Numerical integration and NaN
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I want to calculate the following integral:
integral( pdf(data_one) / pdf(data_two) )
I managed to do this, but I had to resort to a hack to avoid divisions by zero. This hack works but creates deviations that annoy me - what is the proper way to do this?
data_axis = 1:5;
data_one = [1,2,3,4,5];
data_two = [1,1,1,1,5];
% Calculate probability densities
pdf_one = histc(data_one,data_axis); % [0.2,0.2,0.2,0.2,0.2]
pdf_two = histc(data_two,data_axis); % [0.8,0.0,0.0,0.0,0.2]
% HACK: Histc may have generated 0-buckets
%pdf_one = (pdf_one + 0.1);
%pdf_two = (pdf_two + 0.1);
% Normalize the result
pdf_one = pdf_one./sum(pdf_one);
pdf_two = pdf_two./sum(pdf_two);
complicated_integral = trapz(pdf_one./pdf_two);
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Antworten (1)
Walter Roberson
am 11 Mai 2011
Probability Density is not defined for discrete distributions.
Not every probability distribution has a density function: the distributions of discrete random variables do not; nor does the Cantor distribution, even though it has no discrete component, i.e., does not assign positive probability to any individual point.
and
If a probability distribution admits a density, then the probability of every one-point set {a} is zero; the same holds for finite and countable sets.
This implies that you are asking to calculate something which is not defined, or else that your calculation is incorrect, perhaps attempting to sample the counts at a point rather than over a range.
5 Kommentare
Richard Crozier
am 18 Mai 2011
If you know what the underlying form of the pdf of the data from some scientific knowledge of the system from which you have sampled the data, could you not fit the distribution to the datasets and integrate the resulting continuous functions? Of course the accuracy of this is doubtful for some distributions, e.g. an exponential distribution.
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