How to analyze a distribution parameters?

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Avigdor
Avigdor am 1 Apr. 2016
Kommentiert: Avigdor am 2 Apr. 2016
Hi,
I have a distribution of pixels (y) as a function of a parameter (x) where every y value (bin) is the total number of pixels at a given value of x. I'd like to analyze this distribution for mean, median, max, std, etc, and find the corresponding x values for these measurements. Anybody have a way to do this?
Thanks,
Avi
  1 Kommentar
John D'Errico
John D'Errico am 2 Apr. 2016
Please don't indent your paragraphs. That causes the entire paragraph to come out as one long line, which must be scrolled to read it. I removed the indentation for you.

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Akzeptierte Antwort

Roger Stafford
Roger Stafford am 2 Apr. 2016
Bearbeitet: Roger Stafford am 2 Apr. 2016
I fail to see what the difficulty is here. If x is a vector containing all the values of the parameter in question, then y is a vector of all the corresponding pixel counts and y/sum(y) is a vector of the probabilities of each value of x. The mean value of x would then be:
m = sum(x.*y/sum(y))
The standard deviation would be:
s = sqrt(sum((x-m).^2*y/sum(y)))
  3 Kommentare
Roger Stafford
Roger Stafford am 2 Apr. 2016
Bearbeitet: Roger Stafford am 2 Apr. 2016
@Muhammad: The coefficient of determination would depend on what statistical model you are trying to fit.
@Avigdor: As an addendum to the above answer, here is code for finding the statistical median:
[xs,p] = sort(x);
ys = y(p);
c = cumsum(ys/sum(ys));
f = find(c>=1/2,1,'first');
xmed1 = xs(max(f-1,1));
xmed2 = xs(f);
The statistical median is any number between xmed1 and xmed2.
The max is not a statistical entity. Just use matlab's 'max' function on x.
Muhammad Usman Saleem
Muhammad Usman Saleem am 2 Apr. 2016
@Roger: if i am using mean , mean absolute error, correlation and want to find coefficient of determination in matlab then its formula?

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Weitere Antworten (1)

Image Analyst
Image Analyst am 2 Apr. 2016
You can simply do
meany = mean2(y);
mediany = median(y(:));
maxy = max(y(:));
stdDevy = std(y(:));
I don't see how x is involved. Your parameter "x" is basically the gray level and is not involved at all. If you want the range of "x", you can simply do
grayLevelRange = max(y(:)) - min(y(:)) + 1;
  3 Kommentare
Image Analyst
Image Analyst am 2 Apr. 2016
For some arbitrary histogram with bin heights all over the place, no bin will likely have the mean bin count. For example, let's say the bins had counts from 0 to 100,000. And let's say the mean bin height was 23,489.12343. Well, no bin has that since counts must be an integer. So let's say you round to 23,489. Well, there might be 5 or 10 or any number of bins with a count of 23,489. So which one(s) do you want? And why?
Avigdor
Avigdor am 2 Apr. 2016
Hi, the distribution is constructed by counting the number of pixels (y) in discrete grey level ranges (x, bins) over a large range of grey values in some images. We want to find the mean, median of the # of pixels, not so much to know the statistics for pixel count (y), but rather the corresponding grey level/bin for these pixel statistics (x). The distributions are unimodal, and we need the grey level distribution statistics because they relate to some physical parameter of interest. Thanks!

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