How to normalize a histogram such that the area under the probability density function is equal to 1?
My answer to this is the same as in an answer to your earlier question. For a probability density function, the integral over the entire space is 1. Dividing by the sum will not give you the correct density. To get the right density, you must divide by the area. To illustrate my point, try the following example.
[f, x] = hist(randn(10000, 1), 50); % Create histogram from a normal distribution.
g = 1 / sqrt(2 * pi) * exp(-0.5 * x .^ 2); % pdf of the normal distribution
% METHOD 1: DIVIDE BY SUM
figure(1)
bar(x, f / sum(f)); hold on
plot(x, g, 'r'); hold off
% METHOD 2: DIVIDE BY AREA
figure(2)
bar(x, f / trapz(x, f)); hold on
plot(x, g, 'r'); hold off
You can see for yourself which method agrees with the correct answer (red curve).
Another method (more straightforward than method 2) to normalize the histogram is to divide by sum(f * dx) which expresses the integral of the probability density function, i.e.
% METHOD 3: DIVIDE BY AREA USING sum()
figure(3)
dx = diff(x(1:2))
bar(x, f / sum(f * dx)); hold on
plot(x, g, 'r'); hold off
Since 2014b, Matlab has these normalization routines embedded natively in the histogram function (see the help file for the 6 routines this function offers). Here is an example using the PDF normalization (the sum of all the bins is 1).
data = 2*randn(5000,1) + 5; % generate normal random (m=5, std=2)
h = histogram(data,'Normalization','pdf') % PDF normalization
The corresponding PDF is
Nbins = h.NumBins;
edges = h.BinEdges;
x = zeros(1,Nbins);
for counter=1:Nbins
midPointShift = abs(edges(counter)-edges(counter+1))/2;
x(counter) = edges(counter)+midPointShift;
end
mu = mean(data);
sigma = std(data);
f = exp(-(x-mu).^2./(2*sigma^2))./(sigma*sqrt(2*pi));
The two together gives
hold on;
plot(x,f,'LineWidth',1.5)
An improvement that might very well be due to the success of the actual question and accepted answer!
EDIT - The use of hist and histc is not recommended now, and histogram should be used instead. Beware that none of the 6 ways of creating bins with this new function will produce the bins hist and histc produce. There is a Matlab script to update former code to fit the way histogram is called (bin edges instead of bin centers - link). By doing so, one can compare the pdf normalization methods of #abcd (trapz and sum) and Matlab (pdf).
The 3 pdf normalization method give nearly identical results (within the range of eps).
TEST:
A = randn(10000,1);
centers = -6:0.5:6;
d = diff(centers)/2;
edges = [centers(1)-d(1), centers(1:end-1)+d, centers(end)+d(end)];
edges(2:end) = edges(2:end)+eps(edges(2:end));
figure;
subplot(2,2,1);
hist(A,centers);
title('HIST not normalized');
subplot(2,2,2);
h = histogram(A,edges);
title('HISTOGRAM not normalized');
subplot(2,2,3)
[counts, centers] = hist(A,centers); %get the count with hist
bar(centers,counts/trapz(centers,counts))
title('HIST with PDF normalization');
subplot(2,2,4)
h = histogram(A,edges,'Normalization','pdf')
title('HISTOGRAM with PDF normalization');
dx = diff(centers(1:2))
normalization_difference_trapz = abs(counts/trapz(centers,counts) - h.Values);
normalization_difference_sum = abs(counts/sum(counts*dx) - h.Values);
max(normalization_difference_trapz)
max(normalization_difference_sum)
The maximum difference between the new PDF normalization and the former one is 5.5511e-17.
hist can not only plot an histogram but also return you the count of elements in each bin, so you can get that count, normalize it by dividing each bin by the total and plotting the result using bar. Example:
Y = rand(10,1);
C = hist(Y);
C = C ./ sum(C);
bar(C)
or if you want a one-liner:
bar(hist(Y) ./ sum(hist(Y)))
Documentation:
hist
bar
Edit: This solution answers the question How to have the sum of all bins equal to 1. This approximation is valid only if your bin size is small relative to the variance of your data. The sum used here correspond to a simple quadrature formula, more complex ones can be used like trapz as proposed by R. M.
