Editing Negative binomial distribution (section)

==Properties==

===Expectation===

The expected total number of trials needed to see {{mvar|r}} successes is <math>\frac{r}{p}</math>. Thus, the expected number of ''failures'' would be this value, minus the successes:
:<math>
E[\operatorname{NB}(r, p)] = \frac{r}{p} - r = \frac{r(1-p)}{p}
</math>

===Expectation of successes===

The expected total number of failures in a negative binomial distribution with parameters {{math|(''r'', ''p'')}} is {{math|''r''(1 − ''p'')/''p''}}. To see this, imagine an experiment simulating the negative binomial is performed many times. That is, a set of trials is performed until {{mvar|r}} successes are obtained, then another set of trials, and then another etc. Write down the number of trials performed in each experiment: {{math|''a'', ''b'', ''c'', ...}} and set {{math|''a'' + ''b'' + ''c'' + ... {{=}} ''N''}}. Now we would expect about {{math|''Np''}} successes in total. Say the experiment was performed {{mvar|n}} times. Then there are {{math|''nr''}} successes in total. So we would expect {{math|''nr'' {{=}} ''Np''}}, so {{math|''N''/''n'' {{=}} ''r''/''p''}}. See that {{math|''N''/''n''}} is just the average number of trials per experiment. That is what we mean by "expectation". The average number of failures per experiment is {{math|1=''N''/''n'' − ''r'' = ''r''/''p'' − ''r'' = ''r''(1 − ''p'')/''p''}}. This agrees with the mean given in the box on the right-hand side of this page.

A rigorous derivation can be done by representing the negative binomial distribution as the sum of waiting times. Let <math>X_r \sim\operatorname{NB}(r, p)</math> with the convention <math>X</math> represents the number of failures observed before <math>r</math> successes with the probability of success being <math>p</math>. And let <math>Y_i \sim Geom(p)</math> where <math>Y_i</math> represents the number of failures before seeing a success. We can think of <math>Y_i</math> as the waiting time (number of failures) between the <math>i</math>th and <math>(i-1)</math>th success. Thus
:<math>
X_r = Y_1 + Y_2 + \cdots + Y_r.
</math>
The mean is 
:<math>
E[X_r] = E[Y_1] + E[Y_2] + \cdots + E[Y_r] = \frac{r(1-p)}{p},
</math>
which follows from the fact <math>E[Y_i] = (1-p)/p</math>.

=== Variance ===
When counting the number of failures before the {{mvar|r}}-th success, the variance is&nbsp;{{math|''r''(1 − ''p'')/''p''{{sup|2}}}}. 
When counting the number of successes before the {{mvar|r}}-th failure, as in alternative formulation (3) above, the variance is&nbsp;{{math|''rp''/(1 − ''p''){{sup|2}}}}.

===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]].

=== Recurrence relations ===

The following [[recurrence relations]] hold:

For the probability mass function
: <math> \begin{cases}
(k+1) \Pr (X=k+1)-p \Pr (X=k) (k+r)=0, \\[5pt]
\Pr (X=0)=(1-p)^r.
\end{cases}
</math>

For the moments <math>m_k = \mathbb E(X^k),</math>
: <math> m_{k+1} = r P m_k + (P^2 + P) {d m_k \over dP}, \quad P:=(1-p)/p, \quad m_0=1.
</math>

For the cumulants
: <math> \kappa_{k+1} = (Q-1)Q {d \kappa_k \over dQ}, \quad Q:=1/p, \quad \kappa_1=r(Q-1).
</math>