Skip to main content

plotting - Probability density histogram with unequal bin widths



I am confused by output of Histogram and HistogramList for probability density ("PDF") when bin widths are not equal.


According to this and this pages and other sources, density histograms are computed by dividing counts by bar widths and total number of observations. But Histogram obviously uses another algorithm as one can see from the following example:


SeedRandom[1];
data = RandomVariate[NormalDistribution[0, 1], 15];
HistogramList[data, {{-2, 0, 1}}, "PDF"][[2]]
BinCounts[data, {{-2, 0, 1}}]/((Length[data] - 3) Differences[{-2, 0, 1}])


{1/6, 2/3}


{1/6, 2/3}




The outputs are identical when bin counts are divided by total number of observations minus 3 in this case. Why is it? What algorithm Histogram uses for determining this difference (in other cases I got other numbers)?


An addition


Rod has answered the original question but there is another issue: if one gives upper bin boundary equal to the upper datapoint value then this value will be excluded from the histogram. It does not contradict the documentation where stated that {{b1,b2,...}} will use the bins [b1,b2),[b2,b3),.... Here is an illustration:


HistogramList[data, {{-2, 0, Max[data] - 10^-13}}, 
   "PDF"][[2]] - HistogramList[data, {{-2, 0, Max[data]}}, "PDF"][[2]]
HistogramList[data, {{-2, 0, Max[data] + 10^-13}}, "PDF"][[2]] - 
 HistogramList[data, {{-2, 0, Max[data]}}, "PDF"][[2]]



{0, 2.99205*10^-14}


{-(1/105), 0.0123342}



One can see that subtraction of 10^-13 does not alter the result significantly as expected but addition of 10^-13 changes it considerably because now the point Max[data] is included in the histogram. One can check this directly:


HistogramList[data, {{-2, 0, Max[data]}}, "PDF"][[2]]
BinCounts[data, {{-2, 0, Max[data]}}]/ 
  Differences[{-2, 0, Max[data]}]/(Length[data] - 1)

HistogramList[data, {{-2, 0, Max[data] + 10^-13}}, "PDF"][[2]]
BinCounts[data, {{-2, 0, Max[data] + 10^-13}}]/ 

  Differences[{-2, 0, Max[data]}]/Length[data]


{1/7, 0.462531}


{1/7, 0.462531}


{2/15, 0.474865}


{2/15, 0.474865}




Answer



Using SeedRandom[1] you get 3 observations higher than 1. When you use Histogram[data,{{-2,0,1}}] you're excluding those 3 observations...



If you exclude those 3 observations, now your probability (i.e., "PDF") should be based on 12 observations, and not 15...


Comments

Post a Comment

Popular posts from this blog

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

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