin{aligned} E\left\{ \sum _{u \in D_h} \frac{I_h^{A_1}I_{hu}^{A_1}}{n_h^{A_1}d_n^{A_1}} \right\}= & {} E\left( \frac{I_h^{A_q}}{d_n^{A_q}}\right) , \\ \sum _{u \in D_h} E \left( \frac{I_h^{A_1}I_{hu}^{A_1}}{n_h^{A_1}d_n^{A_1}}\right)= & {} E\left( \frac{I_h^{A_1}}{d_n^{A_1}}\right) , \\ H E \left( \frac{I_h^{A_1}I_{h1}^{A_1}}{n_h^{A_1}d_n^{A_1}}\right)= & {} E\left( \frac{I_h^{A_1}}{d_n^{A_1}}\right) , \\ A= & {} \frac{1}{H} E\left( \frac{I_h^{A_1}}{d_n^{A_1}}\right) . \end{aligned}$$
We again observe that \(I_h^{A_1}/d_n^{A_1}\); \(h=1,\ldots , d\), are identically distributed. Let \(B=E(I_h^{A_1}/d_n^{A_1})\); \(h=1,\ldots ,d\). It is then easy to see that \(B= 1/d\). When \(q=2\), the proof of (i) is given in Ozturk (2014). This completes the proofs of (i) and (ii).
Proof of (iii): To simplify the notation, we drop the superscript \(A_1\) in part (iii) and (iv). All random variables in these parts are constructed from design \(A_1\), even though they are not explicitly stated. We first observe that
$$\begin{aligned}&\displaystyle \sum _{u \in D_h} \sum _{v \in D_h} \frac{I_h^2I_{hu}I_{hv}}{d_n^2n_h^2} = \frac{I_h^2}{d_n^2} \\&\displaystyle \sum _{u \in D_h}\sum _{(v \ne u) \in D_h} \frac{I_h^2I_{hu}I_{hv}}{d_n^2n_h^2} + \sum _{u \in D_v} \frac{I_h^2I^2_{hu}}{d_n^2n_h^2} =\frac{I_h^2}{d_n^2}. \end{aligned}$$
Taking the expected value in both sides of the above equation, after some simplification, we obtain
$$\begin{aligned} {\text {Cov}}\left( \frac{I_1I_{11}}{d_nn_1},\frac{I_1I_{12}}{d_nn_1}\right) = -\frac{1}{H-1} {\text {Var}}\left( \frac{I_1I_{11}}{d_nn_1}\right) +{\text {Var}}\left( I_1/d_n\right) \frac{1}{H\left( H-1\right) }. \end{aligned}$$
Proof of (iv): We again observe the following equation
$$\begin{aligned} \sum _{u \in D_h} \sum _{v \in D_{h'}} \frac{I_{h}I_{hu}I_{h'}I_{h'v}}{n_hn_{h'}d_n^2} = \frac{I_hI_{h'}}{d_n^2}. \end{aligned}$$
Taking the expected values in both sides, we write
$$\begin{aligned} H^2E\left( \frac{I_{h}I_{ha}I_{h'}I_{h'b}}{n_hn_{h'}d_n^2}\right)= & {} E(I_hI_{h'}/d_n^2)\nonumber \\ {\text {Cov}}\left( \frac{I_{h}I_{ha}}{d_nn_h}, \frac{I_{h'}I_{h'b}}{d_nn_{h'}}\right) + \frac{1}{d^2H^2}= & {} \frac{1}{H^2}\left\{ {\text {Cov}}(I_h/d_n, I_{h'}/d_n)+\frac{1}{d^2}\right\} \nonumber \\ {\text {Cov}}\left( \frac{I_{h}I_{ha}}{d_nn_h}, \frac{I_{h'}I_{h'b}}{d_nn_{h'}}\right)= & {} -\frac{1}{H^2} {\text {var}}(I_1/d_n); \quad a \in D_h, b \in D_{h'}. \end{aligned}$$
(8)
