Problem using trapz for obtaining probabilities

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Stephan
Stephan 2012-11-8
I use numerical integration to calculate the the integral as seen below. Because this represents a loss distribution, values for P_D should be in the interval [0,1]. The problem is that e.g. P_D(d=0)>10!
How can I fix this problem?
PD = .01;
N = 50;
r = .1;
for d=0:N
f = -10:.01:10;
CPD = normcdf(norminv(PD,0,1),sqrt(r)*f,sqrt(1-r));
l = CPD.^(d).* (1-CPD).^(N-d);
P_D(1,d+1) = nchoosek(N,d) * trapz(f,l);
end

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回答（1 个）

Tom Lane
Tom Lane 2012-11-9
I don't quite understand what you are trying to do. I suggest you consider d=0 and step through your code to see why the first number it calculates is greater than 1.
I am suspicious that
nchoosek(N,d) * CPD.^(d).* (1-CPD).^(N-d);
has the form of a binomial distribution probability function, but you are integrating it over f (CPD is a function of f) in a way I don't understand.
In the first iteration with d=0, try plot(f,l) right before the call to trapz. You are integrating a thing that looks like a cdf rather than a pdf.

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