Proof of Proposition 1
Let \({\nu _{{n_{r}}}} = {\mathbf {H}_{{n_{r}},q}}{x_{n}} - {\mathbf {H}_{{n_{r}},p}}{x_{m}}\) for nr=1,…,Nr. Recall that Nr=1. We note that ν1 is a zero mean complex Gaussian variable with variance:
$$ {\begin{aligned} \sigma_{{\nu_{1}}}^{2} &= \frac{1}{K}\sum\limits_{k = 1}^{K} \left| \sqrt {{G_{q}}\left({\theta_{k}^{t}},{\phi_{k}^{t}}\right)} \exp \left(j{\Omega_{q}}\left({\theta_{k}^{t}},{\phi_{k}^{t}}\right)\right){x_{n}}\right.\\& \left.- \sqrt {{G_{p}}\left({\theta_{k}^{t}},{\phi_{k}^{t}}\right)} \exp \left(j{\Omega_{p}}\left({\theta_{k}^{t}},{\phi_{k}^{t}}\right)\right){x_{m}} \right|^{2} \end{aligned}} $$
(28)
Thus, \({\gamma _{p,q,{x_{m}},{x_{n}}}} = \mid {\nu _{1}}{\mid ^{2}}\) is an exponential random variable whose probability density function is:
$$\begin{array}{*{20}l} f_{x}(x;\lambda) = \lambda\exp(-\lambda x), ~x\geqslant 0 \end{array} $$
with \(\lambda = 1/\sigma _{{\nu _{1}}}^{2}\).
The APEP can then be formulated as follows:
$$\begin{array}{*{20}l} \text{APEP} = \frac{1}{\pi }\int_{0}^{\pi /2} {{\mathbb{E}_{\sigma_{{\nu_{1}}}^{2}}}} \left\{ {{{\left({1 + \frac{{\rho \sigma_{{\nu_{1}}}^{2}}}{{4{{\sin }^{2}}(\vartheta)}}} \right)}^{- 1}}} \right\}{\mkern 1mu} d\vartheta \end{array} $$
(29)
We note that \(\sigma _{{\nu _{1}}}^{2}\) is a random variable that depends on \(\left \{ {{\theta _{k}^{t}},{\phi _{k}^{t}}} \right \}_{k = 1}^{K}\). The expectation can be computed by using the approach introduced in [24], as follows:
$$\begin{array}{*{20}l} \text{APEP} = \frac{1}{\pi }\int_{0}^{\pi /2} {\left({\int_{0}^{\infty} {{M_{S}}} (z){\mkern 1mu} dz} \right)} {\mkern 1mu} d\vartheta \end{array} $$
(30)
where \(S = 1 + \frac {{\rho \sigma _{{\nu _{1}}}^{2}}}{{4{{\sin }^{2}}(\vartheta)}}\), and:
$$\begin{array}{*{20}l} {M_{S}}(z) = \exp (- z){M_{\sigma_{{\nu_{1}}}^{2}}}\left({z\frac{\rho }{{4{{\sin }^{2}}(\vartheta)}}} \right) \end{array} $$
(31)
With the aid of some algebraic manipulations, we obtain the following:
$$ {{}\begin{aligned} &{M_{\sigma_{{\nu_{1}}}^{2}}}\left({z\frac{\rho }{{4{{\sin }^{2}}(\vartheta)}}} \right)\\ &\,=\, {\mathbb{E}_{{\theta_{k}^{t}},{\phi_{k}^{t}}}}\left\{ {\exp \left({ \!- z\frac{\rho }{{4K{{\sin }^{2}}(\vartheta)}}\sum\limits_{k = 1}^{K} {\psi \left({p,q,{x_{m}},{x_{n}},{\theta_{k}^{t}},{\phi_{k}^{t}}} \right)}} \right)} \right\} \end{aligned}} $$
$$ {{}\begin{aligned} &={\mathbb{E}_{{\theta_{k}^{t}},{\phi_{k}^{t}}}}\left\{ {\prod\limits_{k = 1}^{K} {\exp \left({ \,-\, z\frac{\rho }{{4K{{\sin }^{2}}(\vartheta)}}\psi \left({p,q,{x_{m}},{x_{n}},{\theta_{k}^{t}},{\phi_{k}^{t}}} \right)} \right)}} \right\}\\ &\,=\,\prod\limits_{k = 1}^{K} \left(\int_{- \pi }^{\pi} \int_{0}^{\pi} \exp \left(\!- z\frac{\rho }{{4K{{\sin }^{2}}(\vartheta)}}\psi \left(p,q,{x_{m}},{x_{n}},{\theta_{k}^{t}},{\phi_{k}^{t}} \right) \right)\right.\\ & \left. \quad {f_{\theta} }({\theta_{k}^{t}}){f_{\phi} }({\phi_{k}^{t}})d{\theta_{k}^{t}}d{\phi_{k}^{t}} {\vphantom{\int_{- \pi }^{\pi}}}\right) \end{aligned}} $$
(32)
where:
$${{} \begin{aligned} &\psi \left({p,q,{x_{m}},{x_{n}},{\theta_{k}^{t}},{\phi_{k}^{t}}} \right) = \left| \sqrt {{G_{q}}({\theta_{k}^{t}},{\phi_{k}^{t}})} \exp \left(j{\Omega_{q}}\left({\theta_{k}^{t}},{\phi_{k}^{t}}\right)\right){x_{n}} \right. \\ & \left.- \sqrt {{G_{p}}({\theta_{k}^{t}},{\phi_{k}^{t}})} \exp (j{\Omega_{p}}({\theta_{k}^{t}},{\phi_{k}^{t}})){x_{m}} \right|^{2} \end{aligned}} $$
The proof follows from the following identity:
$$ {\begin{aligned} &\int_{- \pi }^{\pi} \int_{0}^{\pi} {f_{\theta} }\left({\theta^{t}}\right){f_{\phi}}({\phi^{t}})d{\theta_{k}^{t}}d{\phi_{k}^{t}} \\& \quad= \int_{- \pi }^{\pi} {f_{\theta} }\left({\theta^{t}}\right)d{\theta_{k}^{t}}\int_{0}^{\pi} {f_{\phi} }\left({\phi^{t}}\right)d{\phi_{k}^{t}} = 1 \end{aligned}} $$
(33)
Proof of Proposition 2
From (29), we obtain, in the high-SNR regime, the following:
$${\begin{aligned} \text{APEP}_{p,q} &= \mathbb{E}_{\sigma^{2}_{\nu_{1}}}\left\{\frac{1}{\pi} \int_{0}^{\pi/2}\left(1+\frac{\rho\sigma^{2}_{\nu_{1}}}{4\sin^{2}(\vartheta)}\right)^{-1}\,d\vartheta \right\} \end{aligned}} $$
$${\begin{aligned} &\overset{\rho \gg 1}{\leq} \mathbb{E}_{\sigma^{2}_{\nu_{1}}}\left\{ \frac{1}{\pi} \int_{0}^{\pi/2}\left(\frac{\rho\sigma^{2}_{\nu_{1}}}{4\sin^{2}(\vartheta)}\right)^{-1}\,d\vartheta \right\} \end{aligned}} $$
$${\begin{aligned} &\,=\, \frac{1}{\rho}\mathbb{E}_{\sigma^{2}_{\nu_{1}}}\left\{\!\frac{1}{\sigma^{2}_{\nu_{1}}} \left(\underbrace{\frac{1}{\pi} \int_{0}^{\pi/2}\left(\frac{1}{4\sin^{2}(\vartheta)}\right)^{-1}\,d\vartheta}_{=1} \right) \right\} \end{aligned}} $$
$$ {\begin{aligned} &= \frac{1}{\rho}\mathbb{E}_{\sigma^{2}_{\nu_{1}}}\left\{\frac{1}{\sigma^{2}_{\nu_{1}}}\right\} \end{aligned}} $$
(34)
The rest of the proof follows by using similar steps as for the proof of Proposition 1.
