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Suppose that $$Y_i \sim 0 \ \ \text{with probability} \ p_i$$ and $$Y_i \sim \text{Poisson}(\lambda_i) \ \ \text{with probability} \ 1-p_i$$

Then why is the same as $$Y_i = 0 \ \ \text{with probability} \ p_i+(1-p_i)e^{\lambda_i}$$ and $$Y=k \ \ \text{with probability} \ (1-p_i)e^{\lambda_i}\lambda_{i}^{k}/k!$$

$Y_i = 0$ means that is can be $0$ independent of the Poisson model or $0$ in the Poisson model? That is why we combined the probabilities? I get the probability for $Y_i = k$.

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