An Adversary-Resilient Doubly Compressed Diffusion LMS Algorithm for Distributed Estimation

Abstract

This paper proposes an adversary-resilient communication-efficient distributed estimation algorithm for time-varying networks. It is a generalization of the doubly compressed diffusion least mean square algorithm that is not adversary-resilient. The major drawback in existing adversary detectors in the literature is that they suggested the detection criterion heuristically. In this paper, an adversary detector is suggested theoretically based on a Bayesian hypothesis test (BHT). It is proved that the test statistics of the detectors is a distance metric compared to a threshold similarly to related papers in the literature. Hence, we prove the validity of the detection criterion based on BHT. The other difficulty encountered in existing works is the determination of thresholds. In this paper, the optimum thresholds are derived in closed form. Since the optimum thresholds need the values of unknown parameters, it is not feasible to derive them. Hence, suboptimal procedures for determining the thresholds are provided. Moreover, the convergence of the mean of the algorithm is investigated analytically. In addition, the Cramer–Rao bound of the problem of distributed estimation based on all node observations in the presence of adversaries is calculated. The simulation results show the effectiveness of the proposed algorithms and demonstrate that the proposed algorithms reach the performance of the algorithm when the adversaries are ideally known in advance, with some delay.

Appendices

Appendix A The Convergence of the Mean Condition

For calculating the sufficient condition of mean convergence of the weight vectors, we shall define some notations. We define $\tilde{\varvec{\psi }}_{k,i}={{\textbf {w}}}_o-\varvec{\psi }_{k,i}$. Then, we collect them in a vector as $\tilde{\varvec{\psi }}_i=\mathrm {col}\{\tilde{\varvec{\psi }}_{1,i},\tilde{\varvec{\psi }}_{2,i},\ldots ,\tilde{\varvec{\psi }}_{N,i}\}$. Let $\mathcal {{\textbf {R}}}_{{{\textbf {u}}}_l,i}={{\textbf {u}}}_{l,i}{{\textbf {u}}}^T_{l.i}$. The other notations are defined as follows:

$$\begin{aligned}&{\mathcal {M}}=\mathrm {diag}\{\mu _1{{\textbf {I}}}_L,\mu _2{{\textbf {I}}}_L,\ldots ,\mu _N{{\textbf {I}}}_L\} \end{aligned}$$

(38)

$$\begin{aligned}&\mathbf {{\mathcal {R}}}_{u,i}=\mathrm {diag}\{\mathbf {\mathcal R}_{u_1,i},\ldots ,{{\mathcal {R}}}_{u_N,i}\} \end{aligned}$$

(39)

$$\begin{aligned}&\mathbf {{\mathcal {C}}}={{\textbf {C}}}\otimes {{\textbf {I}}}_L \end{aligned}$$

(40)

$$\begin{aligned}&{{\mathcal {R}}}_{Q,i}=\mathrm {diag}\Big \{\sum _{l\in \mathcal N_1}c_{l1}{{\textbf {Q}}}_{l,i}\mathcal {{\textbf {R}}}_{{{\textbf {u}}}_l,i},\ldots ,\sum _{l\in \mathcal N_N}c_{lN}{{\textbf {Q}}}_{l,i}\mathcal {{\textbf {R}}}_{{{\textbf {u}}}_l,i}\Big \} \end{aligned}$$

(41)

$$\begin{aligned}&{{\mathcal {R}}}_{\gamma Q,i}=\mathrm {diag}\Big \{\sum _{l\in \mathcal N_1}c_{l1}\gamma _{l1}{{\textbf {Q}}}_{l,i}{{\textbf {R}}}_{{{\textbf {u}}}_l,i},\ldots ,\sum _{l\in \mathcal N_N}c_{lN}\gamma _{lN}{{\textbf {Q}}}_{l,i}{{\textbf {R}}}_{{{\textbf {u}}}_l,i}\Big \} \end{aligned}$$

(42)

$$\begin{aligned}&{{\mathcal {H}}}_i=\mathrm {diag}\{\mathbf {H}_{1,i},\mathbf {H}_{2,i},\ldots ,\mathbf {H}_{N,i}\} \end{aligned}$$

