In Wikipedia's summary on Cramér's $V$, it is mentioned that it can (also) be used in the one-dimensional case as measure of concentration.

  1. What is the formula in this special case?
  2. Is there any reference about this alternative use?

No reference, link or explanation is offered at the article*, so this is guesswork, but I believe this is the intent/reasoning:

*(and as such it should probably be removed from Wikipedia)

1) Cramér's $V$ is a measure of association; it's a function of $\phi$ which is also a measure of association; they cannot really mean that there's a measure of association here.

2) However, Cramér's $V$ can be defined in terms of the statistic for the chi-square test of independence:

$V = \sqrt{ \frac{\chi^2/n}{\min(k - 1,r-1)}}=\sqrt{ \frac{\chi^2}{n(k - 1)}}$ (when $r=1$)

If you place the goodness-of-fit chi-square value in that formula in place of the usual independence chi-square, while you could not (to my mind) reasonably call that Cramer's $V$ (since it doesn't measure association), it would be a perfectly good measure of concentration. It's 0 when the categories are perfectly evenly represented and it's 1 when all the values are in a single category.

| cite | improve this answer | |
  • $\begingroup$ Thanks, Glen. Would you think the probabilities behind the goodness-of-fit chisquared are always to be chosen uniform in such application? $\endgroup$ – Michael M Aug 5 '14 at 6:43
  • 1
    $\begingroup$ If they weren't, it wouldn't be a measure of concentration in the same sense. $\endgroup$ – Glen_b Aug 5 '14 at 7:36
  • 2
    $\begingroup$ +1, it couldn't reasonably be called a measure of association, but I do think it can still be thought of as a measure of the relevant type of effect size for such analyses. Test statistics & p-values conflate magnitude & N to give some sense of the clarity of the result; this formula seems to effectively disentangle the magnitude from N as other measures of effect size do (eg, $d$ & $r$). I think you can also use it to measure the deviation from non-uniform null distributions & it would be doing the same thing, it's just the word "concentration" would no longer characterize it well. $\endgroup$ – gung - Reinstate Monica Aug 9 '14 at 18:25
  • $\begingroup$ @gung +1 I'd agree with everything you have there. $\endgroup$ – Glen_b Aug 9 '14 at 20:56

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.