Evolution of distributed detection performance over balanced binary relay tree networks with bit errors

Abstract

Target detection based on wireless sensor networks can be considered as a distributed binary hypothesis testing problem. In this paper, the evolution of detection error probability with the increase in the network scale is studied for the balanced binary relay tree network with channel noise. Firstly, the iterative expressions of false-alarm probability and missed-detection probability depending on the number of tree network layers are given. Then, the iterative process of two types of error probabilities in the network space is described as a discrete nonlinear switched dynamic system, and the dynamic properties of two types of error probabilities are analyzed in a plane rectangular coordinate system. A globally attractive invariant set of the state of the dynamic system, which is not related to the channel noise, is derived. The switching mode of the system and the total error probability in the invariant set are further analyzed, and a necessary and sufficient convergence condition of the total error probability is provided. Based on this condition, the following detection properties of the network are revealed: (1) as long as the channel bit error probability is not zero, the total error probability does not tend to zero as the network size grows to infinity; (2) when the channel bit error probability is greater than $\frac{2-\sqrt{3}}{2}$, the total error probability will continue to increase with the increase in the network size.

Appendices

Appendices

Proof of Proposition 2

We only need to prove ${B_{1U}}|_{b=0}\subset {B_{1U}}|_{b\ne 0}$ due to the dynamic symmetry of the system.

By the definition of ${B_{1U}}$, the upper boundary of region ${B_{1U}}$ is described by equation of $(1 - (F(b,\alpha ))^2 + (F(b,\beta ))^2 = 1$. For any constant abscissa $\alpha$, this equation defines an implicit function $\beta (b)$, where $b\in [0,\frac{1}{2})$. To prove the proposition, it suffices to prove the positivity of the derivative of this function, i.e., $\beta '_b=\frac{\mathrm{d}\beta }{\mathrm{d}b}>0$. By derivation rules of implicit functions, it is easy to get

$$\begin{aligned} \beta '_b=\frac{(1-F(b,\alpha ))(1-2\alpha )-F(b,\beta )(1-2\beta )}{F(b,\beta )(1-2b)}. \end{aligned}$$

Since $\beta >\alpha$ due to $(\alpha ,\beta )\in U$, we have

$$\begin{aligned} \beta '_b>\frac{(1-2\alpha )(1-F(b,\alpha )-F(b,\beta ))}{F(b,\beta )(1-2b)}. \end{aligned}$$

Now, let us check the intersection point of the upper boundary of ${B_{1U}}$ with the line $\alpha +\beta =1$. Combining their equations yields ${[(1 - 2b)\beta + b]^2} = \frac{1}{2}$, which implies that $\beta = \frac{{{{(\frac{1}{2})}^{\frac{1}{2}}} - b}}{{1 - 2b}} \ge {(\frac{1}{2})^{\frac{1}{2}}}$, or equivalently, $\alpha \le 1-{(\frac{1}{2})^{\frac{1}{2}}}$. It implies that the abscissa of the intersection point of the upper boundary of ${B_{1U}}$ with the line $\alpha +\beta =1$ is less than $\frac{1}{2}$. Considering the fact that $1-2b>0$ by Assumption 2 and $1-F(b,\alpha )-F(b,\beta )\ge 0$ by (3), we get $\beta '_b>0$. The proposition is thus proved.

Proof of Proposition 3

Suppose that $(\alpha _k,\beta _k) \in B_{mU}, m\ge 2$. By definition of $B_{mU}$, we get that

$$\begin{aligned} D^{m-1}(b,(1 - F(b,\alpha _k))+D^{m-1}(b,F(b,\beta _k))<1, \end{aligned}$$

(10)

and

$$\begin{aligned} D^{m-2}(b,(1 - F(b,\alpha _k))+D^{m-2}(b,F(b,\beta _k))\ge 1. \end{aligned}$$

(11)

Note that here we let $D^0(b,y)=y^2$ for simplicity of notation.