(43)

$$\begin{aligned}&{{\mathcal {Q}}}^{'}_i=\mathrm {diag}\Big \{\sum _{l\in \mathcal N_1}c_{l1}({{\textbf {I}}}_L-{{\textbf {Q}}}_{l,i}),\ldots ,\sum _{l\in \mathcal N_N}c_{lN}({{\textbf {I}}}_L-{{\textbf {Q}}}_{l,i})\Big \} \end{aligned}$$

(44)

$$\begin{aligned}&{{\mathcal {Q}}}^{'}_{\gamma ,i}=\mathrm {diag}\Big \{\sum _{l\in \mathcal N_1}c_{l1}\gamma _{l1}({{\textbf {I}}}_L-{{\textbf {Q}}}_{l,i}),\ldots ,\sum _{l\in \mathcal N_N}c_{lN}\gamma _{lN}({{\textbf {I}}}_L-{{\textbf {Q}}}_{l,i})\Big \} \end{aligned}$$

(45)

$$\begin{aligned}&{{\mathcal {R}}}_{u,i}=\mathrm {diag}\{{{\textbf {R}}}_{{{\textbf {u}}}_1,i},{{\textbf {R}}}_{{{\textbf {u}}}_2,i},\ldots ,{{\textbf {R}}}_{{{\textbf {u}}}_N,i}\} \end{aligned}$$

(46)

$$\begin{aligned}&{{\mathcal {Q}}}_i=\mathrm {diag}\{{{\textbf {Q}}}_{1,i},{{\textbf {Q}}}_{2,i},\ldots ,{{\textbf {Q}}}_{N,i}\} \end{aligned}$$

(47)

$$\begin{aligned}&{{\mathcal {F}}}=\mathrm {diag}\Big \{\sum _{l\in \mathcal N_1}a_{l1}\gamma _{l_1,i}({{\textbf {I}}}_L-\mathbf {H}_{l,i}),\ldots ,\sum _{l\in \mathcal N_N}a_{lN}\gamma _{l_N,i}({{\textbf {I}}}_L-\mathbf {H}_{l,i})\Big \} \end{aligned}$$

(48)

$$\begin{aligned}&{{\mathcal {F}}}^{'}=\mathrm {diag}\Big \{\sum _{l\in \mathcal N_1}a_{l1}\gamma _{l_1,i}\mathbf {H}_{l,i},\ldots ,\sum _{l\in \mathcal N_N}a_{lN}\gamma _{l_N,i}\mathbf {H}_{l,i}\Big \} \end{aligned}$$

(49)

$$\begin{aligned}&\Big [{\mathcal R}_{Q(I-H),i}\Big ]_{kl}=c_{lk}{{\textbf {Q}}}_{l,i}\mathcal {{\textbf {R}}}_{u_l,i}({{\textbf {I}}}_L-\mathbf {H}_{k,i}) \end{aligned}$$

(50)

$$\begin{aligned}&\Big [{{\mathcal {R}}}_{\gamma Q(I-H),i}\Big ]_{kl}=c_{lk}\gamma _{lk,i}{{\textbf {Q}}}_{l,i}\mathcal {{\textbf {R}}}_{u_l,i}({{\textbf {I}}}_L-\mathbf {H}_{k,i}) \end{aligned}$$

(51)

$$\begin{aligned}&{{\mathcal {D}}}={{\textbf {D}}}_i\otimes {{\textbf {I}}}_L,\quad d_{kl,i}=\sum _{l\in {\mathcal {N}}}a_{lk}(1-\gamma _{lk,i}) \end{aligned}$$

(52)

$$\begin{aligned}&{{\mathcal {R}}}_{qQ,i}=\mathrm {diag}\Big \{\sum _{l\in \mathcal N_1}c_{l1}{{\textbf {Q}}}_{l,i}\mathcal {{\textbf {R}}}_{{{\textbf {u}}}_l,i}{{\textbf {q}}}_{l,i},\ldots ,\sum _{l\in \mathcal N_N}c_{lN}{{\textbf {Q}}}_{l,i}\mathcal {{\textbf {R}}}_{{{\textbf {u}}}_l,i}{{\textbf {q}}}_{l,i}\Big \} \end{aligned}$$

