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