# how to draw 100 numbers following exponential distribution (erlang) in python by having k and CV and mean exponential distribution?

• how to draw erlang distribution (two parameters k and cv? (k=1,5,10, cv=100%-45%-32%) ) in python by having k and CV and mean exponential distribution=1?

I know numpy.random.exponential(scale=1.0, size=None)¶ by this only has scale but I need two parameters k and cv? (k=1,5,10, cv=100%-45%-32%, )

Guessing, but: the Erlang distribution is a special case of the Gamma (with integer shape parameter).

numpy.random.gamma takes a shape ($$k$$) and a scale ($$\theta$$) parameter. Per Wikipedia, for the Gamma distribution mean=$$k \theta$$, var=$$k \theta^2$$, so CV=$$\sqrt{\textrm{Var}}/m = 1/\sqrt{k}$$. So the bad news is that you can't specify $$k$$ and CV as independent parameters: once you know $$k=1,5,10$$, you already know that the CV is (as you specify) 1, $$1/\sqrt{5}=0.45$$, $$1/\sqrt{10}=0.32$$. The good news is that you can pick the scale ($$\theta$$) parameter any way you want: e.g. you could set $$\theta=1$$ throughout (in which case the mean will equal $$1/\textrm{CV}^2$$), or you can specify the mean $$m$$ and set $$\theta=m/k=m \cdot \textrm{CV}^2$$.

So e.g. numpy.random.gamma(shape=5,size=100) would work (this uses the default scale=1).

If you really want to use numpy.random.exponential to generate 100 k-Erlang deviates, in a marginally less efficient and transparent way, you can generate a 100*k array and sum up the rows, e.g. for $$k=5$$/CV=0.45:

import numpy.random as npr
k = 5
npr.exponential(scale=1,size=(100,k)).sum(axis=1)

• Very similar to @Glen_b's but leaving it anyway since it uses different words so might not be completely redundant. – Ben Bolker Feb 20 '19 at 2:39
• Actually, Ben, I think yours is better. Feel free to edit yours to incorporate anything that was in mine if you think there was anything worth adding (though I don't think there's a need, it looks fine to me). – Glen_b -Reinstate Monica Feb 20 '19 at 2:40
• @BenBolker thanks the only data I saw in the paper is that mean exponential distribution=1? How to add this? Would you please check this link to see what is the input? pdfs.semanticscholar.org/1051/… – user10296606 Feb 21 '19 at 7:35
• don't have time to read in detail. Best guess: they are looking at the flow through k serially linked compartments, each of which have an exponential residence time. The total residence time from start to finish will then be Erlang(k)-distributed ... – Ben Bolker Feb 21 '19 at 14:28
• @BenBolker How to add exponential mean to it? For instance made with exponential mean=1, k=1, and 100 points – user10296606 Feb 23 '19 at 3:11