Secret DMT Analysis for a Multiuser MIMO Fading Channel

Abstract

In wireless communications, eavesdropping is a threat due to the broadcast nature of the channel. Therefore, in addition to the rate and the error probability, the secrecy level should be considered in design and analysis of a system. In this study, by the error probability and the channel outage analysis, we derive the secret diversity-multiplexing tradeoff (DMT) for a multiuser multiple-input multiple-output (MIMO) fading channel. We consider two multiuser schemes, multiple-access channel and user-selection. Secret DMT characterizes the relation of the error probability, the secrecy and the rate of the system at high signal-to-noise ratio (SNR). We quantify how the number of users and antennas at all the nodes affects the secret DMT performance of the system.

Appendices

Appendix A: Proof of Lemma 1

As the error probability is lower bounded by the secrecy outage, it is sufficient to derive the lower bound of the secrecy outage probability at high SNR and get rid of the complexity of exact analysis. First, by the relation of eigenvalues and mutual information of MIMO systems [27], (21) can be written as

$$\begin{aligned} \Pr \left( {\mathcal {O}}^{{\mathcal {S}}}_{s} \right) \doteq \Pr \left( \frac{\Pi _{i=1}^{M}(1+{{\lambda }_{i}}\text {SNR})}{\Pi _{i=1}^{L}(1+{{\mu }_{1,i}}\text {SNR})\ldots \Pi _{i=1}^{L}(1+{{\mu }_{\left| {\mathcal {S}} \right| ,i}}\text {SNR})} <{\text {SNR}^{\left| {\mathcal {S}} \right| {{r}_{s}}}} \right) , \end{aligned}$$

(A1)

where $M=\min {\{\left| {\mathcal {S}} \right| n_t,n_r\}}$, $L=\min {\{n_t,n_e\}}$, $0\le \lambda _1 \le \cdots \le \lambda _M$ and $0\le \mu _{k,1} \le \cdots \le \mu _{k,L}, k=1,\ldots ,\left| {\mathcal {S}} \right|$ are the ordered eigenvalues of ${{\textbf{H}}}_{{\mathcal {S}}}{\textbf{H}}_{{{\mathcal {S}}}}^{\dag }$ and ${{\textbf{G}}}_{k} {\textbf{G}}_{k}^{\dag }$, respectively. We can simplify the denominator at high SNR. First, define the event $E=\left\{ \mu _{k,i}>c| k=1,\ldots ,\left| {\mathcal {S}} \right| , i=1,\ldots ,L \right\}$, where $c>1$ is a constant parameter and the complement of E is denoted by ${\bar{E}}$. We can write

$$\begin{aligned} \Pr \left( {\mathcal {O}}^{{\mathcal {S}}}_{s} \right)&= \Pr ({\mathcal {O}}^{{\mathcal {S}}}_{s}|{ E})\Pr ({E })+\Pr ({\mathcal {O}}^{{\mathcal {S}}}_{s}|{\bar{E}})\Pr ({\bar{E}}) \\&\ge \Pr ({\mathcal {O}}^{{\mathcal {S}}}_{s}|{ E)}\Pr ({E )} \\&\,{\dot{\ge }} \Pr ({\mathcal {O}}^{{\mathcal {S}}}_{s}|{ E)}. \end{aligned}$$

(A2)

The last term is because $\Pr ({E )}$ is constant with respect to SNR at high SNR regime analysis. Then, we can write

$$\begin{aligned} \Pr ({\mathcal {O}}^{{\mathcal {S}}}_{s}|{ E)}&\,{{{\dot{\ge }} }}\Pr \left( \frac{\Pi _{i=1}^{M}(1+{{\lambda }_{i}}\text {SNR})}{{{\left( 1+c\,\text {SNR} \right) }^{L}}\ldots {{\left( 1+c\,\text {SNR} \right) }^{L}}}<{{\text {SNR}}^{\left| {\mathcal {S}} \right| {{r}_{s}}}} \right) \end{aligned}$$

