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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 $\frac{n}{n-1}$?)

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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 $$R_{X,Y}=\frac{\sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)\left(Y_{i}-\bar{Y}\right)}{\sqrt{\sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}}\sqrt{\sum_{i=1}^{n}\left(Y_{i}-\bar{Y}\right)^{2}}}.$$

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  • $\begingroup$ A (a bit similar) question about estimators of $\rho$. $\endgroup$ – ttnphns Jun 28 '16 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. $\qquad$ $\endgroup$ – Michael Hardy Jun 28 '16 at 20:59
  • $\begingroup$ @MichaelHardy: I've corrected that. Thanks for pointing out. $\endgroup$ – Kenny LJ Jun 29 '16 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 Nov 16 '17 at 21:03
  • $\begingroup$ minimum variance unbiased estimator: projecteuclid.org/euclid.aoms/1177706717 $\endgroup$ – Martijn Weterings Aug 25 '18 at 10:00
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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. $\qquad$ $\endgroup$ – Michael Hardy Jun 28 '16 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 Jun 28 '16 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 Jun 28 '16 at 22:07
  • $\begingroup$ For a related question, does anyone know if analogous results exist for any other distributions besides the 2D normal? $\endgroup$ – Riemann1337 Aug 24 '18 at 20:50

protected by whuber Nov 6 '18 at 15:21

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