# How would you find the mean of the zero truncated Poisson distribution? [closed]

Given the probability mass function is,

$$f_T(y)=P(Y=y|Y>0)= \frac{1}{e^\lambda -1} \cdot \frac{\lambda^y}{y!}, y=1,2,3,\dots$$

Where, $$f(y)=\frac{e^{-\lambda}\lambda^y}{y!},y=0,1,..$$

How would you show that mean of this function is, $$\frac{\lambda}{1-e^{-\lambda}}$$

• This is a mixture of a Poisson distribution and an atom at zero with weights $\pi=1-1/(1-e^{-\lambda})$ and $1-\pi,$ respectively . Although one of the mixture weights is negative, that doesn't invalidate the calculations of moments of mixtures as described at stats.stackexchange.com/questions/16608, for instance. Thus, one solution is just to plug in the formula for the expectation of a mixture, which is $\pi(0)+(1-\pi)\lambda,$ giving the desired result. – whuber May 30 at 12:16
• I rolled back your edit. Please do not change the questions after they got already answered, especially when it changes their meaning. – Tim Jun 1 at 11:45

Easy way is directly decomposing the expected value of the original RV:

$$E[Y]=E[Y|Y>0]P(Y>0)+E[Y|Y=0]P(Y=0)=E[Y|Y>0]P(Y>0)$$

$$\lambda=E[Y|Y>0](1-e^{-\lambda})\rightarrow E[Y|Y>0]=\frac{\lambda}{1-e^{-\lambda}}$$

A facile path is noting that y * probability mass function is,

$$y * f_T(y)= y * P(Y=y|Y>0)= y * \frac{1}{e^\lambda -1} \cdot \frac{\lambda^y}{y!}, y=1,2,3,\dots$$

Or:

$$y * f_T(y) = \lambda * \frac{e^{\lambda}}{e^\lambda -1} \cdot e^{-\lambda}\frac{\lambda^{y-1}}{(y-1)!}, y=1,2,3,\dots$$

Where for the original, $$f(y)=\frac{e^{-\lambda}\lambda^y}{y!},y=0,1,..$$

Taking the integral of the PDF part goes to one yielding a mean of $$\lambda * \frac{e^{\lambda}}{e^\lambda -1}$$, the required result.

• This distribution doesn't have a PDF, which makes it difficult to understand what integral you refer to. And what do you mean by "goes to one"?? – whuber May 30 at 12:17
• @whuber It does have a counting density, just not a Lebesgue density, so the „integral“ is a series. – AlexR May 30 at 22:16