why that normalized cross correlation(NCC) lies in $[-1,1]$?

Question

Provided two real number sequences: $a_1,a_2,...,a_n$;$b_1,b_2,...,b_n$, define their means respectively: $$\bar a=\frac{1}{n}\sum_{i=1}^n a_i,\bar b=\frac{1}{n}\sum_{i=1}^n b_i$$ and define their variances and covariance respectively: $$var(a)=\frac{1}{n}\sum_{i=1}^n (a_i-\bar a)^2,var(b)=\frac{1}{n}\sum_{i=1}^n (b_i-\bar b)^2,cov(a,b)=\frac{1}{n}\sum_{i=1}^n (a_i-\bar a)(b_i-\bar b)$$ naturally leads to the definition of normalized cross correlation: $$NCC=\frac{cov(a,b)}{\sqrt{var(a)var(b)}}=\frac{\sum_{i=1}^n(a_i-\bar a)(b_i-\bar b)}{\sqrt{\sum_{i=1}^n (b_i-\bar b)^2 \sum_{i=1}^n (a_i-\bar a)^2}}$$ Now how to show that $NCC$ lies in $[-1,1]$?

Answer 1

Let $x_j:=a_j-\bar a$ and $y_j:=b_j-\bar b$, then your question is equivalent to show that

$$\left(\sum_{j=1}^n x_jy_j\right)^2\le\left(\sum_{j=1}^nx^2_j\right)\left(\sum_{j=1}^ny^2_j\right)=\sum_{j,k=1}^nx^2_jy^2_k$$

Expanding the LHS we have that the above is equivalent to

$$\sum_{j,k=1}^n(x_jy_j)(x_ky_k)\le\sum_{j,k=1}^n x^2_jy_k^2\implies\sum_{j< k}2(x_jx_k)(y_jy_k)\le\sum_{j\neq k}x^2_jy^2_k\\\implies 0\le\sum_{j<k} x^2_jy_k^2-2(x_jy_k)(x_ky_j)+x^2_ky_j^2=\sum_{j< k}(x_jy_k-x_ky_j)^2$$

Answer 2

Take vectors $u=a-\bar a$ and $v=b-\bar b$. Cauchy-Schwarz inequality says that:

$$u.v=u_1v_1+...+u_nv_n\le \sqrt{u_1^2+...+u_n^2}\sqrt{v_1^2+...+v_n^2}$$

which by substitution leads to :

$${\sum_{i=1}^n(a_i-\bar a)(b_i-\bar b)}\le{\sqrt{\sum_{i=1}^n (b_i-\bar b)^2 \sum_{i=1}^n (a_i-\bar a)^2}}$$

or

$$NCC=\frac{cov(a,b)}{\sqrt{var(a)var(b)}}= \frac{\sum_{i=1}^n(a_i-\bar a)(b_i-\bar b)}{\sqrt{\sum_{i=1}^n (b_i-\bar b)^2 \sum_{i=1}^n (a_i-\bar a)^2}}\le1$$