CircularlySymmetric Gaussian random vectors(1)

1、Circularly-Symmctric Gaussian random vectorsRobert G. GallagerJanuary 1, 2008AbstractA number of basic properties about circularly-symmetric Gaussian random vectors are stated and proved here These properties are each probably well known to most researchers who work with Gaussian noise, but I have n

2、ot found them stated together with simple proofs in the literature. They are usually viewed as too advanced or too detailed for elementary texts but are used (correctly or incorrectly) without discussion in more advanced work These results should liavc appeared in Section 7.8.1 of my book, R. G. Gal

3、lager, Principles of Digital Communication/ Cambridge Press, 2008 (PDC08), but I came to understand them only while preparing a solution manual when the book was in the final production stage1 Pseudo-covariance and an exampleLet Z = (Zi, Z、. ，Zn)J be a complex jointly-Gaussian random vector That is，

4、況(ZJ and (ZaJ for 1 k n comprise a set of 2n jointly-Gaussian (real) random variables (rv5s). For a large portion of the situations in which it is useful to view 2n jointly-Gaussian rv5s as a vector of n complex jointly-Gaussian rvs, these vectors have an additional property called circular symmetry

5、 By (lefinition. Z is circularly syninietrie if elZ do not have the same distribution even though, as complex random vectors, they have the same covariance matrixThe definition of circular symmetry seems to capture the intuitive notion of circular syimnetry somewhat better thnii the condition Mz = 0

6、】but the latter is often easier to work with Example 2: Vc now give an example illustrating the importance of the jointly-Gaussian requirement (as opposed to an individually Gaussian requirement) Consider a complex random 2-vector Z for which Z 1) and Z2 = UZ where U is statisticallyindependent of Z

7、i and has possible values 1 with probability 1/2 each Then Z and Z2 are not jointly Gaussian, and in fact 況(ZJ and 況(Z2) have a joint distribution that 1 ri -1is conc(?ntrated on the two ciiagonal axes It is easy to sec that Kz = 0 J and Mz = 0 and thus Z is circularly symmetric The corresponding re

8、al random 4-vector V satisfies Ky = l4, so the rvs are uncorrelatecl However, they are clearly not independent and the probability density does not existThe point of this example is thnt the joint Gaussian i)ropcrty is important in the results that liave been derived The stiibborn believer in an ove

9、rsimplified world might try to think that Examples 1 and 2 rely on the lack of a probability density. This is not true, and each example would lead to the same conclusion if we modified Z by adding a nice random vector Y C(),) This would yield a true probability density and slightly complicate but n

10、ot change the issues discussed.2 Linear transformationsof circularly-symmetrieGaussian vectorsCircular symmetry can also be characterized as the result of a linear transformation (in C) of m iid random variables, each CJV(), 1). Thus let W CJV(), lrn). That is, IVi,. ? Wm are iid and have iid real a

11、nd imaginary parts, thus consisting collectively of 2n iid real rvs, each (0, *) Let A be an arbitrary m by n complex matrix and let Z = A W Then = EA W WW = AA1; Mz = EA W WTAT = 0Thus Z is circularly symmetric and denoted by C;V(0, AA)5As explained in greater detail in Section 7.11.1 of PDC0& the

12、class of covariance nintrices (real or complex) is the same as the class of nonnegative-definite matrices. Each such matrix K can be represented aswhere A is diagonal with nonnegative terins which arc the eigenvalues of K. The columns of Q are orthonormal eigenvectors? of those eigenvalues By choosi

13、ng R = Qx/AQ-13 we see that. Z = R W has the covariance matrix Kz = RRL This means that for any covariance matrix K, there is a matrix R sucli that the random vector Z = RW has covariance K and pseiulo-covciriancc 0. This is suinmcd up in the following tlicor(?m Theorem 2. A necessary and sufficient