The proofs of (v) and (vi) are given in Ozturk (2014) in a slightly different context. \(\square \)
Proof of Theorem 2:
We only consider the unbiasedness of the estimator \(\bar{X}_n^{A_1}\). The proof of the other estimator is similar. To simplify the notation, we again drop the subscript \(A_1\) in \(\bar{X}_n^{A_1}\). Using conditional expectation, we write
$$\begin{aligned} E(\bar{X}_n)= & {} E \left\{ \frac{1}{d_n} \sum _{h=1}^d \frac{I_h}{n_h} \sum _{u \in D_h} \sum _{i=1}^n I(R_{s_i}=u) E(X_{s_i}|R_{s_i}=u)\right\} \\= & {} E \left\{ \frac{1}{d_n} \sum _{h=1}^d \frac{I_h}{n_h} \sum _{u \in D_h} \sum _{i=1}^n I(R_{s_i}=u) \mu _{u:M} \right\} . \end{aligned}$$
Note that \(\sum _{i=1}^n I(R_{s_i}=u)= I_{hu}\) for \(u \in D_h\). Using this notation, the expected value of \(\bar{X}_n\) reduces to
$$\begin{aligned} E(\bar{X}_n)= & {} E \left\{ \frac{1}{d_n} \sum _{h=1}^d \frac{I_h}{n_h} \sum _{u \in D_h} I_{hu} \mu _{u:M} \right\} \end{aligned}$$
(9)
The ratios \( \frac{I_hI_{hu}}{n_hd_n}\); \(u \in D_h\) are identically distributed. From Lemma 1, we have \( E(\frac{I_hI_{hu}}{n_hd_n})= \frac{1}{H} E(\frac{I_h}{d_n})\), for \(u \in D_h\). Using this equality, we write
$$\begin{aligned} E(\bar{X}_n)= & {} E \left\{ \frac{1}{d_n} \sum _{h=1}^d \frac{I_h}{n_h} \sum _{u \in D_h} I_{hu} \mu _{u:M} \right\} \\= & {} E\left\{ \frac{1}{d_n} \sum _{h=1}^d \sum _{u \in D_h}\frac{I_h I_{u}}{n_hd_n} \mu _{u:M} \right\} = \sum _{h=1}^d\sum _{u \in D_h}E\left\{ \frac{I_hI_{h1}}{n_hd_n}\right\} \mu _{u:M} \\= & {} \sum _{h=1}^d \frac{1}{H}\sum _{u \in D_h} E(\frac{I_h}{d_n}) \mu _{u:M} \end{aligned}$$
Again \( \frac{I_h}{d_n}\); \(h=1,\ldots d\), are identically distributed and \(E(\frac{I_1}{d_n}) =1/d\). Hence, we write
$$\begin{aligned} E(\bar{X}_n)= \sum _{h=1}^d \frac{1}{H}\sum _{u \in D_h} E(\frac{I_h}{d_n}) \mu _{u:M}= \frac{1}{dH} \sum _{h=1}^d \sum _{u \in D_h} \mu _{u:M} = \mu . \end{aligned}$$
This completes the proof. \(\square \)
Proof of Theorem 3:
Again to simplify the notation without loss of generality, we write \(\bar{X}_n=\bar{X}_n^{A_1}\) and drop the superscript \(A_1\). Using total variance, we can write
$$\begin{aligned} {\text {Var}}(\bar{X}_n)= {\text {Var}}( E(\bar{X}_n|R))+ E( {\text {Var}}(\bar{X}_n|R)) \end{aligned}$$
From Eq. (9), we write
$$\begin{aligned} E(\bar{X}_n|R)= \sum _{h=1}^d \sum _{u \in D_h} \frac{I_hI_{hu}}{d_nn_h} \mu _{u:M} = \sum _{h=1}^d T_h, \quad T_h=\sum _{u \in D_h} \frac{I_hI_{hu}}{d_nn_h} \mu _{u:M}. \end{aligned}$$
