scipy.stats 中的毛刺分布

Question

I am trying to calculate the mean and std for burr distribution, but I am not quite sure how to input this. The pdf I am using is: f(x) = (alpha*gamma*lambda**alpha*x**(gamma-1))/(lambda+x**gamma)**(alpha+1) from the IFoA Formulae.

I have calculated the parameters to be: alpha = 2.3361635751273977, lambda = 10.596809948869414 and gamma = 0.5 in order to get mean = 500 and std = 600.

Could someone suggest how I should input the data into scipy.stats.burr or scipy.stats.burr12 ?

Answer 1

You need burr12 here, not burr . (The difference is in the sign of the power of x that sits inside another power. Confusingly, it's burr12 that is usually called simply Burr outside of SciPy, not the thing that SciPy calls burr .)

The Burr XII PDF is written in SciPy as c*d*x**(c-1)*(1+x**c)**(-d-1) where c, d are positive shape parameters. Your formula

(alpha*gamma*lamda**alpha*x**(gamma-1)) / (lamda+x**gamma)**(alpha+1)

has lambda in place of 1, so there is some scaling involved. SciPy docs say

burr12.pdf(x, c, d, loc, scale) is identically equivalent to burr12.pdf(y, c, d) / scale with y = (x - loc) / scale .

So, in order for lamda+x**gamma to be a constant multiple of 1 + (x/scale)**gamma , we need scale to be lamda**(1/gamma) . The exponents correspond to SciPy notation as c = gamma and d = alpha . Let's test this:

from scipy.stats import burr12
alpha = 2.3361635751273977
lamda = 10.596809948869414
gamma = 0.5

scale = lamda**(1/gamma)
c = gamma
d = alpha
print(burr12.mean(c, d, loc=0, scale=scale))
print(burr12.std(c, d, loc=0, scale=scale))

which prints

500.0
600.0