Editing Negative binomial distribution (section)

===Relation to the binomial theorem===

Suppose {{mvar|Y}} is a random variable with a [[binomial distribution]] with parameters {{mvar|n}} and {{mvar|p}}.  Assume {{math|1=''p'' + ''q'' = 1}}, with {{math|''p'', ''q'' ≥ 0}}, then

:<math>1=1^n=(p+q)^n.</math>

Using [[Newton's binomial theorem]], this can equally be written as:

:<math>(p+q)^n=\sum_{k=0}^\infty \binom{n}{k} p^k q^{n-k},</math>

in which the upper bound of summation is infinite.  In this case, the [[binomial coefficient]]

: <math>\binom{n}{k} = {n(n-1)(n-2)\cdots(n-k+1) \over k! }.</math>

is defined when {{mvar|n}} is a real number, instead of just a positive integer.  But in our case of the binomial distribution it is zero when {{math|''k'' > ''n''}}.  We can then say, for example

: <math>(p+q)^{8.3}=\sum_{k=0}^\infty \binom{8.3}{k} p^k q^{8.3 - k}.</math>

Now suppose {{math|''r'' > 0}} and we use a negative exponent:

:<math>1=p^r\cdot p^{-r}=p^r (1-q)^{-r}=p^r \sum_{k=0}^\infty \binom{-r}{k} (-q)^k.</math>

Then all of the terms are positive, and the term

:<math>p^r \binom{-r}{k} (-q)^k = \binom{k + r - 1}{k} p^rq^k</math>

is just the probability that the number of failures before the {{mvar|r}}-th success is equal to {{mvar|k}}, provided {{mvar|r}} is an integer.  (If {{mvar|r}} is a negative non-integer, so that the exponent is a positive non-integer, then some of the terms in the sum above are negative, so we do not have a probability distribution on the set of all nonnegative integers.)

Now we also allow non-integer values of {{mvar|r}}.

Recall from above that

:The sum of independent negative-binomially distributed random variables {{math|''r''{{sub|1}}}} and {{math|''r''{{sub|2}}}} with the same value for parameter {{mvar|p}} is negative-binomially distributed with the same {{mvar|p}} but with {{mvar|r}}-value&nbsp;{{math|''r''{{sub|1}} + ''r''{{sub|2}}}}.

This property persists when the definition is thus generalized, and affords a quick way to see that the negative binomial distribution is [[Infinite divisibility (probability)|infinitely divisible]].