Generating a binomial distribution around zero

I'm looking to generate a binomial-esque distribution. I want a binomial distribution but I want it centred around zero (I know this doesn't make much sense with respect to the definition of binomial distributions but still, this is my goal.)

The only way I have found of doing this in python is:

def zeroed_binomial(n,p,size=None):
    return numpy.random.binomial(n,p,size) - n*p

Is there a real name for this distribution? Does this code actually give me what I want (and how can I tell)? Is there a cleaner / nicer / canonical / already implemented way of doing this?

Asked By: jhoyla
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Answer #1:

What you're doing is fine if you want a "discretized" normal distribution centered around 0. If you want integer values, you should round n*p before subtracting.

But the limit of the binomial distribution is just the normal distribution when n becomes large and with p bounded away from 0 or 1. since n*p is not going to be an integer except for certain values, why not just use the normal distribution?

Answered By: Andrew Mao

Answer #2:

The probability distributions implemented in the scipy.stats module allow you to shift distributions arbitrarily by specifying the loc keyword in the constructor. To get a binomial distribution with mean shifted close to 0, you can call

p = stats.binom(N, p, loc=-round(N*p))

(Be sure to use an integer value for loc with a discrete distribution.)

Here's an example:

p = stats.binom(20, 0.1, loc=-2)
x = numpy.arange(-3,5)
bar(x, p.pmf(x))

Edit:

To generate the actual random numbers, use the rvs() method which comes with every random distribution in the scipy.stats module. For example:

>>> stats.binom(20,0.1,loc=-2).rvs(10)
array([-2,  0,  0,  1,  1,  1, -1,  1,  2,  0]) 
Answered By: silvado
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