Derivation of a the log-likelihood for a regression model where the outcome is a mixture between Poisson and a point mass at zero Suppose $ \textbf{Y} = (Y_1, \dots, Y_n)'$ are independent and 
$$\eqalign{
Y_i = 0 & \text{with probability} \ p_i+(1-p_i)e^{-\lambda_i}\\
Y_i = k & \text{with probability} \ (1-p_i)e^{-\lambda_i} \lambda_{i}^{k}/k!
}$$ 
and 
$$\eqalign{
\log(\mathbf{\lambda}) &= \textbf{B} \beta \\
\text{logit}(\textbf{p}) &= \log(\textbf{p}/(1-\textbf{p})) = \textbf{G} \mathbf{\gamma}.
}$$
How do we derive the log-likelihood of $\textbf{Y}$? I know that it is $$L(\gamma,\mathbf{\beta} | \textbf{Y}) = \sum_{y_i=0} \log(e^{G_i \gamma}+\exp(-e^{\textbf{B}_i \mathbf{\beta}})) +\sum_{y_i >0} (y_i \textbf{B}_i \mathbf{\beta}-e^{\textbf{B}_i \mathbf{\beta}})-\sum_{i=1}^{n} \log(1+e^{G_{i} \gamma})-\sum_{y_i >0} \log(y_{i}!)$$
Added. I think that the likelihood function will be $$\prod_{Y_i=0} p_i+(1-p_i)e^{-\lambda_i} \times \prod_{Y_i >0} (1-p_i)e^{-\lambda_i} \lambda_{i}^{k}/k!$$
Then we know that $\lambda = \exp(\textbf{B} \beta)$ and $p = \frac{1}{1+\exp(-G \gamma)}$. So we just substitute these values and take logs?
 A: Your specification of the likelihood in terms of $p_{i}$ and $\lambda_{i}$: 
$$\prod_{i: \ Y_i=0} p_i+(1-p_i) \exp \{-\lambda_i \} \times \prod_{i: \ Y_i >0} (1-p_i)\exp \{-\lambda_i \}  \lambda_{i}^{Y_i}/Y_i!$$
is correct, directly from your model formulation. Taking logs and omitting the constant factorial term, the log-likelihood is 
$$ \sum_{i: \ Y_i=0} \log \left(p_i + (1-p_i)\exp \{-\lambda_i \}  \right) + \sum_{i: \ Y_i >0}\Big( \log(1-p_i) - \lambda_i + Y_i \log(\lambda_i) \Big)  $$
Now using that $p_i = \frac{1}{1 + \exp\{-{\bf G}_{i} {\boldsymbol \gamma}\}}=\frac{\exp\{{\bf G}_{i} {\boldsymbol \gamma}\}}{1+\exp\{{\bf G}_{i} {\boldsymbol \gamma}\}}$ and $\lambda_i = \exp \{ {\bf B}_{i} {\boldsymbol \beta} \}$ and substituting into the equation above and using the fact that $\log(A/B)=\log(A)-\log(B)$ a number of times, the expression above becomes: 
\begin{align*}
& \ \ \ \sum_{i: \ Y_i=0} \log \left( \frac{\exp\{{\bf G}_i {\boldsymbol \gamma}\}}{1 +\exp\{{\bf G}_i {\boldsymbol \gamma}\}} +  \frac{1}{1 + \exp\{{\bf G}_i {\boldsymbol \gamma}\}}
\exp \{ -\exp \{ {\bf B}_i {\boldsymbol \beta} \} \}
 \right) \\
&+ \sum_{i: \ Y_i >0} \log \left( \frac{1}{1 + \exp\{{\bf G}_i {\boldsymbol \gamma}\}} \right) - \exp \{ {\bf B}_i {\boldsymbol \beta} \} + Y_i {\bf B}_i {\boldsymbol \beta} \\
&= \sum_{i: \ Y_i=0} \log \Big(\exp\{{\bf G}_i {\boldsymbol \gamma}\} +
\exp \{ -\exp \{ {\bf B}_i {\boldsymbol \beta} \} \}
 \Big) - \log(1 + \exp\{{\bf G}_i {\boldsymbol \gamma}\}) \\
&+ \sum_{i: \ Y_i >0} \log(1) - \exp \{ {\bf B}_i {\boldsymbol \beta} \} + Y_i {\bf B}_i {\boldsymbol \beta}- \log(1+\exp\{{\bf G}_i {\boldsymbol \gamma}\})\\
&= \sum_{i: \ Y_i=0} \log \Big(\exp\{{\bf G}_i {\boldsymbol \gamma}\} +
\exp \{ -\exp \{ {\bf B}_i {\boldsymbol \beta} \} \}
 \Big) + \sum_{i: \ Y_i >0} \Big( Y_i {\bf B}_i {\boldsymbol \beta} - \exp \{ {\bf B}_i {\boldsymbol \beta} \} \Big) \\
& \ \ \ \ \ \ \ \ - \sum_{i=1}^{n} \log(1+\exp\{{\bf G}_i {\boldsymbol \gamma}\})
\end{align*}
which is exactly the likelihood equation you were trying to derive (without the constant factorial term which is not necessary).   
