'log-likelihood function generated by scipy.stats.rv_continuous.fit
The method scipy.stats.rv_continuous.fit finds the parameters that maximise a log likelihood function which is determined by the input data and the specification of the distribution rv_continuous. For instance, this could be normal or gamma.
The documentation for scipy.stats.rv_continuous.fit doesn't explain how the log-likelihood function is generated and I would like to know how. I need it so I can calculate the value of the log-likelihood at the parameters estimated by fit (i.e. the maximum value).
Solution 1:[1]
Log-likelihood is the logarithm of the probability that a given set of observations is observed given a probability distribution. You can access the value of a probability density function at a point x for your scipy.stats.rv_continuous member using scipy.stats.rv_continuous.pdf(x,params). You would take the product of these values for each member of your data, then take the log of that. For instance:
import numpy as np
from scipy.stats import norm
data = [1,2,3,4,5]
m,s = norm.fit(data)
log_likelihood = np.log(np.product(norm.pdf(data,m,s)))
Solution 2:[2]
While Mark's answer is technically correct, I can only re-iterate nikosd's concern from the comment - taking the product first and then logging it, will in many practical scenarios make your results unusable. If have many (thousands/millions) observations in data, each has a probability of <=1, so your product np.product(norm.pdf(data,m,s)) will be very small and often smaller than numerical precision, making your results unstable/wrong.
Thus the better way - and the reason why log-likelihood is used in the first place - is to first log the individual probabilities np.log(norm.pdf(data,m,s)) and then sum over the resulting vector.
import numpy as np
from scipy.stats import norm
data = [1,2,3,4,5]
m,s = norm.fit(data)
log_likelihood = np.sum(np.log(norm.pdf(data,m,s)))
I thought this was important enough to warrant a separate answer.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Mark Snyder |
| Solution 2 | seulberg1 |