Using the above expression, we obtain
$$\begin{aligned} {\text {Var}}(E(\bar{X}_n|R))= \sum _{h=1}^d {\text {Var}}(T_h) + \sum _{h=1}^d \sum _{h' \ne h} ^d {\text {Cov}}(T_h,T_{h'}) \end{aligned}$$
The variance of \(T_h\) can be expressed as
$$\begin{aligned} {\text {Var}}(T_h)= & {} {\text {Var}}\left\{ \sum _{u \in D_h} \frac{I_hI_{hu}}{d_nn_h} \mu _{u:M}\right\} \\= & {} \sum _{u \in D_h} \mu _{u:M}^2 {\text {Var}}(\frac{I_hI_{hu}}{d_nn_h}) \\&\quad +\sum _{u} \sum _{v} I\{(u \ne v) \in D_h\}\mu _{u:M} \mu _{v:M} {\text {Cov}}\left( \frac{I_hI_{hu}}{d_nn_h}, \frac{I_hI_{hv}}{d_nn_h}\right) \end{aligned}$$
The ratios \(\frac{I_hI_{hu}}{d_nn_h}\); \(u \in D_h\), are identically distributed. The variance of \(T_h\) then reduces to
$$\begin{aligned} {\text {Var}}(T_h)= & {} {\text {Var}}\left( \frac{I_hI_{h1}}{d_nn_h}\right) \sum _{u \in D_h} \mu _{u:M}^2+ {\text {Cov}}\left( \frac{I_hI_{h1}}{d_nn_h}, \frac{I_hI_{h2}}{d_nn_h}\right) \sum _{(u \ne v) \in D_h}\mu _{u:M} \mu _{v:M} \\= & {} {\text {Var}}\left( \frac{I_hI_{h1}}{d_nn_h}\right) \sum _{u \in D_h} \mu _{u:M}^2 +\left\{ \frac{{\text {Var}}(\frac{I_1}{d_n})}{H}-{\text {Var}}(\frac{I_hI_{h1}}{d_nn_h})\right\} \sum _{(u \ne v) \in D_h}\frac{\mu _{u:M} \mu _{v:M}}{H-1} \\= & {} {\text {Var}}\left( \frac{I_hI_{h1}}{d_nn_h}\right) \left\{ \sum _{u \in D_h} \mu _{u:M}^2 - \sum _{(u \ne v) \in D_h}\frac{\mu _{u:M} \mu _{v:M}}{H-1}\right\} \\&\quad + \frac{{\text {Var}}(\frac{I_1}{d_n})}{H(H-1)} \sum _{(u \ne v) \in D_h}\mu _{u:M} \mu _{v:M}. \end{aligned}$$
We now observe that
$$\begin{aligned} \sum _{(u \ne v) \in D_h}\mu _{u:M} \mu _{v:M} =\left( \sum _{u \in D_h} \mu _{u:M}\right) ^2- \sum _{u \in D_h}\mu _{u:M}^2= \mu _h^2- \sum _{u \in D_h}\mu _{u:M}^2, \end{aligned}$$
where \(\mu _h =\sum _{u \in D_h} \mu _{u:M}\). Using the equation above, we write
$$\begin{aligned} {\text {Var}}(T_h)= & {} \frac{{\text {Var}}(\frac{I_hI_{h1}}{d_nn_h})}{H-1}\left\{ H \sum _{u \in D_h}\mu _{u:M}^2 - \mu _h^2 \right\} + \frac{{\text {Var}}(\frac{I_1}{d_n})}{H(H-1)} \left\{ \mu _h^2 -\sum _{u \in D_h}\mu _{u:M}^2 \right\} \nonumber \\= & {} \frac{{\text {Var}}(\frac{I_hI_{h1}}{d_nn_h})}{H-1} H\sum _{u \in D_h}(\mu _{u:M}-\bar{\mu }_h)^2 \nonumber \\&+ \, {\text {Var}}(\frac{I_1}{d_n}) \left\{ \frac{H\bar{\mu }^2_h}{H-1}- \frac{\sum _{u \in D_h} \mu _{u:M}^2}{H(H-1)}\right\} , \end{aligned}$$
(10)
where \(\bar{\mu }_h=\mu _h/H\).
