Minimization of error probability in transceivers - Signal Processing and Optimization for Transceiver Systems - page 522

where the noise variances σ q k are ordered such that

σ q 0 ≤ σ q 1 ≤ σ q M− 1 ... ≤ (16 . 103)

Proof of Eq. (16.100). If K = M then Eq. (16.100) reduces to
M M− 1
=0

σ q

σ q 2 ≥

1 . (16 . 104)

M− 1
=0

Now, from Cauchy-Schwartz inequality we have

σ q 2

M− 1 M− 1 M− 1 M− 1

1 2 σ q = M σ q ≤

=0 =0 =0 =0

which proves (16.104) indeed. Next assume that the power p 0 is such that
K<M. That is, the noise variances are such that Eqs. (16.101) and (16.102)
are true. For simplicity, we first assume that

σ q K = σ q K +1 = ... = σ q M− 1

(16 . 105)

and that Eq. (16.101) holds with equality:

K− 1 K− 1

σ q = σ q k

p 0 + σ q , K ≤ k ≤ M − 1 . (16 . 106)

=0 =0

Substituting into Eq. (16.100) we can rewrite it as

M σ q K K− 1 σ q

=0

K− 1
=0

σ q 2 +( M K ) σ q K K− 1 σ q −

=0

M σ q K K− 1 σ q

σ q − K− 1

=0 =0

−

K ) σ q K 2 ≥

1 ,

K− 1
=0

σ q +( M −

which simplifies to

M σ q K K− 1 σ q

σ q − K− 1

Mσ q K

=0 =0

K− 1
=0

σ q +( M K ) σ q K −

K ) σ q K 2 ≥ 1 .

K− 1
=0

−

σ q +( M −

This can be rewritten as

K ) σ q K + M K− 1 σ q M ( M −

=0

K ) σ q K 2 ≥

1 ,

K− 1
=0

σ q +( M −