I understand the rationale for dividing by $n-1$ when calculating the sample variance, i.e. that if we divide by $n$ we will have an estimate of population variance that is biased to be too low.

Buglear (2013, p. 57) states about the Pearson correlation:

We divide by $n - 1$ for the same reason as we do it when calculating sample standard deviations – it gives us a better estimator of the population equivalent.

[Buglear, J. (2013). Practical Statistics: A Handbook for Business Projects. Kogan Page Publishers]

However, I don't understand why this also applies to correlations. Why is it the case that dividing by $n$ would underestimate the population correlation coefficient?

  • 2
    $\begingroup$ When I compute Pearson correlations, I always divide the numerator and denominator terms by $17$, regardless of the sample size. It works just as well as dividing them both by $n-1$ and is simpler to remember :-). $\endgroup$
    – whuber
    Feb 27, 2016 at 23:49
  • $\begingroup$ @whuber Any reason that you specifically chose 17? $\endgroup$ Nov 16, 2023 at 15:22
  • $\begingroup$ @user16714199 Search our site ;-) $\endgroup$
    – whuber
    Nov 16, 2023 at 16:38

1 Answer 1


We do not need the Bessel correction "-1" to $n$ when we compute correlation, so I think the citated piece is wrong. Let me start by noticing that most of time we compute and use empirical $r$, or the $r$ of the sample, for both describing the sample (the statistic) and the population (the parameter estimate). This is different from variance and covariance coefficients where, typically, we introduce the Bessel correction to distinguish between the statistic and the estimate.

So, consider empirical $r$. It is the cosine similarity of the centered variables ($X$ and $Y$ both were centered): $r= \frac{\sum{X_cY_c}}{\sqrt{\sum X_c^2\sum Y_c^2}}$. Notice that this formula doesn't contain neither $n$ nor $n-1$ at all, we need not to know sample size to obtain $r$.

On the other hand, that same $r$ is also the covariance of the z-standardized variables ($X$ and $Y$ both were centered and then divided by their respective standard deviations $\sigma_x$ and $\sigma_y$): $r= \frac{\sum{X_zY_z}}{n-1}$. I suppose that in your question you are speaking of this formula. That Bessel correction in the denominator, which is called in the formula of covariance to unbias the estimate, - in this specific formula to compute $r$ paradoxically serves to "undo" the unbiasing correction. Indeed, recall that $\sigma_x^2$ and $\sigma_y^2$ had been computed using denominator $n-1$, the Bessel correction. If in the latter formula of $r$ you unwind $X_z$ and $Y_z$, showing how they were computed out of $X_c$ and $Y_c$ using those "n-1"-based standard deviations you'll find out that all "n-1" terms cancel each other from the formula, and you stay in the end with the above cosine formula! The "n-1" in the "covariance formula" of $r$ was needed simply to take off that older "n-1" used.

If we prefer to compute those $\sigma_x^2$ and $\sigma_y^2$ based on denominator $n$ (instead of $n-1$) the formula for yet the same correlation value will be $r= \frac{\sum{X_zY_z}}{n}$. Here $n$ serves to take off that older "n" used, analogously.

So, we needed $n-1$ in the denominator to cancel out the same denominator in the formulas of variances. Or needed $n$ for the same reason in case the variances were computed as biased estimates. Empirical $r$ is itself not based on the information of the sample size.

As for a quest of better population estimate of $\rho$ than the empirical $r$, there we do need corrections, but there exist various approaches and a lot of different alternative formulas, and they use different corrections, usually not $n-1$ one.

  • $\begingroup$ The most standard definition of the sample correlation coefficient is that it's sample covariance divided by the product of two sample standard deviations. You gave several equivalent formulas but not exactly this one. Of course the $n-1$ (or $n$) factor is in both nominator and denominator, so it cancels out as you correctly say. $\endgroup$
    – amoeba
    Feb 27, 2016 at 21:57
  • $\begingroup$ @amoeba, thank you for the caring comment. I didn't specifically discussed in the current answer the "standard" formula $r=cov/\sigma_x\sigma_y$ and that it (obviously) cracks down to the formula of cosine after substitutions, - I had another aim. $\endgroup$
    – ttnphns
    Feb 27, 2016 at 23:33
  • $\begingroup$ BTW, I don't like and usually don't use words "sample covariance" and "sample correlation", the misleading argot. I'd prefer saying "unbiased covariance estimate" and "empirical correlation value". $\endgroup$
    – ttnphns
    Feb 27, 2016 at 23:36
  • $\begingroup$ Not sure I understand your comment about terminology. As far as I understand, "sample covariance" (it can be biased or unbiased, or estimated in any other way) is opposed to "population covariance". So "sample covariance" is an estimate of "population covariance". Why is this misleading? $\endgroup$
    – amoeba
    Feb 28, 2016 at 1:54
  • $\begingroup$ It (with the df="n-1") is misleading because it can be understood as "the covariance of this sample", while in fact this statistic is the "unbiased estimate of the population covariance" given by this sample. A view of any sample statistic is two-fold: either charactarizes just the sample totality, or it serves a possible estimate of the population totality. (Co)variance based on df=n can be labelled both "the sample's" and the "estimate for population". But the one on df=n-1 can hardly be labelled "the sample's" - the Bessel correction was introduced specifically to make an "estimate". $\endgroup$
    – ttnphns
    Feb 28, 2016 at 8:26

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