A Non-Orthogonal Selection Cooperation Protocol with Interference in Multi-Source Cooperative Networks

Abstract

This work proposes a non-orthogonal selection cooperation scheme with interference for multi-source and single destination cooperative networks. In our model, the source nodes can cooperate for one another, i.e., each source node plays the dual role of a source and a relay, thus there is no need for relays. In addition, the source nodes can be continuously transmitted and without dedicated timeslots for cooperative transmissions, which can save system resources and improve spectral efficiency. However, by this way, it will introduce interference with the non-orthogonal transmission mechanism. To overcome this problem, we use general reception scheme in source nodes and successive interference cancellation technology in the destination node, which can reduce the effect of interference effectively. For interference-limited networks, we also derive the theoretical upper bound and lower bound of outage probability of our method. Through the outage probability analysis and comparison, the results show that the spectral efficiency is improved while the system still keeps acceptable transmission reliability.

Appendices

Appendix 1: Proof of Lemma 1 and Its Two Corollaries

It is noted that throughout the derivation, if an exponential random variable with parameter $\lambda$, and its mean value is $1/\lambda$, then the probability density function (PDF)

$$f(x) = \left\{ \begin{array}{ll} \lambda {e^{ - \lambda x}} &\quad x > 0 \\ 0 &\quad x \le 0 \end{array}\right.$$

(37)

Lemma 1

Let X and Y be independent exponential random variables with parameters ${\lambda _1}$ and ${\lambda _2}$, respectively. Let $Z = {k_1}X - {k_2}Y$ $({k_1}> 0, {k_2} > 0)$, then the cumulative distribution function (CDF) of Z $(z > 0)$

$$\begin{aligned} {F_Z}(z) = 1 - {e^{ - \frac{{{\lambda _1}}}{{{k_1}}} \cdot z}}{\left( {1 + \frac{{{\lambda _1}}}{{{\lambda _2}}} \cdot \frac{{{k_2}}}{{{k_1}}}} \right) ^{ - 1}} \end{aligned}$$

(38)

Proof

Let $X' = {k_1}X$, $Y' = {k_2}Y$, then $X'$ and $Y'$ obey exponential distribution with parameters $\frac{{{\lambda _1}}}{{{k_1}}}$ and $\frac{{{\lambda _2}}}{{{k_2}}}$, respectively, and $Z = X' - Y'$. In order to solve ${F_Z}(z)$, we should solve the CDF of the difference between $X'$ and $Y'$, then we have

$$\begin{aligned} {F_Z}(z)& = \Pr \{ Z< z\} \nonumber \\& = \iint \limits _{X' - Y' < Z} {{f_{X'}}(x'){f_{Y'}}(y')dx'dy'} \nonumber \\& = \int _{ - \infty }^{ + \infty } {{f_{Y'}}(y'){F_{X'}}(z + y')dy'} \nonumber \\& = \int _{ - \infty }^{ + \infty } {\frac{{{\lambda _2}}}{{{k_2}}}{e^{ - \frac{{{\lambda _2}}}{{{k_2}}}y}} \cdot (1 - } {e^{ - \frac{{{\lambda _1}}}{{{k_1}}}(z + y)}})dy \end{aligned}$$

(39)

the non-zero range of the integral above

$$\begin{aligned} \left\{ \begin{array}{l} y> 0 \\ z + y> 0 \\ \end{array} \right. \Rightarrow \left\{ \begin{array}{l} y> 0 \\ y > - z \\ \end{array} \right. \end{aligned}$$

(40)

$\square$

Only considering $z > 0$, we have $y > 0$. Then

$$\begin{aligned} {F_Z}(z)= \int _0^{ + \infty } {\frac{{{\lambda _2}}}{{{k_2}}}{e^{ - \frac{{{\lambda _2}}}{{{k_2}}}y}} \cdot (1 - } {e^{ - \frac{{{\lambda _1}}}{{{k_1}}}(z + y)}})dy = 1 - {e^{ - \frac{{{\lambda _1}}}{{{k_1}}} \cdot z}}{\left( {1 + \frac{{{\lambda _1}}}{{{\lambda _2}}} \cdot \frac{{{k_2}}}{{{k_1}}}} \right) ^{ - 1}} \end{aligned}$$

(41)

Corollary 1

Let ${\left| {{\alpha _1}} \right| ^2}$ and ${\left| {{\alpha _2}} \right| ^2}$ be independent exponential random variables with parameters ${\lambda _1}$ and ${\lambda _2}$, respectively. And let $P> 0, R> 0, {N_0} > 0$, then

$$\begin{aligned} \Pr \left\{ {\log \left( {1 + \frac{{P{{\left| {{\alpha _1}} \right| }^2}}}{{P{{\left| {{\alpha _2}} \right| }^2} + {N_0}}}} \right) < R} \right\} = 1 - {e^{ - \frac{{g{N_0}}}{{{P_{}}}} \cdot {\lambda _1}}}\left( {1 + g \cdot \frac{{{\lambda _1}}}{{{\lambda _2}}}} \right) { ^{ - 1}} \end{aligned}$$

