estimation probability over the part of an array

[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!

Answer 1

[Andrei Bobrov]

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]

spasibo za otvet!a vi mojete poyasnit etot kod?mne ne yasno chto meniat' nyjno?

Answer 2

[terance]

still isn't solved((

Answer 3

[Walter Roberson]

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.

[terance]

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)

Answer 4

[terance]

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