Skip to main content

distributions - what is The Formula behind the Kolmogorov-Smirnov test statistics?


I calculate the Kolmogrov-samirnov test statistic for the following data:
R = {0.171434, 0.134263, 0.155931, 0.135479, 0.196356, 0.152357, 0.133084, 0.10537, 0.14654, 0.116676, 0.123145, 0.145377, 0.12366, 0.156681, 0.208564, 0.202139, 0.227931, 0.15622, 0.118042, 0.104006, 0.322, 0.327, 0.331, 0.34, 0.383, 0.454}
with weibull distribution with parameters (3.08, 5.39) and it was
0.226532

but i recalculate it by excel by the same data and same parameters then the result was
0.614
so i want help to know the formula behind the Kolmogorov-Smirnov test statistic in mathematica
the formula behind the Kolmogorov-Smirnov test statistic in excel is
D=max|i/n-CDF|
Thanks



Answer



You have likely made an error in computing the test statistic by not sorting R and by using incorrect limits on empirical CDF values.


R = {0.171434, 0.134263, 0.155931, 0.135479, 0.196356, 0.152357, 
   0.133084, 0.10537, 0.14654, 0.116676, 0.123145, 0.145377, 0.12366, 

   0.156681, 0.208564, 0.202139, 0.227931, 0.15622, 0.118042, 
   0.104006, 0.322, 0.327, 0.331, 0.34, 0.383, 0.454};

dist = WeibullDistribution[3.08, 1/5.39];
n = Length[R];

Max[Abs[Range[0,n-1]/n - CDF[dist, Sort@R]]]

(* 0.226532 *)


This is precisely what we get using the built in test.


KolmogorovSmirnovTest[R, dist, "TestStatistic"]

(* 0.226532 *)

I can replicate what you found in Excel by shifting the empirical values by one and by not sorting R.


Max[Abs[Range[1,n]/n - CDF[dist, R]]]

(* 0.614413 *)


The implementation in Mathematica is more general than this method. If you happen to have ties in the data then you have to actually compute the empirical CDF or they won't be handled properly. For example...


SeedRandom[1];
r2 = RandomChoice[R, n];

Max[Abs[Range[0, n - 1]/n - CDF[dist, r2]]]

(* 0.762458 *)

Which is not equivalent to the correct value...


MaxValue[Abs[CDF[EmpiricalDistribution[r2], x] - CDF[dist, x]], {x}]


(* 0.150597 *)

Comments

Post a Comment

Popular posts from this blog

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
Post a Comment
Read more

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more