(42)

Proof

$$\begin{aligned} Pr \left\{ {\log \left( {1 + \frac{{P{{\left| {{\alpha _1}} \right| }^2}}}{{P{{\left| {{\alpha _2}} \right| }^2} + {N_0}}}} \right)< R} \right\}& = \Pr \left\{ {\frac{{P{{\left| {{\alpha _1}} \right| }^2}}}{{P{{\left| {{\alpha _2}} \right| }^2} + {N_0}}}< g} \right\} \nonumber \\& = \Pr \left\{ {P{{\left| {{\alpha _1}} \right| }^2} - g \cdot P{{\left| {{\alpha _2}} \right| }^2} < g \cdot {N_0}} \right\} \end{aligned}$$

(43)

Let ${k_1} = P$, ${k_2} = g \cdot P$, $Z = g \cdot {N_0}$, according to the result of Lemma 1, we have (42) and the proof is complete. $\square$

Corollary 2

Let ${\left| {{\alpha _1}} \right| ^2}$ and ${\left| {{\alpha _2}} \right| ^2}$ be independent exponential random variables with parameters ${\lambda _1}$ and ${\lambda _2}$, respectively. And let $P> 0, R> 0, {N_0} > 0$, then we can obtain

$$\begin{aligned}&\begin{aligned}&Pr \left\{ {\left( {{{\left| {{\alpha _1}} \right| }^2} \ge {{\left| {{\alpha _2}} \right| }^2}} \right) \cap \left( {\log \left( {1 + \frac{{P{{\left| {{\alpha _1}} \right| }^2}}}{{P{{\left| {{\alpha _2}} \right| }^2} + {N_0}}}} \right) < R} \right) } \right\} \\&= \left\{ \begin{array}{ll} {\left( {1 + \frac{{{\lambda _1}}}{{{\lambda _2}}}} \right) ^{ - 1}} - {e^{ - \frac{{g \cdot {N_0}}}{P} \cdot {\lambda _1}}} \cdot {\left( {1 + g \cdot \frac{{{\lambda _1}}}{{{\lambda _2}}}} \right) ^{ - 1}}, &{} g > 1\\ {\left( {1 + \frac{{{\lambda _1}}}{{{\lambda _2}}}} \right) ^{ - 1}} \cdot \left[ {1 - {e^{ - \left( {{\lambda _1} + {\lambda _2}} \right) \cdot \frac{{g \cdot {N_0}}}{{P(1 - g)}}}}} \right] - {e^{ - {\lambda _1} \cdot \frac{{g \cdot {N_0}}}{P}}} \cdot {\left( {1 + g \cdot \frac{{{\lambda _1}}}{{{\lambda _2}}}} \right) ^{ - 1}}&{}\\ \cdot \left[ {1 - {e^{ - \left( {{\lambda _2} + g \cdot {\lambda _1}} \right) \cdot \frac{{g \cdot {N_0}}}{{P(1 - g)}}}}} \right] , &{} g \le 1\\ \end{array} \right. \end{aligned} \end{aligned}$$

(44)

$$\begin{aligned}&\begin{aligned}&Pr \left\{ \begin{array}{l} \left\{ {{{\left( {{{\left| {{\alpha _1}} \right| }^2}< \left| {{\alpha _2}} \right| } \right) }^2} \cap \left( {\log \left( {1 + \frac{{P{{\left| {{\alpha _2}} \right| }^2}}}{{P{{\left| {{\alpha _1}} \right| }^2} + {N_0}}}} \right)< R} \right) } \right\} \\ \cap \left( {\log \left( {1 + \frac{{P{{\left| {{\alpha _1}} \right| }^2}}}{{P{{\left| {{\alpha _2}} \right| }^2} + {N_0}}}} \right) < R} \right) \\ \end{array} \right\} \\&= \left\{ \begin{array}{ll} {\left( {1 + \frac{{{\lambda _2}}}{{{\lambda _1}}}} \right) ^{ - 1}} - {e^{ - {\lambda _2} \cdot \frac{{g \cdot {N_0}}}{P}}} \cdot {\left( {1 + g \cdot \frac{{{\lambda _2}}}{{{\lambda _1}}}} \right) ^{ - 1}},&{} g > 1 \\ {\left( {1 + \frac{{{\lambda _2}}}{{{\lambda _1}}}} \right) ^{ - 1}}\left[ {1 - {e^{ - ({\lambda _1} + {\lambda _2}) \cdot \frac{{g \cdot {N_0}}}{{P \cdot (1 - g)}}}}} \right] \\ - {e^{ - {\lambda _2} \cdot \frac{{g \cdot {N_0}}}{P}}} \cdot {\left( {1 + g \cdot \frac{{{\lambda _2}}}{{{\lambda _1}}}} \right) ^{ - 1}} \cdot \left[ {1 - {e^{ - ({\lambda _1} + g \cdot {\lambda _2}) \cdot \frac{{g \cdot {N_0}}}{{P \cdot (1 - g)}}}}} \right] , &{} g \leqslant 1 \\ \end{array} \right. \\ \end{aligned} \end{aligned}$$

