Published in IET Signal Processing Received on 23rd May 2011 Revised on 28th April 2012 doi: 10.1049/iet-spr.2011.0199
ISSN 1751-9675
Linear coherent distributed estimation with
cluster-based sensor networks
C.-A. Lin C.-H. Wu
Department of Electrical and Control Engineering, National Chiao Tung University, Hsinchu, Taiwan E-mail: [email protected]
Abstract: The authors consider distributed estimation using sensor network with coherent multiple access channel model and LMMSE fusion rule. The sensors in the network are divided into a number of clusters. Sensors within the same cluster are allowed to collaborate through an amplification matrix to form a message this then transmitted. They formulate the problem of choosing the amplification matrices as an optimal power allocation problem under a total power constraint. The solution gives the optimal amplification matrices as scaled outer products of the observation gain and the channel gain vectors. The authors show that collaboration improves performance and, in simulations, demonstrate that the amount of improvement is closely related to the amount of collaboration.
1 Introduction
Distributed estimation has attracted much attention in signal processing research for sensor networks [1]. In distributed estimation scenario, a certain parameter is measured by spatially distributed sensors and the measurements are sent to a fusion centre (FC) where a final estimate is formed. Owing to energy constraints, power efficiency is an important issue since it is closely related to the network lifetime. To enhance power efficiency, many research works focus on cluster-based sensor networks in which the problem is to efficiently organise sensors into clusters so that network lifetime can be improved[2, 3]. For example, Wimalajeewa and Jayaweera [4] introduced sensor selection schemes to minimise the estimation distortion, whereas Heinzelman et al. [5] developed a communication protocol to save power. Recently, analogue transmission schemes aiming at minimising the estimation distortion by optimally allocating power for each sensor have been studied based on the coherent multiple access channel (MAC) model[6 – 9], the orthogonal MAC model[10 – 15], as well as the hybrid MAC model [16]. Among them, some works consider distributed estimation of a scalar parameter [13, 14] or a vector parameter [9, 10] with spatially correlated sensor observations. The work in [15]
addresses robust estimation that takes account of the uncertainty in the local observing noise variance. Fang and Li [12] considered a cluster-based network architecture in which closely located sensors are able to collaborate to form local messages for transmission through the orthogonal MAC.
In this paper, we consider distributed estimation of a scalar parameter by optimally allocating power based on cluster-based wireless sensor networks with the coherent MAC
model. We assume that the sensors in the network are already divided into a number of clusters. The sensors in the same cluster are allowed to collaborate, while collaboration is prohibited for sensors in different clusters. The collaboration is through an amplification matrix, for each cluster, that forms a message from the measurements for transmission to the FC. In other words, collaboration means the measurements within a cluster are linearly combined locally. At the FC, the parameter is estimated based on the linear minimum mean-squared error (LMMSE) rule. The mean-squared error (MSE) depends on the choice of the amplification matrices. We study the problem of choosing the amplification matrices so that the corresponding MSE is minimised. We formulate the problem as one of optimal power allocation under a total power constraint. The solution shows that the optimal amplification matrices are scaled outer products of the observation gain and the channel gain vectors. For comparison, two special cases are also considered: the full collaboration case, in which all sensors are in the same cluster, and the non-collaboration case, in which each cluster has only one sensor. We show that with the optimal amplification matrices, collaboration indeed improves performance in terms of MSE. We demonstrate through simulation results that the amount of improvement is closely related to the amount of collaboration.
The rest of this paper is organised as follows. In Section 2, we describe the model of cluster-based sensor network and the problem we address. In Section 3, we solve the optimisation problem to obtain the optimal amplification matrices and show that with optimal amplification matrices, collaboration indeed improves performance. In Section 4, simulation results are given to verify the analytical result. Section 5 is a brief conclusion.
2 System model and problem formulation
We consider a wireless sensor network consisting of K spatially deployed sensors for estimating a random source signal u. The sensors in the network are divided into L clusters, as shown in Fig. 1. The lth cluster has Kl sensors
and the measurement at the kth sensor is given by
xl,k = fl,ku+ nl,k, 1≤ l ≤ L, 1 ≤ k ≤ Kl (1)
where fl,kis the observation gain and nl,kis the measurement
noise. In vector form, (1) becomes
xl= flu+ nl, 1≤ l ≤ L (2) where xl= [xl,1· · · xl,K l] T , fl= [fl,1· · · fl,K l] T and nl= [nl,1 · · · nl,Kl]
T. The collaboration between sensors in the lth
cluster is through an amplification matrix Al[ RNl×Kl,
which takes xl [ RKl to form the message vector
Alxl[ RNl. The messages are then sent to the FC and the
signal y received at the FC can be expressed as
y= L l=1 gTlAl(flu+ nl)+ n (3) where gl= [gl,1· · · gl,N l]
Tis the channel gain vector and n is
the additive noise at the receiver. In practice, the sensors which are geographically closely located can compose a cluster. The collaboration between sensors in the same cluster can be implemented by choosing one sensor as the cluster head whose task is to collect and process information sent from other sensors to form a message vector and transmit it to the FC.
