# How to show sample correlation is sample covariance for standardized values?

Given a matrix $$X$$ and the resulting sample correlation matrix $$R$$, consider the standardized observations:

$$\frac{(x_{jk} - \bar x)} {\sqrt{S_{kk}}} \quad k=1,2,...,p \quad j=1,2,...,n$$

Show that these standardized quantities have sample covariance matrix $$R$$.

Sample covariance is defined as:

$$\frac{1}{n} \sum\limits_{j=1}^n{(x_{ji} - \bar x_i)(x_{ji} - \bar x_k)}$$

Sample correlation coefficient:

$$r_{ik}=\frac{S_{ik}} {\sqrt{S_{ii}}\sqrt{S_{kk}}} = \frac{\sum\limits_{j=1}^n(x_{ji}-\bar x_i)(x_{jk}-\bar x_k)}{\sqrt{\sum\limits_{j=1}^n(x_{ji}-\bar x_i)^2\sum\limits_{j=1}^n(x_{jk}-\bar x_k)^2}}$$

It is also known that: $$R (correlation) = D^{-1/2}SD^{-1/2}$$ S = variance-covariance matrix

D = sample standard deviation matrix

How do I show that for standardized quantities sample correlation is just the sample covariance?

Attempted answer: For standardized variable, variance is defined as: $$\frac{1}{n}\sum_j\frac{(x_{jk} - \bar x_k)^2} {S_{kk}}=\frac{S_{kk}}{S_{kk}}=1, \quad k=1,2,...,p,$$

thus D = diagonal matrix with variance satisfies: $$D = I$$ and since $$R (correlation) = D^{-1/2}SD^{-1/2}$$ we can conclude that $$R=D$$

• By computing $D$? – Vincent Guillemot Mar 8 at 10:52
• @VincentGuillemot Sorry I made a typo, should be -1/2 instead of 1/2. – lydias Mar 8 at 13:30
• @VincentGuillemot What if we do not know the last condition? How do we prove based only on knowing it's standardized quantities? – lydias Mar 8 at 13:32

The variance of a standardized variable is $$\frac{1}{n}\sum_j\frac{(x_{jk} - \bar x_k)^2} {S_{kk}}=\frac{S_{kk}}{S_{kk}}=1, \quad k=1,2,...,p,$$ and so is, therefore, its standard deviation. Hence, $$D$$, which is a diagonal matrix with the variances (not the standard deviations, that is given by $$D^{1/2}$$) on the main diagonal, satisfies $$D=I$$, and therefore, $$D^{1/2}=D^{-1/2}=I$$, the identity matrix.
Notice we do not need to subtract the sample average, as a demeaned variable has mean zero by construction: $$\sum_{j=1}^n(x_j-\bar x)=\sum_{j=1}^nx_j-n\bar x=0$$.