Array Gain and Diversity Order of Multiuser MISO Antenna Systems

Abstract

In considering generalized zero-forcing, this paper derives the optimal linear signal processing solution for the multiuser multiple-input single-output (MISO) broadcast fading channels where a single base station is communicating simultaneously to many mobile stations in the same frequency and time slot. We then investigate the array gain, defined as the average of the squared channel response, and the diversity order, defined as the inverse of the normalized variance of the squared channel response, of the optimized system. It is concluded that each user in the multiuser MISO system behaves like a single-user system with (n _T  − M + 1)-th order receiver diversity if there are M simultaneous users and n _T antennas are employed at the base station.

Appendices

Appendix I: Proof of Array Gain

In the optimized MISO antenna system, the transmit antenna weights can be found using (12). It follows that

$$ \left.{\mathbf{g}}_m\right|_{\rm opt}={\mathbf{G}}_{\rm opt}{\mathbf{e}}_m={\mathbf{H}}^{\dag}\left({\mathbf{H}}{\mathbf{H}}^{\dag}\right)^{-1}{\mathbf{e}}_m. $$

(25)

The square of the resultant channel response can be expressed as

$$ \begin{aligned} \beta_m^2&=\frac{ 1}{\|\left.{\mathbf{g}}_m\right|_{\rm opt}\|^2}\\ &=\frac{ 1}{ {\mathbf{e}}_m^{\dag}\left({\mathbf{H}}{\mathbf{H}}^{\dag}\right)^{-1}{\mathbf{e}}_m}\\ &=\frac{ {\tt det}\left[{\mathbf{H}}{\mathbf{H}}^{\dag}\right]}{ {\tt det}\left[{\mathbf{H}}^{(m)-}{\mathbf{H}}^{(m)-\dag}\right]} \end{aligned} $$

(26)

where H ^(m)− agrees with the channel matrix H except that the m-th row corresponding to the channel from the base station to the m-th mobile station is missing, and ${\tt det}[\cdot]$ denotes the determinant of a matrix. Without loss of generality, we shall consider β ²₁ in the following for convenience.

Expanding ${\tt det}[{\mathbf{H}}{\mathbf{H}}^{\dag}]$ , we have

$$ {\tt det}[{\mathbf{H}}{\mathbf{H}}^{\dag}]=\sum_{m=1}^M(-1)^{m-1}{\mathbf{h}}_1{\mathbf{h}}_m^{\dag}\times{\tt det}\left[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}\right] $$

(27)

where ${\mathbf{H}}=[{\mathbf{h}}_1,{\mathbf{h}}_2, \cdots ,\,{\mathbf{h}}_M]$ . As a consequence,

$$ \beta_1^2={\mathbf{h}}_1{\mathbf{h}}_1^{\dag}-\sum_{m=2}^M(-1)^m{\mathbf{h}}_1{\mathbf{h}}_m^{\dag}\times\frac{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}]}{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}]}. $$

(28)

In addition,

$$ \frac{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}]}{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}]}=\frac{ \sum_{i=2}^M(-1)^{i-1}{\mathbf{h}}_i{\mathbf{h}}_1^{\dag}{\tt det}[{\mathbf{M}}_i]}{\sum_{j=2}^M(-1)^{m+j-1}{\mathbf{h}}_j{\mathbf{h}}_m^{\dag}{\tt det}[{\mathbf{M}}_j]} $$

(29)

where M _i is the sub-block matrix of ${\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}$ after removing the i-th row and the m-th column. Then, each term under the summation of (28) can be expressed as

$$ (-1)^m{\mathbf{h}}_1{\mathbf{h}}_m^{\dag}\times\frac{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}]}{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}]}=\frac{\sum_{i=2}^M(-1)^{i-1}{\tt det}[{\mathbf{M}}_i]{\mathbf{h}}_i({\mathbf{h}}_1^{\dag}{\mathbf{h}}_1){\mathbf{h}}_m^{\dag}}{\sum_{j=2}^M(-1)^{j-1}{\tt det}[{\mathbf{M}}_j]{\mathbf{h}}_j{\mathbf{h}}_m^{\dag}}. $$

(30)