In a similar fashion, we can write
$$\begin{aligned} {\text {Cov}}(T_h,T_{h'})= & {} {\text {Cov}}\left( \sum _{u \in D_h} \frac{I_hI_{hu} \mu _{u:M}}{d_nn_h}, \sum _{v \in D_{h'}} \frac{I_{h'}I_{h'v} \mu _{v:M}}{d_nn_{h'}}\right) \\= & {} \sum _{u \in D_h} \sum _{v \in D_{h'}} \mu _{u:M} \mu _{v:M} {\text {Cov}}\left( \frac{I_hI_{hu} }{d_nn_h} , \frac{I_{h'}I_{h'v} }{d_nn_{h'}}\right) \\= & {} {\text {Cov}}\left( \frac{I_hI_{h1} }{d_nn_h} , \frac{I_{h'}I_{h'1}}{d_nn_{h'}}\right) \sum _{u \in D_h} \sum _{v \in D_{h'}} \mu _{u:M} \mu _{v:M} \end{aligned}$$
Note that from Lemma 1, we have
$$\begin{aligned} {\text {Cov}}\left( \frac{I_hI_{h1} }{d_nn_h} , \frac{I_{h'}I_{h'1}}{d_nn_{h'}}\right) =-\frac{1}{H^2(d-1)} {\text {Var}}(\frac{I_1}{d_n}). \end{aligned}$$
Using the above expression, we obtain
$$\begin{aligned} {\text {Cov}}(T_h,T_{h'}) = -\frac{1}{H^2(d-1)} {\text {Var}}(\frac{I_1}{d_n}) \sum _{u \in D_h} \sum _{v \in D_{h'}} \mu _{u:M} \mu _{v:M} \end{aligned}$$
and
$$\begin{aligned} \sum _{h=1}^d \sum _{h' \ne h}^d {\text {Cov}}(T_h,T_{h'})= & {} -\frac{{\text {Var}}(I_1/d_n)}{H^2(d-1)} \left\{ H^2d^2\mu ^2-H^2\sum _{h=1}^d \bar{\mu }^2_h\right\} . \end{aligned}$$
(11)
Combining Eqs. (10) and (11) in \({\text {Var}}(E(\bar{X}_n|R))\), after some simplifications, we obtain
$$\begin{aligned} {\text {Var}}(E(\bar{X}_n|R))= & {} \frac{H{\text {Var}}(\frac{I_1I_{11}}{d_nn_1})}{H-1} \sum _{h=1}^d\sum _{u \in D_h}(\mu _{u:M}-\bar{\mu }_h)^2 \\&+ {\text {Var}}(\frac{I_1}{d_n}) \sum _{h=1}^d \left\{ \frac{H\bar{\mu }^2_h}{H-1}- \frac{\sum _{u \in D_h} (\mu _{u:M}^2}{H(H-1)}\right\} \\&- \frac{{\text {Var}}(I_1/d_n)}{(d-1)} \left\{ d^2\mu ^2-\sum _{h=1}^d \bar{\mu }^2_h\right\} \\= & {} \frac{H}{H-1} {\text {Var}}\left( \frac{I_1I_{11}}{d_nn_1}\right) \sum _{h=1}^d \sum _{u \in D_h}( \mu _{u:M} -\bar{\mu }_h)^2 \\&+\,{\text {Var}}(\frac{I_1}{d_n})\sum _{h=1}^d \bar{\mu }_h^2\left( \frac{H}{H-1}+\frac{1}{d-1}\right) \\&-\,{\text {Var}}(\frac{I_1}{d_n})\left\{ \sum _{h=1}^d \sum _{u \in D_h}\frac{\mu _{u:M}^2}{H(H-1)} +\frac{d^2}{d-1}\mu ^2\right\} . \end{aligned}$$
Note that \( \sum _{u \in D_h}\mu _{u:M}^2= \sum _{u \in D_h}(\mu _{u:M}-\bar{\mu }_h)^2+H\bar{\mu }^2_h\). Using this equality in the above expression and combining similar terms, we write
