estimation probability over the part of an array

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terance
terance 2011-5-24
[EDIT: 20110524 10:11 CDT - reformat - WDR]
Hello!
I have an array of data and I want to estimate the mean probability over the window of this array. I wrote the following code, but it shows me zeros except the last window of an array.
function y=pm(x,length)
for i=1:size(x)-length
m=mean(x(i:i+length));
s=std(x(i:i+length));
k(i)=probability_estimate(x(i:i+length),m,s);
end;
y=k;
end
function y=probability_estimate(x,m,s)
for i=1:length(x)
k(i)=lognpdf(x(i),m,s);
end;
y=mean(k);
end
>>d=pm(x,length);
would you be so kind to explain to me how to correct this code? Thank you for your answers!

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回答（4 个）

Andrei Bobrov
Andrei Bobrov 2011-5-24
function y=probability_estimate(x,m,s)
y=mean(lognpdf(x,m,s));
end
EDIT
idxs = bsxfun(@plus,1:l,(0:length(x)-l)');
outcell = arrayfun(@(y)mean(lognpdf(x(idxs(y,:)),...
mean(x(idxs(y,:))),std(x(idxs(y,:))))),1:size(idxs,1),'un',0);
d = [outcell{:}]';
your "length" -> "l"
with loop:
s1 = size(idxs,1);
d = zeros(s1,1);
for j = 1:s1
y = x(idxs(j,:));
d(j) = mean(lognpdf(y),mean(y),std(y));
end
terance
terance 2011-5-24
still isn't solved((
Walter Roberson
Walter Roberson 2011-5-24
size(x) is a two-element array. i=1:size(x)-length is not going to do what you want. Are you working with an array or with a row vector or with a column vector?
Also, please do not use a variable named "length" as you are very likely to run in to difficulties with the function of that name.
  1 个评论
terance
terance 2011-5-24
I'm working with an 1 dimensional array.
I have a sample, named x(e.x. x=[1 2 3 4 5 6 7]) and I need to estimate the following:
1)I choose th size of the window (length=2,step=1), so I have the following windows: [1 2], [2 3] [3 4] [4 5] [5 6][6 7]
2)over each window, I have to calculate the probability of each window-element and then returnn the mean of the probabilities of each elements in the window (the probability is log-normal, with mean over the window and std over the window)

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terance
terance 2011-5-24
problem is solved, here is the corrected code (thx god for debugging!=))
function y=pm(x,L)
a=length(x);
for i=1:a-L
m=mean(x(i:i+L));
s=std(x(i:i+L));
k(i)=probability_estimate(x(i:i+L),m,s);
end;
y=k;
end
function y=probability_estimate(x,m,s) for z=1:length(x) k(z)=normcdf(x(z),m,s); end; y=mean(k); end

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