14、 condition for a random vector to be a circularly- symmetric jointly- Gaussian random vector is that it has the form Z = AW where W is iid complex Gaussian and A is an arbitrary complex matrixWe now have three equivalent characterizations for circularly-symmetric Gaussian random vectots，first, the d

15、efinition in terins of phase invariance, second, in terins of zero pseudo covariance, and tliir(L in terins of linear transformations of iid Gaussian vectors. One advantage of the third characteriation is that the jointly-Gaussian requirement is automatical ly met, whereas the other two depend on th

16、at, as a separate requirement Another advantage of the third characterization is that the usual motivation for modeling random vectors as circularly syinnietric is tliat they are linear transforinations of essentially iid complex Gaussian random vectors.3 The real and imaginary parts of circularly s

17、ymmet ric random vectorsSince the probability density of a complex random variable or vector is defined in terms of the real and imaginary parts of that variable or vector, we now pause to discuss tliese relationships .Tlie major reason for using complex vector spaces and complex random vectors2A co

18、mplex matrix with orthonormal columns is called a unitary matrix is to avoid all the detail of the real and imaginary parts, but our intuition comes from IR2 and R：In more detail, lri = KzK? implies that 強(Kz)猊(K,) - (Kz)(Kj1) = l and (Kz)(K1) + = 0. With this, it is seen that the product of (6) and

19、 (7) is hn, and the major source of confusion in treating complex rancloni vectors comes from assuming tliat Cn is roughly the same as Rn. This assumption causes additional confusion whon dealing with circular symmetry.Let Z be a random complex n-vector Z = (Z】，Zn)T and let the corresponding real ra

20、ndom 2n-vector V consist of the real and imaginary components of Z. taken in the order 卩= (3?(Zj.，況(ZJ，纨乙)产We start by relating the 2n by 2n real covariance matrix Ky = EV VJ to the n by n matrix Kz = EZZt The (real) covariance matrix K y can tlien be expressed in block form in terms of the (complex

21、) covariance matrix KzE況(Z)況(ZT E5R(Z)3(ZT) _K y =_ E&(Z)猊0) EG(Z)S(ZT) _Assume in what follows that Z is circularly symmetric The pseudo-covariance matrix, Mz is then zero, so expressing (2), in matrix form, we get E3?(Z)Ji(ZT) = *?R(Kz) Using (3)： (4), and (5) in the same way, Ky can be expressed

22、as_烈(Kz)-护(Kz厂Ky =1-23?(Kz)It can then be verified by matrix multiplication3 with (7) that if Kz is non-singular, then Ky1 is also noil-singular and its inverse is given in block form by_ 25?(K?)2&dK J =(8).29(K)2況(K?).We next relate the eigenvalue, eigenvector pairs of Kz to those of Ky. Let Xj be

23、an cigcmvaluo of Kz and let qj = (7ij5 . . QnjV be a corresponding (ngenvector, chosen so that the set of eigenvectors is orthonormal Let紀(g町),综(3(zT). Using this and the expression in (8) for K詁：vJyV = 2zz z(11)Establishing this equality also uses the fact thatis real. We now recall that eacheigenv

24、alue Xj of Kz corresponds to two eigenvalues, both Xj/2, of K y Thusdet(Ky) = 22n(detKz)2(12)Substituting these relations into (10), we get an expression for fz in terms of Kz(13)Note that (10) is valid for any jointly-Gaussian 2n-vector, whereas (11) and (12), and thus (13), depend on circular symn

25、ietiy. Next： suppose ail arbitrary random vectOT Z has the density in (13). We have seen that there is a circularly-symmetrie Gaussian random vector with the covariance matrix Kz, and this also has the density in (13). Since the density fully describes whether Z is circularly-symmet.ric jointly Gaus

26、sian, we conclude that (13) is a necessary- and sufficient condition for a non-singular complex random vector to be circularly-symrnetrie jointly Gaussian. The following theorein summarizes this.Theorem 3. Assume that Z is a complex random n-vector with an arbitrary non-singular covariance Kz Then (

