How to find the mean of a two-part Poisson-based hurdle model 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


$$
\Pr[y_i=j] = \begin{cases} e^{-\theta} &amp; j=0, \\[10pt]
\dfrac{(1-e^{-\theta})e^{-\lambda} \lambda^j}{j!(1-e^{-\lambda})} &amp; j = 1,2,\ldots \end{cases}
$$
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 $x&gt;0$. 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) &amp; = \sum_{x=0}^\infty x\Pr(X=x) \\
&amp; = \sum_{x=0}^\infty x\cdot\frac{\mu^x e^{-\mu}}{x!} \\
&amp; = \sum_{x=1}^\infty x\cdot\frac{\mu^x e^{-\mu}}{x!} &amp; &amp; \text{(starting with $1$ rather than $0$} \\
&amp; &amp; &amp; \phantom{(}\text{since the $0$ term vanishes because} \\
&amp; &amp; &amp; \phantom{(}\text{of the multiplication by $x$)} \\
&amp; =\sum_{x=1}^\infty \frac{\mu^x e^{-\mu}}{(x-1)!} &amp; &amp; \text{(since $x/(x!) = 1/(x-1)!$ if $x\ge1$)} \\[8pt]
&amp; = \mu \sum_{x=1}^\infty \frac{\mu^{x-1} e^{-\mu}}{(x-1)!} &amp; &amp; \text{(factors not depending on the index $x$)} \\
&amp; &amp; &amp; \phantom{(} \text{can be pulled out)} \\
&amp; = \mu \sum_{y=0}^\infty \frac{\mu^y e^{-\mu}}{y!} &amp; &amp; \text{where $y=x-1$} \\
&amp; &amp; &amp; \text{so $y$ goes from $0$ to $\infty$ as $x$ goes from $1$ to $\infty$} \\
&amp; = \mu \cdot 1 &amp; &amp; \text{by line $(2)$, stated above} \\
&amp; = \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
$$
\Pr(X=x) = \begin{cases} e^{-\theta} &amp; \text{if }x=0, \\
(1-e^{-\theta}) \dfrac{\lambda^x e^{-\lambda}}{x!(1-e^{-\lambda})} &amp; \text{if } x\ge 1. \end{cases}
$$
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.