I'm making some modifications to the load testing framework that we're using throughout the company, and this is a question for which I would love to have an answer.
I was under the impression that the following 2 approaches to generating a Poisson distribution would be equivalent, but I'm clearly wrong:
#!/usr/bin/env python
from numpy import average, random, std
from random import expovariate
def main():
for count in 5.0, 50.0:
data = [random.poisson(count) for i in range(10000)]
print 'npy_poisson average with count=%d: ' % count, average(data)
print 'npy_poisson std_dev with count=%d: ' % count, std(data)
rate = 1 / count
data = [expovariate(rate) for i in range(10000)]
print 'expovariate average with count=%d: ' % count, average(data)
print 'expovariate std_dev with count=%d: ' % count, std(data)
if __name__ == '__main__':
main()
This results in output that looks like:
npy_poisson average with count=5: 5.0168
npy_poisson std_dev with count=5: 2.23685443424
expovariate average with count=5: 4.94383067075
expovariate std_dev with count=5: 4.95058985422
npy_poisson average with count=50: 49.9584
npy_poisson std_dev with count=50: 7.07829565927
expovariate average with count=50: 50.9617389096
expovariate std_dev with count=50: 51.6823970228
Why does the standard deviation when I use the built in random.expovariate scale proportionately with number of events in a given interval, while the expovariate std_deviation scales at a rate of log base 10 (count)??
Follow up question: Which one is more appropriate if you're simulating the frequency with which users interact with your service?
Because your assumptions are wrong. The mean / variance of a Poisson distribution are both lambda, hence the stdev is sqrt(lambda). The mean / variance of an exponential distribution are 1/lambda and 1/lambda^2 respectively. So std = sqrt(1/(1/rate)^2) = sqrt(rate^2) = rate which is exactly what you are seeing here.
I'd suggest reading the Wikipedia article on queuing theory for your follow up question.