I'm trying to make sense of a question which uses a zero-inflated poisson model given by:

$$ f(x; \lambda,\omega) = \begin{cases} \omega + (1-\omega)e^{-\lambda} &\mbox{if } x = 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1) \\ \frac{(1-\omega)e^{-\lambda}\lambda^x}{x!} & \mbox{if } x = 1,2,3,\dots \ \ \ \ \ \ \ (2)\end{cases} $$

We are given a table of data, x = 0, 1, 2, 3, 4 and the number of occurrences of each.

In lectures we covered a very similar question, however it used a zero-truncated poisson, given by:

$$ g(x; \theta) = \frac{e^{-\theta}\theta^x}{x!(1-e^{-\theta)}} x = 1,2,3,\dots \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3) $$

Obviously if I equate $(2)$ with $(3)$ I get $(1-\omega) = (1-e^{-\theta})^{-1}$.

Now the question asks me to find the log-likelihood functions for $\lambda$ and $\omega$, then show that the MLE for $\hat{\lambda}$ is given by

$$\frac{\hat{\lambda}}{1-e^{-\hat{\lambda}}} = \bar{x}$$

(Here $\bar{x}$ is the mean for $x_i \neq 0$.)

Now, this is the exact answer we got in lectures - and it's the answer I get here if I ignore (1) in my calculations. If I try to include both (1) and (2) in calculating the MLE, things get pretty messy.

My question is, what happens to (1)? Is there actually a difference between these two distributions? Also, once I calculate $\lambda$ and $\omega$, and use (1) to get a value for the frequency of $x=0$, I get a value very close to zero, whereas the frequency in the original table is 97 (all the other frequencies calculated by the model tally closely, as expected).

One other thing, if you are being kind enough to consider answering my question, is that I have no idea what a link-function is, and presume we are not expected to use it in answering this question.


1 Answer 1


My question is, what happens to (1)? Is there actually a difference between these two distributions?

The Poisson model is helpful when we work with counts, but it is very restrictive: the model says that the mean is equal to the variance. The zero-inflated Poisson is useful when the variance is inflated by a great numbers of zeros. In other hand, the zero-truncated Poisson is applied when we don't have zeros in the data set.

  • 1
    $\begingroup$ Thank you for your answer. However in the example I have from lectures, the zero-truncated Poisson is used for a dataset (the fairly widely used 1992 Leroux and Puterman foetal lamb movements study) which does include lots of zeros. It gives a much better fit than the standard Poisson in this case, for all the non-zero counts. $\endgroup$
    – Lewy
    Apr 10, 2013 at 0:56

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