# The special case of the negative binomial, the geometric and calculation with scipy

### A particular case?

One might consider the geometric distribution is simply a special case of the negative binomial with $N=1$.

We know the geometric provides: "The probability the first occurrence of a success will occur on the $x^{th}$ year". Say for $p=0.1$, $q=1-p$, and $x=8$ this is given by:

$$f_x(8; 0.1) = pq^{x-1} = 0.1 \cdot 0.9^{7}$$

Now, if we want the particular case of the negative binomial, for the first year:

$$f_x(8; 1, 0.1) = \binom{x-1}{k-1}p^kq^{x-k} = \binom{7}{0} 0.1^1 0.9^{7}$$

But when I use scipy I have a slightly different result, and need to reduce my $x$ value by one for the negative binomial:

from scipy.stats import nbinom, geom
N = 1 # first year
p = 0.1 #ten year flood

f_geom = geom(p).pmf
print("""The probability from the geometric distribution
that the first occurrence of a 10-year flood will occur on the
8th year is:
{0:8.5f}""".format(f_geom(8)))

f_nbinom = nbinom(N, p).pmf
print("""The probability from the negative binomial that
a 10-year flood will occur the first time on the
8th year is:
{0:8.5f}""".format(f_nbinom(7)))

Yields:

The probability from the geometric distribution
that the first occurrence of a 10-year flood will occur on the
8th year is:
0.04783

The probability from the negative binomial that
a 10-year flood will occur the first time on the
8th year is:
0.04783

Could someone provide an explanation for this?

• Its simply the number of trials vs number of failures versions of the geometric/ negative binomial. This is explicit at the Wikipedia geometric distribution article Sep 6, 2017 at 0:08