A pretty straightforward question:

Does it make sense to measure Kendall's tau correlation between let's say a vector $a$ with length $n=1000$ and a vector $b$ with $n=200$? Does $n$ have to be the same for both quantities involved?

How the coefficient behaves in the case that they are not?

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    $\begingroup$ It's not straightforward at all. My answer here is that the concept and calculation of correlation requires vectors not only of equal length but also in one-to-one correspondence, i.e. pairing of values is uniquely defined in some natural way, as when measurements are for the same people, places, times, or whatever it is. As it's not defined, there is no question of discussing how Kendall correlation behaves otherwise. $\endgroup$
    – Nick Cox
    Commented Oct 28, 2015 at 15:00
  • $\begingroup$ The deeper question is how do you want to compare your vectors, as there are arbitrarily many ways to do that. You could, for example, construct some kind of outer product with each pairwise comparison forming one entry in a matrix. We need some idea of your purpose to advise well. $\endgroup$
    – Nick Cox
    Commented Oct 28, 2015 at 15:01
  • $\begingroup$ Thanks. You are right, when reading the question is not so straightforward. Let me come with a better explanation in next coupled of days. Thanks for your time anyhow. $\endgroup$
    – Kwnwps
    Commented Oct 28, 2015 at 15:39

1 Answer 1


To me this is a case involving computational complexity as it's applied in computer science. MDL or minimum description length is an information-theoretic metric for comparing these vectors that is rooted in the complexity of the contents of the vectors. MDL basically involves data compression and encryption. Here's the Wiki definition of MDL:

Any set of data can be represented by a string of symbols from a finite (say, binary) alphabet.

And the shortest path to representing that string is the MDL.

There's a good Wiki discussion here:


as well as an excellent, wide-ranging book length discussion in Oded Goldreich's Computational Complexity:


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    $\begingroup$ Can you explain the relation between the question and your answer? $\endgroup$
    – Michael M
    Commented Oct 28, 2015 at 14:59
  • $\begingroup$ @MichaelM Is your question as simple and precise as "do the n's have to be the same for both vectors*? If so, then you are correct that my answer completely misses that point! If not and you are asking how two vectors of differing length and, presumably, content are to be compared, then my answer offers one suggestion as to how to answer that question. $\endgroup$
    – user78229
    Commented Oct 28, 2015 at 15:04

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