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pythonalgorithmnumpyprobabilitybinomial-coefficients
bigTree
bigTree

Reputation: 2173

Computing a binomial probability for huge numbers

I want to compute binomial probabilities on python. I tried to apply the formula:

probability = scipy.misc.comb(n,k)*(p**k)*((1-p)**(n-k))

Some of the probabilities I get are infinite. I checked some values for which p=inf. For one of them, n=450,000 and k=17. This value must be greater than 1e302 which is the maximum value handled by floats.

I then tried to use sum(np.random.binomial(n,p,numberOfTrials)==valueOfInterest)/numberOfTrials

This draws numberOfTrials samples and computes the average number of times the value valueOfInterest is drawn.

This doesn't raise any infinite value. However, is this a valid way to proceed? And why this way wouldn't raise any infinite value whereas computing the probabilities does?

Upvotes: 8

Views: 5451

Answers (4)

Алексей Патрашов
Алексей Патрашов

Reputation: 11

To avoid multiplicity like zero by like infinity use step by step multiplication as this.

def Pbinom(N,p,k):
    q=1-p
    lt1=[q]*(N-k)
    gt1=list(map(lambda x: p*(N-k+x)/x, range(1,k+1)))
    Pb=1.0
    while (len(lt1) + len(gt1)) > 0:
        if Pb>1:
            if len(lt1)>0:
                Pb*=lt1.pop()
            else:
                if len(gt1)>0:
                    Pb*=gt1.pop()
        else:
            if len(gt1)>0:
                Pb*=gt1.pop()
            else:
                if len(lt1)>0:
                    Pb*=lt1.pop()
    return Pb

Upvotes: 1

Bi Rico
Bi Rico

Reputation: 25823

Because you're using scipy I thought I would mention that scipy already has statistical distributions implemented. Also note that when n is this large the binomial distribution is well approximated by the normal distribution (or Poisson if p is very small).

n = 450000
p = .5
k = np.array([17., 225000, 226000])

b = scipy.stats.binom(n, p)
print b.pmf(k)
# array([  0.00000000e+00,   1.18941527e-03,   1.39679862e-05])
n = scipy.stats.norm(n*p, np.sqrt(n*p*(1-p)))
print n.pdf(k)
# array([  0.00000000e+00,   1.18941608e-03,   1.39680605e-05])

print b.pmf(k) - n.pdf(k)
# array([  0.00000000e+00,  -8.10313274e-10,  -7.43085142e-11])

Upvotes: 9

Sudeep Juvekar
Sudeep Juvekar

Reputation: 5068

Work in the log domain to compute combination and exponentiation functions and then raise them to exponent.

Something like this:

combination_num = range(k+1, n+1)
combination_den = range(1, n-k+1)
combination_log = np.log(combination_num).sum() - np.log(combination_den).sum()
p_k_log = k * np.log(p)
neg_p_K_log = (n - k) * np.log(1 - p)
p_log = combination_log + p_k_log + neg_p_K_log
probability = np.exp(p_log)

Gets rid of numeric underflow/overflow because of large numbers. On your example with n=450000 and p = 0.5, k = 17, it returns p_log = -311728.4, i. e., the log of final probability is pretty small and hence underflow occurs while taking np.exp. However, you can still work with log probability.

Upvotes: 7

hivert
hivert

Reputation: 10667

I thing you should do all you computation using logarithms:

from scipy import special, exp, log
lgam = special.gammaln

def binomial(n, k, p):
    return exp(lgam(n+1) - lgam(n-k+1) - lgam(k+1) + k*log(p) + (n-k)*log(1.-p))

Upvotes: 7

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