Using the fact that the channel elements are all independent and identically distributed and assuming that $\hbox{E}[|h^{(m)}_{\ell,k}|^2]=1\,\forall m,k,\ell,$ the correlation matrix of the channel is $\hbox{E}[{\mathbf{h}}_1^{\dag}{\mathbf{h}}_1]={\mathbf{I}}.$ Now, taking the average on both sides of (30), we obtain

$$ {\rm E}\left\{(-1)^m{\mathbf{h}}_1{\mathbf{h}}_m^{\dag}\times\frac{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}]}{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}]}\right\}={\rm E}\left\{\frac{\sum_{i=2}^M(-1)^{i-1}{\tt det}[{\mathbf{M}}_i]{\mathbf{h}}_i({\mathbf{h}}_1^{\dag}{\mathbf{h}}_1){\mathbf{h}}_m^{\dag}}{\sum_{j=2}^M(-1)^{j-1}{\tt det}[{\mathbf{M}}_j]{\mathbf{h}}_j{\mathbf{h}}_m^{\dag}}\right\} $$

(31)

$$ =\hbox{E}\left\{\frac{\sum_{i=2}^M(-1)^{i-1}{\tt det}[{\mathbf{M}}_i]{\mathbf{h}}_i{\rm E}({\mathbf{h}}_1^{\dag}{\mathbf{h}}_1){\mathbf{h}}_m^{\dag}}{\sum_{j=2}^M(-1)^{j-1}{\tt det}[{\mathbf{M}}_j]{\mathbf{h}}_j{\mathbf{h}}_m^{\dag}}\right\} $$

(32)

$$ =1. $$

(33)

From (31) to (32), we have used the statistical independence between h ₁ and ${\mathbf{h}}_m\,(m\ne 1)$ .

As a result,

$$ \begin{aligned} \Upomega_1=\hbox{E}[\beta_1^2]&={\rm E}[{\mathbf{h}}_1{\mathbf{h}}_1^{\dag}]-\sum_{m=2}^M{\rm E}\left\{(-1)^m{\mathbf{h}}_1{\mathbf{h}}_m^{\dag}\times\frac{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}]}{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}]}\right\}\\ &=n_T-M+1. \end{aligned} $$

(34)

For $\Upomega_m\,(m\ne 1),$ the proof is similar.

Appendix II: Proof of Diversity Order

From (15),

$$ \Uppsi_1=\frac{\Upomega_1^2}{{\rm E}[(\beta_1^2-\Upomega_1)^2]}=\frac{\Upomega_1^2}{{\rm E}[\beta_1^4]-\Upomega_1^2}, $$

(35)

and β ⁴₁ can be expressed as

$$ \begin{aligned} \beta_1^4&=({\mathbf{h}}_1{\mathbf{h}}_1^{\dag})^2+2({\mathbf{h}}_1{\mathbf{h}}_1^{\dag})\sum_{m=2}^M(-1)^{m+1}{\mathbf{h}}_1{\mathbf{h}}_m^{\dag}\frac{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}]}{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}]}\\ &+\sum_{m=2}^M\left(\frac{{\mathbf{h}}_1{\mathbf{h}}_m^{\dag}{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}]}{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}]}\right)^2\\ &+\sum_{m,n=2\atop m\ne n}^M\frac{(-1)^{m+n}{\mathbf{h}}_1{\mathbf{h}}_m^{\dag}{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}]{\mathbf{h}}_1{\mathbf{h}}_n^{\dag}{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(n)-\dag}]}{\left({\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}]\right)^2}. \end{aligned} $$

(36)

The expectation of β ⁴₁ can be found by considering the expectation of each term on the right hand side (RHS). It can be shown (see Appendix III) that

$$ \hbox{E}\left\{({\mathbf{h}}_1{\mathbf{h}}_1^{\dag})^2\right\}=n_T^2+n_T; $$

(37)

$$ \hbox{E}\left\{\frac{({\mathbf{h}}_1{\mathbf{h}}_1^{\dag})({\mathbf{h}}_1{\mathbf{h}}_m^{\dag}){\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}]}{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}]}\right\}=(-1)^m(n_T+1); $$

(38)

$$ \hbox{E}\left\{\left(\frac{{\mathbf{h}}_1{\mathbf{h}}_m^{\dag}{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}]}{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}]}\right)^2\right\}=2; $$

