Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Select $n$ numbers from a set $\{1,2,...,U\}$, $y_i$ is the $i$th number selected, and $x_i$ is the rank of $y_i$ in the $n$ numbers. The rank is the order of the a number after the $n$ numbers are sorted in ascending order.

We can get $n$ data points $(x_1, y_1), (x_2, y_2), ..., (x_n, y_n)$, And a best fit line for these data points can be found by linear regression. $r_{xy}$ (correlation coefficient) is the goodness of the fit line, I want to calculate $E(r_{xy})$ or $E(r_{xy}^2)$ (correlation of determination).

share|cite|improve this question
This seems to be the rank correlation: – PEV Apr 12 '11 at 16:16
It seems unlikely that there is a nice formula for $E(r_{xy})$. For instance, when $U=6$ and $n=3$ my calculations give ${3\over 10}+{9\sqrt{21}\over 140} +{2\sqrt{39}\over 65} + {\sqrt{7}\over 28}+{\sqrt{57}\over 76}$. It seems that the average correlation is quite large, usually more than $9/10$. – Byron Schmuland Apr 13 '11 at 15:54

Just a hint, a probability approach:

One can compute $P(x |y) $ : given the value of a extracted number $y=1 \ldots u$, probability that its rank (among the $n$ numbers) is $x=1 \ldots n$.

$\displaystyle P(x |y) = \frac{ {y - 1 \choose x - 1} {u - y \choose n-x} }{ {u-1 \choose n-1} } , \; \; n-u+y \le x \le y $

From this one can (formally or numerically; analytically... I doubt it) compute $E(x|y) = \sum x P(x |y)$

And then we could compute $E(x \; y) = E_y ( y \; E_x ( x | y ) )$

And $Cov(x y) = E(x \; y) - E(x) E(y) = E(x \; y) - \frac{n+1}{2} \frac{u+1}{2}$

share|cite|improve this answer
I have calculated the Rxy using the above formula - the above one is population correlation coefficient, and I also use a random generated data to calculate the sample correlation coefficient. It seems the population correlation coefficient is always smaller than the sample correlation coefficient. – Fan Zhang May 17 '11 at 15:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.