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Question of mean

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Tommy
Tommy el 21 de Mayo de 2012
When I write a program about consumption (C) with 10^5 trajectories, I want to calculate the expectation of them with the command mean(C(:,j)). However, at j=1, all the C are equal to 0.86, but the mean(C(:,1))=0.859999999999891. Would anyone tell me why it happens and how to correct it please? Thank you.

Respuesta aceptada

Tom Lane
Tom Lane el 21 de Mayo de 2012
As Oleg and Titus imply, one idea is to change the test so that it allows for things that differ by a small amount to be treated as roughly equal. Another thing that sometimes works is to refine your answer by another application of mean. This works in some cases but doesn't change the fundamental fact that two theoretically identical things may be slightly different when represented by double precision on a computer. Anyway, here's what I had in mind:
>> x = repmat(0.86,1e5,1);
>> fprintf('%25.18g\n',x(1))
0.85999999999999999
>> m = mean(x); fprintf('%25.18g\n',m)
0.86000000000030574
>> m = m + mean(x-m); fprintf('%25.18g\n',m)
0.85999999999999999

Más respuestas (1)

Titus Edelhofer
Titus Edelhofer el 21 de Mayo de 2012
Hi,
0.86 is not exactly representable in computers double precision representation:
fprintf('%.20f\n', 0.86)
0.85999999999999999000
Titus
  3 comentarios
Tommy
Tommy el 21 de Mayo de 2012
then would you please tell me how to get the right answer please? since I need the correct one to finish the loop of
if C(i,j)>=mean(C(:,j))
THX a lot!
Walter Roberson
Walter Roberson el 21 de Mayo de 2012
Recode your test to
if C(i,j) * size(C,1) >= sum(C(:,j))

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