Proof of Proposition 3
By using the Maclaurin series expansion and keeping the first two dominant terms, we have the following:
$$\begin{array}{*{20}l} &\exp \left(- z\bar \zeta \left({K,\vartheta} \right)\psi \left({p,q,{x_{m}},{x_{n}},{\theta^{t}},{\phi^{t}}} \right) \right) &\\&\quad= 1 - z\bar \zeta \left({K,\vartheta} \right)\psi \left({p,q,{x_{m}},{x_{n}},{\theta^{t}},{\phi^{t}}} \right) + \\ &\qquad\:\:\:\:{\mathcal{O}}\left({z\bar \zeta \left({K,\vartheta} \right)\psi \left({p,q,{x_{m}},{x_{n}},{\theta^{t}},{\phi^{t}}} \right)} \right)\\ &\quad\approx 1 - z\bar \zeta \left({K,\vartheta} \right)\psi \left({p,q,{x_{m}},{x_{n}},{\theta^{t}},{\phi^{t}}} \right) \end{array} $$
(35)
By using this approximation, we have the following:
$$ \begin{aligned} &\int_{0}^{\pi} {\int_{- \pi }^{\pi} {\exp} } \left(- z\bar \zeta \left({K,\vartheta} \right)\psi \left({p,q,{x_{m}},{x_{n}},{\theta^{t}},{\phi^{t}}} \right) \right)\\&\quad{f_{\theta} }\left({\theta^{t}}\right){f_{\phi} }\left({\phi^{t}}\right){\mkern 1mu} d{\theta^{t}}{\mkern 1mu} d{\phi^{t}}\\ &\approx \int_{0}^{\pi} \int_{- \pi }^{\pi} \left({1 - z\bar \zeta \left({K,\vartheta} \right)\psi \left({p,q,{x_{m}},{x_{n}},{\theta^{t}},{\phi^{t}}} \right)} \right)\\&\quad {f_{\theta} }\left({\theta^{t}}\right){f_{\phi} }\left({\phi^{t}}\right){\mkern 1mu} d{\theta^{t}}{\mkern 1mu} d{\phi^{t}} \\ & = 1 - z\bar \zeta \left({K,\vartheta} \right)\int_{0}^{\pi} {\int_{- \pi }^{\pi} {\psi \left({p,q,{x_{m}},{x_{n}},{\theta^{t}},{\phi^{t}}} \right)}}\\&\quad {f_{\theta} }\left({\theta^{t}}\right){f_{\phi} }\left({\phi^{t}}\right){\mkern 1mu} d{\theta^{t}}{\mkern 1mu} d{\phi^{t}} \end{aligned} $$
(36)
Let us consider the following integral:
$$\begin{array}{*{20}l} {} \Theta = \int_{0}^{\pi} {\int_{- \pi }^{\pi} {\psi \left({p,q,{x_{m}},{x_{n}},{\theta^{t}},{\phi^{t}}} \right)}} {f_{\theta} }\left({\theta^{t}}\right){f_{\phi} }\left({\phi^{t}}\right)d{\theta^{t}}d{\phi^{t}} \end{array} $$
(37)
In addition, the following holds true:
$$\begin{array}{*{20}l} {\underset{K \to + \infty}{\lim }} {\left({1 - \frac{{z\frac{\rho }{{4{{\sin }^{2}}(\vartheta)}}\Theta }}{K}} \right)^{K}} = \exp \left({ - z\frac{\rho }{{4{{\sin }^{2}}(\vartheta)}}\Theta} \right) \end{array} $$
(38)
where we used the following notable limit:
$$\begin{array}{*{20}l} {\underset{x \to + \infty}{\lim }} {\left({1 + \frac{k}{x}} \right)^{x}} = {e^{k}}, \end{array} $$
(39)
Therefore, the asymptotic APEP is:
$$\begin{array}{*{20}l} \text{APEP} &\!\leq\! \frac{1}{\pi }\int_{0}^{\pi /2} {\int_{0}^{\infty} {\exp \left({ \,-\, z\left({1 \,+\, \frac{\rho }{{4{{\sin }^{2}}(\vartheta)}}\Theta} \right)} \right)}} dzd\vartheta \notag\\ &{{\stackrel{(a)}{=}} \frac{1}{\pi }\int_{0}^{\pi /2} {{{\left({1 + \frac{\rho }{{{4{\sin }^{2}}(\vartheta)}}\Theta} \right)}^{- 1}}} d\vartheta} \notag\\ &{ {\stackrel{(b)}{=}} \frac{1}{2}\left({1 - \sqrt {\frac{{\rho \Theta }}{{\rho \Theta + 1}}}} \right)} \end{array} $$
(40)
where the first equality (a) comes from the fact that \(\int _{0}^{\infty } {\exp \left ({ - zc} \right) =} \frac {1}{c},c > 0\) and the second equality (b) follows from (5A.4a) in [25]. This concludes the proof.
Proof of Proposition 4
By using steps similar to Proposition 3, the asymptotic APEP can be written as follows:
$$\begin{array}{*{20}l} \text{APEP} &\leq \frac{1}{\pi }\int_{0}^{\pi /2} {{{\left({1 + \frac{\rho }{{{4{\sin }^{2}}(\vartheta)}}\Theta} \right)}^{- 1}}} d\vartheta \notag\\ & {\stackrel{\rho \gg 1}{\le}} \frac{1}{{\rho \Theta }}\left({\underbrace {\frac{1}{\pi }\int_{0}^{\pi /2} {{{\left({\frac{1}{{4{{\sin }^{2}}(\vartheta)}}} \right)}^{- 1}}} {\mkern 1mu} d\vartheta }_{= 1}} \right), \end{array} $$
(41)
which concludes the proof.
Proof of Proposition 5
We first introduce the following lemma [26] for application to hermitian quadratic forms in complex normal variables.
Lemma 1
Let vn(n=1,…,N) be a set of complex Gaussian random variables having zero mean. Let κ, with v=[v1,⋯,vN]T, be an Hermitian quadratic form:
$$\begin{array}{*{20}l} \kappa = \mathbf{v}^{H}\mathbf{I}_{N}\mathbf{v} \end{array} $$
(42)
Its MGF is as follows:
$$\begin{array}{*{20}l} M_{\kappa}(s) = \prod_{n = 1}^{N} (1-s\lambda_{n})^{-1} \end{array} $$
(43)
where λn is the nth eigenvalue of the covariance matrix \(\mathbf {R}_{\mathbf {v}} = \mathbb {E}\{\mathbf {v}\mathbf {v}^{H}\}\).
Proof
See [26]. □
The proof of Proposition 5 can be split in three steps.
Step 1: By using Lemma 1 with N=Nr=2, we have:
$$\begin{array}{*{20}l} \text{APEP} \approx \frac{1}{{{\rho^{2}}}}{\mathbb{E}_{{\theta_{k}^{t}},{\phi_{k}^{t}},{\theta_{k}^{r}},{\phi_{k}^{r}}}}\left\{ {\left({\frac{3}{{{\lambda_{1}}{\lambda_{2}}}}} \right)} \right\} \end{array} $$
(44)
where λ1 and λ2 are eigenvectors of the covariance matrix:
$$\begin{array}{*{20}l} \mathbf{R} = \left[ {\begin{array}{*{20}{c}} {\mathbb{E}\left\{ {\left. {{\nu_{1}}{\nu^{*}}_{1}} \right\}} \right.}&{\mathbb{E}\left\{ {\left. {{\nu_{1}}{\nu^{*}}_{2}} \right\}} \right.}\\ {\mathbb{E}\left\{ {\left. {{\nu_{2}}{\nu^{*}}_{1}} \right\}} \right.}&{\mathbb{E}\left\{ {\left. {{\nu_{2}}{\nu^{*}}_{2}} \right\}} \right.} \end{array}} \right] \end{array} $$
(45)
and R is assumed to be full rank.
It is worth mentioning that λ1 and λ2 depend on the random variables \({{\theta _{k}^{t}},{\phi _{k}^{t}},{\theta _{k}^{r}},{\phi _{k}^{r}}}\). In particular, from Lemma 1, we have:
$$\begin{array}{*{20}l} {} \text{APEP} = \mathbb{E}_{{\theta_{k}^{t}},{\phi_{k}^{t}},{\theta_{k}^{r}},{\phi_{k}^{r}}}\left\{ {\left. {\frac{1}{\pi }\int\limits_{0}^{\pi /2} {\prod\limits_{{n_{r}} = 1}^{2} {{{(1 + \frac{\rho }{{4{{\sin }^{2}}\left(\vartheta \right)}}{\lambda_{n_{r}}})}^{- 1}}} d\vartheta} } \right\}} \right. \end{array} $$
(46)
Also, we have the following:
$$\begin{array}{*{20}l} &\frac{1}{{\left({1 + \frac{\rho }{{4{{\sin }^{2}}\left(\vartheta \right)}}{\lambda_{1}}} \right)\left({1 + \frac{\rho }{{4{{\sin }^{2}}\left(\vartheta \right)}}{\lambda_{2}}} \right)}} \\&= \frac{{{A_{1}}}}{{\left({1 + \frac{\rho }{{4{{\sin }^{2}}\left(\vartheta \right)}}{\lambda_{1}}} \right)}} + \frac{{{A_{2}}}}{{\left({1 + \frac{\rho }{{4{{\sin }^{2}}\left(\vartheta \right)}}{\lambda_{2}}} \right)}} \end{array} $$
(47)
which yields:
$$\begin{array}{*{20}l} {A_{1}} = \frac{{{\lambda_{1}}}}{{{\lambda_{1}} - {\lambda_{2}}}}, \qquad {A_{2}} = - \frac{{{\lambda_{2}}}}{{{\lambda_{1}} - {\lambda_{2}}}} \end{array} $$
(48)
For high SNR, we have:
$$\begin{array}{*{20}l} {\left({1 + \frac{\rho }{{4{{\sin }^{2}}\left(\vartheta \right)}}{\lambda_{n_{r}}}} \right)^{- 1}} &\approx {\left({\frac{\rho }{{4{{\sin }^{2}}\left(\vartheta \right)}}{\lambda_{n_{r}}}} \right)^{- 1}}\\&\quad - {\left({\frac{\rho }{{4{{\sin }^{2}}\left(\vartheta \right)}}{\lambda_{n_{r}}}} \right)^{- 2}} \end{array} $$
(49)
where we have used the second-order Taylor approximation.