$$\begin{aligned} {\text {Var}}(E(\bar{X}_n|R))= & {} \frac{H}{H-1} {\text {Var}}\left( \frac{I_1I_{11}}{d_nn_1}\right) \sum _{h=1}^d \sum _{u \in D_h}\left( \mu _{u:M} -\bar{\mu }_h\right) ^2 \nonumber \\&-\,\frac{1}{H(H-1)}{\text {Var}}\left( \frac{I_1}{d_n}\right) \sum _{h=1}^d \sum _{u \in D_h}\left( \mu _{u:M}-\bar{\mu }_h\right) ^2 \nonumber \\&+{\text {Var}}(\frac{I_1}{d_n}) \left\{ \sum _{h=1}^d\bar{\mu }_h^2\left( \frac{H(d-1)+(H-1)}{(H-1)(d-1)}-\frac{1}{H-1}\right) -\frac{d^2 \mu ^2}{d-1}\right\} \nonumber \\= & {} \sum _{h=1}^d \sum _{u \in D_h}( \mu _{u:M} -\bar{\mu }_h)^2 \left\{ \frac{H}{H-1} {\text {Var}}(\frac{I_1I_{11}}{d_nn_1})- \frac{1}{H(H-1)}{\text {Var}}\left( \frac{I_1}{d_n}\right) \right\} \nonumber \\&+\,\frac{d}{d-1} {\text {Var}}(\frac{I_1}{d_n})\sum _{h=1}^d(\bar{\mu }_h -\mu )^2. \end{aligned}$$
(12)
We now consider \(E({\text {Var}}(\bar{X}_n|R))\). Let
$$\begin{aligned} L_h= \sum _{u \in D_h}\sum _{i=1}^n \frac{I(R_{s_i}=u)}{d_nn_h}X_{s_i}. \end{aligned}$$
The conditional variance of \(\bar{X}_n\) given ranks R can be written as follows
$$\begin{aligned} {\text {Var}}(\bar{X}_n|R)= {\text {Var}}\left( \sum _{h=1}^d L_h\right) = \sum _{h=1}^d{\text {Var}}(L_h) +\sum _{h=1}^d \sum _{h'\ne h}^d {\text {Cov}}(L_h,L_{h'}). \end{aligned}$$
For the joint rankings, the variance of \(L_h\) is given by
$$\begin{aligned} {\text {Var}}(L_h)= \sum _{u \in D_h} \frac{I_h^2I_{hu}^2}{d_n^2n_h^2} \sigma ^2_{u:M}+ \sum _{(u \ne v) \in D_h} \frac{I_h^2 I_{hu}I_{hv}}{d_n^2n_h^2} \sigma _{u,v:M}^2 \end{aligned}$$
From the above equations, we write
$$\begin{aligned} E({\text {Var}}(L_h))= & {} E\left( \frac{I_h^2I_{ha}^2}{d_n^2n_h^2}\right) \sum _{u \in D_h} \sigma ^2_{u:M} +E\left( \frac{I_h^2I_{ha}I_{hb}}{d_n^2n_h^2}\right) \sum _{(u \ne v) \in D_h} \sigma _{u,v:M}^2; \quad (a,b) \in D_h\\= & {} {\text {Var}}\left( \frac{I_hI_{ha}}{d_nn_h}\right) \sum _{u \in D_h} \sigma ^2_{u:M} + {\text {Cov}}\left( \frac{I_hI_{ha}}{d_nn_h},\frac{I_hI_{hb}}{d_nn_h}\right) \sum _{(u \ne v) \in D_h} \sigma _{u,v:M}^2\\&+ \frac{1}{d^2H^2} \sum _{(u , v) \in D_h} \sigma _{u,v:M}^2; \quad (a,b) \in D_h \end{aligned}$$
Using part (iv) of Lemma 1, after some simplification, we obtain
$$\begin{aligned} E({\text {Var}}(L_h))= & {} \frac{\sum _{(u,v) \in D_h}\sigma _{u,v:M}^2}{H-1} \left\{ \frac{H-1}{d^2H^2}+\frac{{\text {Var}}(I_h/d_n)}{H}-{\text {Var}}(\frac{I_hI_{ha}}{d_nn_h}\right\} \\&+ \,\frac{\sum _{u \in D_h} \sigma _{u:M}^2}{H-1} \left\{ H {\text {Var}}(\frac{I_hI_{ha}}{d_nn_h})-{\text {Var}}(I_h/d_n)/H\right\} . \end{aligned}$$
