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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?

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    i would only see sense in such a distribution if p = 0.5 (otherwise, the notion of centered seems far-fetched) Commented Mar 22, 2013 at 14:28
  • Perhaps centred is the wrong word, I mean that the most frequent element should be zero. Commented Mar 22, 2013 at 14:29
  • so basically a binomial such as f(0) = max(f(x)). What are you doing with that ? Commented Mar 22, 2013 at 14:32
  • Testing out a theory I have for my steganography paper :P. Commented Mar 22, 2013 at 14:36
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    n*p is the mean of the binomial distribution. The most frequent element is the mode, and according to wikipedia it is either floor((n + 1)p) or floor((n + 1)p) − 1 Commented Mar 22, 2013 at 15:52

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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?

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This the all true for large N, but the op might want a binomial with a small N, in which case the normal distribution would not be a good approximation.
Good call on the rounding of n*p.
@AndrewMao Could you elaborate a bit on what you mean by rounding $n*p$ before subtracting in order to get "discretized" normal distribution?
@Alex if you want an integer-valued support ..., -2, -1, 0, 1, 2, ... instead of real-valued ones that are oddly shifted like ..., -1.714, -0.714, 0.286, 1.286, .... Of course, the mean of this adjusted distribution will not be 0.
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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))

bar plot

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])
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I'm actually looking for a list of random numbers that conform to a distribution, not just what the distribution looks like. I suppose I could generate the numbers from the distribution but that doesn't feel clean.
@jhoyla Every distribution in scipy.stats of course comes with a method to generate the random numbers. See my edit. Why do you think it doesn't feel clean?

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