Thus, by substituting (48) and (49) in (46), we obtain:
$$ {} \begin{aligned} \text{APEP} &= {\mathbb{E}_{{\theta_{k}^{t}},{\phi_{k}^{t}},{\theta_{k}^{r}},{\phi_{k}^{r}}}}\left\{ \left. \sum\limits_{i = 1}^{2} {A_{i}}\left[ \frac{1}{\pi }\int\limits_{0}^{\pi /2} {{\left({\frac{\rho }{{4{{\sin }^{2}}\left(\vartheta \right)}}{\lambda_{i}}} \right)}^{- 1}}d\vartheta\right. \right. \right.\\ &\quad \left. \left. - \frac{1}{\pi }\int\limits_{0}^{\pi /2} {{{\left({\frac{\rho }{{4{{\sin }^{2}}\left(\vartheta \right)}}{\lambda_{i}}} \right)}^{- 2}}d\vartheta} \right] \right\} \\ &= {\mathbb{E}_{{\theta_{k}^{t}},{\phi_{k}^{t}},{\theta_{k}^{r}},{\phi_{k}^{r}}}}\left\{ \frac{1}{\rho }\left[ \frac{1}{{{\lambda_{i}}}}\frac{1}{\pi }\int\limits_{0}^{\pi /2} {{{\left({\frac{1}{{4{{\sin }^{2}}\vartheta }}} \right)}^{- 1}}d\omega} \right. \right.\\ &\quad \left.\left. - \frac{1}{\rho }\frac{1}{{{\lambda^{2}}_{i}}}\frac{1}{\pi }\int\limits_{0}^{\pi /2} {{{\left({\frac{1}{{4{{\sin }^{2}}\left(\vartheta \right)}}} \right)}^{- 2}}d\vartheta} \right] \right\} \\ & {\stackrel{(a)}{=}} { \mathbb{E}_{{\theta_{k}^{t}},{\phi_{k}^{t}},{\theta_{k}^{r}},{\phi_{k}^{r}}}}\left\{ \frac{1}{\rho }\left(\frac{{{\lambda_{1}}}}{{{\lambda_{1}} - {\lambda_{2}}}}\left[ {\frac{1}{{{\lambda_{1}}}} - \frac{3}{{\rho {\lambda^{2}}_{1}}}} \right] \right.\right.\\ &\quad \left.\left. - \frac{{{\lambda_{2}}}}{{{\lambda_{1}} - {\lambda_{2}}}}\left[ {\frac{1}{{{\lambda_{2}}}} - \frac{3}{{\rho {\lambda^{2}}_{2}}}} \right] \right) \right\} \\ &= {\mathbb{E}_{{\theta_{k}^{t}},{\phi_{k}^{t}},{\theta_{k}^{r}},{\phi_{k}^{r}}}}\left\{ {\frac{1}{{{\rho^{2}}}}\left({\frac{3}{{{\lambda_{1}}{\lambda_{2}}}}} \right)} \right\} \end{aligned} $$
(50)
where (a) follows from:
$$\begin{array}{*{20}l}{} \frac{1}{\pi }\int\limits_{0}^{\pi /2} {{{\left(\! {\frac{1}{{4{{\sin }^{2}}\left(\omega \right)}}}\! \right)}^{- 1}}d\omega} =\! 1, \;\; \frac{1}{\pi }\int\limits_{0}^{\pi /2} {{{\left(\! {\frac{1}{{4{{\sin }^{2}}\left(\omega \right)}}}\! \right)}^{- 2}}d\omega} = 3 \end{array} $$
(51)
Step 2: We compute the explicit expression of the product λ1λ2. To this end, we introduce the following lemma.
Lemma 2
Assume that R, defined in (45), is full rank and has two distinct eigenvalues λ1 and λ2. The product of the two eigenvectors λ1 and λ2 is as follows:
$$ {\begin{aligned} &{\lambda_{1} \lambda_{2}} \,=\, {\mathbb{E}_{{\theta^{t}},{\phi^{t}},{\theta_{k}^{r}},{\phi_{k}^{r}}}}\left({\frac{1}{{{K^{2}}}}\left[ {\sum\limits_{k = 1}^{K} {\sum\limits_{k^{\prime} = 1}^{K} {\!\chi \left({{\theta_{k}}^{t},{\phi_{k}^{t}}} \right)\chi \left({{\theta_{k}}^{\prime{t}},\phi_{k}^{\prime{t}}} \right)}} \Upsilon} \right]} \right) \end{aligned}} $$
(52)
where
$$\begin{array}{*{20}l} \chi \left({{\theta_{k}}^{t},{\phi_{k}^{t}}} \right) = &\sqrt {{G_{q}}\left({{\theta_{k}}^{t},{\phi_{k}^{t}}} \right)} \exp \left({j{\Omega_{q}}\left({{\theta_{k}}^{t},{\phi_{k}^{t}}} \right)} \right) \\&- \sqrt {{G_{p}}\left({{\theta_{k}}^{t},{\phi_{k}^{t}}} \right)} \exp \left({j{\Omega_{p}}\left({{\theta_{k}}^{t},{\phi_{k}^{t}}} \right)} \right) \end{array} $$
(53)
and
$$\begin{array}{*{20}l} \Upsilon =1 - \cos \left(\left\| \mathbf{k} \right\|{d}\left(\sin \left({{\theta^{r}}} \right)\sin \left({{\phi^{r}}} \right) \right.\right. \\ -\left.\left. \sin \left({\theta^{\prime{r}}} \right)\sin \left({\phi^{\prime{r}}} \right) \right) \right) \end{array} $$
(54)
Proof
By definition:
$$\begin{array}{*{20}l} \nu_{1} &= \frac{1}{{\sqrt K }}\sum\limits_{k = 1}^{K} {{\beta_{k}}\chi \left({{\theta_{k}}^{t},{\phi_{k}^{t}}} \right)} \end{array} $$
(55)
$$\begin{array}{*{20}l} {\nu_{2}} &\,=\, \frac{1}{{\sqrt K }}\sum\limits_{k = 1}^{K} {\beta_{k}}{\chi \left({{\theta_{k}}^{t},{\phi_{k}^{t}}} \right)\!\exp \left({j\left\| k \right\|{d}\sin \left({{\theta_{k}^{r}}} \right)\sin \left({{\phi_{k}^{r}}} \right)} \right)} \end{array} $$
(56)
Moreover, it is known that the product of the eigenvalues is equal to the determinant of the covariance matrix (i.e., \(\det (\mathbf {R}) = {\lambda _{1}}{\lambda _{2}}\)). Thus, λ1λ2 can be computed directly from \(\det (\mathbf {R})\) as follows:
$$\begin{array}{*{20}l} \det \left(\mathbf{R} \right) \,=\, \mathbb{E}\left\{ {\left. {{\nu_{1}}{\nu^{*}}_{1}} \right\}} \right.\mathbb{E}\left\{ {\left. {{\nu_{2}}{\nu^{*}}_{2}} \right\}} \right. \,-\, \mathbb{E}\left\{ {\left. {{\nu_{2}}{\nu^{*}}_{1}} \right\}} \right.\mathbb{E}\left\{ {\left. {{\nu_{1}}{\nu^{*}}_{2}} \right\}} \right. \end{array} $$
(57)
By substituting (55) and (56) into (57), we obtain (52) where we have used the identity:
$$\frac{{\exp \left({j\phi} \right) + \exp \left({ - j\phi} \right)}}{2} = \cos \left(\phi \right)$$
and if \(k = k^{\prime }\) we have \( 1 - \exp \left (j\left \| {{\mathbf {k}}} \right \|{d}\left (\sin \left ({{\theta _{k}^{r}}} \right)\sin \left ({{\phi _{k}^{r}}} \right) - \sin \left ({\theta _{k}^{\prime {r}}} \right)\sin \left ({\phi _{k}^{\prime {r}}} \right) \right) \right) = 0\). □
Step 3: We exploit the asymptotic analysis introduced in Proposition 3 to derive a closed-form expression of the APEP. We apply the following formula that is a special case of (40) in [24] for X=λ1λ2:
$$\begin{array}{*{20}l} \mathbb{E}\left({{X^{- 1}}} \right) = \int_{0}^{\infty} {{M_{X}}\left({ - z} \right)dz} \end{array} $$
(58)
where \({M_{X}}\left (z \right) = \mathbb {E}_{X}\left \{\exp \left ({Xz} \right) \right \}\).