(39)

$$ \hbox{E}\left\{\frac{({\mathbf{h}}_1{\mathbf{h}}_m^{\dag}){\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}]({\mathbf{h}}_1{\mathbf{h}}_n^{\dag}){\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(n)-}]}{\left({\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}]\right)^2}\right\}=(-1)^{m+n}. $$

(40)

It follows that

$$ \begin{aligned} {\rm E}[\beta_1^4]&=(n_T^2+n_T)-2(M-1)(n_T+1)+2(M-1)+(M-1)(M-2)\\ &=n_T^2+M^2+3n_T-3M-2Mn_T+2. \end{aligned} $$

(41)

Substituting (41) into (35) and using Ω₁ in Theorem 3.2, we obtain

$$ \Uppsi_1=n_T-M+1 $$

(42)

and this completes the proof of Theorem 3.3.

Appendix III: Proofs of the Results (37)–(40)

Given i.i.d. zero-mean unity-variance complex Gaussian random variables, $h^{(m)}_{\ell,k}$ , for $m=1,\ldots,\,M, \ell=1$ (i.e., single-element mobile receivers), and k = 1,..., n _T , it is easy to check that $\hbox{E}[|h^{(m)}_{\ell,k}|^2]=1$ and $\hbox{E}[|h^{(m)}_{\ell,k}|^4]=2.$ Based on these results, we prove (37)–(40).

3.1 Proof of (37)

Expanding $\hbox{E}\{({\mathbf{h}}_1{\mathbf{h}}_1^{\dag})^2\}$ , we have

$$ \begin{aligned} \hbox{E}\left\{({\mathbf{h}}_1{\mathbf{h}}_1^{\dag})^2\right\}&={\rm E}\left\{\sum_{k=1}^{n_T}|h^{(1)}_{1,k}|^4+\sum_{k,\ell=1\atop k\ne\ell}^{n_T}|h^{(1)}_{1,k}|^2|h^{(1)}_{1,\ell}|^2\right\}\\ &=2n_T+n_T(n_T-1)=n_T^2+n_T. \end{aligned} $$

(43)

$\square$

3.2 Proof of (38)

Using (29), we can write

$$ \begin{aligned} &\quad{\rm E}\left\{\frac{({\mathbf{h}}_1{\mathbf{h}}_1^{\dag})({\mathbf{h}}_1{\mathbf{h}}_m^{\dag}){\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}]}{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}]}\right\}\\ &={\rm E}\left\{(-1)^m\frac{\left[\sum_{i=2}^M(-1)^{i-1}{\tt det}({\mathbf{M}}_i){\mathbf{h}}_i\right]{\mathbf{h}}_1^{\dag}({\mathbf{h}}_1{\mathbf{h}}_1^{\dag})({\mathbf{h}}_1{\mathbf{h}}_m^{\dag})}{\left[\sum_{j=2}^M(-1)^{j-1}{\tt det}({\mathbf{M}}_j){\mathbf{h}}_j\right]{\mathbf{h}}_m^{\dag}}\right\}\\ &={\rm E}\left\{(-1)^m\frac{\left[\sum_{i=2}^M(-1)^{i-1}{\tt det}({\mathbf{M}}_i){\mathbf{h}}_i\right]{\rm E}\left\{{\mathbf{h}}_1^{\dag}({\mathbf{h}}_1{\mathbf{h}}_1^{\dag}){\mathbf{h}}_1\right\}{\mathbf{h}}_m^{\dag}}{\left[\sum_{j=2}^M(-1)^{j-1}{\tt det}({\mathbf{M}}_j){\mathbf{h}}_j\right]{\mathbf{h}}_m^{\dag}}\right\}. \end{aligned} $$

(44)

Now, consider

$$ \hbox{E}\left\{{\mathbf{h}}_1^{\dag}({\mathbf{h}}_1{\mathbf{h}}_1^{\dag}){\mathbf{h}}_1\right\} =\hbox{E}\left\{\left(\sum_{k=1}^{n_T}|h^{(1)}_{1,k}|^2\right)\left[ \begin{array}{llll} |h^{(1)}_{1,1}|^2 & h^{(1)}_{1,2}h^{(1)*}_{1,1}&\cdots & h^{(1)}_{1,n_T}h^{(1)*}_{1,1}\\ h^{(1)}_{1,1}h^{(1)*}_{1,2}&|h^{(1)}_{2,2}|^2 & & \vdots\\ \vdots & & \ddots & \\ h^{(1)}_{1,1}h^{(1)*}_{1,n_T}&\cdots & & |h^{(1)}_{1,n_T}|^2 \end{array}\right]\right\}. $$

