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I have been examining the use of the Point Biserial correlation as a statistic to measure the relationship between a dichotomous variable and a continuous one. Wikipedia et. al. seem to concur that the Point Biserial Correlation is a special case of the Pearson Correlation, but I cannot find a proof for this, algebraic or otherwise, and it is making me wary of using this in the context of the research I am doing (I need to do some statistical confidence testing afterwards). I have tried deriving the truth myself, but have chased everything round in a circle.

Any advice greatly appreciated.

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  • $\begingroup$ Many introductory statistics textbooks provide a reference for the equivalence of various types of correlation coefficients and point out which are not mathematically identical. Some texts provide proofs as well. Webpages can be helpful but are not a substitute for textbooks. $\endgroup$ – Joel W. Jul 2 '14 at 15:16
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    $\begingroup$ @JoelW. - I'm sure that is true. However, having metaphorically chewed my way through a not insignificant number of statistical textbooks and not been able to find mention of the Point Biserial Coefficient, much less a relationship between it and the Pearson Correlation, I decided to ask here- do you have a particular recommendation? $\endgroup$ – Philip Adler Jul 2 '14 at 15:27
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    $\begingroup$ Philip, biserial and point-biserial correlation coefficients are very different. You should choose either to update the title or the text. $\endgroup$ – ttnphns Jul 2 '14 at 15:36
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    $\begingroup$ @ttnphns - so I keep being told- but I need a proof, not an assertion; the formulae I have been given for their calculation appear different, and I cannot make one equate to the other algebraically. $\endgroup$ – Philip Adler Jul 2 '14 at 16:05
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    $\begingroup$ The derivation of the formula for point biserial r from the Pearson product moment correlation formula is given on pages 29-30 in the text Applied Multiple Regression/Correlation Analysis, 3rd ed., by Cohen, Cohen, West, and Aiken (Lawrence Erlbaum). $\endgroup$ – Joel W. Jul 2 '14 at 17:43
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Let the $n$ data consist of $n_0\gt 0$ $(x, 0)$ pairs and $n_1\gt 0$ $(x, 1)$ pairs. Their Pearson correlation coefficient will be the same as the reversed data consisting of corresponding $(0,x)$ and $(1,x)$ pairs. Because there are exactly two distinct values of the first coordinates, the regression line of the reversed data must pass through the mean points $(0,M_0)$ and $(1,M_1)$, whence it has slope $(M_1-M_0)/(1-0) = M_1-M_0$. The correlation coefficient is obtained by standardizing this: it must be multiplied by the standard deviation of the first coordinates and divided by the standard deviation of the second coordinates (the original $x$ values), written $s_n$. The standard deviation of the first coordinates is readily computed from the fact that they consist of $n_0$ zeros and $n_1$ ones; it equals

$$\sqrt{\frac{n_1}{n}\left(1-\frac{n_1}{n}\right)} = \sqrt{\frac{n_0n_1}{n^2}}.$$

Consequently the Pearson correlation coefficient is

$$r = \frac{M_1-M_0}{s_n}\sqrt{\frac{n_0n_1}{n^2}},$$

which is precisely the Wikipedia formula for the point-biserial coefficient.

Figure: x vs. y with linear regression line and mean points shown

The heights of the red dots depict the mean values $M_0$ and $M_1$ of each vertical strip of points. The dashed gray line is the regression line.

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    $\begingroup$ For more on the relationships among correlation coefficients and regression slopes please see the answer at stats.stackexchange.com/a/104577. $\endgroup$ – whuber Jul 2 '14 at 16:32
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    $\begingroup$ Thankyou, it was the reversing of the pairs that I had failed to appreciate. $\endgroup$ – Philip Adler Jul 2 '14 at 17:10
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As Cohen, Cohen, West, and Aiken(2003, p 30) states

sdY is the sample value, which is divided by n rather than n — 1

Thus, when calculating point-biserial-correlation using a formula rather than Pearson, make sure you use the population standard deviation to get what Pearson formula produces. This might be important to conduct significance tests.

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