For \(h \ne h'\), using the similar arguments, we obtain
$$\begin{aligned} E({\text {Cov}}(L_h,L_{h'}))= & {} \sum _{u \in D_h} \sum _{v \in D_{h'}} E\left( \frac{I_hI_{h'}I_{hu}I_{h'v}}{d_n^2n_hn_{h'}}\right) \sigma ^2_{u,v:M} \\= & {} E\left( \frac{I_hI_{h'}I_{ha}I_{h'b}}{d_n^2n_hn_{h'}}\right) \sum _{u \in D_h} \sum _{v \in D_{h'}} \sigma _{u,v:M}^2; \quad a \in D_h, b \in D_{h'} \\= & {} \frac{1}{H^2} \left[ {\text {Cov}}(I_h/d_n,I_{h'}/d_n)+1/d^2\right] \sum _{u \in D_h} \sum _{v \in D_{h'}} \sigma _{u,v:M}^2 \end{aligned}$$
The last equality follows from Eq. (8). Using the equality \({\text {Cov}}(I_1/d_n,I_2/d_n)=-{\text {Var}}(I_1/d_n)/H^2\), we obtain
$$\begin{aligned} E({\text {Cov}}(L_h,L_{h'}))= \frac{1}{H^2d^2 }\sum _{u \in D_h} \sum _{v \in D_{h'}} \sigma _{u,v:M}^2-\frac{{\text {Var}}(I_1/d_n)}{H^2(d-1)} \sum _{u \in D_h} \sum _{v \in D_{h'}} \sigma _{u,v:M}^2 \end{aligned}$$
Combining the variances and covariances, after some simplifications, we obtain
$$\begin{aligned} E({\text {Var}}(\bar{X}_n|R))= & {} \frac{M(N-M)\sigma ^2}{NH^2}\left\{ 1/d^2-{\text {Var}}(I_1/d_n)/(d-1)\right\} \\&+\sum _{h=1}^d \sum _{(u,v) \in D_h}\frac{\sigma _{u,v:M}^2}{H-1} \left\{ \frac{(Hd-1){\text {Var}}(I_1/d_n)}{H^2(d-1)}-{\text {Var}}(\frac{I_1I_{11}}{d_nn_1})\right\} \\&+\sum _{h=1}^d \sum _{u \in D_h}\frac{\sigma ^2_{u:M}}{H(H-1)}\left\{ H^2{\text {Var}}(\frac{I_1I_{11}}{d_nn_1})-{\text {Var}}(I_1/d_n)\right\} . \end{aligned}$$
Finally, the variance of \(\bar{X}_n\) is obtained
$$\begin{aligned} {\text {Var}}(\bar{X}_n)= & {} {\text {Var}}(E(\bar{X}_n|R))+E({\text {Var}}(\bar{X}_n|R)) \\= & {} \sum _{h=1}^d \sum _{u \in D_h}\frac{( \mu _{u:M} -\bar{\mu }_h)^2}{H-1} \left\{ H {\text {Var}}\left( \frac{I_1I_{11}}{d_nn_1}\right) -\frac{{\text {Var}}(\frac{I_1}{d_n})}{H} \right\} \\&+\,\frac{{\text {Var}}(\frac{I_1}{d_n})}{d-1} \sum _{h=1}^d (\bar{\mu }_h-\mu )^2 \\&+\,\frac{M(N-M)\sigma ^2}{NH^2}\left\{ 1/d^2-{\text {Var}}(I_1/d_n)/(d-1)\right\} \\&+\,\sum _{h=1}^d \sum _{(u,v) \in D_h}\frac{\sigma _{u,v:M}^2}{H-1} \left\{ \frac{(Hd-1){\text {Var}}(I_1/d_n)}{H^2(d-1)}-{\text {Var}}\left( \frac{I_1I_{11}}{d_nn_1}\right) \right\} \\&+\,\sum _{h=1}^d \sum _{u \in D_h}\frac{\sigma ^2_{u:M}}{H(H-1)}\left\{ H^2{\text {Var}}\left( \frac{I_1I_{11}}{d_nn_1}\right) -{\text {Var}}(I_1/d_n)\right\} . \end{aligned}$$