(A3a)

$$\begin{aligned}&\,{{{\dot{\ge }} }}\,\Pr \left( \frac{\Pi _{i=1}^{M}(1+{{\lambda }_{i}}\text {SNR})}{{{\text {SNR}}^{L}}\ldots {{\text {SNR}}^{^{L}}}}<{{\text {SNR}}^{\left| {\mathcal {S}} \right| {{r}_{s}}}} \right) \end{aligned}$$

(A3b)

$$\begin{aligned}&\,{{{\dot{\ge }} }}\,\Pr \left( \Pi _{i=1}^{M}(1+{{\lambda }_{i}}\text {SNR})<{{\text {SNR}}^{\left| {\mathcal {S}} \right| ({{r}_{s}}+L)}} \right) \end{aligned}$$

(A3c)

$$\begin{aligned}&\,\,{{\doteq }}\text {SNR}^ {-d_{\left| {\mathcal {S}} \right| n_t,{{n}_{r}}}(\left| {\mathcal {S}} \right| ({{r}_{s}}+L))} \end{aligned}$$

(A3d)

$$\begin{aligned}&\,{{\doteq }}\left\{ \begin{array}{ll} {\text {SNR}^{{-d_{\left| {\mathcal {S}} \right| (n_t-n_e),{{n}_{r}-\left| {\mathcal {S}} \right| n_e}}(\left| {\mathcal {S}} \right| {{r}_{s}})}}},\,\,n_e < \min (n_t,n_r),\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,\,\, n_r > Kn_e, \\ \,\,\,\,\,\,\,\,\,\,\\ 1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text {O.W.},\\ \end{array} \right. \end{aligned}$$

(A3e)

where the term (A3a) has been derived by substituting the minimum of all $\mu _{k,i}$s and (A3b) follows because $(1+c\,\text {SNR})>\text {SNR}$. Expression (A3c) is similar to the outage probability of a MAC system but with the difference that the multiplexing gain has been shifted by the value L which demonstrates the effect of the eavesdropper. Hence, we can apply the analysis of the MIMO system without secrecy [27] to (A3c) which leads to (A3d) and (A3e). According to (16), we can finally conclude

$$\begin{aligned} \Pr (\text {secrecy outage)}&{{{\dot{\ge }} }}\left\{ \begin{array}{ll} {\text {SNR}^{\underset{{\mathcal {S}}}{\mathop {-\min }}{\,d_{\left| {\mathcal {S}} \right| (n_t-n_e),{{n}_{r}-\left| {\mathcal {S}} \right| n_e}}(\left| {\mathcal {S}} \right| {{r}_{s}})}}},\,\,n_e < \min (n_t,n_r),\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, ,\, n_r > Kn_e, \\ \,\,\,\,\,\,\,\,\,\,\\ 1, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text {O.W.}.\\ \end{array} \right. \end{aligned}$$

(A4)

Thus, Lemma 1 is proved.

Appendix B: Proof of Lemma 2

According to the individual secrecy criterion that we discussed in Sect. 2, even if one user can not achieve secrecy, the system is not secure. Therefore, the probability that secrecy is not achieved can be derived as:

$$\begin{aligned} \Pr (\text {secrecy not achieved}) \triangleq \Pr \left( \underset{N \rightarrow +\infty }{\mathop {\lim }}\, \bigcup \limits _{k\in {\mathcal {K}}} \frac{1}{N}H\left( {{W}_{k}}|{{X}_{{{{\mathcal {K}}_{k}}}}},{{Y}_{e}} \right) <{{R}_s} \right) , \end{aligned}$$

(B5)

where N is the number of channel uses and long enough. To analyze this probability, due to union bound, we can write

$$\begin{aligned} \Pr (\text {secrecy not achieved}) \le \Pr \left( \underset{N \rightarrow +\infty }{\mathop {\lim }}\,\frac{1}{N}\sum \limits _{k=1}^{K}H\left( {{W}_{k}}|{{X}_{{{{\mathcal {K}}_{k}}}}},{{Y}_{e}} \right) <{{R}_s} \right) . \end{aligned}$$