By (4), it is easy to get $F(b, \alpha _{k + 1}) = 1 -b - (1 - 2b)(1 - F(b,\alpha _k))^2$, $F(b,\beta _{k + 1}) = (1 - 2b)(F(b,\beta _k))^2 + b$. Hence, from (10) to (11) it follows that

$$\begin{aligned} D^{m-2}(b,(1 \!-\! F(b,\alpha _{k+1}))\!+\!D^{m-2}(b,F(b,\beta _{k+1}))\!<\!1, \end{aligned}$$

and

$$\begin{aligned} D^{m-3}(b,(1 \!-\! F(b,\alpha _{k+1}))\!+\!D^{m-3}(b,F(b,\beta _{k+1}))\!\ge \! 1. \end{aligned}$$

Note that here we assume $D^{-1}(b,F(b,x))=x$ and $D^{-1}(b,(1-F(b,x)))=1-x$ for simplicity of notation. The above two inequalities imply that $({\alpha _{k + 1}},{\beta _{k + 1}}) \in {B_{(m-1)U}}$. Due to the dynamic symmetry of the system, the case of $(\alpha _k,\beta _k) \in B_{mL}$ can be proved similarly.

Proof of Proposition 4

Due to the dynamic symmetry of the system, we just need to prove $U \subseteq \cup _{m = 1}^\infty {B_{mU}}$.

Let us check the intersection point of the upper boundary of ${B_{mU}}$ with the line $\alpha +\beta =1$. The upper boundary equation of ${B_{mU}}$ is given by

$$\begin{aligned} D^{m-1}(b,(1 - F(b,\alpha ))+D^{m-1}(b,F(b,\beta ))=1. \end{aligned}$$

Combining this equation and the line equation $\alpha + \beta = 1$, we get

$$\begin{aligned} D^{m-1}(b,F(b,\beta )) = \frac{1}{2}. \end{aligned}$$

(12)

In Proof of Proposition 2, we have shown that for $m=1$, this equation implies $\beta = \frac{{{{(\frac{1}{2})}^{\frac{1}{2}}} - b}}{{1 - 2b}} \ge {(\frac{1}{2})^{\frac{1}{2}}}$. Using mathematical induction, we can show that Eq. (12) implies

$$\begin{aligned} \beta \ge \lim \limits _{m \rightarrow \infty }\frac{{{{\left( {\frac{1}{2}} \right) }^{\frac{1}{{2m}}}} - b}}{{1 - 2b}} \ge \lim \limits _{m \rightarrow \infty }{\left( {\frac{1}{2}} \right) ^{\frac{1}{{2m}}}}\mathrm{{ = 1}}, \end{aligned}$$

which indicates that the ordinate of the intersection point of the upper boundary of ${B_{mU}}$ with line $\alpha +\beta =1$ is not less than 1 as $m \rightarrow \infty$. This also implies that the ordinate of the intersection point of the upper boundary of ${B_{mU}}$ with the axis $\alpha =0$ is not less than 1. Since the convexity of the upper boundary curve of ${B_{mU}}$ is obvious, the upper triangular of the state space of the system is below this boundary as $m \rightarrow \infty$, i.e., $U\subseteq \bigcup _{m=1}^{\infty }B_{mU}$. Proposition 4 is thus proved.

Proof of Proposition 6

Due to the dynamic symmetry of the system, we just need to prove ${R_U} \subset {B_{\mathrm{{1}}U}} \cup {B_{2U}}$, i.e, to prove the upper boundary of ${B_{2U}}$ is always higher than the upper boundary of ${R_{U}}$. Our proof will use the following lemma which is proved in [15].

Lemma 1

${R_U} \subset {B_{\mathrm{{1}}U}} \cup {B_{2U}}$ holds for $b = 0$.

Since ${R_U}$ is independent of b, Lemma 1 tells us that to prove Proposition 6 it suffices to prove that the upper boundary of ${B_{2U}}$ with $b\ne 0$ is higher than the upper boundary of ${B_{2U}}$ with $b = 0$.

Suppose that $(\alpha _k^0,\beta _k^0)$ is a point on the upper boundary of $B_{2U}$ with $b = 0$, and $(\alpha _k^b,\beta _k^b)$ is a point on the upper boundary of $B_{2U}$ with $b \in (0,\frac{1}{2})$. To prove Proposition 6, it suffices to prove that $\beta _k^b > \beta _k^0$ when $\alpha _k^0 = \alpha _k^b$.