For the proof of \(\sigma ^2_{A_2}\), we first define a random variable \(Z_{R_{s_i},h}= I(R_{s_i} \in D_h) X_{s_i}\). The conditional distribution of \(Z_{R_{s_i},h}\) given that \(R_{s_i} \in D_h\) under design \(A_2\) is given by
$$\begin{aligned} P(Z_{R_{s_i}}=z|R_{s_i} \in D_h) =\frac{1}{H}\sum _{u \in D_h} P(X_{u:M}=z). \end{aligned}$$
Using this conditional distribution, we can easily establish the following equalities
$$\begin{aligned} E(Z_{R_{s_i}}|R_{s_i} \in D_h)= & {} \frac{1}{H} \sum _{u \in D_h} \mu _{u:M} =\bar{\mu }_h \\ V(Z_{R_{s_i}}|R_{s_i} \in D_h)= & {} \frac{1}{H} \sum _{u \in D_h} \sigma ^2_{u:M} +\frac{1}{H} \sum _{u \in D_h} (\mu _{u:M} -\bar{\mu }_h)^2. \end{aligned}$$
Again to simplify the notation we drop the subscript \(A_2\) in \(\bar{X}^{A_2}\). All random variables in the proof of \(\sigma ^2_{A_2}\) are defined based on design \(A_2\) even if it is not explicitly stated. From the total variance, we write
$$\begin{aligned} {\text {Var}}(\bar{X}^{A_2})={\text {Var}}(E(\bar{X}^{A_2}|R))+E({\text {Var}}(\bar{X}^{A_2}|R)). \end{aligned}$$
We first consider
$$\begin{aligned} {\text {Var}}(E(\bar{X}^{A_2}|R))= & {} {\text {var}}\left( \sum _{h=1}^d \frac{I_h}{d_nn_h} \sum _{i=1}^n E(Z_{R_{s_i},h}|R_{s_i} \in D_h)\right) \\= & {} {\text {var}}\left( \frac{I_1}{d_n}\right) \sum _{h=1} \bar{\mu }^2_h+ {\text {Cov}}\left( \frac{I_1}{d_n},\frac{I_2}{d_n}\right) \sum _{h=1} \sum _{h' \ne h} \bar{\mu }_h \bar{\mu }_{h'} \\= & {} {\text {var}}\left( \frac{I_1}{d_n}\right) \sum _{h=1} \bar{\mu }^2_h- \frac{1}{d-1}{\text {Var}}\left( \frac{I_1}{d_n}\right) (d^2\mu ^2-\sum _{h=1}^d \bar{\mu }_h^2 ) \\= & {} \frac{d}{d-1} {\text {Var}}\left( \frac{I_1}{d_n}\right) \sum _{h=1}^d (\bar{\mu }^2_h-\mu )^2 . \end{aligned}$$
In a similar fashion, we can show that
$$\begin{aligned} E({\text {Var}}(\bar{X}^{A_2}|R))= & {} E\left( \sum _{h=1}^d \frac{I_h^2}{d_n^2n_h^2} \sum _{i=1}^n {\text {Var}}(Z_{R_{s_i},h}|R_{s_i} \in D_h)\right) \\= & {} E\left( \frac{I_1^2}{d_n^2n_1}\right) \sum _{h=1}^d {\text {Var}}(Z_{R_{s_1},h}|R_{s_1} \in D_h) \\= & {} E\left( \frac{I_1^2}{d_n^2n_1}\right) \left\{ \frac{1}{H} \sum _{h=1}^d\sigma ^2_h+ \frac{1}{H} \sum _{h=1}^d \sum _{u \in D_h} (\mu _{u:M}-\bar{\mu }_h)^2\right\} . \end{aligned}$$
We complete the proof by combining expressions \({\text {Var}}(E(\bar{X}^{A_2}|R))\) and \(E({\text {Var}}(\bar{X}^{A_2}|R))\).
