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Mean and Variance of Poisson Distribution

If the random variable  
\[X\]
  follows a Poisson distribution,  
\[X \sim Po(\lambda ),\]
  then  
\[P(X=x)=\frac{\lambda^x e^{- \lambda}}{x!}\]
.
The mean or expectation value of  
\[X\]
  is
\[E(X)=\sum^{\infty}_{x=0} x \frac{\lambda^x e^{- \lambda}}{x!} = \lambda \sum^{\infty}_{x=1} x \frac{\lambda^x e^{- \lambda}}{x!} = \lambda \sum^{\infty}_{x1} \frac{\lambda^x e^{- \lambda}}{(x-1)!} = \lambda \sum^{\infty}_{x=0} \frac{\lambda^x e^{- \lambda}}{x!} = \lambda\]
,br /> The variance of  
\[X\]
  is
\[\begin{equation} \begin{aligned} V(X)=E(X^2)-(E(X))^2 &= \sum^{\infty}_{x=0} x^2 \frac{\lambda^x e^{- \lambda}}{x!}- \lambda^2 \\ &= \sum^{\infty}_{x=0} (x(x-1)+x) \frac{\lambda^x e^{- \lambda}}{x!}- \lambda^2 \\ &= \sum^{\infty}_{x=0} x(x-1) \frac{\lambda^x e^{- \lambda}}{x!}+ \sum^{\infty}_{x=0} x \frac{\lambda^x e^{- \lambda}}{x!}- \lambda^2 \\ &= \sum^{\infty}_{x=0} x(x-1) \frac{\lambda^x e^{- \lambda}}{x!}+\lambda- \lambda^2 \\ &= \lambda^2 \sum^{\infty}_{x=2} \frac{\lambda^x e^{- \lambda}}{(x-2)!}+\lambda- \lambda^2 \\ &= \lambda^2 \sum^{\infty}_{x=0} \frac{\lambda^x e^{- \lambda}}{x!}+\lambda- \lambda^2 \\ &= \lambda^2+\lambda- \lambda^2 = \lambda \end{aligned} \end{equation}\]

Notice that the mena and variance are both equal to  
\[\lambda\]
. Equality of mean and variance indicates a possible poisson distribution.

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