For simplicity, we assume that r, b1, and b2 have a mean of zero and a variance of one. From (1), it follows that (6)Cx1x2(A)=E[x1x2]E[x12]E[x22]=E[(R−b1)(R−b2)]E[(R−b1)2]E[(R−b2)2]=E[(Ar−b1)(Ar−b2)]E[(Ar−b1)2]E[(Ar−b2)2](7)=A2−(E[rb1]+E[rb2])A+E[b1b2](A2−2E[rb1]A+1)(A2−2E[rb2]A+1). Thus, Cx1x2(A) is a function of coefficient A, where A > 0. Based on (7), it is easy to see that Cx1x2(A)→1 as A → +∞. Therefore, and as expected when A is large enough, the correlation of two referential signals x1 and x2 will be larger than that of nonreferential signals b1 and b2. In general, it is not easy to discuss the critical points for Cx1x2(A) because these points depend on the values of E[rb1], E[rb2], and E[b1b2]. For simplicity, we further assume E[rb1] = E[rb2] = c. Obviously, –1 ≤ c ≤ 1 and –1 ≤ E[b1b2] ≤ 1 by noting that r, b1, and b2 have a mean of zero and a variance of one. Then, (7) becomes (8)Cx1x2(A)=A2−2cA+E[b1b2]A2−2cA+1=1−1−E[b1b2]A2−2cA+1. Since A2–2cA+1 has a minimum value at A = c, one can see that, when –1 ≤ c ≤ 0, Cx1x2(A) monotonically increases as A(> 0) increases. Fig. 1(A) and (B)(b) show