(B6)

On the other hand, we have

$$\begin{aligned} H\left( {{W}_{k}}|{{X}_{{{{\mathcal {K}}_{k}}}}},{{Y}_{e}} \right)&=H\left( {{W}_{k}}|{{X}_{{{{\mathcal {K}}_{k}}}}} \right) -I\left( {{W}_{k}};{{Y}_{e}}|{{X}_{{{{\mathcal {K}}_{k}}}}} \right) \\&\ge H\left( {{W}_{k}} \right) -I\left( {{X}_{k}};{{Y}_{e}}|{{X}_{{{{\mathcal {K}}_{k}}}}} \right) , \end{aligned}$$

(B7)

where $H\left( {{W}_{k}}|{{X}_{{{{\mathcal {K}}_{k}}}}} \right) = H\left( {{W}_{k}} \right)$ is due to the independency of $W_k$ and ${X}_{{{{\mathcal {K}}_{k}}}}$ and $I\left( {{X}_{k}};{{Y}_{e}}|{{X}_{{{{\mathcal {K}}_{k}}}}} \right) \ge I\left( {{W}_{k}};{{Y}_{e}}|{{X}_{{{{\mathcal {K}}_{k}}}}} \right)$ is due to the data processing inequality [32]. Hence, we can express

$$\begin{aligned}&\Pr \left( \underset{N \rightarrow +\infty }{\mathop {\lim }}\frac{1}{N}H\left( {{W}_{k}}|{{X}_{{{{\mathcal {K}}_{k}}}}},{{Y}_{e}} \right)<{{R}_s} \right) \\&\quad \le \Pr \left( \underset{N \rightarrow +\infty }{\mathop {\lim }} \frac{1}{N}\left( H\left( {{W}_{k}} \right) -I\left( {{X}_{k}};{{Y}_{e}}|{{X}_{{{{\mathcal {K}}_{k}}}}} \right) \right)<{{R}_s}\right) \\&\quad \le \Pr \left( R-{{C}_{e}}<R-\min ({{n}_{t}},{{n}_{e}})\log \text {SNR} \right) \\&\quad \le \Pr \left( {{C}_{e}}>\min ({{n}_{t}},{{n}_{e}})\log \text {SNR} \right) \\&\quad \doteq 0 , \end{aligned}$$

(B8)

where $R= \frac{1}{N} H\left( {{W}_{k}} \right)$ is the rate of each user, $C_e=\frac{1}{N}I\left( {{X}_{k}};{{Y}_{e}}|{{X}_{{{{{\mathcal {K}}_{k}}}}}} \right)$, the capacity of the Eve’s channel follows [22] and $R_{s}=R-\min ({{n}_{t}},{{n}_{e}})\log \text {SNR}$ follows Sect. 3.2. The final result is due to [33], as the maximum degrees of freedom between the source and the eavesdropper is $\min ({{n}_{t}},{{n}_{e}})$. As, the result of (B8) is true $\forall k\in {\mathcal {K}}$, due to (B6), $\Pr (\text {secrecy not achieved})=0$ and therefore, Lemma 2 is proved.

Appendix C: Proof of Lemma 3

To analyze the lower bound of the secrecy outage probability of user-selection, taking the same method as “Appendix A”, ${\log \left( \det \left( {{{\textbf{I}}}_{{n_e}}}+\text {SNR}{{\textbf{G}}_{k^{*}}}{{{\textbf{G}}}_{k^{*}}^{\dag }} \right) \right) }$ can be written in terms of the ordered eigenvalues of ${{\textbf{G}}_{k^{*}}}{{{\textbf{G}}}_{k^{*}}^{\dag }}$, i.e., $0\le \mu _{k^{*},1} \le \cdots \le \mu _{k^{*},L}$ and hence, (36) can be expressed as

$$\begin{aligned} \Pr \left( {\mathcal {O}}^{k^*}_s \right) \doteq \Pr \left( \frac{\det \left( {{{\textbf{I}}}_{{{n_r}}}}+\text {SNR}{{\textbf{H}}_{k^{*}}}{{{\textbf{H}}}_{k^{*}}^{\dag }} \right) }{\Pi _{i=1}^{{L}}(1+{{\mu }_{k^*,i}}\text {SNR}))}<{\text {SNR}}^{Kr_s} \right) . \end{aligned}$$

