In the recently updated BT text, page 42, a pair containing (3,3) is included in discordant. But in GARP's text, it is included in neither category. Which is true? I searched by myself an example of calculation on the web, but I found none. Thanks.
GARP's text (which is, of course, Meissner's book) is incorrect. Please see my note from last year @ https://www.bionicturtle.com/forum/...lls-tau-and-concordant-discordant-pairs.8209/
I wrote to Meissner and he confirmed the typos. The example has 2 concordant pairs and 8 discordant pairs (and there are actually zero "neither"s in Meissner's example), for a Kendall's tau of -0.60 (i.e., -0.20 is incorrect)
David, I understood that the direction of a compared pair of x and y must be same or different for discordance and non-discordance. Then what is neither category? Could you give me a concrete example in the form of (x1, y1) and (x2, y2)?
Hi @Kenji@NNath is correct, and you can see (at my link above) that even the author (Gunter Meissner) himself said the text should be revised to read: But it should be: neither concordant nor discordant if x(t) = x(t)* or if y(t) = y(t)*
I gave you two examples above: {(1,4), (1,3)} or {(1,4), (2,4)}
Take the case of {(1,4), (2,4)}
what is the relationship between x(t) and x(t)*? It is x(t) < x(t)* as 1 < 2 .
In order for the pair to be concordant, also we should see y(t) < y(t)*; i.e., the second y is greater than the first y, just as the second x is greater than the first x
In order for the pair to be discordant, instead we should see y(t) > y(t)*; ie, although the second x is greater than the first x, the second y is less than the first y
But here y(t) = y(t)* as 4 = 4, so neither applies. I hope that helps!
David, thank you very much. Now I understood better.
...I think even if I understand and remember this, since the text itself is incorrect, it will not be picked up in the exam or even if it appears in the exam and others choose incorrect one, it will not be graded since the text itself is wrong..
Correct me if wrong, but this seems like an easy way to calculate Kendall T is:
Maybe it is actually the same method, but it looks easier because one only counts the the numbers higher or lower than then the number of reference.
C(i) is the number of numbers below Y(i) in the list that are higher than Y(i).
D(i) is the number of numbers below Y(i) in the list that are lower than Y(i).
Eg.
Y(1) = 4
In the list below Y(1), {5} is the only number that is higher than 4, so C(1) = 1
In the list below Y(1), {3,1,2} are lower than 4, so D(1) = 3
Y(2) = 5
In the list below Y(2), no number is higher than 5, so C(2) = 0
In the list below Y(2), {3,1,2} are lower than 5 , so D(2) = 3
Y(3) = 3
In the list below Y(3), no number is higher than 3, so C(3) = 0
In the list below Y(3), {1,2} are lower than 3, so D(3) = 2
Y(4) = 1
In the list below Y(4), {2} is higher than 1, so C(3) = 1
In the list below Y(4), no number is lower than 1, so D(4) = 0
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