(45)

$$\begin{aligned}&\begin{aligned}&Pr \left\{ \begin{array}{l} \left\{ {{{\left( {{{\left| {{\alpha _1}} \right| }^2}< \left| {{\alpha _2}} \right| } \right) }^2} \cap \left( {\log \left( {1 + \frac{{P{{\left| {{\alpha _2}} \right| }^2}}}{{P{{\left| {{\alpha _1}} \right| }^2} + {N_0}}}} \right) > R} \right) } \right\} \\ \quad \cap \left( {\log \left( {1 + \frac{{P{{\left| {{\alpha _1}} \right| }^2}}}{{{N_0}}}} \right) < R} \right) \\ \end{array} \right\} \\&= {e^{ - {\lambda _2} \cdot \frac{{g \cdot {N_0}}}{P}}} \cdot {\left( {1 + g \cdot \frac{{{\lambda _2}}}{{{\lambda _1}}}} \right) ^{ - 1}} \cdot \left[ {1 - {e^{ - ({\lambda _1} + g \cdot {\lambda _2}) \cdot \frac{{g \cdot {N_0}}}{P}}}} \right] \\ \end{aligned} \end{aligned}$$

(46)

Proof

Let $X = {\left| {{\alpha _1}} \right| ^2}$, $Y = {\left| {{\alpha _2}} \right| ^2}$, ${p_1}$, ${p_2}$, ${p_3}$ represent the probability to be solved above, respectively, then we obtain

$$\begin{aligned} {p_1}& = \Pr \left\{ {\left( {X \geqslant Y} \right) \cap \left( {PX - g \cdot PY< g \cdot {N_0}} \right) } \right\} \nonumber \\& = \mathop{\mathop{\iint}\limits_{Px - g \cdot Py < g \cdot {N_0}}}\limits_{x \geqslant y} {{f_X}(x){f_Y}(y)dxdy} \end{aligned}$$

(47)

$$\begin{aligned} {p_2}& = \Pr \left\{ \begin{array}{l} \left( {X< Y} \right) \cap \left( {PY - g \cdot PX< g \cdot {N_0}} \right) \cap \left( {PX - g \cdot PY< g \cdot {N_0}} \right) \\ \end{array} \right\} \nonumber \\& = \mathop{\mathop{\mathop{\iint}\limits _{Px - g \cdot Py< g \cdot {N_0}}}\limits_{Py - g \cdot Px< g \cdot {N_0}}}\limits_{x < y} {{f_X}(x){f_Y}(y)dxdy} \end{aligned}$$

(48)

$$\begin{aligned} {p_3}& = \Pr \left\{ \begin{array}{l} \left( {X< Y} \right) \cap \left( {PY - g \cdot PX> g \cdot {N_0}} \right) \cap \left( {X< \frac{{g \cdot {N_0}}}{P}} \right) \\ \end{array} \right\} \nonumber \\&= \mathop{\mathop{\mathop{\iint}\limits _{Py - g \cdot Px > g \cdot {N_0}}}\limits_{x< y}}\limits_{x < \frac{{g \cdot {N_0}}}{P}} {{f_X}(x){f_Y}(y)dxdy}\end{aligned}$$

(49)

To solve the integrals above, we divide them into $g > 1$ and $g \leqslant 1$.

$g > 1$, we have

$$\begin{aligned}&\begin{aligned} {p_1}&= \int _0^{ + \infty } {{f_Y}(y)\left[ {{F_X}\left( {g \cdot y + \frac{{g \cdot {N_0}}}{P}} \right) - {F_X}\left( y \right) } \right] dy} \\&= \int _0^{ + \infty } {\left[ {{\lambda _2} \cdot {e^{ - \left( {{\lambda _1} + {\lambda _2}} \right) y}}{\text { - }}{\lambda _2} \cdot {e^{ - \frac{{g \cdot {N_0}}}{P} \cdot {\lambda _1}}} \cdot {e^{ - \left( {{\lambda _2} + g \cdot {\lambda _1}} \right) y}}} \right] } dy \\&= {\left( {1 + \frac{{{\lambda _1}}}{{{\lambda _2}}}} \right) ^{ - 1}} - {e^{ - \frac{{g \cdot {N_0}}}{P} \cdot {\lambda _1}}} \cdot {\left( {1 + g \cdot \frac{{{\lambda _1}}}{{{\lambda _2}}}} \right) ^{ - 1}} \\ \end{aligned} \end{aligned}$$