(C9)

As in “Appendix A”, we can simplify the denominator at high SNR, i.e., $(1+{{\mu }_{k^*,i}}\text {SNR}) {\dot{\ge }}\, \text {SNR}$. Therefore, (C9) is lower bounded by

$$\begin{aligned} \Pr \left( {\mathcal {O}}^{k^*}_s\right) {\dot{\ge }} \Pr \left( \det \left( {{{\textbf{I}}}_{{{n_r}}}+\text {SNR}{{\textbf{H}}_{k^{*}}}{{{\textbf{H}}}_{k^{*}}^{\dag }}} \right) <{\text {SNR}}^{Kr_s+{L}} \right) . \end{aligned}$$

(C10)

We define the variable $Z_k=\det \left( {{{\textbf{I}}}_{{{n_r}}}+\text {SNR}{{\textbf{H}}_{k}}{{{\textbf{H}}}_{k}^{\dag }}} \right)$ for $k=1,\ldots ,K$. As $k^*=\arg \underset{k}{\mathop {\max }}\, Z_k$, we can write

$$\begin{aligned} \Pr \left( {\mathcal {O}}^{k^*}_s \right)&{\dot{\ge }} \Pr \left( \max Z_k <{\text {SNR}}^{Kr_s+{L}} \right) \\&{\dot{\ge }} \left( F_{Z_k} \left( {\text {SNR}}^{Kr_s+{L}}\right) \right) ^K, \end{aligned}$$

(C11)

where $F_{Z_k}(.)$ is the cumulative distribution function of ${Z_k}$ and the result is due to the ordered statistics [34]. This yields

$$\begin{aligned} F_{Z_k} \left( {\text {SNR}}^{Kr_s+{L}}\right)=\Pr \left( Z_k<{\text {SNR}}^{Kr_s+{L}}\right) \end{aligned}$$

(C12a)

$$\begin{aligned}=\Pr \left( \det \left( {{{\textbf{I}}}_{{{n_r}}}+\text {SNR}{{\textbf{H}}_{k}}{{{\textbf{H}}}_{k}^{\dag }}} \right) <{\text {SNR}}^{Kr_s+{L}}\right) \end{aligned}$$

(C12b)

$$\begin{aligned}\doteq \text {SNR}^{-d_{n_t,{{n}_{r}}}(Kr_s+{L})} \end{aligned}$$

(C12c)

$$\begin{aligned}\doteq \left\{ \begin{array}{ll} \text {SNR}^{-d_{n_t-n_e,{{n}_{r}-n_e}}(Kr_s)},\,\,\,n_e < \min (n_t,n_r),\\ 1, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text {O.W.},\\ \end{array} \right. \end{aligned}$$

(C12d)

where (C12c) and (C12d) follow the same analysis of (A3d) and (A3e). Substituting the result of (C12) in (C11), the lower bound can be expressed as

$$\begin{aligned} \Pr \left( {\mathcal {O}}^{k^*}_s \right) {\dot{\ge }} \text {SNR}^{-Kd_{n_t-n_e,{{n}_{r}-n_e}}(Kr_s)} \,\,\, \,\,\, n_e< \min (n_t,n_r), \end{aligned}$$

(C13)

and if $n_e \ge \min (n_t,n_r)$, $\Pr \left( {\mathcal {O}}^{k^*}_s \right) \doteq 1$. Therefore, Lemma 3 is proved.