[f,x]=hist(data)
The area for each individual bar is height*width. Since MATLAB will choose equidistant points for the bars, so the width is:
delta_x = x(2) - x(1)
Now if we sum up all the individual bars the total area will come out as
A=sum(f)*delta_x
So the correctly scaled plot is obtained by
bar(x, f/sum(f)/(x(2)-x(1)))
The area of abcd`s PDF is not one, which is impossible like pointed out in many comments.
Assumptions done in many answers here
Assume constant distance between consecutive edges.
Probability under pdf should be 1. The normalization should be done as Normalization with probability, not as Normalization with pdf, in histogram() and hist().
Fig. 1 Output of hist() approach, Fig. 2 Output of histogram() approach
The max amplitude differs between two approaches which proposes that there are some mistake in hist()'s approach because histogram()'s approach uses the standard normalization.
I assume the mistake with hist()'s approach here is about the normalization as partially pdf, not completely as probability.
Code with hist() [deprecated]
Some remarks
First check: sum(f)/N gives 1 if Nbins manually set.
pdf requires the width of the bin (dx) in the graph g
Code
%http://stackoverflow.com/a/5321546/54964
N=10000;
Nbins=50;
[f,x]=hist(randn(N,1),Nbins); % create histogram from ND
%METHOD 4: Count Densities, not Sums!
figure(3)
dx=diff(x(1:2)); % width of bin
g=1/sqrt(2*pi)*exp(-0.5*x.^2) .* dx; % pdf of ND with dx
% 1.0000
bar(x, f/sum(f));hold on
plot(x,g,'r');hold off
Output is in Fig. 1.
Code with histogram()
Some remarks
First check: a) sum(f) is 1 if Nbins adjusted with histogram()'s Normalization as probability, b) sum(f)/N is 1 if Nbins is manually set without normalization.
pdf requires the width of the bin (dx) in the graph g
Code
%%METHOD 5: with histogram()
% http://stackoverflow.com/a/38809232/54964
N=10000;
figure(4);
h = histogram(randn(N,1), 'Normalization', 'probability') % hist() deprecated!
Nbins=h.NumBins;
edges=h.BinEdges;
x=zeros(1,Nbins);
f=h.Values;
for counter=1:Nbins
midPointShift=abs(edges(counter)-edges(counter+1))/2; % same constant for all
x(counter)=edges(counter)+midPointShift;
end
dx=diff(x(1:2)); % constast for all
g=1/sqrt(2*pi)*exp(-0.5*x.^2) .* dx; % pdf of ND
% Use if Nbins manually set
%new_area=sum(f)/N % diff of consecutive edges constant
% Use if histogarm() Normalization probability
new_area=sum(f)
% 1.0000
% No bar() needed here with histogram() Normalization probability
hold on;
plot(x,g,'r');hold off
Output in Fig. 2 and expected output is met: area 1.0000.
Matlab: 2016a
System: Linux Ubuntu 16.04 64 bit
Linux kernel 4.6
For some Distributions, Cauchy I think, I have found that trapz will overestimate the area, and so the pdf will change depending on the number of bins you select. In which case I do
[N,h]=hist(q_f./theta,30000); % there Is a large range but most of the bins will be empty
plot(h,N/(sum(N)*mean(diff(h))),'+r')
There is an excellent three part guide for Histogram Adjustments in MATLAB (broken original link, archive.org link),
the first part is on Histogram Stretching.
Related
How can I use Matlab to plot a univariate normal distribution when it has unknown mean but the mean is also normally distributed with known mean of mean and variance of mean?
Eg. N(mean, 4) and mean ~N(2,8)
Using the law of total probability, one can write
pdf(x) = int(pdf(x | mean) * pdf(mean) dmean)
So, we can calculate it in Matlab as follows:
% define the constants
sigma_x = 4;
mu_mu = 2;
sigma_mu = 8;
% define the pdf of a normal distribution using the Symbolic Toolbox
% to be able to calculate the integral
syms x mu sigma
pdf(x, mu, sigma) = 1./sqrt(2*pi*sigma.^2) * exp(-(x-mu).^2/(2*sigma.^2));
% calculate the desired pdf
pdf_x(x) = int(pdf(x, mu, sigma_x) * pdf(mu, mu_mu, sigma_mu), mu, -Inf, Inf);
pdfCheck = int(pdf_x, x, -Inf, Inf) % should be one
% plot the desired pdf (green) and N(2, 4) as reference (red)
xs = -40:0.1:40;
figure
plot(xs, pdf(xs, mu_mu, sigma_x), 'r')
hold on
plot(xs, pdf_x(xs), 'g')
Note that I also checked that the integral of the calculated pdf is indeed equal to one, which is a necessary condition for being a pdf.