(53)

$$\begin{aligned}&{{\mathcal {R}}}_{\gamma qQ,i}=\mathrm {diag}\Big \{\sum _{l\in \mathcal N_1}c_{l1}\gamma _{l1}{{\textbf {Q}}}_{l,i}\mathcal {{\textbf {R}}}_{{{\textbf {u}}}_l,i}{{\textbf {q}}}_{l,i},\ldots ,\nonumber \\&\qquad \qquad \qquad \sum _{l\in {\mathcal {N}}_N}c_{lN}\gamma _{lN}{{\textbf {Q}}}_{l,i}\mathcal {{\textbf {R}}}_{{{\textbf {u}}}_l,i}{{\textbf {q}}}_{l,i}\Big \} \end{aligned}$$

(54)

$$\begin{aligned}&{{\mathcal {A}}}_{q,i}=\mathrm {diag}\Big \{\sum _{l\in \mathcal N_1}c_{l1}({{\textbf {I}}}_L-{{\textbf {Q}}}_{l,i})\mathcal {{\textbf {R}}}_{{{\textbf {u}}}_l,i}{{\textbf {q}}}_{l,i},\ldots , \nonumber \\&\qquad \qquad \qquad \sum _{l\in {\mathcal {N}}_N}c_{lN}({{\textbf {I}}}_L-{{\textbf {Q}}}_{l,i})\mathcal {{\textbf {R}}}_{{{\textbf {u}}}_l,i}{{\textbf {q}}}_{l,i} \Big \} \end{aligned}$$

(55)

$$\begin{aligned}&{{\mathcal {A}}}_{\gamma q,i}=\mathrm {diag}\Big \{\sum _{l\in \mathcal N_1}c_{l1}\gamma _{l1}({{\textbf {I}}}_L-{{\textbf {Q}}}_{l,i})\mathcal {{\textbf {R}}}_{{{\textbf {u}}}_l,i}{{\textbf {q}}}_{l,i},\ldots , \nonumber \\&\qquad \qquad \qquad \sum _{l\in {\mathcal {N}}_N}c_{lN}\gamma _{lN}({{\textbf {I}}}_L-{{\textbf {Q}}}_{l,i})\mathcal {{\textbf {R}}}_{{{\textbf {u}}}_l,i}{{\textbf {q}}}_{l,i} \Big \} \end{aligned}$$

(56)

$$\begin{aligned}&{{\mathcal {S}}}_i=\mathrm {col}\{{{\textbf {u}}}_{1,i}n_{1,i},\ldots ,{{\textbf {u}}}_{N,i}n_{N,i}\} \end{aligned}$$

(57)

After the above definitions, some manipulations show that we have:

$$\begin{aligned} \tilde{{{\textbf {w}}}}_i=({{\textbf {I}}}_{NL}+\mathbf {\mathcal F})\tilde{\varvec{\psi }}_i+\mathbf {\mathcal F}^{'}\tilde{{{\textbf {w}}}}_{i-1}+\mathbf {{\mathcal {D}}}\tilde{{{\textbf {w}}}}_{i-1} \end{aligned}$$

(58)

For investigating the mean convergence, we take expectation from the above equation. So, we reach:

$$\begin{aligned} E\{\tilde{{{\textbf {w}}}}_i\}=E\{({{\textbf {I}}}_{NL}+\mathbf {\mathcal F})\}E\{\tilde{\varvec{\psi }}_i\}+(E\{\mathbf {\mathcal F}^{'}\}+E\{\mathbf {{\mathcal {D}}}\})E\{\tilde{{{\textbf {w}}}}_{i-1}\} \end{aligned}$$

(59)

In the above formula, the term of $E\{\tilde{\varvec{\psi }}_i\}$ is difficult to compute. It needs to calculate a recursion formula for $\tilde{{{\textbf {w}}}}_i$. It is done, and for brevity we omit the details of the derivation. We have:

$$\begin{aligned} \tilde{\varvec{\psi }}_i= & {} \Big ({{\textbf {I}}}_{NL}-\mathbf {{\mathcal {M}}}\mathbf {\mathcal R}_{Q,i}\mathbf {{\mathcal {H}}}_i-\mathbf {{\mathcal {M}}}\mathbf {\mathcal Q}^{'}_i\mathbf {{\mathcal {R}}}_{u,i}-\mathbf {\mathcal M}\mathbf {{\mathcal {R}}}_{Q(I-H),i} \nonumber \\&\quad -\, \mathbf {{\mathcal {M}}}\mathbf {{\mathcal {R}}}_{\gamma Q,i}\mathbf {\mathcal H}_i-\mathbf {{\mathcal {M}}}\mathbf {\mathcal Q}^{'}_{\gamma ,i}\mathbf {{\mathcal {R}}}_{u,i}-\mathbf {\mathcal M}\mathbf {{\mathcal {R}}}_{\gamma Q(I-H),i}\Big )\tilde{{{\textbf {w}}}}_{i-1} \nonumber \\&\quad -\, \Big (\mathbf {{\mathcal {M}}}\mathbf {{\mathcal {C}}}^T\mathbf {\mathcal Q}_i+\mathbf {{\mathcal {M}}}\mathbf {{\mathcal {Q}}}^{'}_i+\mathbf {\mathcal M}\mathbf {{\mathcal {C}}}^T\mathbf {\mathcal Q}_{\gamma ,i}+\mathbf {{\mathcal {M}}}\mathbf {\mathcal Q}^{'}_{\gamma ,i}\Big )\mathbf {{\mathcal {S}}}_i \nonumber \\&\quad -\, \mathbf {{\mathcal {M}}}\mathbf {\mathcal R}_{qQ,i}-\mathbf {{\mathcal {M}}}\mathbf {{\mathcal {R}}}_{\gamma qQ,i}-\mathbf {{\mathcal {M}}}\mathbf {{\mathcal {A}}}_{q,i}-\mathbf {\mathcal M}\mathbf {{\mathcal {A}}}_{\gamma q,i} \end{aligned}$$

(60)

Since we assume that $E\{{{\textbf {q}}}_{k,i}\}=0$ and expectation of noise vectors are zero, the expectations of the above terms in the fourth row of (60) are zero. Then, some calculations lead to the following formula:

$$\begin{aligned} E\{\tilde{\varvec{\psi }}_i\}= & {} \left( {{\textbf {I}}}_{NL}-\frac{MM_{\varDelta }}{L^2}\mathbf {\mathcal M}\mathbf {{\mathcal {R}}}-\left( 1-\frac{M_{\varDelta }}{L}\right) \mathbf {\mathcal M}\mathbf {{\mathcal {R}}}_u \nonumber \right. \\&\left. \quad -\, \frac{M_{\varDelta }}{L}\left( 1-\frac{M}{L}\right) \mathbf {\mathcal M}\mathbf {{\mathcal {C}}}^T\mathbf {\mathcal R}_u-p_a\frac{MM_{\varDelta }}{L^2}\mathbf {{\mathcal {M}}}\mathbf {\mathcal R}-p_a\left( 1-\frac{M_{\varDelta }}{L}\right) \mathbf {{\mathcal {M}}}\mathbf {\mathcal R}_u \nonumber \right. \\&\quad \left. -\, p_a\frac{M_{\varDelta }}{L}\left( 1-\frac{M}{L}\right) \mathbf {\mathcal M}\mathbf {{\mathcal {C}}}^T\mathbf {\mathcal R}_u\right) E\{\tilde{{{\textbf {w}}}}_{i-1}\}=\mathbf {\mathcal B}E\{\tilde{{{\textbf {w}}}}_{i-1}\}, \end{aligned}$$

(61)

where

$$\begin{aligned}&\mathbf {{\mathcal {R}}}_u=E\{\mathbf {\mathcal R}_{u,i}\}=\mathrm {diag}\{\mathbf {\mathcal R}_{u_1},\ldots ,\mathbf {{\mathcal {R}}}_{u_N}\} \end{aligned}$$

(62)

$$\begin{aligned}&\mathbf {{\mathcal {R}}}=\mathrm {diag}\{\mathbf {\mathcal R}_1,\ldots ,\mathbf {{\mathcal {R}}}_N\} \end{aligned}$$

(63)

with

$$\begin{aligned} \mathbf {{\mathcal {R}}}_k=\sum _{l\in {\mathcal {N}}_k}c_{lk}\mathbf {\mathcal R}_{u_l}. \end{aligned}$$

(64)