(45)

Apparently, the expectation of each entry is given by

$$ \begin{aligned} \hbox{E}\left\{h^{(1)*}_{1,i}h^{(1)}_{1,j}\sum_{k=1}^{n_T}| h^{(1)}_{1,k}|^2\right\}&=\left\{\begin{array}{ll} \hbox{E}[|h^{(1)}_{1,i}|^4]+{\rm E}[|h^{(1)}_{1,i}|^2] \sum_{j=1\atop j\ne i}^{n_T}{\rm E}[|h^{(1)}_{1,j}|^2] & i=j\\ 0 & i\ne j \end{array}\right.\\ &=\left\{\begin{array}{ll} n_T+1 & i=j\\ 0 & i\ne j. \end{array}\right. \end{aligned} $$

(46)

Consequently, we have

$$ \hbox{E}\left\{{\mathbf{h}}_1^{\dag}({\mathbf{h}}_1{\mathbf{h}}_1^{\dag}) {\mathbf{h}}_1\right\}=(n_T+1){\mathbf{I}} $$

(47)

so that, (44) yields

$$ \hbox{E}\left\{\frac{({\mathbf{h}}_1{\mathbf{h}}_1^{\dag})({\mathbf{h}}_1{\mathbf{h}}_m^{\dag}){\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}]}{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}]}\right\}=(-1)^m(n_T+1) $$

(48)

and this completes the proof. $\square$

3.3 Proof of (39)

Now, define an n _T -dimensional row vector,

$$ {\mathbf{p}}=[p_1\,p_2\,\cdots]\triangleq\sum_{i=2}^M(-1)^{i-1}{\tt det}({\mathbf{M}}_i){\mathbf{h}}_i $$

(49)

which is independent of h ₁. Using (29),

$$ \hbox{E}\left\{\left(\frac{{\mathbf{h}}_1{\mathbf{h}}_m^{\dag}{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}]}{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}]}\right)^2\right\}={\rm E}\left\{\left(\frac{{\mathbf{ph}}_1^{\dag}{\mathbf{h}}_1{\mathbf{h}}_m^{\dag}}{{\mathbf{ph}}_m^{\dag}}\right)^2\right\}. $$

(50)

Expanding the RHS,

$$ \hbox{E}\left\{\left(\frac{{\mathbf{ph}}_1^{\dag}{\mathbf{h}}_1{\mathbf{h}}_m^{\dag}}{{\mathbf{ph}}_m^{\dag}}\right)^2\right\}= \hbox{E}\left\{\frac{\sum_{i,j,k,\ell=1}^{n_T}p_ih^{(m)*}_{1,j}p_kh^{(m)*}_{1,\ell} \hbox{E}\left\{h^{(1)*}_{1,i}h^{(1)}_{1,j}h^{(1)*}_{1,k}h^{(1)}_{1,\ell}\right\}}{ \sum_{i=1}^{n_T}\left[p_ih^{(m)*}_{1,i}\right]^2+\sum_{i,j=1\atop i\ne j}^{n_T}p_ih^{(m)*}_{1,i}p_jh^{(m)*}_{1,j}}\right\}. $$

(51)

The numerator of (51) can be simplified by partitioning $i,j,k,\ell$ in the following:

$$ \Im_1=\{i=j=k=\ell\}, $$

(52)

$$ \Im_2=\{i=j\ne k=\ell\}, $$

(53)

$$ \Im_3=\{i=\ell\ne j=k\}, $$

(54)

and

$$ \begin{aligned} \Im_4&=\{i=k=\ell\ne j\}\cup\{j=k=\ell\ne i\}\cup\{i=j=k\ne\ell\}\\ &\cup\{i\ne j,k,\ell\}\cup\{j\ne i,k,\ell\}\cup\{k\ne i,j,\ell\}\cup\{\ell\ne i,j,k\}. \end{aligned} $$

(55)

In other words,

$$ \sum_{i,j,k,\ell=1}^{n_T}=\sum_{i,j,k,\ell\in\Im_1}+\sum_{i,j,k,\ell\in\Im_2}+ \sum_{i,j,k,\ell\in\Im_3}+\sum_{i,j,k,\ell\in\Im_4}. $$