(50)

$$\begin{aligned}&\begin{aligned} {p_2}&= \int _0^{ + \infty } {{f_X}(x)\left[ {{F_Y}\left( {g \cdot x + \frac{{g \cdot {N_0}}}{P}} \right) - {F_Y}\left( x \right) } \right] dx} \\&= \int _0^{ + \infty } {\left[ {{\lambda _1} \cdot {e^{ - \left( {{\lambda _1} + {\lambda _2}} \right) x}} - {\lambda _1} \cdot {e^{ - {\lambda _2} \cdot \frac{{g \cdot {N_0}}}{P}}} \cdot {e^{ - \left( {{\lambda _1} + g \cdot {\lambda _2}} \right) \cdot x}}} \right] } dx\\&= {\left( {1 + \frac{{{\lambda _2}}}{{{\lambda _1}}}} \right) ^{ - 1}} - {e^{ - {\lambda _2} \cdot \frac{{g \cdot {N_0}}}{P}}} \cdot {\left( {1 + g \cdot \frac{{{\lambda _2}}}{{{\lambda _1}}}} \right) ^{ - 1}} \\ \end{aligned} \end{aligned}$$

(51)

For ${p_3}$, the integration area of y is $(x + \frac{{g \cdot {N_0}}}{P}, + \infty )$, and the integration area of x is $(0, \frac{{g \cdot {N_0}}}{P})$. Thus, we get

$$\begin{aligned} \begin{aligned} {p_3}&= \int _0^{\frac{{g \cdot {N_0}}}{P}} {{f_X}(x)\int _{g \cdot x + \frac{{g \cdot {N_0}}}{P}}^{ + \infty } {{f_Y}(y)} dxdy} \\&= \int _0^{\frac{{g \cdot {N_0}}}{P}} {{\lambda _1} \cdot {e^{ - {\lambda _1}x}} \cdot {e^{ - {\lambda _2} \cdot \left( {g \cdot x + \frac{{g \cdot {N_0}}}{P}} \right) }}dx} \\&= {e^{ - {\lambda _2} \cdot \frac{{g \cdot {N_0}}}{P}}} \cdot {\left( {1 + g \cdot \frac{{{\lambda _2}}}{{{\lambda _1}}}} \right) ^{ - 1}} \cdot \left[ {1 - {e^{ - ({\lambda _1} + g \cdot {\lambda _2}) \cdot \frac{{g \cdot {N_0}}}{P}}}} \right] \\ \end{aligned} \end{aligned}$$

(52)

$g \leqslant 1$, we have

$$\begin{aligned}&\begin{aligned} {p_1}&= \int _0^{\frac{{g \cdot {N_0}}}{{P(1 - g)}}} {{f_Y}(y)\left[ {{F_X}\left( {g \cdot y + \frac{{g \cdot {N_0}}}{P}} \right) - {F_X}\left( y \right) } \right] dy} \\&= {\left( {1 + \frac{{{\lambda _1}}}{{{\lambda _2}}}} \right) ^{ - 1}} \cdot \left[ {1 - {e^{ - \left( {{\lambda _1} + {\lambda _2}} \right) \cdot \frac{{g \cdot {N_0}}}{{P(1 - g)}}}}} \right] - {e^{ - {\lambda _1} \cdot \frac{{g \cdot {N_0}}}{P}}} \\&\;\;\;\;\cdot {\left( {1 + g \cdot \frac{{{\lambda _1}}}{{{\lambda _2}}}} \right) ^{ - 1}} \cdot \left[ {1 - {e^{ - \left( {{\lambda _2} + g \cdot {\lambda _1}} \right) \cdot \frac{{g \cdot {N_0}}}{{P(1 - g)}}}}} \right] \\ \end{aligned} \end{aligned}$$

(53)

$$\begin{aligned}&\begin{aligned} {p_2} =&{\left( {1 + \frac{{{\lambda _2}}}{{{\lambda _1}}}} \right) ^{ - 1}}\left[ {1 - {e^{ - ({\lambda _1} + {\lambda _2}) \cdot \frac{{g \cdot {N_0}}}{{P \cdot (1 - g)}}}}} \right] - {e^{ - {\lambda _2} \cdot \frac{{g \cdot {N_0}}}{P}}} \\&\cdot {\left( {1 + g \cdot \frac{{{\lambda _2}}}{{{\lambda _1}}}} \right) ^{ - 1}} \cdot \left[ {1 - {e^{ - ({\lambda _1} + g \cdot {\lambda _2}) \cdot \frac{{g \cdot {N_0}}}{{P \cdot (1 - g)}}}}} \right] \\ \end{aligned} \end{aligned}$$

(54)

For ${p_3}$, the integration area of y is $(g \cdot x + \frac{{g \cdot {N_0}}}{P}, + \infty )$, and the integration area of x is $(0, \frac{{g \cdot {N_0}}}{P})$. Although the integration area has been changed, the integration interval is the same, and the result is also same with $g > 1$. $\square$

Combining (50) and (53), we get (44). And combining (51) and (54), we get (45). According to (52) and the integration interval is consistent in both cases, we get (46). Then we obtain the desired results.