Table 1 Final MSD of various diffusion algorithms

Then, replacing (61) into (59) and with some calculations, we reach to

$$\begin{aligned} E\{\tilde{{{\textbf {w}}}}_i\}= & {} \Big (1+p_a(1-\frac{M}{L})\Big )\mathbf {\mathcal B}E\{\tilde{{{\textbf {w}}}}_{i-1}\} \nonumber \\&\quad +\, \Big (p_a\frac{M}{L}+(1-p_a)\Big )E\{\tilde{{{\textbf {w}}}}_{i-1}\}=(\mathbf {\mathcal B}^{'}+\mathbf {{\mathcal {Y}}})E\{\tilde{{{\textbf {w}}}}_{i-1}\}, \end{aligned}$$

(65)

where $\mathbf {\mathcal B}^{'}=\Big (1+p_a(1-\frac{M}{L})\Big )\mathbf {{\mathcal {B}}}$ and $\mathbf {{\mathcal {Y}}}=[p_a\frac{M}{L}+(1-p_a)]{{\textbf {I}}}_N$. Then, (65) can be written in the following recursive form

$$\begin{aligned} E\{\tilde{{{\textbf {w}}}}_i\}=(\mathbf {{\mathcal {B}}}^{'}+\mathbf {\mathcal Y})E\{\tilde{{{\textbf {w}}}}_{i-1}\} \end{aligned}$$

(66)

Therefore, similar to [18], the proposed AR-DC-DLMS asymptotically converges in the mean toward ${{\textbf {w}}}_o$ if, and only if $\rho (\mathbf {{\mathcal {B}}}^{'}+\mathbf {{\mathcal {Y}}})<1$ where $\rho (.)$ stands for the spectral radius of the matrix argument. From matrix algebra, we have $\rho ({{\textbf {X}}})\le ||{{\textbf {X}}}||$ for any induced norm. So, we have:

$$\begin{aligned} \rho (\mathbf {{\mathcal {B}}}^{'}+\mathbf {\mathcal Y})\le ||\mathbf {{\mathcal {B}}}^{'}+\mathbf {\mathcal Y}||_{b,\infty }\le \mathrm {max}||\Big [\mathbf {\mathcal B}^{'}+\mathbf {{\mathcal {Y}}}\Big ]_{kl}|| \end{aligned}$$

(67)

where $||.||_{b,\infty }$ is the block maximum norm. Deducing from (67), we will have:

$$\begin{aligned} \rho (\mathbf {{\mathcal {B}}}^{'}+\mathbf {\mathcal Y})\le & {} \mathrm {max}_{k,l}||{{\textbf {I}}}_L-\mu _k\left[ \frac{MM_{\varDelta }}{L^2}+\left( 1-\frac{M_{\varDelta }}{L}\right) \mathbf {\mathcal R}_{u_k}\nonumber \right. \\&\left. \quad +\, \frac{M_{\varDelta }}{L}\left( 1-\frac{M}{L}\right) c_{lk}\mathbf {\mathcal R}_{u_l}-p_a[\frac{MM_{\varDelta }}{L^2}\mathbf {\mathcal R}_k+\left( 1-\frac{M_{\varDelta }}{L}\right) \mathbf {{\mathcal {R}}}_{u_k} \nonumber \right. \\&\left. \quad +\, \frac{M_{\varDelta }}{L}\left( 1-\frac{M}{L}\right) c_{lk}\mathbf {\mathcal R}_{u_l}]+p_a\frac{M}{L}+(1-p_a)\right] ||<1 \end{aligned}$$

(68)

Similar to [18], as a linear combination with positive coefficients of positive-definite matrices $\mathbf {{\mathcal {R}}}_k$, $\mathbf {{\mathcal {R}}}_{u_k}$, and $\mathbf {{\mathcal {R}}}_{u_l}$, the matrix in square brackets on the RHS of (68) is positive definite. Then, the condition in right side of (68) holds if (29) is satisfied. Then, the $\lambda _{\mathrm {max},k}$ is given by (30) and the proof is completed.

