Regions that the IQS is Suboptimal

After identifying the best IQS for a given hypothesis distribution, we next test whether the NQS can improve the best IQS or not. As aforementioned, it would be easier to disprove the optimality of the IQS rather than proving its optimality because we only need to show the existence of an NQS that outperforms the best IQS.

To achieve this goal, we choose to replace one of the sensors’ LRQ by a distinct one. By this, we may possibly identify some area on the hypothesis distribution domain, which the NQS outperforms the IQS. However, we found that such areas may be changed for different number of sensors n. We then ask ourselves whether there exists a specific area that the NQS is always better than the IQS for most n. For this purpose, we overlap our experimental results for different n and look for the common area that the NQS (or more specifically, the near-IQS) is better. We found some desired common areas:

Segment 1: (n − 1) type 1 + 1 type 2 : P1 = ²₃,₆₄⁵ ≤ P2 < ¹₆ Segment 2: (n − 1) type 1 + 1 type 3 : P1 = ²₃,¹₆ < P2 ≤ ₁₉₂⁴⁹ Segment 3: (n − 1) type 2 + 1 type 1 : P2 = ²₃,₆₄⁵ ≤ P1 < ¹₆ Segment 4: (n − 1) type 2 + 1 type 3 : P2 = ²₃,¹₆ < P1 ≤ ₁₉₂⁴⁹

Segment 5: (n − 1) type 3 + 1 type 1 : P1+ P2 = ³₄,²¹₆₄ ≤ P1 < ³₈,³₈ < P2 ≤ ²⁷₆₄ Segment 6: (n − 1) type 3 + 1 type 2 : P1+ P2 = ³₄,³₈ < P1 ≤ ²⁷₆₄,²¹₆₄ ≤ P2 < ³₈

(3.5)

Now take the first segment as an example. We have already known that the type-1 IQS is the best IQS in the specified ranges of P1 and P2. We then replace one of the local quantizer to a type-2 quantizer ˆg. For reader’s convenience, the post-quantization hypothesis distributions of the type-1 and type-2 quantizers are tabulated in Table 3.2.

Table 3.2:

u 0 1

P¯g(u) P1 1 − P1

P¯g(u) ¹₃ ²₃

u 0 1

Pˆg(u) P2 1 − P2

Qgˆ(u) ¹₃ ²₃

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Using the same technique introduced in Chapter 2, we derive:

γn

Taking the distributions in Table 3.2 into (3.6), we can rewrite (3.6) as γn of (1/3, 64/113) and (64/113, 64/79). So the critical l lies in

1

and

Similar derivation can be applied to Fig. 3.2(b). The two ranges for Fig. 3.2(b) are (1/3, 128/193) and (128/193, 128/191), where ˆγ(π) is larger than ¯γ(π) in the former range but is smaller than ¯γ(π) in the latter range. This results in the derivation that

1

A) Line Segment 1

We now give a general result for line segment 1. For P1 = 2/3, the two gamma functions are given by:

and for P2 < 1 − P1 = 1/3,

γ(π) in the former range but is smaller than ¯γ(π) in the latter range. This results in the derivation that Suppose n = 2k is even. Therefore, if

k −1 always outperform γn,1. Since Figure 3.1 indicates that γn,1 is the optimal IQS between the two margins of γ_n,1 = γ_n,2 and γ_n,1 = γ_n,3 for ¹₃ ≤ P₁ ≤ 1 and 0 ≤ P₂ ≤ ¹₃, we can then decide numerically that for the line segment decided by P₁ = ²₃ and ₆₄⁵ ≤ P₂ < ¹₆, the IQS is only suboptimal.

Note that when P2 = ¹₆, no integer satisfies both (3.12) and (3.13); hence, using (n − 1) type 1 + 1 type 2 quantizers have the same performance as γn,1. In such case, we can no longer claim that the IQS is only suboptimal.

B) Line Segment 2

We can similarly examine our result regarding line segment 2.

Again, for P1 = 2/3, the two gamma functions are given by:

¯

γ(π) in the former range but is smaller than ¯γ(π) in the latter range. This results in the

derivation that Suppose n = 2k is even. Therefore, if

k − 1 quantizers always outperform γn,1. Since Figure 3.1 indicates that γn,1 is the optimal IQS between the two margins of γn,1 = γn,2 and γn,1 = γn,3 for ¹₃ ≤ P1 ≤ 1 and 0 ≤ P2 ≤ ¹₃, we can then decide numerically that for the line segment decided by P1 = ²₃ and ¹₆ < P2 < ₁₉₂⁴⁹, the IQS is only suboptimal.

Note that when P2 = ¹₆, no integer satisfies both (3.14) and (3.15); hence, using (n − 1) type 1 + 1 type 3 quantizers have the same performance as γn,1. In such case, we can no longer claim that the IQS is only suboptimal.