Appendix 2: Proof of Lemma 2

Lemma 2

Let ${\left| {{\alpha _1}} \right| ^2},{\left| {{\alpha _2}} \right| ^2}, \ldots ,{\left| {{\alpha _N}} \right| ^2}$ be independent exponential random variables with parameters ${\lambda _1},{\lambda _2}, \ldots ,{\lambda _N}$, respectively. Let ${\left| {{\alpha _s}} \right| ^2}$ be an independent exponential random variable with parameter ${\lambda _s}$ and ${\left| {{\alpha _b}} \right| ^2} = \max \{ {\left| {{\alpha _1}} \right| ^2},{\left| {{\alpha _2}} \right| ^2}, \ldots ,{\left| {{\alpha _N}} \right| ^2}\}$, $P> 0, R> 0, {N_0} > 0$, then we can obtain

$$\begin{aligned} \begin{aligned}&Pr \left\{ {\left( {{{\left| {{\alpha _b}} \right| }^2} \geqslant {{\left| {{\alpha _s}} \right| }^2}} \right) \cap \left( {\log \left( {1 + \frac{{P{{\left| {{\alpha _b}} \right| }^2}}}{{P{{\left| {{\alpha _s}} \right| }^2} + {N_0}}}} \right) < R} \right) } \right\} \\&= \left\{ \begin{array}{ll} \sum \limits _{m = 1}^N {\mathop{\mathop{\sum}\limits_{{S_m} \subseteq \{ 1,2, \ldots ,N\}}}\limits_ {\left| {{S_m}} \right| = m} {{{( - 1)}^m}} \cdot \left[ \begin{array}{l} {e^{ - \frac{{g \cdot {N_0}}}{P} \cdot \sum \nolimits _{j \in {S_m}} {{\lambda _j}} }} \cdot {\left( {1 + g \cdot \frac{{\sum \nolimits _{j \in {S_m}} {{\lambda _j}} }}{{{\lambda _s}}}} \right) ^{ - 1}} \\ - {\left( {1 + \frac{{\sum \nolimits _{j \in {S_m}} {{\lambda _j}} }}{{{\lambda _s}}}} \right) ^{ - 1}} \\ \end{array} \right] } &{} g > 1 \\ \sum \limits _{m = 1}^N {\mathop{\mathop{\sum}\limits_{{S_m} \subseteq \{ 1,2, \ldots ,N\}}}\limits_ {\left| {{S_m}} \right| = m} {{{( - 1)}^m} \cdot \left\{ {{e^{ - \frac{{g \cdot {N_0}}}{P} \cdot \sum \nolimits _{j \in {S_m}} {{\lambda _j}} }} \cdot {{\left( {1 + g \cdot \frac{{\sum \nolimits _{j \in {S_m}} {{\lambda _j}} }}{{{\lambda _s}}}} \right) }^{ - 1}} } \right. } } \\ \cdot \left[ {1 - {e^{ - \left( {{\lambda _s} + g \cdot \sum \nolimits _{j \in {S_m}} {{\lambda _j}} } \right) \cdot \frac{{g \cdot {N_0}}}{{P(1 - g)}}}}} \right] &{} g \leqslant 1 \\ \left. { - {{\left( {1 + \frac{{\sum \nolimits _{j \in {S_m}} {{\lambda _j}} }}{{{\lambda _s}}}} \right) }^{ - 1}} \cdot \left[ {1 - {e^{ - \left( {{\lambda _s} + \sum \nolimits _{j \in {S_m}} {{\lambda _j}} } \right) \cdot \frac{{g \cdot {N_0}}}{{P(1 - g)}}}}} \right] } \right\} \\ \end{array}\right. \\ \end{aligned} \end{aligned}$$

(55)

Proof

Set $X = {\left| {{\alpha _b}} \right| ^2}$, $Y = {\left| {{\alpha _s}} \right| ^2}$, p represents the probability to be solved, then we have

$$\begin{aligned} p& = \Pr \left\{ {\left( {X \geqslant Y} \right) \cap \left( {PX - g \cdot PY< g \cdot {N_0}} \right) } \right\} \nonumber \\& = \mathop{\mathop{\iint} \limits_{Px - g \cdot Py < g \cdot {N_0}}}\limits_{x \geqslant y} {{f_X}(x){f_Y}(y)dxdy} \end{aligned}$$