For any point $(\alpha _k^b,\beta _k^b)$ on the upper boundary of $B_{2U}$, since $[(1 - 2b)(1 - F(b,\alpha _k^b))^2 + b]^2+ [(1 - 2b)(F(b,\beta _k^b))^2 + b]^2 = 1$, using Eq. (4) we get $(1 - F(b,\alpha _{k + 1}^b))^2 + (F(b,\beta _{k + 1}^b))^2 = 1$. This implies that $(\alpha _{k+1}^b,\beta _{k+1}^b)$ lies on the upper boundary of $B_{1U}$. For $b = 0$, the upper boundary equation of the region ${B_{1U}}$ is ${\left( {1 - \alpha } \right) ^2} + {\beta ^2} = 1$. Based on this boundary equation, we define a function $\beta = G_0(\alpha )$. Obviously, ${G_0}(\alpha )$ is monotony increasing. For $b \in \left( {0,\frac{1}{2}} \right)$, the upper boundary equation of the region ${B_{1U}}$ is $(1-F(b,\alpha ))^2 + F(b,\beta )^2 = 1$. Based on this boundary equation, we define another function $\beta = G_b(\alpha )$. Therefore, for $(\alpha _k^0,\beta _k^0)$ on the upper boundary of $B_{2U}$ with $b = 0$ and $(\alpha _k^b,\beta _k^b)$ on the upper boundary of $B_{2U}$ with $b \in (0,\frac{1}{2})$, we know that $\beta _{k + 1}^0 = G_0(\alpha _{k + 1}^0)$ and $\beta _{k_+ 1}^b = G_b(\alpha _{k+ 1}^b)$.

Since $F(b,\alpha )>\alpha$ holds for any $b \in (0,\frac{1}{2})$ and $\alpha _k^0 = \alpha _k^b$ as supposed, we have $F(b,\alpha _k^b) > \alpha _k^0$. Therefore,

$$\begin{aligned} \left\{ \begin{array}{l} \alpha _{k + 1}^0 = 1 - (1 - \alpha _k^0)^2\\ \alpha _{k+ 1}^b = 1 - (1 - F(b,\alpha _k^b))^2 \end{array} \right. \Rightarrow \alpha _{k+1}^b > \alpha _{k + 1}^0. \end{aligned}$$

Then, by Proposition 2 we have

$$\begin{aligned} G_b(\alpha _{k+1}^b)> G_0(\alpha _{k + 1}^b)> G_0(\alpha _{k+1}^0). \end{aligned}$$

The above discussion has shown that $\beta _{k + 1}^b > \beta _{k + 1}^0$ under the condition $\alpha _k^0 = \alpha _k^b$, Therefore, from the iteration formula $\beta _{k+1}=(F(b,\beta _k))^2$ it follows that

$$\begin{aligned} \left\{ \begin{array}{l} \beta _{k + 1}^0 = (\beta _k^0)^2\\ \beta _{k + 1}^b = (F(b,\beta _k^b))^2 \end{array} \right. \Rightarrow F(b,\beta _k^b)> \beta _k^0 \Rightarrow \beta _k^b > \frac{\beta _k^0 - b}{1 - 2b}. \end{aligned}$$

Similar to what we have shown in Proof of Proposition 4, the ordinate of the intersection point of the upper boundary with the line $\alpha +\beta =1$ is greater than 1/2. Hence,

$$\begin{aligned} \frac{\mathrm{d}(\frac{{{\beta _k^0} - b}}{{1 - 2b}})}{{\mathrm{d} b}} = \frac{{2{\beta _k^0} - 1}}{{{{(1 - 2b)}^2}}}>0 \end{aligned}$$

for all $\beta _k^0 \in [0.5,1]$, which implies that

$$\begin{aligned} {\inf \limits _{b\in (0,0.5)}(\frac{{{\beta _k^0} - b}}{{1 - 2b}})} = {\beta _k^0} \end{aligned}$$

for all $\beta _k^0 \in [0.5,1]$. Thus, $\beta _k^b > \beta _k^0$. Proposition 6 is proved.