(56)

To proceed, we obtain the expected value in the numerator of (51) for index values in the above four sets. It can be easily shown that

$$ \hbox{E}\left\{h^{(1)*}_{1,i}h^{(1)}_{1,j}h^{(1)*}_{1,k}h^{(1)}_{1,\ell}\right\}= \left\{\begin{array}{ll} \hbox{E}\left\{|h^{(1)}_{1,i}|^4\right\}=2 & i,j,k,\ell\in\Im_1,\\ \hbox{E}\left\{|h^{(1)}_{1,i}|^2\right\}{\rm E}\left\{|h^{(1)}_{1,k}|^2\right\}=1 & i,j,k,\ell\in\Im_2,\\ \hbox{E}\left\{|h^{(1)}_{1,i}|^2\right\}{\rm E}\left\{|h^{(1)}_{1,j}|^2\right\}=1 & i,j,k,\ell\in\Im_3,\\ 0 & i,j,k,\ell\in\Im_4. \end{array}\right. $$

(57)

As a result,

$$ \hbox{E}\left\{\left(\frac{{\mathbf{h}}_1{\mathbf{h}}_m^{\dag}{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}]}{{\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}]}\right)^2\right\}= \hbox{E}\left\{\frac{\sum_{\Im_1}2\left[p_ih^{(m)*}_{1,i}\right]^2+\sum_{\Im_2,\Im_3} p_ih^{(m)*}_{1,i}p_jh^{(m)*}_{1,j}} {\sum_{i=1}^{n_T}\left[p_ih^{(m)*}_{1,i}\right]^2+\sum_{i,j=1\atop i\ne j}^{n_T}p_ih^{(m)*}_{1,i}p_jh^{(m)*}_{1,j}}\right\}=2. $$

(58)

3.4 Proof of (40)

Similar to (49), we define another n _T -dimensional row vector as

$$ {\mathbf{q}}=[q_1\,q_2\,\cdots]\triangleq\sum_{i=2}^M(-1)^{i-1}{\tt det}({\mathbf{N}}_i){\mathbf{h}}_i $$

(59)

where N _i is the sub-block matrix of ${\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}$ after removing the i-th row and n-th column. As can be noted, q is also independent of h ₁. Then, we can write

$$ \hbox{E}\left\{\frac{({\mathbf{h}}_1{\mathbf{h}}_m^{\dag}){\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}]({\mathbf{h}}_1{\mathbf{h}}_n^{\dag}){\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(n)-}]}{\left({\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}]\right)^2}\right\}=(-1)^{m+n}{\rm E}\left\{\frac{({\mathbf{ph}}_1^{\dag}{\mathbf{h}}_1{\mathbf{h}}_m^{\dag})({\mathbf{qh}}_1^{\dag}{\mathbf{h}}_1{\mathbf{h}}_n^{\dag})}{({\mathbf{ph}}_m^{\dag})({\mathbf{qh}}_n^{\dag})}\right\}. $$

(60)

The RHS is further expanded as

$$ (-1)^{m+n}\hbox{E}\left\{\frac{({\mathbf{ph}}_1^{\dag}{\mathbf{h}}_1{\mathbf{h}}_m^{\dag})({\mathbf{qh}}_1^{\dag}{\mathbf{h}}_1{\mathbf{h}}_n^{\dag})}{({\mathbf{ph}}_m^{\dag})({\mathbf{qh}}_n^{\dag})}\right\}=(-1)^{m+n} \hbox{E}\left\{\frac{\sum_{i,j,k,\ell=1}^{n_T}p_ih^{(m)*}_{1,j}q_kh^{(n)*}_{1,\ell} \hbox{E}\left\{h^{(1)*}_{1,i}h^{(1)}_{1,j} h^{(1)*}_{1,k}h^{(1)}_{1,\ell}\right\}}{\sum_{i=1}^{n_T}p_ih^{(m)*}_{1,i}q_ih^{(n)*}_{1,i}+ \sum_{i,j=1\atop i\ne j}^{n_T}p_ih^{(m)*}_{1,i}q_jh^{(n)*}_{1,j}}\right\}. $$

(61)