Then, we obtain:
$$ {\begin{aligned} {\mathbb{E}_{X}}\left({\exp \left({Xz} \right)} \right) &= {\mathbb{E}_{{\theta^{t}},{\phi^{t}},{\theta^{r}},{\phi^{r}}}}\left(\prod\limits_{k = 1}^{K} {\prod\limits_{k^{\prime} = 1}^{K} {\exp} } \left(\frac{1}{{{K^{2}}}}{\chi}\left({{\theta_{k}}^{t},{\phi_{k}^{t}}} \right)\right.\right.\\ &\quad \left.\left. {\chi}\left({{\theta_{k}}^{\prime{t}},\phi_{k}^{\prime{t}}} \right) \Upsilon z {\vphantom{\prod\limits_{k = 1}}}\right) \right)\\ &=\prod\limits_{k = 1}^{K} \prod\limits_{k^{\prime} = 1}^{K} \left(\int_{- \pi }^{\pi}\! \int_{- \pi }^{\pi}\! \int_{- \pi }^{\pi}\! \int_{- \pi }^{\pi}\! \int_{0}^{\pi}\! \int_{0}^{\pi}\!\! \int_{0}^{\pi}\! \int_{0}^{\pi} \exp\! \left(\! \frac{1}{{{K^{2}}}}\!{\chi}\left(\! {{\theta_{k}}^{t},{\phi_{k}^{t}}} \!\right)\!{\chi} \right.\right.\\ & \left.\left({{\theta_{k}}^{\prime{t}},\phi_{k}^{\prime{t}}} \right) \Upsilon z {\vphantom{\int_{- \pi }^{\pi}}}\right){f_{{\phi^{t}}_{k^{\prime}}}}{f_{{\phi^{t}}_{k}}}{f_{{\phi^{r}}_{k^{\prime}}}}{f_{{\phi^{r}}_{k}}}{f_{{\theta^{t}}_{k^{\prime}}}}{f_{{\theta^{t}}_{k}}}{f_{{\theta^{r}}_{k^{\prime}}}}{f_{{\theta^{r}}_{k}}}\\ &\left.\quad{\phi_{k^{\prime}}}^{t}d{\phi_{k}}^{t}d{\phi_{r^{\prime}}}^{r}d{\phi_{k}}^{r} d{\theta_{k}}^{\prime{t}}d{\theta_{k}}^{t}d{\theta_{r}}^{\prime{r}}d{\theta_{k}}^{r} {\vphantom{\int_{- \pi }^{\pi}}}\right)\\ &\,=\,\left(\! \int_{- \pi }^{\pi}\! \int_{- \pi }^{\pi}\! \int_{- \pi }^{\pi}\! \int_{- \pi }^{\pi}\! \int_{0}^{\pi}\!\! \int_{0}^{\pi}\!\! \int_{0}^{\pi}\!\! \int_{0}^{\pi} \!\exp\! \left(\! {\frac{1}{{{K^{2}}}}{\chi}\!\left(\! {{\theta^{t}},{\phi^{t}}}\! \right)\chi\!\left(\! {\theta^{\prime{t}}}\!,\phi^{\prime{t}} \right)\!\Upsilon z}\! \right)\right.\\ &\left.\quad{f_{{\phi^{t}}}}{f_{\phi^{\prime{t}}}}{f_{{\phi^{r}}}}{f_{\phi^{\prime{r}}}}{f_{{\theta^{t}}}}{f_{\theta^{\prime{t}}}}{f_{{\theta^{r}}}}{f_{\theta^{\prime{r}}}}d{\phi^{t}}d\phi^{\prime{t}}d{\phi^{r}}d\phi^{\prime{r}} d{\theta^{t}}d\theta^{\prime{t}}d{\theta^{r}}d\theta^{\prime{r}} {\vphantom{\int_{- \pi }^{\pi}}}\right)^{{K^{2}}} \end{aligned}} $$
(59)
where the last equality follows from the assumption of independent and identically distributed random variables.
Moreover, by using the following approximation:
$$\begin{array}{*{20}l} &\exp \left({\frac{1}{{{K^{2}}}}{\chi}\left({{\theta^{t}},{\phi^{t}}} \right)\chi\left({\theta^{\prime{t}},\phi^{\prime{t}}} \right)\Upsilon z} \right) \approx 1 \\ & \qquad+ \frac{1}{{{K^{2}}}}{\chi}\left({{\theta^{t}},{\phi^{t}}} \right)\chi\left({\theta^{\prime{t}},\phi^{\prime{t}}} \right)\Upsilon z \end{array} $$
(60)
in (59) and using the notable limit in (39), we obtain the following:
$$ {\begin{aligned} &{\mathbb{E}_{X}}\left({\exp \left({Xz} \right)} \right) {\stackrel{K \to \infty}{=}} \\ &\exp \left(z\int_{- \pi }^{\pi} \int_{- \pi }^{\pi} \int_{- \pi }^{\pi} \int_{- \pi }^{\pi} \int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\pi} \int_{0}^{\pi} \left({{\chi}\left({{\theta^{t}},{\phi^{t}}} \right)\chi\left({\theta^{\prime{t}},\phi^{\prime{t}}} \right)\Upsilon} \right)\right.\\ &\quad\left.{f_{{\phi^{t}}}}{f_{\phi{\prime{t}}}}{f_{{\phi^{r}}}}{f_{\phi^{\prime{r}}}}{f_{{\theta^{t}}}}{f_{\theta^{\prime{t}}}}{f_{{\theta^{r}}}}{f_{\theta^{\prime{r}}}}d{\phi^{t}}d \phi^{\prime{t}}d{\phi^{r}}d\phi^{\prime{r}} d{\theta^{t}}d\theta^{\prime{t}}d{\theta^{r}}d\theta^{\prime{r}} {\vphantom{\int_{- \pi }^{\pi}}}\right) \end{aligned}} $$
(61)
Finally, substituting (61) in (58), and then (58) in (44), we conclude the proof by using the following result:
$$ {}\begin{aligned} & \int_{- \pi }^{\pi} \int_{0}^{\pi} \int_{- \pi }^{\pi} \int_{0}^{\pi} \left(1 - \cos \left(\left\| \mathbf{k} \right\|{d}\left(\sin \left({{\theta^{r}}} \right)\sin \left({{\phi^{r}}} \right)\right.\right.\right.\\ &\, \left.\left.\left.- \sin \left({\theta^{\prime{r}}} \right)\sin \left({\phi^{\prime{r}}} \right) \right) \right) \right) {f_{{\phi^{r}}}}{f_{\phi^{\prime{r}}}}{f_{{\theta^{r}}}}{f_{\theta^{\prime{r}}}}d{\phi^{r}}d{\theta^{r}}d \phi^{\prime{r}}d \theta^{\prime{r}} = \\ &1 - {\left({\int_{- \pi }^{\pi} {\int_{0}^{\pi} {\cos \left({\left\| \mathbf{k} \right\|{d}\left({\sin \left({{\theta^{r}}} \right)\sin \left({{\phi^{r}}} \right)} \right)} \right)} {f_{{\phi^{r}}}}{f_{{\theta^{r}}}}d{\phi^{r}}d{\theta^{r}}}} \right)^{2}}\\ &\!\!\!\!\,\quad - {\left({\int_{- \pi }^{\pi} {\int_{0}^{\pi}\! {\sin \left({\left\| \mathbf{k} \right\|{d}\left({\sin \left({{\theta^{r}}} \right)\sin \left({{\phi^{r}}} \right)} \right)} \right){f_{{\phi^{r}}}}{f_{{\theta^{r}}}}d{\phi^{r}}d{\theta^{r}}}} } \right)^{2}} \end{aligned} $$
(62)
Proof of Theorem 1
The proof of Theorem 1 generalizes the steps of Proposition 5 as follows.