In this paper, we assume that (i) E[u] ¼ 0 and E[u2]= s2u, (ii) the measurement noises are zero-mean and mutually uncorrelated, specifically E[nl] ¼ 0, E[nln
T l]= s
2 nIKl and
E[nlnTm]= 0Kl×Km for l = m, (iii) E[n] ¼ 0 and E[n
2
]= s2n,
(iv) the source signal, the measurement noises, and the receiver noise are uncorrelated, that is, E[unl] ¼ 0,
E[un] ¼ 0 and E[nnl] ¼ 0, and (v) the observation gain
vectors fland the channel gain vectors glare known to the FC.
For a given set of amplification matrices Al, the LMMSE
estimate of u using the received signal y in (3) is[17, p. 382] ˆ u =E[uy] E[ y2]y = s2u L l=1gTlAlfl s2 u L l=1gTlAlfl 2 + s2 n L l=1gTlAlATlgl+ s2n y (4)
and the corresponding MSE is J = E[(u − ˆu )2]= s2u− (E[uy])2 E[ y2] = 1 s2 u + L l=1gTlAlfl 2 s2 n L l=1gTlAlATlgl+ s2n −1 (5)
The problem is to minimise the MSE in (5) by choosing optimal amplification matrices Al under a total power constraint. The total transmitted power of the L clusters is L
l=1E[xTlA T
lAlxl]. Hence if P is the amount of power that
the clusters together can used, then we have the following constraint L l=1 tr(E[AlxlxTlATl])= L l=1 tr(s2uAlflfTlATl + s2nAlATl)≤ P (6) where tr(.) denotes the trace of a matrix and we use E[xlxTl]= s2uffT+ s2nIK
l. From (5) and (6), the
optimisation problem under consideration can be written as min Al,1≤l≤LJ subject to L l=1 tr(s2uAlflfTlATl + s2nAlATl)≤ P (7) where J is given in (5).
Remarks: We had assumed that the measurement noises are mutually uncorrelated across all sensors. If the measurement noises are correlated within the same cluster but uncorrelated across different clusters, the problem can still be formulated in the same form as (7). To see this, suppose E[nlnTl]= Rn
l, where Rnl = R
T nl [ R
Kl×Kl is positive
definite and E[nlnTm]= 0K
l×Km for l = m. Let
Rn
l = UnlLnlU
T
nl be the eigenvalue decomposition with
Ln l = diag(s 2 nl,1, . . . , s 2 nl,Kl) . 0, where diag(x1, . . . , xM) is
a diagonal matrix whose mth diagonal element is xm.
By setting A˜l= AlUn lL 1/2 nl and ˜fl= L −1/2 nl U T nlfl, the
corresponding optimisation problem has the same form as (7) with Al, fland s2n replaced by ˜Al, ˜fland 1, respectively.
3 Optimal amplification matrices
In this section, we consider the solution of the optimisation problem (7) with the goal of obtaining a closed form expression for the optimal amplification matrices Al. We
first make the following observations:
1. In problem (7), if the ‘≤ ’ sign in the constraint is replaced by the ‘¼’ sign, the solution does not change. Hence we could consider the optimisation problem with equality constraint. The argument is as follows. Since the constraint function is quadratic in the elements of Al, if a set of Al is such that
strict inequality holds, we can equally scale up each Al so
that equality holds. In addition, if we equally scale up each Al, we obtain a lower function value of J because in (5)
the second term inside the parentheses becomes larger. Consequently, with optimal Al, the inequality constraint
must be active. Fig. 1 Cluster-based sensor network with coherent MAC
2. Consider the optimal MSE in (7), say, J∗ as a function of the power P, then J∗is a strictly decreasing function of P, that is, if P2. P1, then J∗(P2) , J∗(P1). The argument is similar:
if the power level increases, we can equally scale up Al to
obtain a lower value of J and thus a lower value of optimal MSE J∗ can be obtained.
