Fit Gaussian mixture model with weighted observations
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Wolfgang Schwanghart on 23 Nov 2019
Answered: Omkar Mulekar on 5 Jun 2020
Hi everyone, looking at the help of fitgmdist, I cannot see that there is the possibility to weight observations. Is there a reason? Many functions of the Statistics and Machine Learning toolbox support weights. Does anyone have an idea how to include weights, or can anyone point me to an alternative?
Adam Danz on 26 Nov 2019
If you end up giving that a try, keep in mind that the weights must be converted to integers and depending on how that's carried out, it could vastly increase the number of data points. Feel free to pull me in if you decide to go down this route and get stuck.
In a sense, by duplicating the values of the data being fit, you are strengthening their representation in the fit and that's kind of like weighting.
Kaashyap Pappu on 26 Nov 2019
The function fitgmdist fits a distribution to a given data set. This data set generally has points belonging to the same class therefore the ‘weight’ parameter is not needed, since you are essentially just fitting a distribution model to given data.
Functions such as fitcknn, fitcsvm have weights because those are classification models. Weights become essential when data from multiple classes is present for training, but there is a class imbalance, that is data points for each class are not in equal proportion. To account for this imbalance, weights are used and are essential input arguments.
Hope this helps!
Jeff Miller on 26 Nov 2019
It's not exactly clear (to me either) what it means to weight the different observations in this context, but maybe you have something like this in mind:
You have observations X(1:n) with weights W(1:n). Let sumW = sum(W).
Make a new dataset Y with (say) 10000 observations consisting of
round(W(1)/sumW*10000) copies of X(1)
round(W(2)/sumW*10000) copies of X(2)
etc--that is, round(W(i)/sumW*10000) copies of X(i)
Now use fitgmdist with Y. Every Y value will be weighted equally, but the different X's will have weights approximately proportional to their original W values--because their numbers will be in those proportions.
I hope that is clear.
Omkar Mulekar on 5 Jun 2020
There seems to be an answer in this paper:
They talk about a couple of methods for EM using weighted data. See if it's useful for you!
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