Channel capacity - Signal Processing and Optimization for Transceiver Systems - page 230

circularly symmetric, then the expression on the right-hand side of Eq. (6.57)
does not describe its pdf; we can only use the original form (6.54).

Derivation of Eq. (6.57) . We now show that Eq. (6.54) can be rewritten as Eq.
(6.57) when x − m x is circularly symmetric. Let A = A re + j A im be an arbitrary
N × N matrix and define

B = A re

.

− A im

A im A re

With the 2 N 1 real vector u defined in terms of the complex vector x as in Eq.

(6.53), we can verify the following identity by direct substitution:

×

u

T

Bu =Re[ x † Ax ] . (6 . 58)

Identifying B with 0 . 5 C − 1
uu

and using Eqs. (6.44) and (6.45), it therefore follows

that

0 . 5 u

T

C − 1

uu u = x † C − 1 xx x .

(6 . 59)

The same holds if we replace u and x by the zero-mean versions u − m and x − m x .
Next, from Eq. (6.46) we have det [2 C uu ]=[det C xx ] 2 . Using this and Eq. (6.59)
in Eq. (6.54) we obtain Eq. (6.57).

What we have shown is that if x is Gaussian (i.e., u is Gaussian as in Eq. (6.54))
and if x − m x is circularly symmetric, then the pdf of x can be expressed as in Eq.
(6.57). Conversely, suppose we have a complex random vector x whose pdf can
bewrittenasinEq. (6.57). Doesthismeanthat x − m x is circularly symmetric
Gaussian? Yes, indeed; the expression (6.57) defines C xx , and hence defines P
and Q via (6.56). Using these, define a matrix C uu as in Eq. (6.55). Then
Eqs. (6.59) and (6.46) are true (because these are mere algebraic identities), so
Eq. (6.57) can be written as in Eq. (6.54) proving that u is real Gaussian with
covariance (6.55), i.e., x − m x is circularly symmetric!
Summarizing, the pdf of a complex random vector x can be described by (6.57)
if and only if x − m x is circularly symmetric Gaussian. If x − m x is Gaussian
but not circularly symmetric we have to use the original form (6.54) to describe
the pdf.

Example 6.3: Complex scalar Gaussian

Consider again a zero-mean scalar complex random variable x = x re + jx im
as in Ex. 6.2. We say x is Gaussian if f u ( u )isasinEq. (6.54). When x is
circularly symmetric and Gaussian, Eq. (6.54) reduces to (6.57):

e − | x | 2

1
πσ x

σ x

f u ( u )= f x ( x )= (6 . 60)

and furthermore

σ x re = σ x im = σ x / 2

(6 . 61)