7
$\begingroup$

There are several measures of association (or contingency or correlation) between two binary random variables $X$ and $Y$, among others

I wonder how the following number $\kappa$ relates to known measures, if it is statistically interesting, and under which name it is (possibly) discussed:

$$\kappa = 1 - \frac{2}{N}|X \triangle Y|$$

with $|X \triangle Y|$ the number of samples having property $X$ or property $Y$ but not both (exclusive OR, symmetric difference), $N$ the total number of samples. Like the phi coefficient, $\kappa = ± 1$ indicates perfect agreement or disagreement, and $\kappa = 0$ indicates no relationship

$\endgroup$

2 Answers 2

11
$\begingroup$

Using a,b,c,d convention of the 4-fold table, as here,

               Y
             1   0
            -------
        1  | a | b |
     X      -------
        0  | c | d |
            -------
a = number of cases on which both X and Y are 1
b = number of cases where X is 1 and Y is 0
c = number of cases where X is 0 and Y is 1
d = number of cases where X and Y are 0
a+b+c+d = n, the number of cases.

substitute and get

$1-\frac{2(b+c)}{n} = \frac{n-2b-2c}{n} = \frac{(a+d)-(b+c)}{a+b+c+d}$ = Hamann similarity coefficient. Meet it e.g. here. To cite:

Hamann similarity measure. This measure gives the probability that a characteristic has the same state in both items (present in both or absent from both) minus the probability that a characteristic has different states in the two items (present in one and absent from the other). HAMANN has a range of −1 to +1 and is monotonically related to Simple Matching similarity (SM), Sokal & Sneath similarity 1 (SS1), and Rogers & Tanimoto similarity (RT).

You might want to compare the Hamann formula with that of phi correlation (that you mention) given in a,b,c,d terms. Both are "correlation" measures - ranging from -1 to 1. But look, Phi's numerator $ad-bc$ will approach 1 only when both a and d are large (or likewise -1, if both b and c are large): product, you know... In other words, Pearson correlation, and especially its dichotomous-data hypostasis, Phi, is sensitive to the symmetry of marginal distributions in the data. Hamann's numerator $(a+d)-(b+c)$, having sums in place of products, is not sensitive to it: either of two summands in a pair being large is enough for the coefficient to attain close to 1 (or -1). Thus, if you want a "correlation" (or quasi-correlation) measure defying marginal distributions shape - choose Hamann over Phi.

Illustration:

Crosstabulations:
        Y
X    7     1
     1     7
Phi = .75; Hamann = .75

        Y
X    4     1
     1    10
Phi = .71; Hamann = .75
$\endgroup$
5
  • $\begingroup$ Is Hamann similarity widely known and accepted as an interesting measure? $\endgroup$ Commented Jan 16, 2017 at 19:42
  • 1
    $\begingroup$ How can I answer? How much widely/accepted will suffice? :-) It is sure less known than phi correlation or Jaccard similarity. Still, it is sometimes used. Google it to see... One its important property is that it is monotonical equivalent of... (see the citation). $\endgroup$
    – ttnphns
    Commented Jan 16, 2017 at 19:47
  • $\begingroup$ Sorry for my naive question, and thanks for your informative answer:-) $\endgroup$ Commented Jan 16, 2017 at 19:51
  • $\begingroup$ Can you give me a hint, under which typical circumstances I might want a "correlation defying marginal distributions shape" and choose Hamann, and under which circumstances I might want a "correlation NOT defying marginal distributions shape" and choose Phi? $\endgroup$ Commented Jan 17, 2017 at 9:56
  • $\begingroup$ Hans, if you are speaking about scientific fields or aims where we might want to use one over the other - why not ask that as a separate question? Because more people might come to answer. $\endgroup$
    – ttnphns
    Commented Jan 17, 2017 at 17:39
4
$\begingroup$

Hubalek, Z. Coefficients of association and similarity, based on binary (presence-absence) data: an evaluation (Biol. Rev., 1982) reviews and ranks 42 different correlation coefficients for binary data. Only 3 of them meet basic statistical desiderata. Unfortunately, the issue of PRE (proportionate reduction of error) interpretation is not discussed. For the following contingency table:

        present  absent

present    a       b

absent     c       d

the association measure $r$ should meet the following obligatory conditions:

  1. $r(J,K) \le r(J,J) \quad\forall J, K$

  2. $\min(r)$ should be at $a = d = 0$ and $\max(r)$ at $b = c = 0$

  3. $r(J,K) = r(K,J) \quad \forall K,J$

  4. discrimination between positive and negative association

  5. $r$ should be linear with $\sqrt{\chi^2}$ for both subsets $ad-bc < 0 $ and $ad-bc >= 0$ (note that $\chi^2$ violates condition 4)

and ideally the following non-obligatory:

  • range of $r$ should be either $\left\{ -1 \dots +1 \right\}$, $\left\{0 \dots +1 \right\}$, or $\left\{0 \dots \infty \right\}$

  • $r(b=c=0) > r(b = 0 \veebar c = 0)$

  • $r(a=0) = min(r)$ (stricter than 2) above)

  • $r(a+1)-r(a) = r(a+2)-r(a+1)$

  • $r(a=0,b,c,d), r(a=1,b-1,c-1,d+1), r(a=2,b-2,c-2,d+2)\ldots$ should be smooth

  • homogeneous distribution of $r$ in permutation sample

  • random samples from population with known $a,b,c,d$: $r$ should show little variability even in small samples

  • simplicity of calculation, low computer time

All conditions are met by Jaccard $\left( \frac{a}{a+b+c} \right)$, Russel & Rao $\left( \frac{a} {a+b+c+d} \right)$ (both range $\left\{0 \dots +1 \right\}$) and McConnaughey $\left( \frac{a^2 - bc}{(a+b) \times (a+c)}\right)$ (range $\left\{ -1 \dots +1 \right\}$)

$\endgroup$
3
  • $\begingroup$ This would be easier to read if you could edit to use $\LaTeX$ notation. I do a small part to show how. $\endgroup$ Commented Aug 25, 2017 at 16:59
  • $\begingroup$ Please merge your two answers here: edit one of them by adding contents of the other, then delete one. $\endgroup$
    – ttnphns
    Commented Aug 25, 2017 at 16:59
  • $\begingroup$ by your commands ;-) $\endgroup$ Commented Aug 27, 2017 at 7:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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