(56)

where

$$\begin{aligned} {f_Y}(y)& = {\lambda _s}{e^{ - {\lambda _s}y}} \end{aligned}$$

(57)

$$\begin{aligned} {F_X}(x)& = \Pr \{ X< x\} \nonumber \\& = \Pr \left\{ {{X_1}< x,{X_2}< x, \ldots ,{X_N} < x} \right\} \nonumber \\& = \prod \limits _{i = 1}^N {(1 - {e^{ - {\lambda _i}x}})} = 1 \nonumber \\& \quad + \sum \limits _{m = 1}^N {\mathop{\mathop{\sum}\limits_{{S_m} \subseteq \{ 1,2, \ldots ,N\}}}\limits_ {\left| {{S_m}} \right| = m} {{{( - 1)}^m}} } \cdot {e^{ - x \cdot \sum \nolimits _{j \in {S_m}} {{\lambda _j}} }} \end{aligned}$$

(58)

For the last step in (58), in order to expand ${F_X}(x)$, we use the following formula

$$\begin{aligned} \prod \limits _{i = 1}^K {(1 - {a_i}) = 1 + \sum \limits _{l = 1}^K {\mathop{\mathop{\sum}\limits_{{S_l} \subseteq \{ 1,2, \ldots ,K\}}}\limits_ {\left| {{S_l}} \right| = l} {{{( - 1)}^l}\prod \limits _{j \in {S_l}} {{a_j}} } } } \end{aligned}$$

(59)

Next, we divide (56) into $g > 1$ and $g \leqslant 1$. $\square$

$g > 1$, the integration area of x is $(y, g \cdot y + \frac{{g \cdot {N_0}}}{P})$, and the integration area of y is $(0, + \infty )$. Then we have

$$\begin{aligned} \begin{aligned} p&= \int _0^{ + \infty } {{f_Y}(y)\left[ {{F_X}\left( {g \cdot y + \frac{{g \cdot {N_0}}}{P}} \right) - {F_X}\left( y \right) } \right] dy} \\&= \int _0^{ + \infty } {\left\{ {{\lambda _s}{e^{ - {\lambda _s}y}} \cdot \sum \limits _{m = 1}^N {\mathop{\mathop{\sum}\limits_{{S_m} \subseteq \{ 1,2, \ldots ,N\}}}\limits_ {\left| {{S_m}} \right| = m} {{{( - 1)}^m} \cdot \left[ \begin{array}{l} {e^{ - \left( {g \cdot y + \frac{{g \cdot {N_0}}}{P}} \right) \cdot \sum \nolimits _{j \in {S_m}} {{\lambda _j}} }} \\ - {e^{ - y\sum \nolimits _{j \in {S_m}} {{\lambda _j}} }} \\ \end{array} \right] } } } \right\} } dy \\&= \sum \limits _{m = 1}^N {\mathop{\mathop{\sum}\limits_{{S_m} \subseteq \{ 1,2, \ldots ,N\}}}\limits_ {\left| {{S_m}} \right| = m} {{{( - 1)}^m} \cdot \left[ \begin{array}{l} {e^{ - \frac{{g \cdot {N_0}}}{P} \cdot \sum \nolimits _{j \in {S_m}} {{\lambda _j}} }} \cdot {\left( {1 + g \cdot \frac{{\sum \nolimits _{j \in {S_m}} {{\lambda _j}} }}{{{\lambda _s}}}} \right) ^{ - 1}} \\ - {\left( {1 + \frac{{\sum \nolimits _{j \in {S_m}} {{\lambda _j}} }}{{{\lambda _s}}}} \right) ^{ - 1}} \\ \end{array} \right] } } \\ \end{aligned} \end{aligned}$$

(60)

$g \leqslant 1$, the integration area of x is $(y, g \cdot y + \frac{{g \cdot {N_0}}}{P})$, and the integration area of y is $(0, \frac{{g \cdot {N_0}}}{{P(1 - g)}})$. Thus, we just resolve the integral of y in (60), then we obtain

