Bayesian Statistics - Zero Inflated Binomial Model - Calculate Posterior Conditional Distribution

Looking at solving question a) as attached. Even if is not fully solved, some help on where to begin would also be appreciated.

Thanks

Answer

You want to calculate:
$$P(z_i=1\mid x_i,\omega,p)$$
the definition of the $z_i$ is so that $z_i=0\implies x_i=0$. So if $x_i\neq0$ you must have that $z_i=1$, i.e.:
$$P(z_i=1\mid x_i\neq0, \omega,p)=1$$
and, as the text remarks, the onlything to calculate is
$$P(z_i=1\mid x_i=0, \omega,p)$$
First we apply Bayes Theorem to get:
$$P(z_i=1\mid x_i=0,\omega,p) = \frac{P(x_i=0\mid z_i=1, p,\omega)\cdot P(z_i=1\mid p,\omega)}{P(x_i=0\mid p,\omega)}$$
We know that $(x_i|z_i=1)\sim BIN(7;p)$, so $P(x_i=0\mid z_i=1,p,\omega) = (1-p)^7$. We further calculate:
$$P(x_i=0\mid p,\omega) = P(x_i=0\mid z_i=1,p,\omega)\cdot P(z_i=1\mid p,\omega)+P(x_i=0\mid z_i=0,p,\omega)\cdot P(z_i=0\mid p,\omega)$$
which we evaluate to:
$$P(x_i=0\mid p,\omega) = (1-p)^7\cdot \omega+1\cdot(1-\omega)$$
This leaves us with:
$$P(z_i=1\mid x_i=0,p,\omega)=\frac{(1-p)^7\cdot \omega}{(1-p)^7\cdot \omega + (1-\omega)}$$

And in total:

$$P(z_i=1\mid x_i=0,\omega) = \int_0^1 P(z_i=1\mid x_i=0, p,\omega) \underline{p}(p)\,dp$$
where $\underline p(p) = \dfrac{p^{a-1}\cdot (1-p)^{b-1}}{B(a,b)}$ is the Beta distribution. This doesn't look like it would be any fun to evaluate - the text probably would accept this expression as an answer.

The answer is accepted.