A Note on the Bivariate Distribution Representation of Two Perfectly

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Let’s further assume that random variables X and Y are identically distributed, with mean values of equal or opposite signs. We state then that the joint probability distribution fX,Y (x, y) can be compactly represented as follows f (x)δ(x y) if ρ = 1 and µ = µ , f (x, y)= X X Y (4) X,Y 0 ∓ if ρ = ±1 and µ = ±µ =0, ± X ∓ Y 6 where the upper and the lower signs should be evaluated separately, δ(.) is the δ-function or Dirac’s δ-function and f (x)δ(x X ∓ y)= fY (y)δ(y x). Indeed, for perfectly∓ correlated random variables ρ = 1 we have from (1) that ± σ2 = E[xy] µ2, (5) ± ∓ where the (+) and ( ) signs on the left hand side of (4) correspond to ρ =1 and ρ = 1, respectively. On the righthand side − − the (+) and ( ) signs correspond to opposite sign mean values (µ = µX = µY ) and equal sign mean values (µ = µX = µY ), respectively. − − From (2) we have that σ2 = E[x2] µ2. (6) − Consider now E[(x ky)2]=0. Thus, solving for k and combining (5) and (6) gives − E[xy] E2[xy] E2[x2] (7a) k = ± 2 − pE[x ] σ2 µ2 ( σ2 µ2)2 (σ2 + µ2)2 (7b) = ± ± ± ± 2 ± 2 − , p σ + µ where we have used our assumption of identically distributed variables. A straightforward analysis of (7b) leads to three possibilities: (I), k =1 if ρ =1 and µ = µ = µ =0 or µ = µ = µ =0, (II), k = 1 if ρ = 1 and µ = µ = µ =0 X Y X Y 6 − − X Y or µ = µX = µY = 0, and (III), there is no (real) solution if ρ = 1 and µ = µX = µY = 0 or if ρ = 1 and µ = µ = µ =0− . Hence,6 summarizing, we ﬁnd that − 6 − X Y 6 1 if ρ = 1 and µ = µ = µ , k = X Y (8) ± if ρ = ±1 and µ = µ = ±µ =0. {∅} ± X ∓ Y 6 Clearly, k = 1 in the ﬁrst case implies a linear, deterministic relationship between random variables X and Y with probability one. Thus, we± can write f (y x)= δ(y x) and f (x y)= δ(x y), (9) Y |X | ∓ X|Y | ∓ since δ(y x)= δ(x y). Now, inserting (8) into (9) gives ± ± f (x, y)= f (x)δ(y x)= f (y)δ(x y). (10) X,Y X ∓ Y ∓ On the other hand, k = implies in the second case that there exist no linear, deterministic relationship between random variables X and Y and therefore{∅} a perfect correlation between the variables cannot be observed. Hence, in this case

fX,Y (x, y)=0. (11) This ends our derivation. Let us consider bivariate distributions of identically distributed random variables (with mean values of equal or opposite signs) as previously. We further assume they also depend on the correlation coefficient ρ as a parameter, and are well-defined functions for 0 ρ < 1. However, such distributions are in general not well-defined for ρ = 1 since terms of the form (1 ρ)−1, (1 ρ≤)− |1/|2, (1 ρ2)−1 and/or (1 ρ2)−1/2 may appear in their arguments. Examples| | of such bivariate distributions are± the Gaussian± (for both± complex and real± variables) and the family of distributions derived from the complex Gaussian multivariate distribution , [3], such as the bivariate Rice , [4], the bivariate Nakagami, [5], and the bivariate Weibull distributions , [6]. Henceforth, we denote such a bivariate distribution as fX,Y (x, y; ρ) to make a distinction between the joint distribution and bivariate distribution, i.e., we explicitly point out the dependence on ρ for 0 ρ < 1. Therefore, we can write ≤ | | f (x, y)= f (x, y; ρ) if 0 ρ < 1. (12) X,Y X,Y ≤ | | However, as we have shown above the singularities can be represented through the δ-function when ρ =1 and µX = µY , or ρ = 1 and µ = µ or ρ = 1 and µ = µ =0. Thus − X − Y ± X Y fX,Y (x, y) = lim fX,Y (x, y; ρ)= fX (x)δ(x y). (13) ρ→±1 ∓ Hence, Dirac’s δ-function has the following representation f (x, y; ρ) δ(x y) = lim X,Y . (14) ∓ ρ→±1 f ( y) X ± 3

