Are Easter days normally distributed?

Question

How can Mathematica be used to check if Easter days are normally distributed?

By definition, Easter day is the first Sunday after the first full moon after the spring equinox (so between March 23d and April 25th).

This is a self-answered question but of course the purpose is to discover new ideas/commands/etc. so feel free to post your own answer!

Answer 1

Instead of FindFit[], DistributionFitTest[] is the function to use to test the distribution followed by your data. Using easter[] from this answer, generate the number of days Easter is counted from March 23 of that same year:

data = Table[DayCount[DateObject[{k, 3, 23}], DateObject[easter[k]]], {k, 1700, 2018}];

Then,

hh = DistributionFitTest[data // N, Automatic, "HypothesisTestData"];

See the test results:

hh["TestDataTable", All]

The $p$-values of all except one test are quite tiny. More starkly put,

hh["TestConclusion", All]
   {"The null hypothesis that the data is distributed according to the
     NormalDistribution[x, y] is rejected at the 5 percent level based on
     the Anderson-Darling test.",

    ...

    "The null hypothesis that the data is distributed according to the
     NormalDistribution[x, y] is rejected at the 5 percent level based on
     the Shapiro-Wilk test."}

where I have omitted some of the output.

Answer 2

First, we need to get the data, here from the website tlarsen2.tripod.com:

data = Import["http://tlarsen2.tripod.com/thomaslarsen/easterdates.html", "Table"];
dates = Reverse /@ SortBy[Partition[Flatten@Select[data, 
         MemberQ[#, "April"] || MemberQ[#, "March"] &], 3], Last] /. 
    "April" -> 4 /. "March" -> 3;
dates = Table[date[[1 ;; 2]]~Join~{ToExpression@
      StringReplace[date[[3]], LetterCharacter -> ""]}, {date, dates}];
(* {{1700, 4, 11}, {1701, 3, 27}, {1702, 4, 16}, ... *)

Then, convert the dates into the number of days from the first possible date (March 23d), and use Tally to count:

monthsDays = dates[[All, {3, 2}]];
days = If[#[[2]] == 4, #[[1]] + 31 - 22., (#[[1]] - 22.)] & /@ monthsDays;
dist = Sort@Tally[days];

That's the date (counted in days after March 23d) as a function of the year (counted from 1700).

ListPlot[days]

Then, we can fit a Gaussian curve on the distribution (note: to see how to do this properly, check J.M.'s answer)

model[x_] = ampl Evaluate[PDF[NormalDistribution[mu, sigma], x]];
fit = FindFit[dist, model[x], {ampl, mu, sigma}, x]
Show[ListPlot[dist], 
 Plot[model[x] /. fit, {x, 0, 35}, PlotStyle -> Red]]

The result is not really well approximated by a Gaussian.

Answer 3

This is code to compute Easter Sunday for each year (the "Computus"), from Gauss, in the proleptic Gregorian calendar:

computusGauss[year_Integer] := 
  Module[{a, b, c, d, e, f, g, h, i, j, k, month, day},
   a = Mod[year, 19];
   {b, c} = QuotientRemainder[year, 100];
   {d, e} = QuotientRemainder[b, 4];
   f = Quotient[8 b + 13, 25];
   g = Mod[19 a + b - d - f + 15, 30];
   {h, i} = QuotientRemainder[c, 4];
   j = Quotient[a + 11 g, 319];
   k = Mod[2 e + 2 h - i - g + j + 32, 7];
   month = Quotient[g - j + k + 90, 25];
   day = Mod[g - j + k + month + 19, 32];
   {year, month, day}
   ];

The result is periodic with period 5700000 years.

Compute a full period:

data = computusGauss /@ Range[5700000];

Project the results to a common year (it's irrelevant which one you choose):

data[[All, 1]] = 2018;

A date histogram shows this trapezoidal structure, between March 22 and April 25, both included:

DateListStepPlot[Tally[data]]

Answer 4

An article from the Irish Astronomical Journal, "The Frequency Distribution of the Dates of Easter", explains a way to find the frequency of occurrence for each date without computing Easter for the 5,700,000 years in the period. Here's code for the method.

{p, q, r} = {27550, 27075, 26600};
freq = {p, 2 q, 3 q, 4 p, 5 r, 6 p, 7 r, 7 p, 7 q, 7 q, 7 p, 7 r, 7 p,
    7 r, 7 p, 7 q, 7 q, 7 p, 7 r, 7 p, 7 r, 7 p, 7 q, 7 q, 7 p, 7 r, 
   7 p, 141/19 r, 8 p, 7 q, 6 q, 5 p, 4 r, 3 p, 30/19 r};
hist = Partition[Riffle[DayRange["Mar 22", "Apr 25"], freq], 2];