Step 1: From Lemma 1, we have:
$$\begin{array}{*{20}l}{} \text{APEP} = {\mathbb{E}_{{\theta_{k}^{t}},{\phi_{k}^{t}},{\theta_{k}^{r}},{\phi_{k}^{r}}}}\left\{ {\left. {\frac{1}{\pi }\int\limits_{0}^{\pi /2} {\prod\limits_{n = 1}^{{N_{r}}} {{{(1 + \frac{\rho }{{4{{\sin }^{2}}\left(\vartheta \right)}}{\lambda_{n}})}^{- 1}}} d\vartheta} } \right\}} \right. \end{array} $$
(63)
Then, we need to identify the coefficients \(\left \{ {{A_{{n_{r}}}}} \right \}_{{n_{r}} = 1}^{{N_{r}}}\) so that the following is satisfied:
$$\begin{array}{*{20}l} \frac{1}{{\prod\limits_{{n_{r}} = 1}^{{N_{r}}} {\left({1 + \frac{\rho }{{4{{\sin }^{2}}\left(\omega \right)}}{\lambda_{n_{r}}}} \right)} }} = \sum\limits_{{n_{r}} = 1}^{{N_{r}}} {\frac{{{A_{n_{r}}}}}{{\left({1 + \frac{\rho }{{4{{\sin }^{2}}\left(\omega \right)}}{\lambda_{n_{r}}}} \right)}}} \end{array} $$
(64)
To this end, we can exploit the general formula of partial fraction decomposition [pp. 66–67], [27]:
$$\begin{array}{*{20}l} {A_{n_{r}}} = \frac{{{\lambda_{n_{r}}}^{{N_{r}} - 1}}}{{\prod\limits_{{n_{r}} \ne l = 1}^{{N_{r}}} {\left({{\lambda_{n_{r}}} - {\lambda_{l}}} \right)} }} \end{array} $$
(65)
for l,nr=1,…,Nr.
Thus, we have:
$$ {\begin{aligned} \text{APEP} &\!\approx\! {\mathbb{E}_{_{{\theta^{t}},{\phi^{t}},{\theta^{r}},{\phi^{r}}}}}\!\left\{\!\! {\left. {\frac{1}{\pi }\!\int\limits_{0}^{\pi /2} {\sum\limits_{{n_{r}} = 1}^{{N_{r}}}\! {{A_{n}}\!\left[\! {\sum\limits_{m = 1}^{{N_{r}}}\!\! {{{\left(\! { - 1} \right)}^{m + 1}}{{\left(\!\! {\frac{\rho }{{4{{\sin }^{2}}\left(\vartheta \right)}}{\lambda_{{n_{r}}}}}\!\! \right)}^{\! - m}}}}\!\! \right]} d\vartheta} }\! \right\}} \right. \\ &\!= {\mathbb{E}_{{\theta^{t}},{\phi^{t}},{\theta^{r}},{\phi^{r}}}}\left\{ {\sum\limits_{{n_{r}} = 1}^{{N_{r}}} {{A_{n}}\left[ {\sum\limits_{m = 1}^{{N_{r}}} {{{\left({ - 1} \right)}^{m + 1}}\frac{1}{\pi }\int_{0}^{\pi /2}}}\right.}}\right.\\ &\quad\left.{{\left.{{{{\left({\frac{\rho }{{4{{\sin }^{2}}\left(\vartheta \right)}}{\lambda_{{n_{r}}}}}\! \right)}^{\! - m}}d\vartheta} }\! {\vphantom{\sum\limits_{m = 1}^{{N_{r}}}}}\right]}}\! {\vphantom{\sum\limits_{{n_{r}} = 1}^{{N_{r}}}}}\right\} \\ &\!= {\mathbb{E}_{_{{\theta^{t}},{\phi^{t}},{\theta^{r}},{\phi^{r}}}}}\left\{ {\sum\limits_{{n_{r}} = 1}^{{N_{r}}} {A_{n}}\left[ {\sum\limits_{m = 1}^{{N_{r}}} {{{\left({ - 1} \right)}^{m + 1}}{{\left({\rho {\lambda_{{n_{r}}}}} \right)}^{- m}}}}\right.}\right.\\ &\quad\left.{{\left.{{\left({\frac{1}{\pi }\int_{0}^{\pi /2} {{{\left({\frac{1}{{4{{\sin }^{2}}\left(\vartheta \right)}}} \right)}^{- m}}d\vartheta} } \right)}} \right]}} \right\} \end{aligned}} $$
(66)
In the high SNR regime, we can use the Nrth-order Taylor approximation as follows:
$$ {\begin{aligned} {\left({1 + \frac{\rho }{{4{{\sin }^{2}}\left(\vartheta \right)}}{\lambda_{{n_{r}}}}}\! \right)^{- {N_{r}}}} &\approx \sum\limits_{m = 1}^{{N_{r}}} {{{\left({ - 1} \right)}^{m + 1}}{{\left({\frac{\rho }{{4{{\sin }^{2}}\left(\vartheta \right)}}{\lambda_{{n_{r}}}}}\! \right)}^{- m}}}\\ &\quad + \mathcal{O}\left({{{\left({ - 1} \right)}^{{N_{r}} + 1}}{{\left({\frac{\rho }{{4{{\sin }^{2}}\left(\vartheta \right)}}{\lambda_{{n_{r}}}}} \right)}^{- ({N_{r}} + 1)}}} \right) \end{aligned}} $$
(67)
We note that the following holds true:
$$ {\begin{aligned} \frac{1}{\pi }\!\! \int\limits_{0}^{\pi /2}\! {{{\left(\! {\frac{1}{{4{{\sin }^{2}}\!\left(\vartheta \right)}}}\! \right)}^{\! - m}}\! d\vartheta} &= \frac{1}{2}\! \left(\! {\!\begin{array}{*{20}{c}} {2m}\\ m \end{array}}\! \right)\\ &\quad +\! \sum\limits_{k = 0}^{m - 1}\! {{{\left(\! { - 1}\! \right)}^{m - k}}2\!\left(\! {\begin{array}{*{20}{c}} {2n}\\ k \end{array}}\! \right)\!\frac{{\sin \left({\pi\! \left({m - k} \right)} \right)}}{{2\pi \left({m - k} \right)}}} \end{aligned}} $$
(68)
where we have used the following identity [27, p. 31, eq. (1320,1)]:
$$ {}\begin{aligned} {\sin^{2m}}\left(\vartheta \right) &= \frac{1}{{{2^{2m}}}}\left\{\! {\sum\limits_{k = 0}^{m - 1} {{\left({ - 1} \right)}^{m - k}}2\left({\begin{array}{*{20}{c}} {2m}\\ k \end{array}} \right)\!\cos \left({2\left({m - k} \right)\vartheta} \right)}\right.\\ &\quad\left.{+ \left({\begin{array}{*{20}{c}} {2m}\\ m \end{array}}\! \right)} \right\} \end{aligned} $$
(69)
Let us define:
$$\begin{array}{*{20}l} {\alpha_{m}} = \left({\frac{1}{\pi }\int_{0}^{\pi /2} {{{\left({\frac{1}{{4{{\sin }^{2}}\left(\vartheta \right)}}} \right)}^{- m}}d\vartheta} } \right) \end{array} $$
(70)
Then, we have the following:
$$ {\begin{aligned} \text{APEP} &\approx {\mathbb{E}_{{\theta_{k}^{t}},{\phi_{k}^{t}},{\theta_{k}^{r}},{\phi_{k}^{r}}}}\left\{ {\sum\limits_{{n_{r}} = 1}^{{N_{r}}} {{A_{{n_{r}}}}\left[ {\sum\limits_{m = 1}^{{N_{r}}} {{{\left({ - 1} \right)}^{m + 1}}{{\left({\rho {\lambda_{{n_{r}}}}} \right)}^{- m}}{\alpha_{m}}}}\! \right]}}\! \right\} \\ &= {\mathbb{E}_{{\theta_{k}^{t}},{\phi_{k}^{t}},{\theta_{k}^{r}},{\phi_{k}^{r}}}}\left\{ {\sum\limits_{{n_{r}} = 1}^{{N_{r}}} {{A_{{n_{r}}}}\left[ {{{\left({ - 1} \right)}^{{N_{r}} + 1}}{{\left({\rho {\lambda_{{n_{r}}}}} \right)}^{- {N_{r}}}}{\alpha_{{N_{r}}}}}\right.}}\right.\\ &\quad\left.{\left.{ + \sum\limits_{m = 1}^{{N_{r}} - 1} {{{\left({ - 1} \right)}^{m + 1}}{{\left({\rho {\lambda_{{n_{r}}}}} \right)}^{- m}}{\alpha_{m}}}} \right]} \right\} \\ &= {\mathbb{E}_{{\theta_{k}^{t}},{\phi_{k}^{t}},{\theta_{k}^{r}},{\phi_{k}^{r}}}}\left\{ {\sum\limits_{{n_{r}} = 1}^{{N_{r}}} {{A_{{n_{r}}}}{{\left({ - 1} \right)}^{{N_{r}} + 1}}{{\left({\rho {\lambda_{{n_{r}}}}} \right)}^{- {N_{r}}}}{\alpha_{{N_{r}}}}}}\right.\\ &\quad\left.{+ \sum\limits_{n = 1}^{{N_{r}}} {{A_{n}}\left[ {\sum\limits_{m = 1}^{{N_{r}} - 1} {{{\left({ - 1} \right)}^{m + 1}}{{\left({\rho {\lambda_{{n_{r}}}}} \right)}^{- m}}{\alpha_{m}}}} \right]}} \right\} \end{aligned}} $$
(71)
By induction, the following can be proved:
$$\begin{array}{*{20}l} \sum\limits_{{n_{r}} = 1}^{{N_{r}}} {{A_{{n_{r}}}}\left[ {\sum\limits_{m = 1}^{{N_{r}} - 1} {{{\left({ - 1} \right)}^{m + 1}}{{\left({\rho {\lambda_{{n_{r}}}}} \right)}^{- m}}{\alpha_{m}}}} \right]} = 0 \end{array} $$
(72)
and
$$\begin{array}{*{20}l} \sum\limits_{{n_{r}} = 1}^{{N_{r}}} {{A_{{n_{r}}}}{{\left({ - 1} \right)}^{{N_{r}} + 1}}{{\left({\rho {\lambda_{{n_{r}}}}} \right)}^{- {N_{r}}}}{\alpha_{{N_{r}}}}} = \frac{{{\alpha_{{N_{r}}}}}}{{{\rho^{{N_{r}}}}}}\frac{1}{{\prod\limits_{n = 1}^{{N_{r}}} {{\lambda_{n}}} }} \end{array} $$
(73)
Step 2: Let \(X = \prod \limits _{{n_{r}} = 1}^{{N_{r}}} {{\lambda _{{n_{r}}}}}\). By using steps similar to those of Lemma 2, we obtain the following:
$$\begin{array}{*{20}l} \det \left({{\mathbf{R}_{\mathbf{v}}}} \right) &= \frac{1}{{{K^{{N_{r}}}}}}\left(\sum\limits_{{k_{1}} = 1}^{K} \ldots \sum\limits_{{k_{{N_{r}}}} = 1}^{K} \left(\prod\limits_{i = 1}^{{N_{r}}} {\chi}\left({{\theta_{{k_{i}}}}^{t},\phi_{{k_{i}}}^{t}} \right)\right. \right. \\& \left. \left. F\left({\theta_{{k_{1}}}^{r}, \ldots,\theta_{{k_{{N_{r}}}}}^{r},\phi_{{k_{1}}}^{r}, \ldots,\phi_{{k_{{N_{r}}}}}^{r}} {\vphantom{\sum\limits_{{k_{1}} = 1}}}\right) \right) \right) \end{array} $$
(74)
where \(F\left ({\theta _{{k_{1}}}^{r}, \ldots,\theta _{{k_{{N_{r}}}}}^{r},\phi _{{k_{1}}}^{r}, \ldots,\phi _{{k_{{N_{r}}}}}^{r}} \right)\).
Step 3: The APEP is then the following:
$$ {}{\begin{aligned} &\mathbb{E}_{X}\left\{ {\exp \left({Xz} \right)} \right\}\\ &={\mathbb{E}_{{\theta^{t}},{\phi^{t}},{\theta^{r}},{\phi^{r}}}}\left\{ \exp\!\! \left(\!\! \frac{z}{{{K^{{N_{r}}}}}}\!\!\left(\sum\limits_{{k_{1}} = 1}^{K} \ldots\! \sum\limits_{{k_{{N_{r}}}} = 1}^{K} \left[\! \left(\prod\limits_{i = 1}^{{N_{r}}} {{\chi}\left({{\theta_{{k_{i}}}}^{t},\phi_{{k_{i}}}^{t}} \right)} \!\right)\right.\right. \right.\right.\\& \left.\left.\left.\left. F\left({\theta_{{k_{1}}}^{r}, \ldots,\theta_{{k_{{N_{r}}}}}^{r},\phi_{{k_{1}}}^{r}, \ldots,\phi_{{k_{{N_{r}}}}}^{r}} \right) \vphantom{{\sum\limits_{{k_{1}} = 1}^{K}}}\right] \right) \right) \right\}\\ &={\mathbb{E}_{{\theta^{t}},{\phi^{t}},{\theta^{r}},{\phi^{r}}}}\left\{ \prod\limits_{{k_{1}} = 1}^{K} \ldots \prod\limits_{{k_{{N_{r}}}} = 1}^{K} \exp \left(\frac{z}{{{K^{{N_{r}}}}}}\left[ {\prod\limits_{i = 1}^{{N_{r}}} {{\chi}\left({{\theta_{{k_{i}}}}^{t},\phi_{{k_{i}}}^{t}} \right)}} \right]\right.\right. \\& \left. \left. F\left({\theta_{{k_{1}}}^{r}, \ldots,\theta_{{k_{{N_{r}}}}}^{r},\phi_{{k_{1}}}^{r}, \ldots,\phi_{{k_{{N_{r}}}}}^{r}} \right) {\vphantom{\prod\limits_{{k_{{N_{r}}}} = 1}^{K}}}\right) \right\} \\ &=\prod\limits_{{k_{1}} = 1}^{K} \ldots \prod\limits_{{k_{{N_{r}}}} = 1}^{K} {\mathbb{E}_{{\theta^{t}},{\phi^{t}},{\theta^{r}},{\phi^{r}}}}\left\{\exp \left(\frac{z}{{{K^{{N_{r}}}}}}\left[ {\prod\limits_{i = 1}^{{N_{r}}} {{\chi}\left({{\theta_{{k_{i}}}}^{t},\phi_{{k_{i}}}^{t}} \right)}} \right]\right. \right. \\& \left. \left. F\left({\theta_{{k_{1}}}^{r}, \ldots,\theta_{{k_{{N_{r}}}}}^{r},\phi_{{k_{1}}}^{r}, \ldots,\phi_{{k_{{N_{r}}}}}^{r}} \right) {\vphantom{\prod\limits_{i = 1}^{{N_{r}}}}}\right) \right\} \\ &=\left[{\mathbb{E}_{{\theta^{t}},{\phi^{t}},{\theta^{r}},{\phi^{r}}}}\left\{ \exp \left(\frac{z}{{{K^{{N_{r}}}}}}\left[ {\prod\limits_{i = 1}^{{N_{r}}} {{\chi^{i}}\left({{\theta^{i,t}},{\phi^{i,t}}} \right)}} \right] \right.\right.\right. \\& \left.\left. \left. F\left({{\theta^{1,r}}, \ldots,{\theta^{{N_{r}},r}},{\phi^{1,r}}, \ldots,{\phi^{{N_{r}},r}}} \right) {\vphantom{\prod\limits_{i = 1}^{{N_{r}}}}}\right) \right\} \right]^{{N_{r}}} \end{aligned}} $$
(75)
By using the following approximation:
$$ {\begin{aligned} {}&\exp\!\left(\! {\frac{z}{{{K^{{N_{r}}}}}}\!\left[ {\prod\limits_{i = 1}^{{N_{r}}} \!{{\chi}\!\left({{\theta^{i,t}},{\phi^{i,t}}} \right)}} \!\!\right]\!F\!\left({{\theta^{1,r}}, \!\ldots,{\theta^{{N_{r}},r}},{\phi^{1,r}}, \ldots,{\phi^{{N_{r}},r}}} \right)}\! \right) \approx \end{aligned}} $$
(76)
$$ \begin{aligned} {}&1 + \frac{z}{{{K^{{N_{r}}}}}}\left[ {\prod\limits_{i = 1}^{{N_{r}}} {{\chi}\left({{\theta^{i,t}},{\phi^{i,t}}} \right)}}\! \right]\!F\left({{\theta^{1,r}}, \ldots,{\theta^{{N_{r}},r}},{\phi^{1,r}}, \ldots,{\phi^{{N_{r}},r}}} \right) \end{aligned} $$
(77)
we have:
$$ {}{\begin{aligned} \mathbb{E}\left\{ {\exp \left({Xz} \right)} \right\}&=\left[ {\mathbb{E}_{{\theta^{t}},{\phi^{t}},{\theta^{r}},{\phi^{r}}}}\left\{1 + \frac{z}{{{K^{{N_{r}}}}}}\left[ {\prod\limits_{i = 1}^{{N_{r}}} {{\chi}\left({{\theta^{i,t}},{\phi^{i,t}}} \right)}} \right]\right. \right. \\& \left. \left. F\left({{\theta^{1,r}}, \ldots,{\theta^{{N_{r}},r}},{\phi^{1,r}}, \ldots,{\phi^{{N_{r}},r}}} \right) {\vphantom{\prod\limits_{i = 1}^{{N_{r}}}}}\right\} \right]^{{N_{r}}}\\ &=\left[ 1 + \frac{z}{{{K^{{N_{r}}}}}}{\mathbb{E}_{{\theta^{t}},{\phi^{t}},{\theta^{r}},{\phi^{r}}}}\left\{ \left[ {\prod\limits_{i = 1}^{{N_{r}}} {{\chi}\left({{\theta^{i,t}},{\phi^{i,t}}} \right)}} \right] \right. \right. \\& \left. \left. F\left({{\theta^{1,r}}, \ldots,{\theta^{{N_{r}},r}},{\phi^{1,r}}, \ldots,{\phi^{{N_{r}},r}}} \right) \prod\limits_{i = 1}^{{N_{r}}}\right\} \right]^{{N_{r}}}\\ &\approx \exp \left(z{\mathbb{E}_{{\theta^{t}},{\phi^{t}},{\theta^{r}},{\phi^{r}}}}\left\{ \left[ {\prod\limits_{i = 1}^{{N_{r}}} {{\chi}\left({{\theta^{i,t}},{\phi^{i,t}}} \right)}} \right]\right. \right. \\& \left. \left. F\left({{\theta^{1,r}}, \ldots,{\theta^{{N_{r}},r}},{\phi^{1,r}}, \ldots,{\phi^{{N_{r}},r}}} \right) \prod\limits_{i = 1}^{{N_{r}}}\right\} \right) \end{aligned}} $$
(78)
Thus, we obtain:
$$ {\begin{aligned} &\frac{{{\alpha_{{N_{r}}}}}}{{{\rho^{{N_{r}}}}}}\int_{0}^{\infty} {{M_{X}}\left({ - z} \right)dz} = \\ &\frac{{{\alpha_{{N_{r}}}}}}{{{\rho^{{N_{r}}}}}}\int_{0}^{\infty} \exp \left(z{\mathbb{E}_{{\theta^{t}},{\phi^{t}},{\theta^{r}},{\phi^{r}}}}\left\{\left[ \prod\limits_{i = 1}^{{N_{r}}} {{\chi}\left({{\theta^{i,t}},{\phi^{i,t}}} \right)} \right]\right. \right. \\ & \left. \left. F\left({{\theta^{1,r}}, \ldots,{\theta^{{N_{r}},r}},{\phi^{1,r}}, \ldots,{\phi^{{N_{r}},r}}} \right) {\vphantom{\prod\limits_{i =1}^{{N_{r}}}}}\right\} \right)dz \\ &= \frac{{{\alpha_{{N_{r}}}}}}{{{\rho^{{N_{r}}}}{\mathbb{E}_{{\theta^{t}},{\phi^{t}},{\theta^{r}},{\phi^{r}}}}\left\{ {\left[ {\prod\limits_{i = 1}^{{N_{r}}} {{\chi}\left({{\theta^{i,t}},{\phi^{i,t}}} \right)}} \right]F\left({{\theta^{1,r}}, \ldots,{\theta^{{N_{r}},r}},{\phi^{1,r}}, \ldots,{\phi^{{N_{r}},r}}} \right)} \right\}}} \\ &= \frac{{{\alpha_{{N_{r}}}}}}{{{\rho^{{N_{r}}}}{\mathbb{E}_{{\theta^{t}},{\phi^{t}}}}\left\{ {\left[ {\prod\limits_{i = 1}^{{N_{r}}} {{\chi}\left({{\theta^{i,t}},{\phi^{i,t}}} \right)}} \right]} \right\}{\mathbb{E}_{{\theta^{r}},{\phi^{r}}}}\left\{{F\left({{\theta^{1,r}}, \ldots,{\theta^{{N_{r}},r}},{\phi^{1,r}}, \ldots,{\phi^{{N_{r}},r}}} \right)} \right\}}}\\ &= \frac{{{\alpha_{{N_{r}}}}}}{{{\rho^{{N_{r}}}}{{\left[ {{\mathbb{E}_{{\theta^{t}},{\phi^{t}}}}\left\{ {\chi\left({{\theta^{t}},{\phi^{t}}} \right)} \right\}} \right]}^{{N_{r}}}}{\mathbb{E}_{{\theta^{r}},{\phi^{r}}}}\left\{ {F\left({{\theta^{1,r}}, \ldots,{\theta^{{N_{r}},r}},{\phi^{1,r}}, \ldots,{\phi^{{N_{r}},r}}} \right)} \right\}}} \end{aligned}} $$
(79)
This concludes the proof.
Proof of Proposition 5
From Theorem 1, we can obtain (21). We need to prove how to obtain the explicit form of the function F(·). If Nr=3, the determinant of the covariance matrix R can be computed as follows:
$$ {} \begin{aligned} &\det \left({{\mathbf{R}}} \right) = \\ &\mathbb{E}\left\{ {\left. {{\nu_{1}}{\nu^{*}}_{1}} \right\}} \right.\mathbb{E}\left\{ {\left. {{\nu_{2}}{\nu^{*}}_{2}} \right\}} \right.\mathbb{E}\left\{ {\left. {{\nu_{3}}{\nu^{*}}_{3}} \right\}} \right. + \mathbb{E}\left\{ {\left. {{\nu_{1}}{\nu^{*}}_{2}} \right\}} \right.\mathbb{E}\left\{ {\left. {{\nu_{2}}{\nu^{*}}_{3}} \right\}} \right.\\& \mathbb{E}\left\{ {\left. {{\nu_{3}}{\nu^{*}}_{1}} \right\}} \right. + \mathbb{E}\left\{ {{\nu_{1}}{\nu^{*}}_{3}} \right\}\mathbb{E}\left\{ {{\nu_{2}}{\nu^{*}}_{1}} \right\}\mathbb{E}\left\{ {{\nu_{3}}{\nu^{*}}_{2}} \right\} - \mathbb{E}\left\{ {{\nu_{1}}{\nu^{*}}_{3}} \right\}\\& \mathbb{E}\left\{ {{\nu_{2}}{\nu^{*}}_{2}} \right\} \mathbb{E}\left\{ {{\nu_{3}}{\nu^{*}}_{1}} \right\} - \mathbb{E}\left\{ {\left. {{\nu_{1}}{\nu^{*}}_{2}} \right\}} \right.\mathbb{E}\left\{ {\left. {{\nu_{2}}{\nu^{*}}_{1}} \right\}} \right.\mathbb{E}\left\{ {\left. {{\nu_{3}}{\nu^{*}}_{3}} \right\}} \right.\\ & - \mathbb{E}\left\{ {\left. {{\nu_{1}}{\nu^{*}}_{1}} \right\}} \right.\mathbb{E}\left\{ {\left. {{\nu_{2}}{\nu^{*}}_{3}} \right\}} \right.\mathbb{E}\left\{ {\left. {{\nu_{3}}{\nu^{*}}_{2}} \right\}} \right. \end{aligned} $$
(80)
We note that:
$$\begin{array}{*{20}l} &\mathbb{E}\left\{ {\left. {{\nu_{u}}{\nu^{*}_{v}}} \right\}} \right. =\notag\\&\quad\!\! { \frac{1}{K}\sum\limits_{k = 1}^{K} {{\chi\left({{{\theta^{t}_{k}}},{{\phi^{t}_{k}}}} \right)}} \exp \left({j\left({u - v} \right)\left\| \mathbf{k} \right\|{d}\sin \left({{\theta_{k}^{r}}} \right)\sin \left({{\phi_{k}^{r}}} \right)} \right)} \end{array} $$
(81)
for u,v=1,…,Nr. Then, from (80) and (81), we have:
$$\begin{array}{*{20}l} &\det(\mathbf{R})=\frac{1}{{{K^{3}}}}\sum\limits_{{k_{1}} = 1}^{K} {\sum\limits_{{k_{2}} = 1}^{K} {\sum\limits_{{k_{3}} = 1}^{K} {{\chi}\left({{\theta_{{k_{1}}}}^{t},\phi_{{k_{1}}}^{t}} \right)}} } {\chi}\left({{\theta_{{k_{2}}}}^{t},\phi_{{k_{2}}}^{t}} \right)\notag\\&\qquad{\chi}\left({{\theta_{{k_{3}}}}^{t},\phi_{{k_{3}}}^{t}} \right)F\left({\theta_{{k_{1}}}^{r},\theta_{{k_{2}}}^{r},\theta_{{k_{3}}}^{r},\phi_{{k_{1}}}^{r},\phi_{{k_{2}}}^{r},\phi_{{k_{3}}}^{r}} \right) \end{array} $$
