My code is as follows:
from scipy.stats import pareto
values=[0.001,0.1,0.0000000000000000000000000000000019,2.,3.,4.,5.,6.,7.,8.,9.,20.,40.,80.,160.]
for b in values:
par = pareto.rvs(b,loc=0,scale=1, size=1)
print(par[0])
This is the output:
2.4710696399005953e+244
4457.919153166827
inf
1.074371220057211
1.813450580334762
1.129136768422894
1.0493631839305024
1.0741561671750386
1.1357715848324075
1.3165771298337223
1.000756684137126
1.0001559590523508
1.0831688633464842
1.0001668159923398
1.006863649128853
/usr/local/lib/python3.6/dist-packages/scipy/stats/_continuous_distns.py:5414: RuntimeWarning: overflow encountered in power
return pow(1-q, -1.0/b)
This is not what I expected. I thought a pareto distribution gives me values near a given value with a 80/20 chance of lying below/over the given value. I also experimented a bit with the variables loc and scale but did not come to a good result.
Can you please point out where I am wrong in my assumptions or the way I use the scipy pareto module?
I ended up generating a pareto value for 1 and multiplying this with the given value. Parameter scale I set to 0.2.