Introduce $Z_{ij}\in\{0,1\}$, where $Z_{ij}=1$ means that the observation came from the structural-zero component. Conditional on the parameters,
$$
\Pr(Z_{ij}=1\mid Y_{ij})=
\begin{cases}
\displaystyle\frac{\pi_j}{\pi_j+(1-\pi_j)e^{-\alpha_i\beta_j}},&Y_{ij}=0,\\
0,&Y_{ij}>0.
\end{cases}
$$
Given $Z$, the nonstructural observations are independent Poisson variables. Using shape-rate parameterization and the stated unit-rate priors, the remaining Gibbs updates are
$$
\alpha_i\mid-\sim\operatorname{Gamma}\left(
1+\sum_j(1-Z_{ij})Y_{ij},
1+\sum_j(1-Z_{ij})\beta_j
\right),
$$
$$
\beta_j\mid-\sim\operatorname{Gamma}\left(
1+\sum_i(1-Z_{ij})Y_{ij},
1+\sum_i(1-Z_{ij})\alpha_i
\right),
$$
$$
\pi_j\mid-\sim\operatorname{Beta}\left(
1+\sum_iZ_{ij},
1+n-\sum_iZ_{ij}
\right).
$$
Alternating these four standard-distribution updates defines the requested <Gibbs sampler> for the <Zero-inflated Poisson distribution> posterior.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 216 / 5 / Solution

The mass function is the mixture
$$
p_{\pi,\lambda}=(1-\pi)\delta_0
+\pi\operatorname{Poisson}(\lambda),
$$
so it is nonnegative and sums to one. Its first two moments are
$$
\mathbb E[Y]=\pi\lambda,
\qquad
\mathbb E[Y^2]=\pi(\lambda+\lambda^2).
$$
Therefore
$$
\operatorname{Var}(Y)
=\pi\lambda+\pi(1-\pi)\lambda^2.
$$
For $0<\pi<1$ this is strictly larger than the mean, so the <Zero-inflated Poisson distribution> models overdispersion. At $\pi=1$ it reduces to an ordinary Poisson distribution, and at $\pi=0$ it is degenerate at zero.

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 2 / 28J / b / Solution

{c}

A zero-inflated Poisson distribution is a mixture of a point mass at zero and a <Poisson distribution>. If the Poisson component has weight $\pi$ and mean $\lambda$, then its mean is $\pi\lambda$ and its variance is $\pi\lambda+\pi(1-\pi)\lambda^2$.

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 Incoming links: Zero-inflated Poisson distribution