Proof That the Poisson Distribution is a Probablity Distribution

In order for a function to be a probability distribution, it must sum or integrate over the possible values of the random variable(s) to 1.
A random variable  
\[X\]
  is said to be modelled by a Poisson distribution  
\[Po( \lambda)\]
  if  
\[X \sim \frac{e^{- \lambda} \lambda^x} {x!}\]

\[\begin{equation} \begin{aligned} \sum_{x=0}^{\infty} \frac{e^{- \lambda} \lambda^x} {x!} &= e^{- \lambda} \sum_{x=0}^{\infty} \frac{ \lambda^x } {x!} \\ &= e^{- \lambda} e^{ \lambda} =1 \end{aligned} \end{equation}\]
 

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