Uncorrected sample standard deviation in correlation coefficient There seems to be a common use of the uncorrected sample standard deviation in calculating the correlation coefficient:


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*https://www.experts-exchange.com/articles/2728/Covariance-and-Correlation-in-MS-Access.html

*https://www.mrexcel.com/forum/microsoft-access/128620-using-access-find-correlation-between-data-columns.html
I'm wondering if I am missing something about the apparent preference of using the corrected sample standard deviation:


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*https://en.wikipedia.org/wiki/Correlation_and_dependence

*https://en.wikipedia.org/wiki/Standard_deviation#Uncorrected_sample_standard_deviation
I've posted this to:


*

*Stack Exchange

*sci.stat.math
 A: It makes no difference whatsoever, provided you compute consistently.
Suppose you define your version of the variance of a batch of data $X = x_1, x_2, \ldots, x_n$, with mean $\bar x = (x_1 + x_2 + \cdots + x_n)/n$, to be
$$\text{var}_f(X) = f(n)\left((x_1-\bar x)^2 + (x_2 - \bar x)^2 + \cdots + (x_n - \bar x)^2\right)$$
where $f:\mathbb{N}\to (0,\infty)$ is a personalized function giving some positive value $f(n)$ for each positive integer $n$. Some people might use $f(n) = 1/n$, others might use $f(n) = 1/(n-1)$, and others might use something else altogether such as $f(n)=\Gamma((n-1)/2)^2/(2\Gamma(n/2)^2)$.  It doesn't matter, because as we have seen you must use the same function $f$ to compute the covariance of any batch of paired data $(X,Y) = (x_1,y_1), (x_2,y_2), \ldots, (x_n,y_n)$:
$$\text{cov}_f(X,Y) = f(n)\left((x_1-\bar x)(y_1-\bar y) + (x_2-\bar x)(y_2-\bar y) + \cdots + (x_n-\bar x)(y_n-\bar y)\right).$$
Consequently your personal correlation coefficient will be
$$\eqalign{
\rho_f(X,Y) &= \frac{\text{cov}_f(X,Y)}{\sqrt{\text{var}_f(X)}\sqrt{\text{var}_f(Y)}} \\
&= \frac{f(n)\sum_i(x_i-\bar x)(y_i-\bar y)}{\sqrt{f(n)\sum_i(x_i-\bar x)^2}\sqrt{f(n)\sum_i(y_i-\bar y)^2}}\\
&= \frac{\sum_i(x_i-\bar x)(y_i-\bar y)}{\sqrt{\sum_i(x_i-\bar x)^2}\sqrt{\sum_i(y_i-\bar y)^2}}.
}$$
Since the factor $f(n)/(\sqrt{f(n)}\sqrt{f(n)}) = 1$ disappears, your correlation coefficient will be the same as anyone else's.  As a bonus, now you know you don't have to compute $f$.
