Truncated Poisson Distribution at X=0

Truncated Poisson Distribution (at $X=0$)

A discrete random variable $X$ is said to have truncated Poisson distribution (at $X=0$) if its probability mass function is given by

$$ \begin{equation*} P(X=x)= \left\{ \begin{array}{ll} \frac{e^{-\lambda}\lambda^x}{(1-e^{-\lambda})x!}, & \hbox{$x=1,2, \ldots$;} \\ & \hbox{$\lambda>0$}\\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$

Proof

The probability mass function of Poisson distribution $P(\lambda)$ is

$$ \begin{equation*} P(X=x)= \left\{ \begin{array}{ll} \frac{e^{-\lambda}\lambda^x}{x!}, & \hbox{$x=0,1,\ldots $;} \\ & \hbox{$\lambda>0$}\\ 0, & \hbox{Otherwise.} \end{array} \right. \end{equation*} $$

As $P(X=x)$ is a probability mass function, $\sum_{x=0}^n P(X=x) = 1$.

$$ \begin{eqnarray*} & & \sum_{x=0}^\infty P(X=x) = 1 \\ &\Rightarrow & \sum_{x=0}^\infty \frac{e^{-\lambda}\lambda^x}{x!}= 1\\ &\Rightarrow & \frac{e^{-\lambda}\lambda^0}{0!} +\sum_{x=1}^\infty \frac{e^{-\lambda}\lambda^x}{x!} = 1\\ &\Rightarrow & e^{-\lambda} +\sum_{x=1}^\infty \frac{e^{-\lambda}\lambda^x}{x!} = 1\\ &\Rightarrow & \sum_{x=1}^\infty \frac{e^{-\lambda}\lambda^x}{x!} = 1-e^{-\lambda}\\ &\Rightarrow & \frac{1}{1-e^{-\lambda}} \sum_{x=1}^\infty \frac{e^{-\lambda}\lambda^x}{x!} = 1\\ &\Rightarrow & \sum_{x=1}^\infty \frac{e^{-\lambda}\lambda^x}{(1-e^{-\lambda})x!} = 1\\ &\Rightarrow & \sum_{x=1}^\infty P^\prime(X=x) = 1 \end{eqnarray*} $$

where $P^\prime(X=x) =\frac{e^{-\lambda}\lambda^x}{(1-e^{-\lambda})x!}$, $x=1,2, \ldots $, $\lambda>0$. The probability function $P^\prime (X=x)$ is the probability mass function (pmf) of truncated Poisson (at $X=0$) distribution.

Mean of Truncated Poisson Distribution

The mean of truncated Poisson distribution at $X=0$ is

$$ \begin{aligned} E(X) = \frac{\lambda }{(1-e^{-\lambda})}. \end{aligned} $$

proof

The mean of truncated Poisson distribution at $X=0$ is

$$ \begin{eqnarray*} \mu_1^\prime = E(X) &=& \sum_{x=1}^\infty x\cdot P(X=x)\\ &=& \sum_{x=1}^\infty x\cdot\frac{e^{-\lambda}\lambda^x}{(1-e^{-\lambda})x!} \\ &=& \frac{\lambda e^{-\lambda}}{(1-e^{-\lambda})}\sum_{x=1}^\infty \frac{\lambda^{(x-1)}}{(x-1)!} \\ &=& \frac{\lambda e^{-\lambda}}{(1-e^{-\lambda})}e^{\lambda}\\ &=& \frac{\lambda }{(1-e^{-\lambda})}. \end{eqnarray*} $$

Variance of Truncated Poisson Distribution

The variance of truncated Poisson distribution at $X=0$ is

$$ \begin{aligned} V(X) &=\frac{\lambda^2+\lambda}{(1-e^{-\lambda})} - \frac{\lambda^2 }{(1-e^{-\lambda})^2} \end{aligned} $$

Proof

To calculate variance, let us calculate $E[X(X-1)]$.

$$ \begin{eqnarray*} E[X(X-1)]&=& \sum_{x=1}^\infty x(x-1)P(X=x)\\ &=& \sum_{x=1}^\infty x(x-1)\cdot\frac{e^{-\lambda}\lambda^x}{(1-e^{-\lambda})x!} \\ &=&\frac{e^{-\lambda}\lambda^2}{(1-e^{-\lambda})}\sum_{x=2}^\infty \frac{\lambda^{x-2}}{(x-2)!} \\ &=&\frac{e^{-\lambda}\lambda^2}{(1-e^{-\lambda})}e^\lambda\\ &=&\frac{\lambda^2}{(1-e^{-\lambda})}. \end{eqnarray*} $$

Therefore

$$ \begin{aligned} \mu_2^\prime &= E[X(X-1)]+E(X)\\ &=\frac{\lambda^2}{(1-e^{-\lambda})} +\frac{\lambda }{(1-e^{-\lambda})}\\ &=\frac{\lambda^2+\lambda}{(1-e^{-\lambda})} \end{aligned} $$

The variance of truncated Poisson distribution at $X=0$ is

$$ \begin{eqnarray*} \mu_2 &=& \mu_2^\prime - {\mu_1^\prime}^2\\ & = &\frac{\lambda^2+\lambda}{(1-e^{-\lambda})} - \frac{\lambda^2 }{(1-e^{-\lambda})^2}. \end{eqnarray*} $$

VRCBuzz co-founder and passionate about making every day the greatest day of life. Raju is nerd at heart with a background in Statistics. Raju looks after overseeing day to day operations as well as focusing on strategic planning and growth of VRCBuzz products and services. Raju has more than 25 years of experience in Teaching fields. He gain energy by helping people to reach their goal and motivate to align to their passion. Raju holds a Ph.D. degree in Statistics. Raju loves to spend his leisure time on reading and implementing AI and machine learning concepts using statistical models.

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