$$\begin{aligned} \begin{aligned} p&= \sum \limits _{m = 1}^N {\mathop{\mathop{\sum}\limits_{{S_m} \subseteq \{ 1,2, \ldots ,N\}}}\limits_ {\left| {{S_m}} \right| = m}} {{{( - 1)}^m} \cdot \left\{ {{e^{ - \frac{{g \cdot {N_0}}}{P} \cdot \sum \nolimits _{j \in {S_m}} {{\lambda _j}} }} \cdot {{\left( {1 + g \cdot \frac{{\sum \nolimits _{j \in {S_m}} {{\lambda _j}} }}{{{\lambda _s}}}} \right) }^{ - 1}}} \right. } \\&\quad \cdot \left[ {1 - {e^{ - \left( {{\lambda _s} + g \cdot \sum \nolimits _{j \in {S_m}} {{\lambda _j}} } \right) \cdot \frac{{g \cdot {N_0}}}{{P(1 - g)}}}}} \right] \\&\quad \left. { - {{\left( {1 + \frac{{\sum \nolimits _{j \in {S_m}} {{\lambda _j}} }}{{{\lambda _s}}}} \right) }^{ - 1}} \cdot \left[ {1 - {e^{ - \left( {{\lambda _s} + \sum \nolimits _{j \in {S_m}} {{\lambda _j}} } \right) \cdot \frac{{g \cdot {N_0}}}{{P(1 - g)}}}}} \right] } \right\} \\ \end{aligned} \end{aligned}$$

(61)

Combining (60) and (61) completes the proof.

Appendix 3: Proof of Lemma 3

Lemma 3

Let ${\left| {{\alpha _1}} \right| ^2},{\left| {{\alpha _2}} \right| ^2}, \ldots ,{\left| {{\alpha _N}} \right| ^2}$ be independent exponential random variables with parameters ${\lambda _1},{\lambda _2}, \ldots ,{\lambda _N}$, respectively. Let ${\left| {{\alpha _s}} \right| ^2}$ be an independent exponential random variable with parameter ${\lambda _s}$ and ${\left| {{\alpha _b}} \right| ^2} = \max \{ {\left| {{\alpha _1}} \right| ^2},{\left| {{\alpha _2}} \right| ^2}, \ldots ,{\left| {{\alpha _N}} \right| ^2}\}$, $P> 0, R> 0, {N_0} > 0$, then we can obtain

$$\begin{aligned} \begin{aligned}&Pr \left\{ \begin{array}{l} \left( {{{\left| {{\alpha _b}} \right| }^2}< {{\left| {{\alpha _s}} \right| }^2}} \right) \cap \left( {\log \left( {1 + \frac{{P{{\left| {{\alpha _s}} \right| }^2}}}{{P{{\left| {{\alpha _b}} \right| }^2} + {N_0}}}} \right)< R} \right) \\ \cap \left( {\log \left( {1 + \frac{{P{{\left| {{\alpha _b}} \right| }^2}}}{{P{{\left| {{\alpha _s}} \right| }^2} + {N_0}}}} \right) < R} \right) \\ \end{array}\right\} \\&= \left\{ \begin{array}{l} \sum \limits _{m = 1}^N {\mathop{\mathop{\sum}\limits_{{S_m} \subseteq \{ 1,2, \ldots ,N\}}}\limits_ {\left| {{S_m}} \right| = m} {{{( - 1)}^m} \cdot \left[ \begin{array}{l} {e^{ - {\lambda _s} \cdot \frac{{g \cdot {N_0}}}{P}}} \cdot {\left( {1 + g \cdot \frac{{{\lambda _s}}}{t}} \right) ^{ - 1}} \\ - {\left( {1 + \frac{{{\lambda _s}}}{t}} \right) ^{ - 1}} \\ \end{array} \right] } }, g > 1 \\ \sum \limits _{m = 1}^N {\mathop{\mathop{\sum}\limits_{{S_m} \subseteq \{ 1,2, \ldots ,N\}}}\limits_ {\left| {{S_m}} \right| = m} {{{( - 1)}^m} \cdot \left\{ {{e^{ - {\lambda _s} \cdot \frac{{g \cdot {N_0}}}{P}}} \cdot {{\left( {1 + g \cdot \frac{{{\lambda _s}}}{t}} \right) }^{ - 1}} \;\;\;\;\;\;\; g \leqslant 1 } \right. } } \\ \left. { \cdot \left[ {1 - {e^{ - \left( {g \cdot {\lambda _s} + t} \right) \cdot \frac{{g \cdot {N_0}}}{{P(1 - g)}}}}} \right] - {{\left( {1 + \frac{{{\lambda _s}}}{t}} \right) }^{ - 1}} \cdot \left[ {1 - {e^{ - \left( {{\lambda _s} + t} \right) \cdot \frac{{g \cdot {N_0}}}{{P(1 - g)}}}}} \right] } \right\} \\ \end{array} \right. \\ \end{aligned} \end{aligned}$$

(62)

where

$$\begin{aligned} t = \sum \limits _{j \in {S_m}} {{\lambda _j}} \end{aligned}$$

(63)

