Reputation: 173
I've been trying to implement the skewed generalized t distribution in Python to model some financial returns. I based my code on formulas found on Wikipedia, and I used the Beta distribution from scipy.
from scipy.special import beta
import numpy as np
from math import sqrt
def sgt(x, params):
# This function accepts an array of 5 parameters [mu, sigma, lambda, p, q]
mu, sigma, lam, p, q = params
v = (q**(-1/p)) / (sqrt((3*lam*lam + 1)*beta(3/p, q-2/p)/beta(1/p, q) - 4*lam*lam*(beta(2/p, q-1/p)/(beta(1/p, q)))**2))
m = 2*v*sigma*lam*q**(1/p)*beta(2/p, q - 1/p) / beta(1/p, q)
fx = p / (2*v*sigma*(q**(1/p))*beta(1/p, q)*((abs(x-mu+m)**p/(q*(v*sigma)**p*(lam*np.sign(x-mu+m)+1)**p + 1)+1)**(1/p + q)))
return fx
Now, the function seems to work perfectly fine for some sets of parameters, but terribly for other sets of parameters.
For example:
dx = 0.001
x_axis = np.arange(-10, 10, dx)
ok_parameters = [0, 2, 0, 3, 8]
bad_parameters = [0, 2, 0, 1.05, 2.1]
ok_distribution = sgt(x_axis, ok_parameters)
bad_distribution = sgt(x_axis, bad_parameters)
If I try to compute the integrals of those two numbers:
a = np.sum(ok_distribution*dx)
b = np.sum(bad_distribution*dx)
I obtain the results a = 1.0013233154393804 and b = 2.2799746093533346. Now, in theory both of these should be 1, but I assume since I approximated the integral the value won't always be exactly 1. In the second case however I don't understand why the value is so high.
Does anyone know what the issue is?
These are the graphs of the ok distribution (blue) and bad distribution (orange)
Upvotes: 4
Views: 1759
Reputation: 782
I believe there was just a typo (though I couldn't exactly find where) in your definition sgt
. Here is an implementation that works.
%matplotlib inline
import matplotlib.pyplot as plt
from scipy.special import beta
import numpy as np
from math import sqrt
from typing import Union
from scipy import integrate
# Generalised Student T probability Distribution
def generalized_student_t(x:Union[float, np.ndarray], mu:float, sigma:float,
lam:float, p:float, q:float) \
-> Union[float, np.ndarray]:
v = q**(-1/p) * ((3*lam**2 + 1)*(beta(3/p, q - 2/p)/beta(1/p,q)) - 4*lam**2*(beta(2/p, q - 1/p)/beta(1/p,q))**2)**(-1/2)
m = 2*v*sigma*lam*q**(1/p)*beta(2/p,q - 1/p)/beta(1/p,q)
fx = p / (2*v*sigma*q**(1/p)*beta(1/p,q)*(abs(x-mu+m)**p/(q*(v*sigma)**p)*(lam*np.sign(x-mu+m)+1)**p + 1)**(1/p + q))
return fx
def plot_cdf_pdf(x_axis:np.ndarray, pmf:np.ndarray) -> None:
"""
Plot the PDF and CDF of the array returned from the function.
"""
fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(12, 6))
ax1.plot(x_axis, pmf)
ax1.set_title('PDF')
ax2.plot(x_axis, integrate.cumtrapz(x=x_axis, y=pmf, initial = 0))
ax2.set_title('CDF')
pass
dx = 0.0001
x_axis = np.arange(-10, 10, dx)
# Create the Two
distribution1 = generalized_student_t(x=x_axis, mu=0, sigma=1, lam=0, p=2, q=100)
distribution2 = generalized_student_t(x=x_axis, mu=0, sigma=2, lam=0, p=1.05, q=2.1)
plot_cdf_pdf(x_axis=x_axis, pmf=distribution1)
plot_cdf_pdf(x_axis=x_axis, pmf=distribution2)
We can also check that the integral of the PDFs are 1
integrate.simps(x=x_axis, y = distribution1)
integrate.simps(x=x_axis, y = distribution2)
We can see the results of the integral are 0.99999999999999978 and 0.99752026308335162. The reason they are not exactly 1 is due the CDF being defined as integral from -infinity to infinity of the PDF.
Upvotes: 5