I want to plot a gamma distribution with alpha = 29 (the scale) and beta = 3 (the size). In other words, I want to plot the pdf for Gamma(29,3). How do I do this if according to the documentation, the python gamma function only has parameters a and x and the size parameter doesn't exist?
I thought loc was beta, but I think it's actually offset, so the code below is wrong...
import numpy as np
import scipy.stats as stats
from matplotlib import pyplot as plt
x = np.linspace (0, 100, 200)
y1 = stats.gamma.pdf(x, a=29, loc=3) #a is alpha, loc is beta???
plt.plot(x, y1, "y-", label=(r'$\alpha=29, \beta=3$'))
plt.ylim([0,0.08])
plt.xlim([0,150])
plt.show()
According to the documentation, you want to use the scale parameter (theta), but since you are defining beta, which is the inverse of theta, then you pass scale with the value of 1/beta, which in your example would be 1/3 or 0.33333.
Therefore, try:
y1 = stats.gamma.pdf(x, a=29, scale=0.33333)
As @Hielke replied, as far as explained in scipy.stats 1.4.1 documentation it seems that the scalar parameter is equal to beta. Indeed, the function originally developped is :
gamma.pdf(x, a) = x^(a-1) * exp(-x) / gamma(a)
If one replaces x by a combination of the two optional parameters loc and scale as :
x = (y - loc) / scale
One should have :
gamma.pdf(x, a) = (y - loc)^(a-1) * exp( -(y - loc)/scale ) / (scale^(a-1) * gamma(a))
If you take loc = 0 then you recognized the expression of the Gamma distribution as usually defined. You multiply by the inverse of scale and you can conclude that scale = beta in this function and loc is an offset.
Actually I have tried to detail the documentation explaination :
Specifically, gamma.pdf(x, a, loc, scale) is identically equivalent to gamma.pdf(y, a) / scale with y = (x - loc) / scale.
This is not strictly answer for question, but this comes up when searching equivalent for R qgamma with alpha and beta parameters in Python. So just a note:
R:
qgamma(0.025, 5, 41.3)
Python:
from scipy.stats import gamma
gamma.ppf(0.025, 5, scale=1/41.3)