\(\square \)
Proof of Theorem 4:
We first consider the expected value of \(T_1\)
$$\begin{aligned} E(W_1)= & {} d\sum _{h=1}^d\sum _{i=1}^n \sum _{j\ne i}^nE\left( \frac{I_h^*}{ d_n^*n_h(n_h-1)}\right) E\left\{ (Z_{R_{s_i},h}-Z_{R_{s_j},h})^2|(R_{s_i},R_{s_j})\in D_h\right\} \\= & {} d \sum _{h=1}^dE(\frac{I^*_h}{d_n^*})E\left\{ (Z_{R_{s_1},h}-Z_{R_{s_2},h})^2 |(R_{s_i},R_{s_j})\in D_h\right\} \\= & {} 2dE(\frac{I^*_1}{d_n^*}) \sum _{h=1}^d {\text {var}}\left( Z_{R_{s_1},h}|R_{s_1}\in D_h\right) = 2 \sum _{h=1}^d {\text {var}}\left( Z_{R_{s_1},h}|R_{s_1}\in D_h\right) \end{aligned}$$
In a similar fashion, expected value of \(T_2\) is given by
$$\begin{aligned} E(W_2)= & {} \sum _{h=1}^d \sum _{h'\ne h}^d \sum _{i=1} \sum _{j=1}^n E\left( \frac{I_hI_{h'}}{d_n^2n_hn_{h'}}\right) E\left\{ (Z_{R_{s_i},h}-Z_{R_{s_j},h'})^2|(R_{s_i},R_{s_j})\in D_h\right\} \\= & {} E\left( \frac{I_1I_2}{d_n^2}\right) \sum _{h=1}^d \sum _{h' \ne h}^d E\left\{ (Z_{R_{s_1},h}-Z_{R_{s_2},h'})^2|(R_{s_1},R_{s_2})\in D_h\right\} \\= & {} E\left( \frac{I_1I_2}{d_n^2}\right) \left\{ 2(d-1) \sum _{h=1}^d {\text {Var}}\left( Z_{R_{s_1},h}|R_{s_1}\in D_h\right) + 2(d-1) \sum _{h=1}^d (\bar{\mu }_h-\mu )^2\right\} \\&- 2 E\left( \frac{I_1I_2}{d_n^2}\right) \sum _{h=1}\sum _{h'\ne h}(\bar{\mu }_h-\mu )(\bar{\mu }_h'-\mu ) \\= & {} E\left( \frac{I_1I_2}{d_n^2}\right) \left\{ 2(d-1) \sum _{h=1}^d {\text {Var}}(Z_{R_{s_1},h}|R_{s_1}\in D_h)+ 2d \sum _{h=1}^d (\bar{\mu }_h-\mu )^2\right\} \end{aligned}$$
We now combine \(E(W_1)\) and \(E(W_2)\) in \(\hat{\sigma }^2_{A_2}\) to write
$$\begin{aligned} E(\hat{\sigma }^2_{A_2})= & {} E(W_1/2)\left\{ E(\frac{I_1^2}{d_n^2n_1})-{\text {Var}}(\frac{I_1}{d_n})\right\} +\frac{E(W_2/2){\text {Var}}(I_1/d_n)}{E(\frac{I_1I_2}{d_n^2})(d-1)} \\= & {} \left\{ \sum _{h=1}^d {\text {var}}\left( Z_{R_{s_1},h}|R_{s_1}\in D_h\right) \right\} \left\{ E(\frac{I_1^2}{d_n^2n_1})-{\text {Var}}(\frac{I_1}{d_n})\right\} \\&+ \left\{ \sum _{h=1}^d {\text {var}}(Z_{R_{s_1},h}|R_{s_1}\in D_h)+\frac{d}{d-1} \sum _{h=1}^d(\bar{\mu }_h-\mu )^2\right\} {\text {Var}}(I_1/d_n) \\= & {} E\left( \frac{I_1^2}{d_n^2n_1}\right) \sum _{h=1}^d {\text {var}}\left( Z_{R_{s_1},h}|R_{s_1}\in D_h\right) +\frac{d}{d-1} {\text {Var}}(I_1/d_n)\bar{\tau }^2 \\= & {} \frac{d}{d-1} {\text {Var}}(I_1/d_n)\bar{\tau }^2 +\frac{1}{H}E\left( \frac{I_1^2}{d_n^2n_1}\right) \left\{ \sum _{h=1}^d \sigma _h^2 +\sum _{h=1}^d \tau ^2_h \right\} . \end{aligned}$$
This completes the proof. \(\square \)
