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.