3. Since the function J∗(P) is 1 – 1 and decreasing, the inverse function P( J∗) is also 1 – 1 and decreasing. Hence instead of finding the matrices Al that minimise J in (5) under an
equality constraint on power level, we can find the matrices Al that minimise the power level subject to an equality
constraint on MSE. If the constraint value on MSE is such that the resulting minimum power level matches the given value P in (7), the corresponding matrices Al are the
optimal ones we set out to find. We thus consider the following optimisation problem
min Al,1≤l≤L L l=1tr(s 2 uAlflfTlAlT+ s2nAlATl) subject to 1 s2 u + L l=1gTlAlfl 2 s2 n L l=1gTlAlATlgl+ s2n −1 = J∗ ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ (8) where 0 , J∗ ≤ s2u. We note that both the objective function and the constraint function in (8) are quadratic in the elements of Al. This problem is considerably easier to solve than the
original one (7). The main result based on solving (8) is in the following proposition whose proof is given in Appendix 1. Proposition 1: Consider the sensor network model described by (2) and (3). Suppose the total transmitted power from all sensors is no greater than P, then using the LMMSE estimator, the optimal amplification matrix of the lth cluster is given by Aoptl = L i=1 fi 2g i 2 (s2u fi 2+ s2 n) f2i −1 P f2l g lf T l, l= 1, . . . , L (9) where fi= s2n(s2u fi2+ s2n)+ s2ngi2P andx = xTx √ , and the corresponding minimum MSE is
JM= 1 s2 u +L l=1 fl2gl2P s2 n(s2u fl2+ sn2)+ s2ngl2P −1 (10)
The optimal amplification matrix Aoptl is a rank one matrix, which is a scaled outer product of gland fl. As expected the
optimal MSE JM decreases as P increases. Moreover, as
P 1, we have lim P1JM= s2u 1+ (s2 u/s2n) L l=1 fl2 (11)
The limit does not go to zero but approaches a finite value which depends on the signal-to-noise ratio s2uLl=1 fl2/s2n, since the measured signal flu+ nl is amplified by Al,
1≤ l ≤ L.
For comparison, we consider two special cases: L ¼ 1 and L ¼ K. When L ¼ 1, there is full collaboration among the K sensors. The observation gain is f [ RK and the channel
gain is g [ RN, N≤ K. With the optimal amplification matrix Aopt [ RN×K given by (9), the minimum MSE in (10) becomes JC= 1 s2 u + f 2g2P s2 n(s2u f 2+ s2n)+ s2ng2P −1 (12) When L ¼ K, each sensor is a cluster and no collaboration between sensor exists. The scalar observation gains and channel gains are respectively fk and gk, 1≤ k ≤ K. With
the K scalar amplification gains given by (9), the minimum MSE becomes JN= 1 s2 u +K k=1 fk2g2kP s2 n(s2u fk2+ sn2)+ s2ng2kP −1 (13)
To compare the performance of the general case and the two special cases, we assume Nl¼ Kl, 1≤ l ≤ L, in (10) and
N ¼ K in (12), that is, the number of measurements is equal to the number of transmitters in each cluster. Hence the observation gain and the channel gain vectors can be written as f = [ fT1fT2· · · fTL]T= [ f1f2· · · fK]T and
g= [gT1gT2· · · gTL]T = [g1g2· · · gK]T, respectively, where
fl, gl[ RKl and K
1+ · · · + KL¼ K. In terms of the
MSE, it is not unexpected that collaboration improves performance. Indeed, we have the following proposition. Proposition 2: The minimum MSEs JMin (10), JC in (12),
and JNin (13) satisfy
JC≤ JM ≤ JN (14)
The proof of Proposition 2 is based on the following lemma. Lemma 1: For x= [xT1x2T· · · xTL]T [ Rn and y= [ yT1yT2· · · yTL]T [ Rn, where xiand yiare non-zero vectors of
dimension≥1, we have x2y2 x2+ y2≥ L i=1 xi 2y i 2 xi2+ yi2 (15)
Proof: Please see Appendix 2.
We now establish Proposition 2. Let x= s2 u f , y= (s2 n/s2n)P g, xi= s2 u fi and yi= (s2 n/s2n)P gi. Then by Lemma 1, we obtain f 2g2 s2 u f 2+ (s2n/s2n)Pg2+ s2n ≥ L i=1 fi2gi2 s2 u fi2+ (s2n/s2n)Pgi2+ s2n
and thus JC−1 ≥ JM−1, or equivalently, JC≤ JM. The second inequality in (14) follows similarly: apply Lemma 1 to each xi, yi, and their respective scalar components, and their
sums give the desired inequality.