The green plot is the requested pdf. The red plot is added as reference and represents the pdf for a constant mean (equal to the average mean).
I'am new in matlab programming,I should Write a script to generate a random sequence (x1,..., XN) of size N following the normal distribution N (0, 1) and calculate the empirical mean mN and variance σN^2
Then,I should plot them:
this is my essai:
function f = normal_distribution(n)
x =randn(n);
muem = 1./n .* (sum(x));
muem
%mean(muem)
vaem = 1./n .* (sum((x).^2));
vaem
hold on
plot(x,muem,'-')
grid on
plot(x,vaem,'*')
NB:those are the formules that I have used:
I have obtained,a Figure and I don't know if is it correct or not ,thanks for Help
From your question, it seems what you want to do is calculate the mean and variance from a sample of size N (nor an NxN matrix) drawn from a standard normal distribution. So you may want to use randn(n, 1), instead of randn(n). Also as #ThP pointed out, it does not make sense to plot mean and variance vs. x. What you could do is to calculate means and variances for inceasing sample sizes n1, n2, ..., nm, and then plot sample size vs. mean or variance, to see them converge to 0 and 1. See the code below:
function [] = plotMnV(nIter)
means = zeros(nIter, 1);
vars = zeros(nIter, 1);
for pow = 1:nIter
n = 2^pow;
x =randn(n, 1);
means(pow) = 1./n * sum(x);
vars(pow) = 1./n * sum(x.^2);
end
plot(1:nIter, means, 'o-');
hold on;
plot(1:nIter, vars, '*-');
end
For example, plotMnV(20) gave me the plot below.
For this given example:
a=rand(100,1)
ecdf(a)
The obtained plot will have on x proportion of bin values, and on y the proportion of points. I wanted to add a line y=0.5 (50% percent of values) and when cross the line in the plot gather and shows the predicted x value.
It is possible but it suppresses my knowledge.
Thanks in advance
Not sure I interpret your question correctly - but could it be as simple as
a=rand(100,1)
ecdf(a)
hold on
plot([0 1],[1 1]*0.5);
This adds a line at a height of 0.5 from 0 to 1 (which I believe are the limits of the plot that ecdf produced for you).
If you want to find the point where these two lines intersect, you need to obtain the points in the plot using a different form of the ecdf function:
a = rand(100,1)
[f x] = ecdf(a);
figure
plot(x, f); % now you have to make the plot yourself...
hold all
plot(x, 0.5 * ones(size(x))); % add the line at y=0.5
title 'cumulative probability of rand()'
xest = interp1(f, x, 0.5); % interpolate - find x where f would be 0.5
fprintf(1, 'The intercept is at x=%.2f\n", xest);
I am trying to use Matlab's nlinfit function to estimate the best fitting Gaussian for x,y paired data. In this case, x is a range of 2D orientations and y is the probability of a "yes" response.
I have copied #norm_funct from relevant posts and I'd like to return a smoothed, normal distribution that best approximates the observed data in y, and returns the magnitude, mean and SD of the best fitting pdf. At the moment, the fitted function appears to be incorrectly scaled and less than smooth - any help much appreciated!
x = -30:5:30;
y = [0,0.20,0.05,0.15,0.65,0.85,0.88,0.80,0.55,0.20,0.05,0,0;];
% plot raw data
figure(1)
plot(x, y, ':rs');
axis([-35 35 0 1]);
% initial paramter guesses (based on plot)
initGuess(1) = max(y); % amplitude
initGuess(2) = 0; % mean centred on 0 degrees
initGuess(3) = 10; % SD in degrees
% equation for Gaussian distribution
norm_func = #(p,x) p(1) .* exp(-((x - p(2))/p(3)).^2);
% use nlinfit to fit Gaussian using Least Squares
[bestfit,resid]=nlinfit(y, x, norm_func, initGuess);
% plot function
xFine = linspace(-30,30,100);
figure(2)
plot(x, y, 'ro', x, norm_func(xFine, y), '-b');
Many thanks
If your data actually represent probability estimates which you expect come from normally distributed data, then fitting a curve is not the right way to estimate the parameters of that normal distribution. There are different methods of different sophistication; one of the simplest is the method of moments, which means you choose the parameters such that the moments of the theoretical distribution match those of your sample. In the case of the normal distribution, these moments are simply mean and variance (or standard deviation). Here's the code:
% normalize y to be a probability (sum = 1)
p = y / sum(y);
% compute weighted mean and standard deviation
m = sum(x .* p);
s = sqrt(sum((x - m) .^ 2 .* p));
% compute theoretical probabilities
xs = -30:0.5:30;
pth = normpdf(xs, m, s);
% plot data and theoretical distribution
plot(x, p, 'o', xs, pth * 5)
The result shows a decent fit:
You'll notice the factor 5 in the last line. This is due to the fact that you don't have probability (density) estimates for the full range of values, but from points at distances of 5. In my treatment I assumed that they correspond to something like an integral over the probability density, e.g. over an interval [x - 2.5, x + 2.5], which can be roughly approximated by multiplying the density in the middle by the width of the interval. I don't know if this interpretation is correct for your data.
