<p>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 $\dfrac{e^{-\mu} \mu^0}{0!} = e^{-\mu}.$</p>

<p><b>Postscript:</b> 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.</p>

<p>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 $\mu$.</p>

<p><b>Postscript:</b></p>

<p>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 <b>weighted average</b> of</p>

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

<p>The first has expected value $1$.</p>

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

<p>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.</p>

CC BY-SA 3.0
2016-11-04T14:23:03.860Z
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?
MHall
CC BY-SA 3.0
2016-11-04T16:21:30.727Z
@MHall : I've added a postscript to the answer explaining how to find the mean. $\qquad$
Michael Hardy
CC BY-SA 3.0
2016-11-04T16:26:04.860Z
Now I see you've edited the question since I wrote all this.  Maybe I'll be back$\,\ldots\ldots\qquad$
Michael Hardy
CC BY-SA 3.0
2016-11-04T01:28:09.333Z
2016-11-04T18:19:44.770Z
2016-11-04T18:19:44.770Z
Michael Hardy
<p>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?</p>

<p>EDIT: this is the distribution I am given


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



<econometrics>
How to find the mean of a two-part Poisson-based hurdle model
MHall
# How to find the mean of a two-part Poisson-based hurdle model

(Asked by: username_0 on 2016-11-03)

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

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

**Tags:** econometrics


---

## 1 answers

### Answer by username_1 on 2016-11-04. Score: 2 | Accepted answer

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.

> **Comments:**
> **username_0**: 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?  
> **username_1**: @username_0 : I've added a postscript to the answer explaining how to find the mean. $\qquad$  
> **username_1**: Now I see you've edited the question since I wrote all this.  Maybe I'll be back$\,\ldots\ldots\qquad$  

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