Suppose you have two traits that are correlated in a given population, like a person's BMI and their blood pressure. And let's say I want to estimate the probability that in a randomly-selected pair of people drawn from this population, the one that has the higher BMI will also have a higher blood pressure. If I know the pearson correlation coefficient r (or equivalent r^2, the proportion of variance in blood pressure that's explained by BMI), can this be used to get that probability? If not, could I do so using some other correlation measure, or by making some simplifying assumption about the functional relationship between them (say, assuming blood pressure in each individual is a linear function of BMI along with several other independent variables) and/or each one's individual distribution (say, assuming both BMI and blood pressure are normally distributed)?

  • $\begingroup$ If this is what you're interested in you could just estimate this directly from the data by using Kendall's $\tau$ for example. $\endgroup$ – dsaxton Mar 8 '16 at 22:39

No - knowing the correlation (and even linear regression formula) between two traits is not enough to predict the probability that a higher BMI will have a higher blood pressure.

See Anscombe's quartet for a visual example of four dissimilar distributions with identical correlations and fitted linear regression lines to see where making probability predictions based upon the correlation can lead you astray.

If you make simplifying assumptions: i.e., a linear relationship between BMI and blood pressure and normal distributions then yes, you could construct prediction intervals for new measurements using the least squares equation.

However when working with real-world data I would advise avoiding assumptions about the data distribution. A better alternative would be to use bootstrapping to estimate the cumulative distribution function.


If two traits have known correlation, can you predict probability they'll “align” for a random pair?

It depends on which population correlation you look at.

For the Pearson correlation you mention ($\rho$), the answer is "no", at least not without additional assumptions. (RobertF's answer is correct)

If instead you know the population Kendall correlation (Kendall's tau, here denoted $\tau_K$) in a continuous bivariate distribution then the answer is actually yes.

The population Kendall correlation is the difference between the probability of a concordant pair and the probability of a discordant pair:

$$\tau_K = p_C-p_D$$

(the sample Kendall correlation is similarly the difference in sample proportions of concordant and discordant pairs).

Since in continuous bivariate populations $p_C+p_D=1$, if you know $\tau_K$ you can calculate $p_C$:

$\tau_K = p_C-p_D$ $ = p_C-(1-p_C)$ $ = 2p_C-1$

Hence $p_C = \frac12(\tau_K+1)$, a nice simple result.

While $\tau_K$ determines the probability you ask for (at least in the continuous case), the relationship between $\rho$ and $\tau_K$ depends on the structure of the bivariate relationship between the variables (i.e. the copula).

If you assume bivariate normality, then you could work out the (nonlinear) connection between $\tau_K$ and $\rho$. In fact this is a well-known result; we have:

$$\tau_K = \frac{2}{\pi}\arcsin(\rho)$$

- see sec 5.3.2 of Embrechts et al. (2005) [1], which result can also be found in various places -- for example in Meyer (2009) [2]. So in that case

$$p_C = \frac{\arcsin(\rho)}{\pi}+\frac12\,.$$

(However, an assumption of bivariate normality would seem dubious for BMI and blood pressure)

This relationship between $\tau_K$ and $\rho$ actually holds for elliptical distributions more generally. See for example Lindskog, McNeil, & Schmock (2003)[3]. However, again, this assumption for BMI and blood pressure may be dubious -- for example, both measures in practice tend to be right-skew.

[1] Embrechts, P., Frey, R., McNeil, A.J. (2005),
Quantitative Risk Management: Concepts, Techniques, Tools,
Princeton series in Finance, Princeton University Press

[2] Meyer, C. (2009),
The Bivariate Normal Copula,
arXiv:0912.2816v1[math.PR] pdf (December 15)

[3] Lindskog, F., McNeil, A.J., Schmock, U., (2003),
"Kendall’s tau for elliptical distributions"
in: Credit Risk; Measurement, Evaluation and Management, ed. G. Bol et al.,
Contributions to Economics, Physica-Verlag Heidelberg, pp.149–156.
(or see http://www.macs.hw.ac.uk/~mcneil/ftp/KendallsTau.pdf)

  • 2
    $\begingroup$ I'm approving RobertF's answer since it came first and addressed most of what I was asking, but I appreciate this additional information. My layman's impression is that most of the empirical statistical studies I've seen in fields like medicine/psychology/sociology only use Pearson's r or r^2 to describe correlations--if that's true I wonder why Kendall's measure isn't reported more often alongside it, it would seem to have some usefulness in interpreting the results (especially when summarizing for a non-specialist audience, since the idea of picking a random pair is intuitive). $\endgroup$ – Hypnosifl Mar 8 '16 at 23:22
  • $\begingroup$ Indeed that's an excellent question to ponder -- and one might well ask a similar question in relation to a number of other nonparametric quantities of similar interpretability and simplicity. [At least the Kendall correlation is widely used in work with copulas, both the theory and practice.] $\endgroup$ – Glen_b Mar 8 '16 at 23:25

I recommend increasing the variables you are measuring. Age, gender, location etc. weight them in your formula to lower the probability of false negatives. Maximize your ROC curve. It would be interesting to see a model that keeps the same correlation given datasets over different decades.


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