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Negative binomial distribution
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===Poisson distribution=== Consider a sequence of negative binomial random variables where the stopping parameter {{mvar|r}} goes to infinity, while the probability {{mvar|p}} of success in each trial goes to one, in such a way as to keep the mean of the distribution (i.e. the expected number of failures) constant. Denoting this mean as {{mvar|Ξ»}}, the parameter {{mvar|p}} will be {{math|1=''p'' = ''r''/(''r'' + ''Ξ»'')}} : <math> \begin{align} \text{Mean:} \quad & \lambda = \frac{(1-p)r}{p} \quad \Rightarrow \quad p = \frac{r}{r+\lambda}, \\ \text{Variance:} \quad & \lambda \left( 1 + \frac{\lambda}{r} \right) > \lambda, \quad \text{thus always overdispersed}. \end{align} </math> Under this parametrization the probability mass function will be : <math> f(k; r, p) = \frac{\Gamma(k+r)}{k!\cdot\Gamma(r)}(1-p)^k p^r = \frac{\lambda^k}{k!} \cdot \frac{\Gamma(r+k)}{\Gamma(r)\;(r+\lambda)^k} \cdot \frac{1}{\left(1+\frac{\lambda}{r}\right)^r} </math> Now if we consider the limit as {{math|''r'' β β}}, the second factor will converge to one, and the third to the exponent function: : <math> \lim_{r\to\infty} f(k; r, p) = \frac{\lambda^k}{k!} \cdot 1 \cdot \frac{1}{e^\lambda}, </math> which is the mass function of a [[Poisson distribution|Poisson-distributed]] random variable with expected value {{mvar|Ξ»}}. In other words, the alternatively parameterized negative binomial distribution [[convergence in distribution|converges]] to the Poisson distribution and {{mvar|r}} controls the deviation from the Poisson. This makes the negative binomial distribution suitable as a robust alternative to the Poisson, which approaches the Poisson for large {{mvar|r}}, but which has larger variance than the Poisson for small {{mvar|r}}. : <math> \operatorname{Poisson}(\lambda) = \lim_{r \to \infty} \operatorname{NB} \left(r, \frac{r}{r + \lambda}\right). </math>
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