(82)
where:
$$ {\begin{aligned} &F\left({\theta_{{k_{1}}}^{r},\theta_{{k_{2}}}^{r},\theta_{{k_{3}}}^{r},\phi_{{k_{1}}}^{r},\phi_{{k_{2}}}^{r},\phi_{{k_{3}}}^{r}} \right) =\\ &\left[\! \begin{array}{l} 1\\ +\exp\! \left({ \!- j\left\| k \right\|{d}\left[ \!{\sin \left({\theta_{{k_{1}}}^{r}} \right)\sin\! \left({\phi_{{k_{1}}}^{r}}\! \right) \,+\, \sin\! \left({\theta_{{k_{2}}}^{r}} \!\right)\sin\! \left({\phi_{{k_{2}}}^{r}}\! \right)\! -\! 2\sin\! \left({\theta_{{k_{3}}}^{r}} \right)\!\sin\! \left({\phi_{{k_{3}}}^{r}} \!\right)} \!\right]} \right)\\ + \exp\! \left({ \!- j\left\| k \right\|{d}\left[ \!{2\sin \left({\theta_{{k_{1}}}^{r}} \!\right)\sin\! \left({\phi_{{k_{1}}}^{r}} \!\right) \,-\, \sin\! \left({\theta_{{k_{2}}}^{r}} \!\right)\sin\! \left({\phi_{{k_{2}}}^{r}} \!\right) \,-\, \sin\! \left({\theta_{{k_{3}}}^{r}}\! \right)\sin\! \left({\phi_{{k_{3}}}^{r}} \right)}\! \right]} \right)\\ - \exp \left({ - j\left\| k \right\|{d}\left[ \!{2\sin\! \left({\theta_{{k_{1}}}^{r}} \right)\sin\! \left({\phi_{{k_{1}}}^{r}} \right) - 2\sin\! \left({\theta_{{k_{3}}}^{r}} \right)\sin\! \left({\phi_{{k_{3}}}^{r}} \right)} \right]} \right)\\ - \exp \left({ - j\left\| k \right\|{d}\left[ \!{\sin\! \left({\theta_{{k_{1}}}^{r}} \right)\sin\! \left({\phi_{{k_{1}}}^{r}} \right) - \sin\! \left({\theta_{{k_{2}}}^{r}} \right)\sin\! \left({\phi_{{k_{2}}}^{r}} \right)} \right]} \right)\\ - \exp \left({ - j\left\| k \right\|{d}\left[ \!{\sin\! \left({\theta_{{k_{2}}}^{r}} \right)\sin\! \left({\phi_{{k_{2}}}^{r}} \right) - \sin\! \left({\theta_{{k_{3}}}^{r}} \right)\sin\! \left({\phi_{{k_{3}}}^{r}} \right)} \right]} \right) \end{array} \right] \end{aligned}} $$
(83)
which can be further simplified as follows:
$$ {\begin{aligned} &F\left({\theta_{{k_{1}}}^{r},\theta_{{k_{2}}}^{r},\theta_{{k_{3}}}^{r},\phi_{{k_{1}}}^{r},\phi_{{k_{2}}}^{r},\phi_{{k_{3}}}^{r}} \right) =\\ &\left[\! \begin{array}{l} 1\\ + \cos \left({ \,-\, \left\| k \right\|{d}\left[ \!{\sin\! \left({\theta_{{k_{1}}}^{r}} \!\right)\sin\! \left({\phi_{{k_{1}}}^{r}} \right) \,+\, \sin\! \left({\theta_{{k_{2}}}^{r}} \!\right)\sin \!\left({\phi_{{k_{2}}}^{r}}\! \right)\! -\! 2\sin \!\left({\theta_{{k_{3}}}^{r}} \right)\sin \left({\phi_{{k_{3}}}^{r}} \right)} \right]} \right)\\ + \cos \left({ - \left\| k \right\|{d}\left[ {2\sin\! \left({\theta_{{k_{1}}}^{r}} \right)\!\sin\! \left({\phi_{{k_{1}}}^{r}}\! \right)\! -\! \sin \left({\theta_{{k_{2}}}^{r}} \!\right)\!\sin \!\left({\phi_{{k_{2}}}^{r}} \right) \,-\, \sin\! \left({\theta_{{k_{3}}}^{r}}\! \right)\!\sin\! \left({\phi_{{k_{3}}}^{r}}\! \right)} \right]} \right)\\ - \cos \left({ - \left\| k \right\|{d}\left[ {2\sin\! \left({\theta_{{k_{1}}}^{r}} \right)\sin\! \left({\phi_{{k_{1}}}^{r}} \right) - 2\sin\! \left({\theta_{{k_{3}}}^{r}} \right)\sin\! \left({\phi_{{k_{3}}}^{r}} \right)} \right]} \right)\\ - \cos \left({ - \left\| k \right\|{d}\left[ {\sin\! \left({\theta_{{k_{1}}}^{r}} \right)\sin\! \left({\phi_{{k_{1}}}^{r}} \right) - \sin\! \left({\theta_{{k_{2}}}^{r}} \right)\sin\! \left({\phi_{{k_{2}}}^{r}} \right)} \right]} \right)\\ - \cos \left({ - \left\| k \right\|{d}\left[ {\sin\! \left({\theta_{{k_{2}}}^{r}} \right)\sin\! \left({\phi_{{k_{2}}}^{r}} \right) - \sin\! \left({\theta_{{k_{3}}}^{r}} \right)\sin\! \left({\phi_{{k_{3}}}^{r}} \right)} \right]} \right) \end{array}\! \right] \end{aligned}} $$
(84)
where we have used similar observations as for Nr=2.
Now, we need to compute \(\mathbb {E}_{{\theta ^{r}},{\phi ^{r}}}\left ({F\left ({{\theta ^{1,r}},{\theta ^{2,r}},{\theta ^{3,r}},{\phi ^{1,r}},{\phi ^{2,r}},{\phi ^{3,r}}} \right)} \right)\). Let us define the following:
$$\begin{array}{*{20}l} \begin{array}{l} a = \left\| k \right\|{d}\sin \left({\theta_{{k_{1}}}^{r}} \right)\sin \left({\phi_{{k_{1}}}^{r}} \right)\\ b = \left\| k \right\|{d}\sin \left({\theta_{{k_{2}}}^{r}} \right)\sin \left({\phi_{{k_{2}}}^{r}} \right)\\ c = \left\| k \right\|{d}\sin \left({\theta_{{k_{3}}}^{r}} \right)\sin \left({\phi_{{k_{3}}}^{r}} \right) \end{array} \end{array} $$
(85)
By applying the following trigonometric identities:
$$ {} \begin{aligned} &\cos \left({a + b - 2c} \right) = \cos \left(a \right)\cos \left(b \right)\cos \left({2c} \right)\\& \quad - \sin \left(a \right)\sin \left(b \right)\cos \left({2c} \right) + \sin \left(a \right)\cos \left(b \right)\sin \left({2c} \right)\\&\quad + \sin \left(b \right)\cos \left(a \right)\sin \left({2c} \right) \\ &\cos \left({c + b - 2a} \right) = \cos \left(c \right)\cos \left(b \right)\cos \left({2a} \right)\\&\quad - \sin \left(c \right)\sin \left(b \right)\cos \left({2a} \right) + \sin \left(c \right)\cos \left(b \right)\sin \left({2a} \right)\\ &\quad + \sin \left(b \right)\cos \left(c \right)\sin \left({2a} \right) \end{aligned} $$
(86)
and noting that a,b, and c are independent, we eventually obtain the desired result after some algebraic manipulations.