Is it true that $R_{X,Y}$ is an unbiased estimator for $\rho_{X,Y}$? That is, $$\mathbf{E}\left[R_{X,Y}\right]=\rho_{X,Y}?$$

If not, what is an unbiased estimator for $\rho_{X,Y}$? (Perhaps there is a standard unbiased estimator that's used? Also, is it analogous to the unbiased sample variance, where we simply make the simple adjustment of multiplying the biased sample variance by $\frac{n}{n-1}$?)


The population correlation coefficient is defined as $$\rho_{X,Y}=\frac{\mathbf{E}\left[\left(X-\mu_{X}\right)\left(Y-\mu_{Y}\right)\right]}{\sqrt{\mathbf{E}\left[\left(X-\mu_{X}\right)^{2}\right]}\sqrt{\mathbf{E}\left[\left(Y-\mu_{Y}\right)^{2}\right]}},$$ while the sample correlation coefficient is defined as $$R_{X,Y}=\frac{\sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)\left(Y_{i}-\bar{Y}\right)}{\sqrt{\sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}}\sqrt{\sum_{i=1}^{n}\left(Y_{i}-\bar{Y}\right)^{2}}}.$$

  • $\begingroup$ A (a bit similar) question about estimators of $\rho$. $\endgroup$
    – ttnphns
    Commented Jun 28, 2016 at 9:19
  • $\begingroup$ The question "what is the unbiased estimator" presupposes that there is one and that there is only one. A priori, there doesn't appear to be any reason to think that. $\qquad$ $\endgroup$ Commented Jun 28, 2016 at 20:59
  • $\begingroup$ @MichaelHardy: I've corrected that. Thanks for pointing out. $\endgroup$
    – user46481
    Commented Jun 29, 2016 at 3:22
  • $\begingroup$ Just stumbled upon this thread, and I think this might be an interesting read sciencedirect.com/science/article/pii/S0167715298000352 (I haven't yet read it myself tbh) $\endgroup$
    – martn
    Commented Nov 16, 2017 at 21:03
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    $\begingroup$ minimum variance unbiased estimator: projecteuclid.org/euclid.aoms/1177706717 $\endgroup$ Commented Aug 25, 2018 at 10:00

1 Answer 1


This is not an easy question but some expressions are available. If you are talking about the Normal distribution in particular, then the answer is NO! We have

$$\mathbb{E} \widehat{\rho} = \rho \left[1 - \frac{\left(1-\rho^2 \right)}{2n} + O\left( \frac{1}{n^2} \right) \right]$$

as seen in Chapter 2 of Lehmann's Theory of Point Estimation. There are infinitely many terms in the expression above but we are essentially considering terms of equal or lower order than $n^{-2}$ negligible.

This formula shows that the sample correlation coefficient is only unbiased for $\rho = 0$, i.e. independence, as one would expect. It is also unbiased for the degenerate cases with $|\rho| = 1$, but that is not very interesting. In general cases the bias will be of order $\frac{1}{n}$ but quite small for all reasonable sample sizes.

In Normal distributions the sample correlation coefficient is the mle, which means that it is asymptotically unbiased. You can also see that from the above formula as $\mathbb{E} \widehat{\rho} \to \rho$. Note that this already follows from the boundedness and the consistency of the sample correlation coefficient through the bounded convergence theorem.

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    $\begingroup$ There may be infinitely many terms in the expression above, but "infinite terms" would be there are some terms, each of which is infinite. $\qquad$ $\endgroup$ Commented Jun 28, 2016 at 21:00
  • $\begingroup$ Suppose all points in a bivariate population lie on a straight line with nonzero slope. Then all points in any sample do so too. I conjecture therefore that if population correlation has absolute value $|\rho| = 1$ so also sample correlation $|r| \equiv 1$. $\endgroup$
    – Nick Cox
    Commented Jun 28, 2016 at 22:05
  • $\begingroup$ @NickCox That's true, in the degenerate case the sample correlation coefficient would return $|1|$ with no estimation error. $\endgroup$
    – JohnK
    Commented Jun 28, 2016 at 22:07
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    $\begingroup$ For a related question, does anyone know if analogous results exist for any other distributions besides the 2D normal? $\endgroup$ Commented Aug 24, 2018 at 20:50
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    $\begingroup$ What if you consider $\rho^2$ and $R^2$ instead of $\rho$ and $R$? For instance you can find unbiased estimators for the variance and covariance, while the estimate for the standard-deviation you obtain from the unbiased variance is not generally unbiased. $\endgroup$
    – Diger
    Commented Apr 27, 2021 at 20:26