4 Numerical results
In this section, we use numerical simulations to verify the analytical result established in Section 3. In all simulations, the random parameters, u, nl, k and n, are zero-mean
Gaussian, and we assume that s2u= s2n= 1. The observation
gains fl,k are assumed to be uniformly distributed in the
interval [0.5, 1]. The channel gains are taken as cgd 23.5
, where d is uniformly drawn from the interval [1, 10] and cg¼ 22.6 is a normalisation constant to make E[ gl,n] ¼ 1
as in[7].
We first consider the effect of different numbers of transmitters N, where N≤ K, in the full collaboration case. We set K ¼ 10 and s2n = 0.4. InFig. 2, we plot the average
MSE against N with power levels P ¼ 0, 5 and 10 dB. We note that as N increases, the MSEs decrease; also, large power levels result in smaller MSEs.
In all the simulations to follow, the number of sensors and the number of transmitters are set equal. Fig. 3 shows the average MSE against P for the full collaboration and non-collaboration cases with different observation noises, s2n= 0.4 and s2n= 0.8. We set K ¼ 20. For s2n= 0.4, the case with full collaboration performs better than the non-collaboration case. Moreover, as the transmitted power increases, the MSEs for both two cases decrease. In fact, from (11), these two cases approach identical MSE as Fig. 2 MSE of full collaboration case with different numbers of transmitters
Fig. 4 MSE of full collaboration and non-collaboration cases with different power levels
Fig. 5 MSEs for Kl¼ 1, 4, 8, and K with different number of
sensors
Fig. 6 Comparison of the coherent MAC model to that of the orthogonal MAC model
Fig. 3 MSE of full collaboration and non-collaboration cases with different power levels
P 1. We also see that the MSE of the case with s2n= 0.4 is smaller than that of the case with s2n= 0.8, that is, a large
signal-to-noise ratio results in a good performance.
Fig. 4shows the comparison for the full collaboration case and two non-collaboration cases. The first non-collaboration case uses the optimal power allocation scheme and the second case uses equal power allocation scheme, in which the amplification gains are chosen as ak =
P/K √
, 1≤ k ≤ K. We set K ¼ 50 and s2n= 0.4. Clearly, optimal power allocation improves performance over equal power allocation. The reduction in MSE by full collaboration with optimal power allocation is about 10 dB compared with the equal power allocation scheme.
We now consider two multiple cluster cases: in case 1, each cluster consists of 4 sensors, and in case 2, each cluster consists of eight sensors. Hence for a fixed number of sensors K, case 1 has K/4 clusters and case 2 has K/8 clusters. We compare their performance with the full collaboration and non-collaboration cases. We set P ¼ 0 dB and s2n= 0.4.Fig. 5shows that MSE of case 2 is less than
that of case 1 since for a fixed K, case 2 has a smaller number of clusters and thus more collaboration among sensors. That the full collaboration has the lowest MSE and the non-collaboration case has the highest MSE is as predicted by (14).
For comparison, we also simulate the scheme proposed in
[12] based on the orthogonal MAC model, where the measurement vector for lth cluster is xl¼ flu+ nl which
then transmits to the lth receiver through a diagonal channel gain matrix Dl after multiplying by an amplification matrix
Al; at the FC, the received signal vector from the lth cluster
is yl¼ DlAlxl+ nl, l ¼ 1, . . ., L, where the additive noise
nl is assumed to be E[nl] ¼ 0, E[nln T l]= s 2 nIKl, E[nln T j]=
0Kl×Kl for j = l. After collecting L signal vectors at
the FC, the LMMSE fusion rule is used for estimating the source signal. The performance comparison for the orthogonal and coherent MAC models is plotted in Fig. 6, in which we take P ¼ 10 dB, Kl¼ 3 for all clusters, and
s2n= 1. We see that the MSE of the coherent MAC model performs better than that of the orthogonal MAC model. This is because by using the orthogonal MAC model, the number of receiver noises increases as the number of clusters increases. However, by using the coherent MAC model, there is only one receiver noise regardless of the number of clusters.
