# Talk:Rank correlation

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I don't understand what this is all about. Thanks. 205.228.73.12 11:27, 3 October 2007 (UTC)

I agree that a lack of an example makes the content difficult to understand. I have added a new section on a rank correlation measure known as the rank-biserial correlation. I also worked through an example, so perhaps this will be easier to understand than the earlier sections. --Friend of facts2 (talk) 17:18, 8 October 2015 (UTC)

## When?

Are there situations where Spearman's ρ is more suitabke than Kendall's τ, or vice-versa? How can we choose which one to use? I have no clue about it, but it would be useful to know that. Calimo (talk) 13:18, 23 January 2009 (UTC)

I agree. It is useless to list two options if no information is supplied allowing to somehow distinguish them at a glance without clicking individually each one and see for each one an explanation ignoring the other. 212.198.146.203 (talk) 06:47, 30 March 2009 (UTC)

## Kendall (1944) reference

The reference Kendall (1944) in the beginning of section Rank_correlation#General_correlation_coefficient is unavailable. Sieste (talk) 15:53, 25 January 2016 (UTC)

## General Correlation Coefficient, notation

It would be much clearer, here and below, if the notation specified that when a sum iterates over ${\displaystyle i}$ and ${\displaystyle j}$, we exclude the elements where ${\displaystyle i=j}$. Perhaps

${\displaystyle \sum _{i,j=1;i\neq j}^{n}{a_{ij}b_{ij}}}$

(although it would be nicer if the lower limits were stacked on two lines).

(I haven't addressed this, though I have made a few smaller changes for clarity.)

Eac2222 (talk) 14:25, 1 August 2016 (UTC)

## Proof in the case of Kendall's τ

I believe an expert should look at this. The proof may be missing information, or incorrect.

If we have ${\displaystyle a_{ij}=\operatorname {sgn}(r_{j}-r_{i})}$, then don't we also need to define ${\displaystyle s_{i}}$ as the rank of the ${\displaystyle i}$th member according to the ${\displaystyle y}$-quality, and define ${\displaystyle b_{ij}=\operatorname {sgn}(s_{j}-s_{i})}$?

(I haven't addressed this, though I have made a few smaller changes for clarity.)

Eac2222 (talk) 14:25, 1 August 2016 (UTC)