Proof

Let $X = {\left| {{\alpha _b}} \right| ^2}$, $Y = {\left| {{\alpha _s}} \right| ^2}$, p represents the probability to be solved, then we have

$$\begin{aligned} \begin{aligned} p&= \Pr \left\{ \begin{array}{l} \left( {X< Y} \right) \cap \left( {PY - g \cdot PX< g \cdot {N_0}} \right) \cap \left( {PX - g \cdot PY< g \cdot {N_0}} \right) \\ \end{array} \right\} \\&= \mathop{\mathop{\mathop{\iint}\limits_{Px - g \cdot Py< g \cdot {N_0}}}\limits_{Py - g \cdot Px< g \cdot {N_0}}}\limits_{x < y} {{f_X}(x){f_Y}(y)dxdy} \\ \end{aligned} \end{aligned}$$

(64)

where

$$\begin{aligned} {F_Y}(y)& = 1 - {e^{ - {\lambda _s}y}} \end{aligned}$$

(65)

$$\begin{aligned} {f_X}(x)& = \sum \limits _{m = 1}^N {\mathop{\mathop{\sum}\limits_{{S_m} \subseteq \{ 1,2, \ldots ,N\}}}\limits_ {\left| {{S_m}} \right| = m} {{{( - 1)}^{m + 1}} \cdot t} } \cdot {e^{ - x \cdot t}} \end{aligned}$$

(66)

Next, we divide (64) into $g > 1$ and $g \leqslant 1$.

$g > 1$ the integration area is shown in Fig. 11a. We can see from Fig. 11a, the integration area of x is $(0, + \infty )$, and the integration area of y is $(x, g \cdot x + \frac{{g \cdot {N_0}}}{P})$. Then we have

$$\begin{aligned} \begin{aligned} p&= \int _0^{ + \infty } {{f_X}(x)\int _x^{g \cdot x + \frac{{g \cdot {N_0}}}{P}} {{f_Y}(y)} dxdy} \\&= \int _0^{ + \infty } {{f_X}(x)\left[ {{F_Y}\left( {g \cdot x + \frac{{g \cdot {N_0}}}{P}} \right) - {F_Y}\left( x \right) } \right] dx} \\&= \sum \limits _{m = 1}^N {\mathop{\mathop{\sum}\limits_{{S_m} \subseteq \{ 1,2, \ldots ,N\}}}\limits_ {\left| {{S_m}} \right| = m} {{{( - 1)}^m} \cdot \left[ {{e^{ - {\lambda _s} \cdot \frac{{g \cdot {N_0}}}{P}}} \cdot {{\left( {1 + g \cdot \frac{{{\lambda _s}}}{t}} \right) }^{ - 1}} - {{\left( {1 + \frac{{{\lambda _s}}}{t}} \right) }^{ - 1}}} \right] } } \\ \end{aligned} \end{aligned}$$

(67)

$g \leqslant 1$, the integration area is shown in Fig. 11b. We an see from Fig. 11b, the integration area of y is $(x, g \cdot x + \frac{{g \cdot {N_0}}}{P})$, and the integration area of x is $(0, \frac{{g \cdot {N_0}}}{{P(1 - g)}})$. Thus, we just resolve the integral of x in (67), then we obtain

$$\begin{aligned} \begin{aligned} p&= \int _0^{\frac{{g \cdot {N_0}}}{{P(1 - g)}}} {{f_X}(x)\int _x^{g \cdot x + \frac{{g \cdot {N_0}}}{P}} {{f_Y}(y)} dxdy} \\&= \int _0^{\frac{{g \cdot {N_0}}}{{P(1 - g)}}} {{f_X}(x)\left[ {{F_Y}\left( {g \cdot x + \frac{{g \cdot {N_0}}}{P}} \right) - {F_Y}\left( x \right) } \right] dx} \\&= \sum \limits _{m = 1}^N {\mathop{\mathop{\sum}\limits_{{S_m} \subseteq \{ 1,2, \ldots ,N\}}}\limits_ {\left| {{S_m}} \right| = m} {{{( - 1)}^m} \cdot \left\{ {{e^{ - {\lambda _s} \cdot \frac{{g \cdot {N_0}}}{P}}} \cdot {{\left( {1 + g \cdot \frac{{{\lambda _s}}}{t}} \right) }^{ - 1}}} \right. } } \\&\quad\left. { \cdot \left[ {1 - {e^{ - \left( {g \cdot {\lambda _s} + t} \right) \cdot \frac{{g \cdot {N_0}}}{{P(1 - g)}}}}} \right] - {{\left( {1 + \frac{{{\lambda _s}}}{t}} \right) }^{ - 1}} \cdot \left[ {1 - {e^{ - \left( {{\lambda _s} + t} \right) \cdot \frac{{g \cdot {N_0}}}{{P(1 - g)}}}}} \right] } \right\} \end{aligned} \end{aligned}$$

(68)

$\square$

Combining (67) and (68) completes the proof.