Appendix B Calculating the Fisher-Information Matrix

For calculating the FIM, the total likelihood can be computed as

$$\begin{aligned} p({{\textbf {X}}}|{{\textbf {w}}})=\prod _{k=1}^Np({{\textbf {x}}}_k|{{\textbf {w}}})=\prod _{k=1}^Np_{{{\textbf {z}}}_k}({{\textbf {x}}}_k-{{\textbf {U}}}_k{{\textbf {w}}}). \end{aligned}$$

(69)

Since ${{\textbf {z}}}_k=[z_{k,1},z_{k,2},\ldots ,z_{k,T}]^T$ with $z_{k,i}={{\textbf {u}}}^T_{k,i}\tilde{{{\textbf {q}}}}_{k,i}+n_{k,i}$, the $z_{k,i}$ is Gaussian with zero mean and variance $\sigma ^2_{z_{k,i}}=E(z^2_{k,i})$. Since $n_{k,i}$ and $\tilde{{{\textbf {q}}}}_{k,i}$ are assumed to be independent and uncorrelated with mean zero, we have $\sigma ^2_{z_{k,i}}=E\{({{\textbf {u}}}^T_{k,i}\tilde{{{\textbf {q}}}}_{k,i})^2\}+E\{n^2_{k,i}\}$. This would be equal to $\sigma ^2_{z_{k,i}}=E\{{{\textbf {u}}}^T_{k,i}\tilde{{{\textbf {q}}}}_{k,i}\tilde{{{\textbf {q}}}}^T_{k,i}{{\textbf {u}}}_{k,i}\}+\sigma ^2_{k,n}$ where $\sigma ^2_{k,n}$ is the variance of noise at node k. Some simple calculations lead to the following formula

$$\begin{aligned} \sigma ^2_{z_{k,i}}={{\textbf {u}}}^T_{k,i}E\{\tilde{{{\textbf {q}}}}_{k,i}\tilde{{{\textbf {q}}}}^T_{k,i}\}{{\textbf {u}}}_{k,i}+\sigma ^2_{k,n}=p_a\sigma ^2_q||{{\textbf {u}}}_{k,i}||^2+\sigma ^2_{k,n}, \end{aligned}$$

(70)

where it is assumed the elements of $\tilde{{{\textbf {q}}}}_{k,i}$ are uncorrelated. Since the elements of ${{\textbf {z}}}_k$ are independent of each other, the vector ${{\textbf {z}}}_k$ is Gaussian with zero mean and diagonal covariance matrix equal to ${{\textbf {P}}}_k=\mathrm {diag}(\sigma ^2_{z_{k,i}})$. So, from (69), we can write the log-likelihood as $\log p({{\textbf {X}}}|{{\textbf {w}}})=\sum _{k=1}^N \Big [-\frac{I}{2}\log (2\pi )-\frac{1}{2}\sum _i\log (\sigma ^2_{z_{k,i}})-\frac{1}{2}\Big ({{\textbf {x}}}_k-{{\textbf {U}}}_k{{\textbf {w}}}\Big )^T{{\textbf {P}}}_k^{-1}\Big ({{\textbf {x}}}_k-{{\textbf {U}}}_k{{\textbf {w}}}\Big )\Big ]$. So, the partial derivative will be equal to $\frac{\partial \log p({{\textbf {X}}}|{{\textbf {w}}})}{\partial w_j}=-\frac{1}{2}\sum _{k=1}^N\frac{\partial }{\partial w_j}\Big [{{\textbf {e}}}_k^T{{\textbf {P}}}_k^{-1}{{\textbf {e}}}_k\Big ]$ where ${{\textbf {e}}}_k={{\textbf {x}}}_k-{{\textbf {U}}}_k{{\textbf {w}}}$. We can write $B={{\textbf {e}}}_k^T{{\textbf {P}}}_k^{-1}{{\textbf {e}}}_k=\sum _{i=1}^I P^{-1}_{k,ii}e^2_{k,i}$. So, the partial derivative is equal to $\frac{\partial B}{\partial w_j}=\sum _{i=1}^I 2P^{-1}_{ii}e_{k,i}\frac{\partial e_{k,i}}{\partial w_j}$. Also, we have $\frac{\partial e_{k,i}}{\partial w_j}=-U_{k,i,l}$. Taking the second partial derivative and doing some simple manipulations, we reach $F_{l,j}=-E\{\frac{\partial ^2\log p({{\textbf {X}}}|{{\textbf {w}}})}{\partial w_lw_j}\}=-\sum _{k=1}^N\sum _{i=1}^T P^{-1}_{k,ii}U_{k,i,j}U_{k,i,l}$, where this formula leads to (35).