= Solution
Let $I_0=\{i:y_i=0\}$ and $I_+=\{i:y_i>0\}$. Under the proposed <zero-inflated negative binomial model>, <independence> gives the <likelihood function>
$$
\boxed{L(\pi,\beta;r)=\prod_{i\in I_0}\left[\pi+(1-\pi)\left(\frac{r}{r+e^{x_i^T\beta}}\right)^r\right]\prod_{i\in I_+}(1-\pi)f_{\rm NB}(y_i;r,e^{x_i^T\beta}).}
$$
The positive-count factor uses the ordinary, untruncated <negative binomial distribution>. With $f_i=f_{\rm NB}(y_i;r,\lambda_i)$, the <log-likelihood> is
$$
\ell=\sum_{i\in I_0}\log[\pi+(1-\pi)f_i]+\sum_{i\in I_+}[\log(1-\pi)+\log f_i].
$$
For latent indicators $a_i$, the complete-data <log-likelihood> is
$$
\ell_c=\sum_i\{a_i\log\pi+(1-a_i)[\log(1-\pi)+\log f_i]\},
$$
where $a_i=1$ is allowed only when $y_i=0$. This separation is what makes the <expectation-maximization algorithm> convenient.
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