Finally to see quantitatively the relation between collaboration and MSE, we consider a network with 30 sensors and P ¼ 0 dB. We perform ten simulations with the number of cluster ranging from 4 to 9. The number of sensors in each cluster is randomly chosen from 1 to 10. In each case, we count the total number of entries in the amplification matrices Al. For example, in the first case
there are four clusters, the numbers of sensors in the clusters are respectively 9, 9, 9 and 3, and the number of entries is 92+ 92+ 92+ 32¼ 252. Table 1 shows the number of cluster, the number of entries, and the corresponding MSE for each case. From the table, we see that the MSE decreases as the number of entries increases.
For comparison, the MSE for the two special cases are respectively JN¼ 0.1689 and JC¼ 0.0392.
5 Conclusion
We study optimal collaboration for distributed estimation in cluster-based wireless sensor network. We show that the optimal amplification matrix of each cluster is a rank one matrix obtained as a scaled outer product of the observation gain and the channel gain vectors. We also show that with optimal amplification matrices, estimation performance is improved compared with the non-collaboration case. We demonstrate, through simulation results, that the amount of improvement is closely related to the amount of collaboration.
6 Acknowledgment
We thank the reviewers for their helpful suggestions that improve the paper. Research sponsored by National Science Council under grant NSC 97-2221-E009-046-MY3.
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Table 1 Different number of sensors in clusters
K ¼ 30 P ¼ 0 dB
clusters, L 4 5 5 6 6 6 7 7 8 9
number of entries 252 226 218 218 200 184 184 166 162 124
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8 Appendix 1: Proof of Proposition 1
Write ATl = [al,1al,2· · · al,N l], where al,j[ R Kl. Define slack variables tl= gTlAlfl=Nl n=1gl,na T l,nfl, 1≤ l ≤ L. The
problem (8) is rewritten as (see (16)) The Lagrangian function for (16) is
L(al,n, tl, ll, l0)= L l=1 s2u Nl n=1 (aTl,nfl)2+ s2n Nl n=1 (aTl,nal,n) + L l=1 ll tl− Nl n=1 gl,naTl,nfl + l0 s 2 n L l=1 Nl n=1 gl,naTl,n Nl m=1 gl,mal,m + s2 n− 1 J∗− 1 s2 u −1 L l=1 tl 2⎤ ⎦
where ll, l0[ R, and the associated necessary conditions for
optimality are ∂L ∂al,n= 2a T l,n(s2u flfTl + sn2IKl)− llgl,nfTl + 2l0s2ngl,ngTlAl= 0T1×Kl, 1≤ n ≤ Nl, 1≤ l ≤ L ⇒ 2Al(s2uflfTl + s2nIKl)− llglf T l + 2l0s 2 nglg T lAl= 0Nl×Kl, 1≤ l ≤ L (17) ∂L ∂tl= ll− 2l0 1 J∗− 1 s2 u −1 L i=1 ti = 0, 1≤ l ≤ L (18) ∂L ∂ll= tl− g T lAlfl= 0, 1 ≤ l ≤ L (19) ∂L ∂l0= s 2 n L l=1 gTlAlATlgl+ s2n− 1 J∗− 1 s2 u −1 L l=1 tl 2 = 0 ⇒ s2 n L l=1 gTlAlATlgl+ s2n= 1 J∗− 1 s2 u −1 L l=1 tl 2 (20) It follows from (18) that l1¼ . . . ¼ lL. Let ll¼ l, ∀l. It
follows from (17) that
Al+ l0s2nglgTlAl(s2uflfTl + s2nIK l) −1 =l 2glf T l(s2uflfTl + s2nIKl) −1 ⇒ Al+ l0glg T lAl− s2ul0tl s2 n+ s2u fl2 glfTl = l 2(s2 n+ s2u fl2) glfTl ⇒ Al= s2ul0tl s2 n+ s2u fl2 (IN l + l0glg T l)−1glfTl + l 2(s2 n+ s2u fl2) (INl+ l0glgTl)−1glfTl ⇒ Al= s2ul0tl (s2 n+ s2u fl2)(1+ l0gl2) glfTl + l 2(s2 n+ s2u fl2)(1+ l0gl2) glfTl (21)
