Peacock in his paper on 'Two-dimensional goodness-of-fit testing in astronomy', http://adsabs.harvard.edu/full/1983MNRAS.202..615P, described the issue with the non-uniqueness of CDF in higer (N >= 2) dimensions.

While its definition is N=1 is unambiguous, "in N dimensions there are 2^N - 1 independent ways of defining a cumulative probability distribution." I can imagine two ways to define it in N=2 as shown in the figure

enter image description here

but accordingly to Peacock there should be 3 ways to do it. I have the feeling I am on a wrong track here so would be very grateful for any hints.

UPDATE & CORRECTIONS In fact I have found at least 4 ways to define a CDF, in part thanks to @jbowman comment, see figure (with corrected values for CDFs). Does that mean that Peacock's formula is incorrect? enter image description here

  • 2
    $\begingroup$ Take your first one and reverse the 0.4 and 0.6 areas by reversing which line to the two points defining those areas is "on top" (unclear, I admit, but hopefully you'll figure it out!) $\endgroup$
    – jbowman
    Nov 14, 2019 at 0:01
  • 1
    $\begingroup$ The correct number is $2^n,$ not $2^n-1.$ Even in one dimension there are two equally valid definitions. So what? They are all equivalent. Peacock is clear about this: see his discussion on p. 617. $\endgroup$
    – whuber
    Nov 14, 2019 at 0:44
  • $\begingroup$ @whuber I cannot see in Peacock's paper where he corrects himself about the 2^N - 1 statement. He later talks about quadrants (defined for each data point) in which he calculates the probabilities in order to find the point where one observes the largest difference. Also other sources talk about '2^N-1', see Lopes et al. bura.brunel.ac.uk/bitstream/2438/1166/1/acat2007.pdf or Wikipedia en.wikipedia.org/wiki/… $\endgroup$
    – mjs
    Nov 14, 2019 at 10:05
  • $\begingroup$ @whuber as you can see in the EDIT I have found 2^N ways to define the CDF, it seems Peacock's formula is wrong. $\endgroup$
    – mjs
    Nov 14, 2019 at 11:19
  • 2
    $\begingroup$ Note the emphasis on "$2^n-1$ independent ways" in these works. For continuous distributions, the sum of all $2^n$ different possible CDFs is constantly $1:$ a linear dependence. I believe that's what these authors may be accounting for. $\endgroup$
    – whuber
    Nov 14, 2019 at 14:46


Your Answer

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