poisson distribution question 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 $Y_i = k$.
 A: Imagine that you play the following game.  You flip a coin that has probability $p$ of landing heads, and probability $1-p$ of landing tails.  
If the result is head, you win $0$ dollars.  If the result is tail, then you play a Poisson game with parameter $\lambda$, that is, a game in which you win $k$ dollars with probability $\frac{e^{-\lambda}\lambda^k}{k!}$.
Let random variable $Y$ denote your winnings in the game described above. We want to find the probability distribution of $Y$.
Let's first deal with $P(Y=0)$.  You can win $0$ dollars in two ways: (i) the coin flip results in head or (ii) the coin flip results in tail, but the Poisson distribution subgame results in $0$. The probability of (i) is $p$. To find the probability of (ii), note that we must get tail (probability $1-p$) and the Poisson subgame must give result $0$ (probability $\frac{e^{-\lambda}\lambda^0}{0!}$, or more simply $e^{-\lambda}$). Thus
$$P(Y=0)=p+(1-p)e^{-\lambda}.$$
Next we deal with $P(Y=k)$ where $k\gt 0$.  To get $k$ dollars, we must have gotten tail on the coin toss (probability $1-p$ and gotten result $k$ in the Poisson subgame (probability $\frac{e^{-\lambda}\lambda^k}{k!}$). So if $k \gt 0$, then
$$P(Y=k)=(1-p)\frac{e^{-\lambda}\lambda^k}{k!}.$$
A: I'm not entirely sure what is meant by a random variable following a distribution with a certain probability. But here is my interpretation. Let $X$ be a random variable such that $P(X=0)=p$ and $P(X=1)=1-p$ for some $p>0$, and let $Z\sim \text{po}(\lambda)$, $\lambda>0$, such that $Z$ and $X$ are independent Then $Y=ZX$ has the property that you ask for. Then 
$$
\{Y=0\}=\{X=0\}\cup(\{Z=0\}\cap\{X=1\})
$$
and hence 
$$
P(Y=0)=P(X=0)+P(X=1,Z=0)=P(X=0)+P(X=1)P(Z=0)
$$
due to independence. Using the definitions of $X$ and $Z$ we obtain
$$
P(Y=0)=p+(1-p)\frac{\lambda^0}{0!}e^{-\lambda}=p+(1-p)e^{-\lambda}.
$$