where in the second equation we use the matrix inversion lemma [18, p. 45] and tl= gTlAlfl. Substituting (21) into (19), we obtain tl= lgl2 fl2/(2wl), where wl= s2u fl2+ s2n+ l0s2ngl2, and thus from (21), we obtain Al = l/(2wl)glfTl. From (18) and tl= lgl2 fl2/2wl, we have 1 J∗− 1 s2 u =L i=1 l0gi2fi2 wi (22) Substituting tl= lgl2 fl2/(2wl), Al= l/(2wl)glfTl, and (22) into (20), we have l 2= L i=1 gi 2 f i 2 (s2u fi2+ s2n) wi −1 snl0 (23) min al,n,tl L l=1 s 2 u Nl n=1(a T l,nfl)2+ s2n Nl n=1(a T l,nal,n) subject to Nl n=1gl,n aTl,nfl= tl, 1≤ l ≤ L s2nL l=1 Nl n=1gl,na T l,n Nl m=1gl,mal,m + s2 n= 1 J∗− 1 s2 u −1 L l=1tl 2 ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ (16)
From (23) and Al = l/(2wl)glfTl, the minimum total power can be written as follows:
Pmin= L l=1 tr(s2uAlflf T lA T l + s 2 nAlA T l)= l0s 2 n It follows that l0= Pmin/s2n (24)
and thus from (22), we obtain 1 J∗− 1 s2 u = L i=1 Pmingi2 fi2 s2 n(s2u fl2+ sn2)+ Pmins2ngl2 (25)
Equation (25) gives the relation between the achieved minimum power and the constraint J∗ on MSE. The optimal amplification matrices are Al= l/(2wl)glfTl, where l and wl depends on Pmin through (24). In view of observation
(iii) in Section 3, if we set Pmin¼ P, the corresponding
MSE is given in (10) and the corresponding amplification
matrices is in (9). A
9 Appendix 2: Proof of Lemma 1
We first show that for x= [x1· · · xn]T[ Rn and y= [y1· · · yn]T[ Rn, the following inequality holds
x2y2 x2+ y2≥ n i=1 x2iy2i x2 i + y2i (26) or equivalently, n i=1x2i n j=1y2j n j=1(x2j + y2j) ≥ n i=1 x2iy 2 i n k=i(x 2 k+ y 2 k) n k=1(x2k+ y2k) ⇔ n i=1 x2i n j=1 y2j n k=1 (x2k+ y2k) − n j=1 (x2j + y2j) n i=1 x2iy2i n k=i (x2k+ y2k) ≥ 0 The left-hand side of the above inequality can be written as follows: (see equation at the bottom of the page)
Thus, we obtain (26). Now, let x= [xT1· · · xTL]T= [x1· · · xn]T andxl2 = ˜x2l, then we have
x2=n i=1 x2i = L l=1 xl 2 =L l=1 ˜x2l = ˜x2 (27)
where ˜x= [˜x1· · · ˜xL]T. By the same way, we have
y2= n i=1 y2i = L l=1 yl2= L l=1 ˜y2l = ˜y2 (28) From (26) – (28), we have x2y2 x2+ y2= ˜x2˜y2 ˜x2+ ˜y2 ≥ L l=1 ˜x2l˜y2l ˜x2l + ˜y2l =L l=1 xl2yl2 xl2+ yl2
and the result follows. A
n i=1 n j=1 x2iy2j n k=1 (x2k+ y2k)− n i=1 n j=1 x2iy2i(x2j + y2j) n k=i (x2k+ y2k) = n i=1 n j=i x2iy2j n k=1 (x2k+ y2k)− n i=1 x2iy2i n k=1 (x2k+ y2k)− n i=1 n j=i x2iy2i(x2j + y2j) n k=i (x2k+ y2k) = n i=1 n j=i x2iy2j n k=1 (x2k+ y2k)− n i=1 n j=i x2iy2i(x2j + y2j) n k=i (x2k+ y2k) = n i=1 n j=i x2iy2j(x2i + y2i)− n i=1 n j=i x2iy2i(x2j + y2j) n k=i (x2k+ y2k)= n i=1 n j=i x2i(x2iy2j − x2jy2i) n k=i (x2k+ y2k) = n i=1 n j.i x2i(x2iy2j − x2jy2i) n k=i (x2k+ y2k)+ n i=1 n j,i x2i(x2iy2j − x2jy2i) n k=i (x2k+ y2k) = n i=1 n j.i x2i(x2iy2j − x2jy2i) n k=i (x2k+ y2k)+ n i=1 n j.i x2j(x2jy2i − x2iy2j) n k=j (x2k+ y2k) = n i=1 n j.i x2i(x2j + y2j)(xi2y2j − x2jy2i)− x2j(x2i + y2i)(xi2y2j − x2jy2i) n k=i, j (x2k+ y2k) = n i=1 n j.i (x2iy2j − x2jy2i)2 n k=i, j (x2k+ y2k)≥ 0
