Relationship Between Gamma Distribution and Poisson Distribution
Theorem
For every natural number $k$, 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.
Explanation
- For $k, \theta > 0$, the continuous probability distribution $\Gamma ( k , \theta )$ with the following probability density function is called the gamma distribution. $$ f(x) = {{ 1 } \over { \Gamma ( k ) \theta^{k} }} x^{k - 1} e^{ - x / \theta} \qquad , x > 0 $$
- For $\lambda > 0$, the discrete probability distribution $\text{Poi} ( \lambda )$ with the following probability mass function is called the Poisson distribution. $$ p(x) = {{ e^{-\lambda} \lambda^{x} } \over { x! }} \qquad , x = 0 , 1 , 2, \cdots $$
This equality shows that the cumulative distribution functions of the gamma distribution and the Poisson distribution are related to each other. Considering that the gamma distribution has a relationship with the exponential distribution, one can guess that this is quite plausible.
Proof
We 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 that $\displaystyle \int_{\mu}^{\infty} { { z^{N-1} e^{-z} } \over { \Gamma (N) } } dz = \sum_{x=0}^{N-1} { { {\mu}^{x} e^{-\mu} } \over {x!} }$ holds when $k=N$, by integration by parts, $$ \begin{align*} \int_{\mu}^{\infty} { { z^{N-1} e^{-z} } \over \Gamma (N) } dz =& \int_{\mu}^{\infty} { { z^{N-1} e^{-z} } \over { (N-1)! } } dz \\ =& \left[ { { z^{N} e^{-z} } \over { N! } } \right] _{\mu} ^{\infty} - \int_{\mu}^{\infty} - { { z^{N} e^{-z} } \over { N! } } dz \\ =& - { { {\mu}^{N} e^{-\mu} } \over { N! } } + \int_{\mu}^{\infty} { { z^{N} e^{-z} } \over { \Gamma (N+1) } } dz \\ =& \sum_{x=0}^{N-1} { { {\mu}^{x} e^{-\mu} } \over {x!} } \end{align*} $$ Rearranging 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, the following holds for every natural number $k$. $$ \int_{\mu}^{\infty} { { z^{k-1} e^{-z} } \over { \Gamma (k) } } dz = \sum_{x=0}^{k-1} { { {\mu}^{x} e^{-\mu} } \over {x!} } $$
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