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Can anyone show how the expected value and variance of the zero inflated Poisson, with probability mass function

$$ f(y) = \begin{cases} \pi+(1-\pi)e^{-\lambda}, & \text{if }y=0 \\ (1-\pi)\frac{\lambda^{y}e^{-\lambda}}{y!}, & \text{if }y=1,2.... \end{cases} $$

where $\pi$ is the probability that the observation is zero by a binomial process and $\lambda$ is the mean of the Poisson, is derived?

The result is expected value $\mu =(1-\pi)\lambda$ and the variance is $\mu+ \frac{\pi}{1-\pi}\mu^{2}$.

ADD: I am looking for a process. For example, can you use a moment generating function? Ultimately I'd like to see how to do this to better understand the zero inflated gamma and other, as well.

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    $\begingroup$ It seems you know a model for how such a probability distribution would arise. Can you use that to help you? $\endgroup$ – cardinal Nov 20 '11 at 4:08
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Method 0: The lazy statistician.

Note that for $y \neq 0$ we have $f(y) = (1-\pi) p_y$ where $p_y$ is the probability that a Poisson random variable takes value $y$. Since the term corresponding to $y = 0$ does not affect the expected value, our knowledge of the Poisson and the linearity of expectation immediately tells us that $$ \mu = (1-\pi) \lambda $$ and $$ \mathbb E Y^2 = (1-\pi) (\lambda^2 + \lambda) \> . $$

A little algebra and the identity $\mathrm{Var}(Y) = \mathbb E Y^2 - \mu^2$ yields the result.

Method 1: A probabilistic argument.

It's often helpful to have a simple probabilistic model for how a distribution arises. Let $Z \sim \mathrm{Ber}(1-\pi)$ and $Y \sim \mathrm{Poi}(\lambda)$ be independent random variables. Define $$ X = Z \cdot Y \>. $$ Then, it is easy to see that $X$ has the desired distribution $f$. To check this, note that $\renewcommand{\Pr}{\mathbb P}\Pr(X = 0) = \Pr(Z=0) + \Pr(Z=1, Y=0) = \pi + (1-\pi) e^{-\lambda}$ by independence. Similarly $\Pr(X = k) = \Pr(Z=1, Y=k)$ for $k \neq 0$.

From this, the rest is easy, since by the independence of $Z$ and $Y$, $$ \mu = \mathbb E X = \mathbb E Z Y = (\mathbb E Z) (\mathbb E Y) = (1-\pi)\lambda \>, $$ and, $$ \mathrm{Var}(X) = \mathbb E X^2 - \mu^2 = (\mathbb E Z)(\mathbb E Y^2) - \mu^2 = (1-\pi)(\lambda^2 + \lambda) - \mu^2 = \mu + \frac{\pi}{1-\pi}\mu^2 \> . $$

Method 2: Direct calculation.

The mean is easily obtained by a slight trick of pulling one $\lambda$ out and rewriting the limits of the sum. $$ \mu = \sum_{k=1}^\infty (1-\pi) k e^{-\lambda} \frac{\lambda^k}{k!} = (1-\pi) \lambda e^{-\lambda} \sum_{j=0}^\infty \frac{\lambda^j}{j!} = (1-\pi) \lambda \> . $$

A similar trick works for the second moment: $$ \mathbb E X^2 = (1-\pi) \sum_{k=1}^\infty k^2 e^{-\lambda} \frac{\lambda^k}{k!} = (1-\pi)\lambda e^{-\lambda} \sum_{j=0}^\infty (j+1) \frac{\lambda^j}{j!} = (1-\pi)(\lambda^2 + \lambda) \>, $$ from which point we can proceed with the algebra as in the first method.


Addendum: This details a couple tricks used in the calculations above.

First recall that $\sum_{k=0}^\infty \frac{\lambda^k}{k!} = e^\lambda$.

Second, note that $$ \sum_{k=0}^\infty k \frac{\lambda^k}{k!} = \sum_{k=1}^\infty k \frac{\lambda^k}{k!} = \sum_{k=1}^\infty \frac{\lambda^k}{(k-1)!} = \sum_{k=1}^\infty \frac{\lambda \cdot \lambda^{k-1}}{(k-1)!} = \lambda \sum_{j=0}^\infty \frac{\lambda^j}{j!} = \lambda e^{\lambda} \>, $$ where the substitution $j = k-1$ was made in the second-to-last step.

In general, for the Poisson, it is easy to calculate the factorial moments $\mathbb E X^{(n)} = \mathbb E X(X-1)(X-2)\cdots(X-n+1)$ since $$ e^\lambda \mathbb E X^{(n)} = \sum_{k=n}^\infty k(k-1)\cdots(k-n+1) \frac{\lambda^k}{k!} = \sum_{k=n}^\infty \frac{\lambda^n \lambda^{k-n}}{(k-n)!} = \lambda^n \sum_{j=0}^\infty \frac{\lambda^j}{j!} = \lambda^n e^\lambda \>, $$ so $\mathbb E X^{(n)} = \lambda^n$. We get to "skip" to the $n$th index for the start of the sum in the first equality since for any $0 \leq k < n$, $k(k-1)\cdots(k-n+1) = 0$ since exactly one term in the product is zero.

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  • $\begingroup$ Cardinal, this is fantastic. Would you mind giving a quick detail on pulling out the $\lambda$? My summation is <very> rusty. Thanks! $\endgroup$ – B_Miner Nov 20 '11 at 5:44
  • $\begingroup$ Thanks again for this. This may be an easy question, but what happens to the top part the pdf (when y=0) $\pi+(1-\pi)e^{-\lambda}$ why doesnt' it get included in the calculation for $\mu$? $\endgroup$ – B_Miner Nov 20 '11 at 13:45
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    $\begingroup$ Recall the definition of the expected value for a discrete random variable: $\mu = \mathbb E Y = \sum_{y=0}^\infty y \mathbb P(Y = y)$. So for $y = 0$, the term in the expected value is $0 \cdot (\pi + (1-\pi)e^{-\lambda}) = 0$. $\endgroup$ – cardinal Nov 20 '11 at 14:32

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