📂Probability Distribution

The Relationship between Gamma Distribution and Poisson Distribution

Theorem

For all natural numbers $k$, the following holds. $$ \int_{\mu}^{\infty} { { z^{k-1} e^{-z} } \over { \Gamma (k) } } dz = \sum_{x=0}^{k-1} { { {\mu}^{x} e^{-\mu} } \over {x!} } $$

---

• $\Gamma$ is the gamma function.

Description

• $k, \theta > 0$ For the following probability density function for continuous probability distribution $\Gamma ( k , \theta )$ is called Gamma Distribution. $$ f(x) = {{ 1 } \over { \Gamma ( k ) \theta^{k} }} x^{k - 1} e^{ - x / \theta} \qquad , x > 0 $$
• $\lambda > 0$ For the following probability mass function for discrete probability distribution $\text{Poi} ( \lambda )$ is called Poisson Distribution. $$ p(x) = {{ e^{-\lambda} \lambda^{x} } \over { x! }} \qquad , x = 0 , 1 , 2, \cdots $$

This equation shows that there is a relation between the cumulative probability distribution functions of gamma distribution and Poisson distribution. This is plausible given that gamma distribution has a relationship with exponential distribution.

Proof

Use mathematical induction.

When $k=1$ $$ \int_{\mu}^{\infty} { { z^{0} e^{-z} } \over { \Gamma (0) } } dz = e^{-\mu} = \sum_{x=0}^{0} { { {\mu}^{x} e^{-\mu} } \over {x!} } $$ Assuming $k=N$ when $\displaystyle \int_{\mu}^{\infty} { { z^{N-1} e^{-z} } \over { \Gamma (N) } } dz = \sum_{x=0}^{N-1} { { {\mu}^{x} e^{-\mu} } \over {x!} }$ holds according to integration by parts $$ \begin{align*} \int_{\mu}^{\infty} { { z^{N-1} e^{-z} } \over \Gamma (N) } dz =& \int_{\mu}^{\infty} { { z^{N-1} e^{-z} } \over { (N-1)! } } dz \\ =& \left[ { { z^{N} e^{-z} } \over { N! } } \right] _{\mu} ^{\infty} - \int_{\mu}^{\infty} - { { z^{N} e^{-z} } \over { N! } } dz \\ =& - { { {\mu}^{N} e^{-\mu} } \over { N! } } + \int_{\mu}^{\infty} { { z^{N} e^{-z} } \over { \Gamma (N+1) } } dz \\ =& \sum_{x=0}^{N-1} { { {\mu}^{x} e^{-\mu} } \over {x!} } \end{align*} $$ Reorganizing the last two lines gives $$ \begin{align*} \int_{\mu}^{\infty} { { z^{N} e^{-z} } \over { \Gamma (N+1) } } dz =& { { {\mu}^{N} e^{-\mu} } \over { N! } } + \sum_{x=0}^{N-1} { { {\mu}^{x} e^{-\mu} } \over {x!} } \\ =& \sum_{x=0}^{N} { { {\mu}^{x} e^{-\mu} } \over {x!} } \end{align*} $$ And, by mathematical induction, for all natural numbers $k$, the following holds. $$ \int_{\mu}^{\infty} { { z^{k-1} e^{-z} } \over { \Gamma (k) } } dz = \sum_{x=0}^{k-1} { { {\mu}^{x} e^{-\mu} } \over {x!} } $$

■