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).

  • $\begingroup$ This seems to be the rank correlation: en.wikipedia.org/wiki/Rank_correlation $\endgroup$ Apr 12 '11 at 16:16
  • $\begingroup$ 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 $9/10$. $\endgroup$
    – user940
    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 $Cov(x y) = E(x \; y) - E(x) E(y) = E(x \; y) - \frac{n+1}{2} \frac{u+1}{2}$

  • $\begingroup$ 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. $\endgroup$
    – Fan Zhang
    May 17 '11 at 15:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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