Circularly Symmetric Gaussian Random Vectors Saravanan Vijayakumaran sarva@ee.iitb.ac.in Department of Electrical Engineering Indian Institute of Technology Bombay October 1, 2013 1 / 11
Circularly Symmetric Gaussian Random Vectors Definition (Complex Gaussian Random Variable) If X and Y are jointly Gaussian random variables, Z = X + jY is a complex Gaussian random variable. Definition (Complex Gaussian Random Vector) If X and Y are jointly Gaussian random vectors, Z = X + j Y is a complex Gaussian random vector. Definition (Circularly Symmetric Gaussian RV) A complex Gaussian random vector Z is circularly symmetric if e j φ Z has the same distribution as Z for all real φ . If Z is circularly symmetric, then • E [ Z ] = E [ e j φ Z ] = e j φ E [ Z ] = ⇒ E [ Z ] = 0. • cov ( e j φ Z ) = E [ e j φ Z e − j φ Z H ] = cov ( Z ) . (See footnote) • Define pseudocovariance of Z as E [ ZZ T ] . E [ ZZ T ] = E [ e j φ Z e j φ Z T ] = e 2 j φ E [ ZZ T ] = ⇒ E [ ZZ T ] = 0 cov ( e j φ Z ) = cov ( Z ) holds for any zero mean Z (even non-circulary symmetric Z ) 2 / 11
Circular Symmetry for Random Variables • Consider a circularly symmetric complex Gaussian RV Z = X + jY • E [ Z ] = 0 = ⇒ E [ X ] = 0 and E [ Y ] = 0 ⇒ E [ ZZ T ] = E [ Z 2 ] = 0 = • Pseudocovariance zero = ⇒ var ( X ) = var ( Y ) , cov ( X , Y ) = 0 • If Z is a circularly symmetric complex Gaussian random variable, its real and imaginary parts are independent and have equal variance 3 / 11
Complex Gaussian Random Vector PDF • The pdf of a complex random vector Z is the joint pdf of its real and imaginary parts i.e. the pdf of � X � ˜ Z = Y • For a complex Gaussian random vector, the pdf is given by 1 � − 1 � m ) T C − 1 p ( z ) = p (˜ 2 (˜ z − ˜ Z (˜ z − ˜ z ) = exp m ) ˜ 1 ( 2 π ) n ( det ( C ˜ Z )) 2 m = E [˜ where ˜ Z ] and � C X � C XY C ˜ Z = C YX C Y � ( X − E [ X ])( X − E [ X ]) T � C X = E � ( Y − E [ Y ])( Y − E [ Y ]) T � C Y = E � ( X − E [ X ])( Y − E [ Y ]) T � C XY = E � ( Y − E [ Y ])( X − E [ X ]) T � = C YX E 4 / 11
Circularly Symmetry and Pseudocovariance • Covariance of Z = X + j Y � ( Z − E [ Z ])( Z − E [ Z ]) H � C Z = E = C X + C Y + j ( C YX − C XY ) • Pseudocovariance of Z = X + j Y is � ( Z − E [ Z ])( Z − E [ Z ]) T � E = C X − C Y + j ( C XY + C YX ) • Pseudocovariance is zero = ⇒ • C X = C Y , C XY = − C YX • C Z = 2 C X + 2 j C YX � T for zero pseudocovariance is • The covariance of ˜ � Z = X Y � C X � C X � 1 − 1 C XY � − C YX � 2 Re ( C Z ) 2 Im ( C Z ) � C ˜ Z = = = 1 1 C YX C Y C YX C X 2 Im ( C Z ) 2 Re ( C Z ) 5 / 11
Circularly Symmetry and Pseudocovariance Theorem A complex Gaussian vector is circularly symmetric if and only if its mean and pseudocovariance are zero. Proof. • The forward direction was shown in the first slide. • For reverse direction, assume Z is a complex Gaussian vector with zero mean and zero pseudocovariance. � T • The pdf of Z is the the pdf of ˜ � Z = X Y � � 1 − 1 z T C − 1 p ( z ) = p (˜ 2 ˜ Z ˜ z ) = exp z ˜ 1 ( 2 π ) n ( det ( C ˜ Z )) 2 where � 1 − 1 � 2 Re ( C Z ) 2 Im ( C Z ) C ˜ Z = 1 1 2 Im ( C Z ) 2 Re ( C Z ) • We want to show that e j φ Z has the same pdf as Z 6 / 11
Circularly Symmetry and Pseudocovariance Proof Continued. • e j φ Z has zero mean and zero pseudocovariance • cov ( e j φ Z ) = cov ( Z ) • If Z is a complex Gaussian vector, e j φ Z is a complex Gaussian vector (Assignment 4) • Let U = e j φ Z and ˜ � T � Re ( U ) Im ( U ) U = • The pdf of U is the the pdf of ˜ U 1 � − 1 � u T C − 1 p ( u ) = p (˜ 2 ˜ U ˜ u ) = exp u ˜ 1 ( 2 π ) n ( det ( C ˜ U )) 2 where � 1 � 1 − 1 − 1 � � 2 Re ( C U ) 2 Im ( C U ) 2 Re ( C Z ) 2 Im ( C Z ) C ˜ U = = 1 1 1 1 2 Im ( C U ) 2 Re ( C U ) 2 Im ( C Z ) 2 Re ( C Z ) • U has the same pdf as Z 7 / 11
PDF of Circularly Symmetric Gaussian Vectors • The pdf of a complex Gaussian vector Z = X + j Y is the pdf of ˜ � T Z = � X Y � � 1 − 1 m ) T C − 1 p ( z ) = p (˜ 2 (˜ z − ˜ Z (˜ z − ˜ z ) = exp m ) ˜ 1 ( 2 π ) n ( det ( C ˜ Z )) 2 • If Z is circularly symmetric, the pdf is given by 1 � � − z H C − 1 p ( z ) = π n det ( C Z ) exp Z z We write Z ∼ CN ( 0 , C Z ) 8 / 11
Projection of Complex WGN is Circularly Symmetric • Let ψ 1 ( t ) , ψ 2 ( t ) , . . . , ψ K ( t ) be a complex orthonormal basis • Let n ( t ) be complex white Gaussian noise with PSD 2 σ 2 • Consider the projection of n ( t ) onto the orthonormal basis � n , ψ 1 � . N = . . � n , ψ K � • N is circularly symmetric and its pdf is −� n � 2 � � 1 1 � � − n H C − 1 p ( n ) = π K det ( C N ) exp N n = ( 2 πσ 2 ) K exp 2 σ 2 • If Y = s i + N , then the pdf of Y is −� y − s i � 2 1 � � p ( y ) = ( 2 πσ 2 ) K exp 2 σ 2 9 / 11
Reference Circularly Symmetric Gaussian Random Vectors Robert G. Gallager http://www.rle.mit.edu/rgallager/documents/ CircSymGauss.pdf 10 / 11
Thanks for your attention 11 / 11
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