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The Relationship between Gamma Distribution and Poisson Distribution 📂Probability Distribution

The Relationship between Gamma Distribution and Poisson Distribution

Theorem

For all natural numbers kk, the following holds. ∫μ∞zk−1e−zΓ(k)dz=∑x=0k−1μxe−μx! \int_{\mu}^{\infty} { { z^{k-1} e^{-z} } \over { \Gamma (k) } } dz = \sum_{x=0}^{k-1} { { {\mu}^{x} e^{-\mu} } \over {x!} }

Description

  • k,θ>0k, \theta > 0 For the following probability density function for continuous probability distribution Γ(k,θ)\Gamma ( k , \theta ) is called Gamma Distribution. f(x)=1Γ(k)θkxk−1e−x/θ,x>0 f(x) = {{ 1 } \over { \Gamma ( k ) \theta^{k} }} x^{k - 1} e^{ - x / \theta} \qquad , x > 0
  • λ>0\lambda > 0 For the following probability mass function for discrete probability distribution Poi(λ)\text{Poi} ( \lambda ) is called Poisson Distribution. p(x)=e−λλxx!,x=0,1,2,⋯ 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=1k=1 ∫μ∞z0e−zΓ(0)dz=e−μ=∑x=00μxe−μx! \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=Nk=N when ∫μ∞zN−1e−zΓ(N)dz=∑x=0N−1μxe−μx!\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 ∫μ∞zN−1e−zΓ(N)dz=∫μ∞zN−1e−z(N−1)!dz=[zNe−zN!]μ∞−∫μ∞−zNe−zN!dz=−μNe−μN!+∫μ∞zNe−zΓ(N+1)dz=∑x=0N−1μxe−μx! \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 ∫μ∞zNe−zΓ(N+1)dz=μNe−μN!+∑x=0N−1μxe−μx!=∑x=0Nμxe−μx! \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 kk, the following holds. ∫μ∞zk−1e−zΓ(k)dz=∑x=0k−1μxe−μx! \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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