C) Line Segment 3

We now prove the validity of line segment 3. For P2 = 2/3, the two gamma functions are

given by:

γ(π) in the former range but is smaller than ¯γ(π) in the latter range. This results in the derivation that

Suppose n = 2k is even. Therefore, if (n − 1) type 2 + 1 type 1 quantizers always outperform γn,2. Since Figure 3.1 indicates that γn,2 is the optimal IQS between the two margins of γn,1= γn,2 and γn,2 = γn,3 for 0 ≤ P1 ≤ ¹₃ and ¹₃ ≤ P2 ≤ 1, we can then decide numerically that for the line segment decided by P2 = ²₃ and ₆₄⁵ ≤ P1 < ¹₆, the IQS is only suboptimal.

Note that when P1 = ¹₆, no integer satisfies both (3.16) and (3.17); hence, using (n − 1) type 2 + 1 type 1 quantizers have the same performance as γn,2. In such case, we can no longer claim that the IQS is only suboptimal.

D) Line Segment 4

We now proceed to prove the validity of line segment 4. For P2 = 2/3, the two gamma functions are given by:

¯

It can be verified that _3P12²₊₂ < _4−3P¹ ₁₂; hence, than ¯γ(π) in the former range but is smaller than ¯γ(π) in the latter range. This results in the derivation that Suppose n = 2k is even. Therefore, if

k − 1 (n − 1) type 2 + 1 type 3 quantizers always outperform γn,2. Since Figure 3.1 indicates that γ_n,2 is the optimal IQS between the two margins of γ_n,1= γ_n,2 and γ_n,2 = γ_n,3 for 0 ≤ P₁ ≤ ¹₃ and ¹₃ ≤ P₂ ≤ 1, we can then decide numerically that for the line segment decided by P₂ = ²₃ and ¹₆ ≤ P1 < ₁₉₂⁴⁹, the IQS is only suboptimal.

Note that when P1 = ¹₆, no integer satisfies both (3.18) and (3.19); hence, using (n − 1) type 2 + 1 type 3 quantizers have the same performance as γn,2. In such case, we can no longer claim that the IQS is only suboptimal.

E) Line Segment 5

Our next proof is for line segment 5. For P1+ P2 = 3/4, the two gamma functions are

Hence, to check wether the IQS is suboptimal, we examine:

8

17 < 1

1 + (⁸₉)^l(⁴₃)^n−l−1 < 1 3P₁− 1, or equivalently,

log₂(3P₁) + (1 − n) log₂(⁴₃)

log₂(²₃) < l < log₂(⁹₈) + (1 − n) log₂(⁴₃)

log₂(²₃) = αn − 1. (3.20) where α = ^2−log_1−log²⁽³⁾

2(3) ≈ 0.7095 is an irrational number.

Now, if we place probability mass 1/N at points {αn − ⌊αn⌋}^N_n=1 to form a probability measure µn, then by the theory about the so-called uniformly distributed modulo 1 [6], we have µn converges in distribution to µ, where µ is a uniform distribution over (0, 1]. This indicates as long as

P₁ < 3 8,

there exists integer l satisfying (3.20) for infinitely many n. Hence, we conclude that when 0 ≤ P1 < ⁸₃ subject to P1 + P2 = ³₄, using (n − 1) type 3 + 1 type 1 quantizers always outperform γn,3. Since Figure 3.1 indicates that γn,3 is the optimal IQS between the two margins of γn,1 = γn,3 and γn,2 = γn,3, we can then decide numerically that for the line segment decided by P₁+ P₂ = ³₄, ²¹₆₄ ≤ P₁ < ³₈ and ³₈ < P₂ ≤ ²⁷₆₄, the IQS is suboptimal infinitely often in n.

Note that when P1 = ³₈, no integer satisfies (3.20); hence, using (n − 1) type 3 + 1 type 1 quantizers have the same performance as γn,3. In such case, we can no longer claim that the IQS is only suboptimal.

F) Line Segment 6

We now turn to the last proof about line segment 6. For P12 = P1 + P2 = 3/4, the two

gamma functions are given by:

Hence, to examine whether the IQS is suboptimal or not, we continue to derive 8

where α = ^2−log_1−log²⁽³⁾

2(3) ≈ 0.7095 is an irrational number.

Again, if we place probability mass 1/N at points {αn − ⌊αn⌋}^N_n=1 to form a probability measure µn, then by the theory about the so-called uniformly distributed modulo 1 [6], we have µn converges in distribution to µ, where µ is a uniform distribution over (0, 1]. This indicates as long as

P2 < 3 8,

there exists integer l satisfying (3.21) for infinitely many n. Hence, we conclude that when 0 ≤ P₂ < ⁸₃ subject to P₁ + P₂ = ³₄, using (n − 1) type 3 + 1 type 2 quantizers always outperform γn,3. Since Figure 3.1 indicates that γn,3 is the optimal IQS between the two margins of γn,1 = γn,3 and γn,2 = γn,3, we can then decide numerically that for the line segment decided by P1+ P2 = ³₄, ³₈ ≤ P1 < ²⁷₆₄ and ²¹₆₄ ≤ P2 < ³₈, the IQS is suboptimal infinitely often in n.

Note that when P2 = ³₈, no integer satisfies (3.21); hence, using (n − 1) type 3 + 1 type 2 quantizers have the same performance as γn,3. In such case, we can no longer claim that the IQS is only suboptimal.