# Conditional density following Gamma distribution

We know that if a random variable say x ~ Gamma(a, b), then its probability density function is $$\propto x^{a-1} exp^{-bx}$$.

In a Bayesian hierarchical model, for example

$$Z_1, \cdots, Z_n |\theta \sim iid \; Gamma(r, \theta)$$, (r known)\$ ;

$$\theta \sim \; Gamma(a, b)$$,

the full conditional distribution of $$\theta$$ is

$$p(\theta | Z_{1:n}) \propto p(\theta, Z_{1:n}) \propto p(Z_{1:n}|\theta) \times p(\theta)$$,

by $$Z_{1:n} | \theta \sim iid \; Gamma(r, \theta)$$, I think $$p(Z_{1:n}|\theta)$$ shall be $$\Pi_i Z_i^{r-1} exp^{- \theta Z_i}$$, but the reading material suggests

$$p(Z_{1:n}|\theta) = \Pi_i \theta^{r} exp^{-\theta Z_i}$$,

I'm not sure I fully understand why it writes $$\theta^{r}$$ instead of $$Z_i^{r-1}$$ in front of $$exp^{-\theta Z_i}$$.

Can anyone explain? Thanks.

The full joint density of $$Z_{1:n}$$ conditional on $$\theta$$ is $$p(z_{1:n}|\theta)= \prod_{i=1}^n \frac{\theta^r}{\Gamma(r)} z_i^{r-1} \exp(-\theta z_i) \,.$$ If we consider this as a function of $$z_{1:n}$$, the term $$\theta^{r}/\Gamma(r)$$ is just a normalising constant and can be dropped.
As you point out, the conditional distribution of $$\theta$$ given $$z_{1:n}$$ (i.e. the posterior of $$\theta$$ given the observed sample) is $$p(\theta|z_{1:n}) \propto p(\theta) p(z_{1:n}|\theta) = p(\theta) \prod_{i=1}^n \frac{\theta^r}{\Gamma(r)} z_i^{r-1} \exp(-\theta z_i)$$ This is a function of $$\theta$$, so we need to keep the terms involving $$\theta$$ (such as $$\theta^{r}$$), but we can drop the other terms (such as $$z_i^{r-1}$$). We can thus write $$p(\theta|z_{1:n}) \propto p(\theta) \prod_{i=1}^n \theta^r \exp(-\theta z_i) \propto \theta^{a-1} \exp(-b\theta)\,\, \theta^{nr} \exp(-\theta n \bar{z})$$ which we identify as another Gamma density.