Proof of Theorem 5

By the dynamic symmetry of the system, the proof is just provided for the region U.

Firstly, we prove the part on the relationship of ${M_U}$ and ${R_U}$. The upper boundary equation of ${M_U}$ is $\beta = \alpha + \frac{{2b}}{{{{\left( {1 - 2b} \right) }^2}}}$, and the upper boundary equation of ${R_U}$ is $\beta = 2\sqrt{\alpha }- \alpha$. Let us define function

$$\begin{aligned} P(\alpha )= & {} \alpha + \frac{{2b}}{{{{\left( {1 - 2b} \right) }^2}}} - \left( {2\sqrt{\alpha }- \alpha } \right) \\= & {} 2\left[ {\frac{b}{{{{\left( {1 - 2b} \right) }^2}}} - \sqrt{\alpha }+ \alpha } \right] \end{aligned}$$

for $\alpha \in \left[ {0,\frac{1}{4}} \right]$. The derivative of this function is $P'(\alpha ) = 2 - \frac{1}{{\sqrt{\alpha }}} \le 0$. Therefore, ${P_{\max }}(\alpha ) = P(0) \ge 0$,${P_{\min }}(\alpha ) = P(\frac{1}{4})$, where

$$\begin{aligned} \left\{ \begin{array}{lll} P\left( {\frac{1}{4}} \right) < 0, &{}\mathrm{if} &{} b \in \left[ {0,\frac{{2 - \sqrt{3} }}{2}} \right) ,\\ P\left( {\frac{1}{4}} \right) = 0, &{} \mathrm{if} &{} b = \frac{{2 - \sqrt{3} }}{2},\\ P\left( {\frac{1}{4}} \right) > 0, &{} \mathrm{if} &{} b \in \left( {\frac{{2 - \sqrt{3} }}{2},\frac{1}{2}} \right) . \end{array} \right. \end{aligned}$$

Therefore, when $b \in \left[ {0,\frac{{2 - \sqrt{3} }}{2}} \right)$, ${R_U}\backslash {M_U} \ne \emptyset$, ${M_U}\backslash {R_U} \ne \emptyset$ hold; when $b = \frac{{2 - \sqrt{3} }}{2}$, ${R_U} \subset \overline{M_U}$ holds, and the upper boundary of ${M_U}$ and the upper boundary of ${R_U}$ have only one intersection point $\left( \frac{1}{4},\frac{3}{4}\right)$; when $b \in \left( {\frac{{2 - \sqrt{3} }}{2},\frac{1}{2}} \right)$, ${R_U} \subset {M_U}$ holds, and the upper boundary of ${M_U}$ and the upper boundary of ${R_U}$ have no intersection.

Next, we prove the part on the relationship of ${M_U}$ and ${B_{1U}}$.

The upper boundary equation of ${M_U}$ can be also written as $F(b,\beta ) = F(b,\alpha ) + \frac{{2b}}{{1 - 2b}}$, and the upper boundary equation of ${B_{1U}}$ is $F(b,\beta ) = \sqrt{2F(b,\alpha ) - (F(b,\alpha ))^2}$. Let us introduce a variable $t=F(b,\beta )$ and define function

$$\begin{aligned} Q(y) = y + \frac{{2b}}{{1 - 2b}} - \sqrt{2y - y^2} \end{aligned}$$

for $y \in \left[ {b,\frac{{2 - \sqrt{2} }}{2}} \right]$. The derivative of this function is $Q'(y) = 1 - \frac{{1 - y}}{{\sqrt{2y - y^2}} } \le 0$. Therefore,

$$\begin{aligned} Q_{\max }(y) = Q(b) = b + \frac{2b}{1 - 2b} - \sqrt{2b - b^2}. \end{aligned}$$

Obviously, when $b \in \left[ 0,\frac{2 -\sqrt{3}}{2}\right)$, ${Q_{\max }}(y) \le 0$. Hence, the upper boundary of ${B_{1U}}$ must be higher than the upper boundary of ${M_U}$. The proof is completed.