Your data follow a Gaussian curve and you describe them as probabilities. Are these numbers (y) your raw data – or did you generate them from e.g. a histogram over a larger data set? If the latter, the estimate of the distribution parameters could be improved by using the original full data.
How to normalize a histogram such that the area under the probability density function is equal to 1?
My answer to this is the same as in an answer to your earlier question. For a probability density function, the integral over the entire space is 1. Dividing by the sum will not give you the correct density. To get the right density, you must divide by the area. To illustrate my point, try the following example.
[f, x] = hist(randn(10000, 1), 50); % Create histogram from a normal distribution.
g = 1 / sqrt(2 * pi) * exp(-0.5 * x .^ 2); % pdf of the normal distribution
% METHOD 1: DIVIDE BY SUM
figure(1)
bar(x, f / sum(f)); hold on
plot(x, g, 'r'); hold off
% METHOD 2: DIVIDE BY AREA
figure(2)
bar(x, f / trapz(x, f)); hold on
plot(x, g, 'r'); hold off
You can see for yourself which method agrees with the correct answer (red curve).
Another method (more straightforward than method 2) to normalize the histogram is to divide by sum(f * dx) which expresses the integral of the probability density function, i.e.
% METHOD 3: DIVIDE BY AREA USING sum()
figure(3)
dx = diff(x(1:2))
bar(x, f / sum(f * dx)); hold on
plot(x, g, 'r'); hold off
Since 2014b, Matlab has these normalization routines embedded natively in the histogram function (see the help file for the 6 routines this function offers). Here is an example using the PDF normalization (the sum of all the bins is 1).
data = 2*randn(5000,1) + 5; % generate normal random (m=5, std=2)
h = histogram(data,'Normalization','pdf') % PDF normalization
The corresponding PDF is
Nbins = h.NumBins;
edges = h.BinEdges;
x = zeros(1,Nbins);
for counter=1:Nbins
midPointShift = abs(edges(counter)-edges(counter+1))/2;
x(counter) = edges(counter)+midPointShift;
end
mu = mean(data);
sigma = std(data);
f = exp(-(x-mu).^2./(2*sigma^2))./(sigma*sqrt(2*pi));
The two together gives
hold on;
plot(x,f,'LineWidth',1.5)
An improvement that might very well be due to the success of the actual question and accepted answer!
EDIT - The use of hist and histc is not recommended now, and histogram should be used instead. Beware that none of the 6 ways of creating bins with this new function will produce the bins hist and histc produce. There is a Matlab script to update former code to fit the way histogram is called (bin edges instead of bin centers - link). By doing so, one can compare the pdf normalization methods of #abcd (trapz and sum) and Matlab (pdf).
The 3 pdf normalization method give nearly identical results (within the range of eps).
TEST:
A = randn(10000,1);
centers = -6:0.5:6;
d = diff(centers)/2;
edges = [centers(1)-d(1), centers(1:end-1)+d, centers(end)+d(end)];
edges(2:end) = edges(2:end)+eps(edges(2:end));
figure;
subplot(2,2,1);
hist(A,centers);
title('HIST not normalized');
subplot(2,2,2);
h = histogram(A,edges);
title('HISTOGRAM not normalized');
subplot(2,2,3)
[counts, centers] = hist(A,centers); %get the count with hist
bar(centers,counts/trapz(centers,counts))
title('HIST with PDF normalization');
subplot(2,2,4)
h = histogram(A,edges,'Normalization','pdf')
title('HISTOGRAM with PDF normalization');
dx = diff(centers(1:2))
normalization_difference_trapz = abs(counts/trapz(centers,counts) - h.Values);
normalization_difference_sum = abs(counts/sum(counts*dx) - h.Values);
max(normalization_difference_trapz)
max(normalization_difference_sum)
The maximum difference between the new PDF normalization and the former one is 5.5511e-17.
