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If I have a hurdle model in which the distribution is one Poisson distribution for observations $=0$ and another distribution for observations not equal to zero, how do I go about finding the mean of such a distribution?

EDIT: this is the distribution I am given

$$ \Pr[y_i=j] = \begin{cases} e^{-\theta} & j=0, \\[10pt] \dfrac{(1-e^{-\theta})e^{-\lambda} \lambda^j}{j!(1-e^{-\lambda})} & j = 1,2,\ldots \end{cases} $$

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Maybe I'm not understanding your question. Poisson distributions with expected values $\lambda$ and $\mu$ are given by $$ \Pr(X=x) = \frac{\lambda^x e^{-\lambda}}{x!}\quad\text{ and }\quad \Pr(X=x) = \frac{\mu^x e^{-\mu}}{x!} \quad \text{ for } x=0,1,2,3,\ldots. $$ Suppose the second one holds for $x>0$. It follows that $$ \Pr(X=0) = 1 - \underbrace{\sum_{x=1}^\infty \frac{\mu^x e^{-\mu}}{x!}}_\text{starting at 1, not 0} = e^{-\mu}. $$ Since the sum of the probabilities must be $1$, the probability that $X=0$ can only be $1$ minus the sum of the other probabilities, and that is $\dfrac{e^{-\mu} \mu^0}{0!} = e^{-\mu}.$

Postscript: In comments I am asked how to find the mean of this distribution. Remember that $$ \sum_{x=0}^\infty \frac{\mu^x}{x!} = e^\mu, \tag 1 $$ and from that we get the sum of the probabilities, $$ \sum_{x=0}^\infty \frac{\mu^x e^{-\mu}}{x!} = 1 \tag 2 $$ We will use the sum $(2)$ in finding the mean.

The mean is \begin{align} \operatorname{E}(X) & = \sum_{x=0}^\infty x\Pr(X=x) \\ & = \sum_{x=0}^\infty x\cdot\frac{\mu^x e^{-\mu}}{x!} \\ & = \sum_{x=1}^\infty x\cdot\frac{\mu^x e^{-\mu}}{x!} & & \text{(starting with $1$ rather than $0$} \\ & & & \phantom{(}\text{since the $0$ term vanishes because} \\ & & & \phantom{(}\text{of the multiplication by $x$)} \\ & =\sum_{x=1}^\infty \frac{\mu^x e^{-\mu}}{(x-1)!} & & \text{(since $x/(x!) = 1/(x-1)!$ if $x\ge1$)} \\[8pt] & = \mu \sum_{x=1}^\infty \frac{\mu^{x-1} e^{-\mu}}{(x-1)!} & & \text{(factors not depending on the index $x$)} \\ & & & \phantom{(} \text{can be pulled out)} \\ & = \mu \sum_{y=0}^\infty \frac{\mu^y e^{-\mu}}{y!} & & \text{where $y=x-1$} \\ & & & \text{so $y$ goes from $0$ to $\infty$ as $x$ goes from $1$ to $\infty$} \\ & = \mu \cdot 1 & & \text{by line $(2)$, stated above} \\ & = \mu. \end{align} So the mean is $\mu$.

Postscript:

Your new revision of the question isn't quite consistent with your statement that "the distribution is one Poisson distribution for observations $=0$ and another distribution for observations not equal to zero". You have $$ \Pr(X=x) = \begin{cases} e^{-\theta} & \text{if }x=0, \\ (1-e^{-\theta}) \dfrac{\lambda^x e^{-\lambda}}{x!(1-e^{-\lambda})} & \text{if } x\ge 1. \end{cases} $$ Thus you have a weighted average of

  • The distribution of a random variable constantly equal to $1$, and
  • The conditional distribution of a Poisson-distributed random variable given that it is $\ge 1.$

The first has expected value $1$.

The second has expected value $$ \frac 1 {1- e^{-\lambda}} \sum_{x=1}^\infty \cdots = \frac 1 {1-e^{-\lambda}} \left( (-e^{-\lambda}\cdot0) + \sum_{x=0}^\infty \cdots \right) = \frac 1 {1-e^{-\lambda}} \cdot ( \lambda ). $$

After that, use the fact that the expected value of the weighted average of two distributions is the weight average of their two expected values.

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  • $\begingroup$ Thanks! I did not understand that initially. However, once I have the distributions set equal to one, how do I find the mean of this distribution? $\endgroup$ – MHall Nov 4 '16 at 14:23
  • $\begingroup$ @MHall : I've added a postscript to the answer explaining how to find the mean. $\qquad$ $\endgroup$ – Michael Hardy Nov 4 '16 at 16:21
  • $\begingroup$ Now I see you've edited the question since I wrote all this. Maybe I'll be back$\,\ldots\ldots\qquad$ $\endgroup$ – Michael Hardy Nov 4 '16 at 16:26

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