27、13) is a necessary and sufficient condition for Z to be CA/0 Kz)If Kz is singular, then one or more components of Z arc deterministic linear combinations of the other components, so the most, convenient way of specifying Z is by first removing components of Z that are deterministic linear combinatio

28、ns of the remaining components and using (13) for tlie remaining componentsAs witli real jointly-Gaussian random vectors，it is often more insightful to express the probability d(?nsity in terms of the eigenvalues and cigciifimctions of Kz Letting 入，An be the eigenvalues of Kz (repeated as necessary)

29、 and qi， 、qn be the corresponding eigenfunctions, we can use (6) to express Kz asSubstituting this into (13)7rn det(Kz) 卩9Expressing qz as the projection of z on q” i.e., as z, q and recalling that (let Kz = Yj Xj、this bccoinosn(14)fz(z) = H 灰 exp (-|z,如)|2入)3This is the density of n independent cir

30、cularly-symmetric Gaussian random variables. (Z.gi，. qu) with variances 入1，An respectively. In other words, expressing Z in the orthonormal basis q13 ,Qn, the variables in this basis are independent, circularly-syminctric Gaussian random variables with variances 入i，：入介 This is the same as the analog

31、ous result for jointly-Gaussian real random vectors which says that there is always an orthonormal basis in which the variables are Gaussian and independent. This analogy forms the simplest, way to (sort of) visualize circularly-symmetric Gaussian vectors - they have the same kind of elliptical syin

32、nietry as the real case, except that here, each complex random variable is also circularly symmetric. It should be clear that (14) is also an if-and-only-if condition for circularly-symmetric joint.ly-Gaussian random vectors5 Linear functionals of circularly-symmetrie random vectorsLet Z CN Kz) If s

33、ome other random vector Y can be expressed as Y = BZ, then Y is also a circularly-symmetric jointly-Gaussian random vector. To see this, represent Z as Z = A W where W CJV(OJ)/Then Y = BA W, so Y CV()?Thishelps show why circular symmetry is important - it is invariant to linear transformations. If B

34、 is 1 by n (i.e., if it is a row vector 6T) then Y = bJZ is a complex rv. Such tvs are called linear functionals of Z Thus all linear functionals of a circularly-symmetrie jointly Gaussian random vector are circularly-symmetric Gaussian tvs.Conversely, we now want to show that if all linear function

35、als of a complex random vector Z are circularly-symmetric Gauss诅n，then Z must also be circularly-symmetric： and jointly-Gaussian The question of being jointly Gaussian can be separated from that of being circularly symmctrie. Thus assume that for all complex n-vcctors the complex rv b Z is complex G

36、aussian. Looking at V and b as real 2zi vectors, V = (脫0),，脫(Zn),m(Zi),. ,?(ZJ 沁1卢=(況(),.,況低),3). ,3(6n)T,we see that rT V is a Gaussian random variable for all b Since r can be arbitrarily chosen by choosing the real and imaginary components of b, we sec that all real linear functionals of V are Ga

37、ussian. It is known for real random vectors (see for example Section 7.3.6 of PDC08) that V is jointly Gaussian if all its linear functionals are Gaussian random variables. Thus Z is complex jointly Gaussian if all its linear functionals are complex Gaussian random variables.We could now show that Z

38、 is also circnlarlv svnimctrie and jointly Gaussian if bTZ is circularly-symmetric Gaussian for all b, but it is just as easy, and yields a slightly stronger result, to show that Z C人厂(0, Kz) if Z is jointly Gaussian and, in addition, the limited set of linear functionals Zj + Zk is circularly symmetric Gaussian for all j, k If Z)+ Z) is circularly syininetric for all 久 then EZj = 0. so that the main diagonal of Mz is zero. If in addition, Z)+ Z& is circularly syinirietric, then E(Zj + Z2 = 0. But since EZJ = EZl = 0, 2EZjZj = 0