How to fit an uniform distribution to a histogram?-CodePudding

I have a set of data that is generated from an uniform distribution. Now I want to fit the corresponding histogram to an uniform distribution, such that there is a 'ㄇ' shape of line plotted on that histogram. I tried to fit it by using the MATLAB built-in function histfit, but there is no such an option of uniform distribution for histfit. How can I do it?

data = unifrnd(-100,100,1000,1);

%% MATLAB built-in function: 'histfit'
figure(1);
hh = histfit(data); % No options for 'histfit' to fit data to an uniform distribution

%% Manually fitting a histogram to an uniform distribution
figure(2);
numBars = length(hh(1).XData);
histogram(data, numBars);
% TODO: How to do next to plot a line that fits the data to an uniform distribution?

CodePudding user response：

Since a uniform distribution is just 1/(b-a) for a region [a, b), you can define a function that calculates this

x = -200:200;
y = unifdist(x, -100, 100);

figure;
plot(x, y)

function ret = unifdist(x, a, b)
  ret = ones(length(x), 1)/(b-a);
  ret(x < a | x >= b) = 0;
end

There's probably a much simpler and faster way to do this, but it works.

CodePudding user response：

As an alternative to @CrisLuengo's answer, you could use the Method of Moments to estimate the parameters a,b of the uniform distribution U[a,b]. Following equations are sufficient to solve for the parameters. The MoM just tells us to equate the (sample-) mean and variance to the distribution mean and variance:

mean(samples) = (a b)/2,    variance(samples) = (b-a)^2/12

This results in a, b = mean(samples) - sqrt(3 * variance(samples)). In MATLAB you can compute this as follows:

m = mean(data);
v = var(data);
a = m - sqrt(3*v);
b = m   sqrt(3*v);

To plot this you can just define vector

x = linspace(-lower_limit, upper_limit, number_of_points);
y = (a < x) .* (x < b);
plot(x, y, '-r');

CodePudding user response：

min(data) and max(data) give an estimate of the two parameters of the uniform distribution, assuming data is uniformly distributed.

Note that this is a biased estimator, see here for a correction to remove the bias (one of the answers considers the case where the lower bound is not 0). [Link thanks to @flawr.]