how to estimate cdf from ksdensity pdf

I have a quick question about ksdensity. For a given variable I derive distribution by binning into a specified number of bins, calculating the number of samples in each bin. For example, I used 10 bins to get the pdfs

To sample distributions, those pdf's are converted to cdf's, I draw random values from 0-1 so sample that cdf at bin center i. This allows to only get back the values at each bin center which for contionous variable is not what I want. Thus I then sample uniformatly from i-1 to i+1.

I was wondering if I can used ksdensity to do this as the more robust soluton. So essentially finding CDF from PDF that was estimated using Kernel Desnity?

Thank you!

Answer 1

[Star Strider]

There is an option for ksdensity to return the cdf:

x = [randn(30,1); 5+randn(30,1)];
[f,xi] = ksdensity(x);

figure
plot(xi,f)
grid
title('PDF')

[f,xi] = ksdensity(x, 'Function','cdf');

figure
plot(xi,f)
grid
title('CDF')

Experiment with your own distribution. This is discontinuous, so you may need to use the interp1 function for intermediate values.

[jenka]

Is there a sampling function that would allow me to sample from that cdf? Thank you so much this is very useful:)

[Star Strider]

My pleasure.

I am not certain that it is possible to create random numbers from an empirical distribution. Looking through the documentation, you would have to fit a specific distribution to your data with the fitdist (link) function. If that provided an acceptable approximation, you could then generate random numbers from it using the random function. I am not aware that any other option exists.

[jenka]

actually, I think this might work:

x = [randn(30,1); 5+randn(30,1)];
[f,xi] = ksdensity(x);
% Generate uniform random number between 0 and 1
ur =  rand(m,n);
% Interpolate ur from empirical cdf and extraplolate for out of range
% values.
xr = interp1(xi,f,ur,[],'extrap');

[Star Strider]

O.K. I am not certain what m and n are, and without more information, I continue to guess at the appropriate solution. I alluded (and linked) to interp1 earlier.

You can calculate the probabilities, then weight them appropriately.

Example —

rn = repelem(xi, fix(f/min(f)));                    % Generate Random Values

figure
histogram(rn, numel(f))                             % Plot Histogram

That will reproduce the PDF of the distribution.

Experiment to get the result you want.