Assume now that Xc C and Yc C are two random variables defined on some probability space. Thus, the Pearson product-moment correlation∈ coefficient∈ between them is defined as, [3] ∗ ∗ E[(xc µXc )(yc µYc )] ρc = − − , (15) σXc σYc ∗ where E[.] denotes the expected value operator, (.) denotes complex-conjugate, µXc = E[xc] and µYc = E[xc] denote the mean values, and σXc and σYc denote the standard deviations, i.e. σ = E[ x µ 2] and σ = E[ y µ 2]. (16) Xc | c − Xc | Yc | c − Yc | ∗ R We further assume that ρc = ρc , i.e., ρc p. In this case if Xc and Yc pare identically distributed with mean values of equal or opposite signs, then the following holds∈ true ρ =1 ρ = 1 if x = x , y = y , (17) | c| ⇒ ± | c| ±| c| where, the upper and the lower signs should be evaluated separately. The converse, i.e., ρ = 1 ρc =1, doesn’t hold in general. This follows from the definition of the correlation coefficients (1) and (15). | | ⇒ | |

III. ANEXAMPLE: THE BIVARIATE RICE DISTRIBUTION Consider the bivariate Rice distribution of two identically distributed random variables X R and Y R ∈ ∈ 2 2 2 2K (1+K)(x +y ) (1 + K) xy − 1+ρ − − 2 c 2β(1 ρc ) (18) fX,Y (x, y; ρc)= 2 2 e 2πβ (1 ρc) 2π − ρc(1+K)xy cos θ 2 2 − 2 2K(1 + K)(x + y +2xy cos θ) β(1 ρc ) e I0 2 dθ, × s (1 + ρc) β ! Z0

x2 y2 where β = 2 = 2 is the average power of, e.g., the signal received at one antenna, K is the power of the line-of-sight(LOS) component to the scattered power and ρc is the correlation coefficient (15), and I0(.) is the modified Bessel function of 0th order. We see from (18) that fX,Y (x, y; ρc) is not well-defined for ρc = 1, so it should be evaluate in the limit ρc 1 to obtain a value if it exists. ± → Evaluating the limit by regrouping terms and using the Algebraic Limit Theorem gives

−K− (1+K)xy (1 + K)2xye 2β (19) lim fX,Y (x, y; ρc)= 2 G1(x, y) ρc→1 2β 2π K(1 + K)(x2 + y2 +2xy cos θ) G2(x,y,θ)I0 dθ, × s 2β ! Z0 where (1+K)(x−y)2 − − 2 e 2β(1 ρc ) (20) G1(x, y) = lim 2 1/2 , ρc→1 √π(1 ρ ) − c ρc(1+K)xy(1−cos θ) − − 2 e β(1 ρc ) (21) G2(x,y,θ) = lim 2 1/2 , ρc→1 √π(1 ρ ) − c 0 The limits in (20) and (21) are of the form 0 , which can be resolved by noticing that Dirac’s-δ can be represented in terms of the limit 2 − z e ǫ2 δ(z) = lim . (22) ǫ→0 √πǫ Thus, G (x, y)= δ (1 + K)/(2β)(x y) , (23) 1 − and p G (x,y,θ)= δ xy(1 + K)(1 cos θ)/β . (24) 2 − Let’s ﬁrst consider (23), which by applying the followingp property of Dirac’s δ-function δ(x x ) δ(g(x)) = − n , (25) dg(xn) n X dx

where the xn satisfy g(xn)=0, is reduced to 2β G1(x, y)= δ(x y), (26) r1+ K − Similarly, applying (25) to (24) we identify g(θ)= ((1 + K)xy(1 cos θ))/β, (27) − with the modulus of the ﬁrst derivative given by p

dg(θ) (1 + K)xy θ = cos , (28) dθ s 2β 2

where we have used some standard mathematical identities and operations. Further, it is straightforward to see that the zeros of (27) are given by θn =2πn. Thus, ∞ 2β G (x,y,θ)= δ(θ 2πn), (29) 2 (1 + K)xy − s n=0 X The shifting property of Dirac’s δ-function

f(x)δ(x x )dx = f(x ), (30) − 0 0 DZ where is the domain of integration and x0 . Thus, inserting (26) and (29) into (18) and applying (30) we obtain the ﬁnal resultD ∈ D

lim fX,Y (x, y; ρc)= δ(x y) (31) ρc→1 −

2 x(1 + K) −K− (1+K)x 2K(1 + K) e 2β I0 x . × β s β !

IV. CONCLUSIONS The Pearson product-moment correlation coefﬁcient ρ plays a fundamental role in the analysis of the performance of diversity systems for wireless communications. We have shown that for bivariate distributions belonging to the Gaussian family, e.g., Rayleigh, Ricean, Nakagami and Weibull, and perfectly correlated random variables they can be represented by Dirac’s δ- function if the corresponding linear dependence exists. The presented result can be used in the analytical and numerical analysis of the diversity performance of perfectly correlated diversity branches. In addition, for nearly perfectly correlated results can be obtained by expanding the joint probability distribution around ρ 1 or ρ 1. Finally, we believe that for the sake of completeness the limiting cases corresponding to ρ = 1 and/or ρ ≈= 1 should≈ − also be given when providing the joint probability distribution of two random variables. −

ACKNOWLEDGMENTS This work has been partly supported by the EU FP6 “GAWIND” project and the EU FP7 IAPP “IAPP@RANPLAN” project.

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