# 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.

- answered
- 345 views
- $14.76