Using (57), we can rewrite the numerator of RHS in (61) as

$$ 2\sum\limits_{i=1}^{n_T}p_ih^{(m)*}_{1,i}q_ih^{(n)*}_{1,i}+\sum_{i,j=1\atop i\ne j}^{n_T}p_ih^{(m)*}_{1,i} q_jh^{(n)*}_{1,j}+\sum_{i,j=1\atop i\ne j}^{n_T}p_ih^{(m)*}_{1,j}q_jh^{(n)*}_{1,i}. $$

(62)

Therefore,

$$ \begin{aligned} &\quad(-1)^{m+n}{\rm E}\left\{\frac{({\mathbf{ph}}_1^{\dag}{\mathbf{h}}_1{\mathbf{h}}_m^{\dag})({\mathbf{qh}}_1^{\dag}{\mathbf{h}}_1{\mathbf{h}}_n^{\dag})}{({\mathbf{ph}}_m^{\dag})({\mathbf{qh}}_n^{\dag})}\right\}\\ &=(-1)^{m+n}\left[1+{\rm E}\left\{\frac{\sum_{i=1}^{n_T}p_ih^{(m)*}_{1,i}q_ih^{(n)*}_{1,i}+ \sum_{i,j=1\atop i\ne j}^{n_T}p_ih^{(m)*}_{1,j}q_jh^{(n)*}_{1,i}}{\sum_{i=1}^{n_T}p_ih^{(m)*}_{1,i}q_ih^{(n)*}_{1,i}+ \sum_{i,j=1\atop i\ne j}^{n_T}p_ih^{(m)*}_{1,i}q_jh^{(n)*}_{1,j}}\right\}\right]\\ &=(-1)^{m+n}\left[1+{\rm E}\left\{\frac{({\mathbf{ph}}_n^{\dag})({\mathbf{qh}}_m^{\dag})}{({\mathbf{ph}}_m^{\dag})({\mathbf{qh}}_n^{\dag})}\right\}\right]. \end{aligned} $$

(63)

Note that

$$ \begin{aligned} {\mathbf{ph}}_n^{\dag}&=\sum_{i=2}^M(-1)^{i-1}{\tt det}({\mathbf{M}}_i){\mathbf{h}}_i{\mathbf{h}}_n^{\dag}\\ &={\tt det}\left[\begin{array}{lllllll} {\mathbf{h}}_2{\mathbf{h}}_n^{\dag} & {\mathbf{h}}_2{\mathbf{h}}_2^{\dag} & {\mathbf{h}}_2{\mathbf{h}}_3^{\dag} & \cdots & {\mathbf{h}}_2{\mathbf{h}}_n^{\dag} & \cdots & {\mathbf{h}}_2{\mathbf{h}}_{n_T}^{\dag}\\ {\mathbf{h}}_3{\mathbf{h}}_n^{\dag} & {\mathbf{h}}_3{\mathbf{h}}_2^{\dag} & {\mathbf{h}}_3{\mathbf{h}}_3^{\dag} & \cdots & {\mathbf{h}}_3{\mathbf{h}}_n^{\dag} & \cdots & {\mathbf{h}}_3{\mathbf{h}}_{n_T}^{\dag}\\ \vdots & & \ddots & & \vdots & \cdots & \vdots\\ {\mathbf{h}}_{n_T}{\mathbf{h}}_n^{\dag} & {\mathbf{h}}_{n_T}{\mathbf{h}}_2^{\dag} & & \cdots & {\mathbf{h}}_{n_T}{\mathbf{h}}_n^{\dag} & & {\mathbf{h}}_{n_T}{\mathbf{h}}_{n_T}^{\dag} \end{array}\right] \end{aligned} $$

(64)

and because the first column is the same as the (n + 1)-th column, this is a singular matrix and the determinant is thus zero. Same is also true for ${\mathbf{qh}}_m^{\dag}$ . As a result, we conclude that

$$ \hbox{E}\left\{\frac{({\mathbf{h}}_1{\mathbf{h}}_m^{\dag}){\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(m)-\dag}]({\mathbf{h}}_1{\mathbf{h}}_n^{\dag}){\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(n)-}]}{\left({\tt det}[{\mathbf{H}}^{(1)-}{\mathbf{H}}^{(1)-\dag}]\right)^2}\right\}=(-1)^{m+n}. $$

(65)

□