1
$\begingroup$

I am trying to calculate Kendall’s tau coefficient for example given by Scipy in python. It is for tied ranks with tau-b version. The tau should be -0.47140452079103173 but I get a different result. I use formulas in "Handbook of Parametric and Nonparametric. Statistical Procedures. David J. Sheskin. Chapman & Hall/CRC." page 900.

$$x = [12, 2, 1, 12, 2]$$ $$y = [1, 4, 7, 1, 0]$$

rank the data: $$rank x = [4.5, 2.5, 1, 4.5, 2.5]$$ $$rank y = [2.5, 4, 5, 2.5, 1]$$

dimensions: $n=5, m=2 $ $$rank x+rank y = [7, 6.5, 6, 7, 3.5]$$ $$T=7+6.5+6+7+3.5=30$$ $$U=7^2+6.5^2+6^2+7^2+3.5^2=188.5$$

calculate the tie correlation(${\displaystyle\sum_{i=1}^m}{\displaystyle\sum_{a=1}^s}(t^3-t)$):

in x there are 2 tied groups each with 2 members: $(2^3-2)+(2^3-2)=12$

in y there is 1 tied group with 2 members: $(2^3-2)= 6$

total tie correlations: $12+6=18$

$$W=\frac{12U-3m^2n(n+1)^2}{m^2n(n^2-1)-m{\displaystyle\sum_{i=1}^m}{\displaystyle\sum_{a=1}^s}(t^3-t)}=\frac{12(188.5)-3\times 2^2\times 5(5+1)^2}{2^2\times5(5^2-1)-2(18)}=0.22973$$

Sorry for the low-quality question. there are not many numeric questions for tau-b. I hope this question can be helpful also for others who review python's example.

$\endgroup$
  • 1
    $\begingroup$ Scipy may use an adjusted algorithm for the calculation. Have you taken a look at the code? $\endgroup$ – Jon Feb 10 '17 at 21:48
  • $\begingroup$ You say you used the " formulas in "Handbook of Parametric and Nonparametric. ". Are these all listed out? Otherwise, readers may not know which formulas/algorithms you are referring to. $\endgroup$ – Jon Feb 10 '17 at 21:49
  • $\begingroup$ Also, there appear to be multiple algorithms for Kendall's tau: en.wikipedia.org/wiki/Kendall_rank_correlation_coefficient. It'll be useful to know which Scipy is using. $\endgroup$ – Jon Feb 10 '17 at 21:51
  • $\begingroup$ @jon the formula is the one written in the question. $\endgroup$ – Woeitg Feb 10 '17 at 22:40
3
$\begingroup$

The formula you are using is calculating Kendall's W, also known as Kendall's coefficient of concordance. Kendall's W not the same as Kendall's tau-b.

To calculate the Kendall tau-b for the given data set, you can use the formula in the Wikipedia page. I count $n_0=10$, $n_1=2$, $n_2=1$, $n_c=2$, $n_d=6$, so that $$ \tau_B=\frac{ 2-6}{\sqrt{(10-2)(10-1)}}=-\frac4{\sqrt{72}}=-.4714045. $$

EDIT: How to calculate $n_1$? We see that $x$ has two groups of ties, namely $\{2,2\}$ and $\{12,12\}$, each with two ties per group, so $t_1=2$ and $t_2=2$ and $$n_1:=\frac12\sum_it_i(t_i-1)=\frac12(2\cdot1)+\frac12(2\cdot1)=2.$$ As for $n_2$, we see that $y$ has one group of ties, namely $\{1,1\}$, so $u_1=2$ and $$n_2:=\frac12\sum_ju_j(u_j-1)=\frac12(2\cdot1)=1.$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ thanks for the answer. I will take a look and then accept it. about the calculation of U, I used ranks not vectors itself. $\endgroup$ – Woeitg Feb 10 '17 at 22:44
  • $\begingroup$ @Woeitg You are right about the ranks. I can't believe I missed that! $\endgroup$ – grand_chat Feb 10 '17 at 22:45
  • $\begingroup$ well, actually I did that mistake in the first place and then modify the question. maybe you saw it before modification. Now I add more modification to clarify it. $\endgroup$ – Woeitg Feb 10 '17 at 22:48
  • $\begingroup$ @Woeitg Right, I recall you got a value of $W$ that was much higher than 1. $\endgroup$ – grand_chat Feb 10 '17 at 22:50
  • $\begingroup$ Can you add steps for calculating n1 and n2? number of ties in x is 2 (there are two group of similar values) and 1 (there is one group of similar value). feed it in formula gives n1=(2(2-1)/2)+(2(2-1)/2+0+(2(2-1)/2)+(2(2-1)/2=4 $\endgroup$ – Woeitg Feb 10 '17 at 23:31

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.