hist can not only plot an histogram but also return you the count of elements in each bin, so you can get that count, normalize it by dividing each bin by the total and plotting the result using bar. Example:
Y = rand(10,1);
C = hist(Y);
C = C ./ sum(C);
bar(C)
or if you want a one-liner:
bar(hist(Y) ./ sum(hist(Y)))
Documentation:
hist
bar
Edit: This solution answers the question How to have the sum of all bins equal to 1. This approximation is valid only if your bin size is small relative to the variance of your data. The sum used here correspond to a simple quadrature formula, more complex ones can be used like trapz as proposed by R. M.
[f,x]=hist(data)
The area for each individual bar is height*width. Since MATLAB will choose equidistant points for the bars, so the width is:
delta_x = x(2) - x(1)
Now if we sum up all the individual bars the total area will come out as
A=sum(f)*delta_x
So the correctly scaled plot is obtained by
bar(x, f/sum(f)/(x(2)-x(1)))
The area of abcd`s PDF is not one, which is impossible like pointed out in many comments.
Assumptions done in many answers here
Assume constant distance between consecutive edges.
Probability under pdf should be 1. The normalization should be done as Normalization with probability, not as Normalization with pdf, in histogram() and hist().
Fig. 1 Output of hist() approach, Fig. 2 Output of histogram() approach
The max amplitude differs between two approaches which proposes that there are some mistake in hist()'s approach because histogram()'s approach uses the standard normalization.
I assume the mistake with hist()'s approach here is about the normalization as partially pdf, not completely as probability.
Code with hist() [deprecated]
Some remarks
First check: sum(f)/N gives 1 if Nbins manually set.
pdf requires the width of the bin (dx) in the graph g
Code
%http://stackoverflow.com/a/5321546/54964
N=10000;
Nbins=50;
[f,x]=hist(randn(N,1),Nbins); % create histogram from ND
%METHOD 4: Count Densities, not Sums!
figure(3)
dx=diff(x(1:2)); % width of bin
g=1/sqrt(2*pi)*exp(-0.5*x.^2) .* dx; % pdf of ND with dx
% 1.0000
bar(x, f/sum(f));hold on
plot(x,g,'r');hold off
Output is in Fig. 1.
Code with histogram()
Some remarks
First check: a) sum(f) is 1 if Nbins adjusted with histogram()'s Normalization as probability, b) sum(f)/N is 1 if Nbins is manually set without normalization.
pdf requires the width of the bin (dx) in the graph g
Code
%%METHOD 5: with histogram()
% http://stackoverflow.com/a/38809232/54964
N=10000;
figure(4);
h = histogram(randn(N,1), 'Normalization', 'probability') % hist() deprecated!
Nbins=h.NumBins;
edges=h.BinEdges;
x=zeros(1,Nbins);
f=h.Values;
for counter=1:Nbins
midPointShift=abs(edges(counter)-edges(counter+1))/2; % same constant for all
x(counter)=edges(counter)+midPointShift;
end
dx=diff(x(1:2)); % constast for all
g=1/sqrt(2*pi)*exp(-0.5*x.^2) .* dx; % pdf of ND
% Use if Nbins manually set
%new_area=sum(f)/N % diff of consecutive edges constant
% Use if histogarm() Normalization probability
new_area=sum(f)
% 1.0000
% No bar() needed here with histogram() Normalization probability
hold on;
plot(x,g,'r');hold off
Output in Fig. 2 and expected output is met: area 1.0000.
Matlab: 2016a
System: Linux Ubuntu 16.04 64 bit
Linux kernel 4.6
For some Distributions, Cauchy I think, I have found that trapz will overestimate the area, and so the pdf will change depending on the number of bins you select. In which case I do
[N,h]=hist(q_f./theta,30000); % there Is a large range but most of the bins will be empty
plot(h,N/(sum(N)*mean(diff(h))),'+r')
There is an excellent three part guide for Histogram Adjustments in MATLAB (broken original link, archive.org link),
the first part is on Histogram Stretching.