# Finding the full conditonal distribution when there are multiple distributions involved

6 neighboring countries have the following disease instances: $y = (y_1, y_2,...,y_n)$ with a population of $x = (x_1, x_2,...,x_n)$.

The following model and prior distributions are considered:

$y_i|\theta_i,p_i \sim \text{Poisson}(\theta_i x_i)$

$\theta_i | \alpha, \beta \sim \text{gamma}(\alpha, \beta)$

$\alpha \sim \text{gamma}(1,1)$

$\beta \sim \text{gamma}(10,1)$

a) Find the full conditional rate $p(\theta_i | \theta_{-i}, \alpha, \beta, x, y)$

b) Find the posterior distribution.

Attempt:

a) For finding the conditional rate with two variables, I would use Bayes' theory. I am not sure if this applies with multiple distributions.

$$p(\theta_i | \theta_{-i}, \alpha, \beta, x, y) = \frac{P(\theta_i \bigcap \theta_{-i} \bigcap \alpha \bigcap \beta \bigcap x \bigcap y)}{P( \theta_{-i}, \alpha, \beta, x, y)}$$

$$= \frac{P(\theta_{-i}, \alpha, \beta, x, y | \theta_i)P(\theta_i)}{\sum_{i=1}^6 P(\theta_{-i}, \alpha, \beta, x, y | \theta_i)P(\theta_i)}$$

b) The posterior probability is the (prior)x(likelihood). So this would be $$\text{Poisson}(\theta_i x_i) \times L(\theta_i x_i)$$

I'm not sure how to do the pdf of a Poisson variable as it is variable. The likelihood function is $L(\theta_i y_i) = \frac{\theta_i^{\sum_{i=1}^n y_i} e^{-n \theta_i}}{y_1!,y_2!,..,y_n!}$

• First off, I assume the first two distributions you write are independent over $i$? For part (a), Bayes theorem applies to multiple variables - let $x$ be a vector and $y$ be a vector, then $p(y|x) = p(x,y)/p(x)$. Note that $p(y|x)$, $p(x,y)$ and $p(x)$ are all joint probability distributions over the relevant variables. For part(b), your expression of the likelihood is incorrect - your likelihood statement is $y_i| \text{some vars} \sim \text{Poisson}(\theta_i x_i)$ - i.e. it is a distribution for $y_i$, and note that $\theta_i \times x_i$ is simply the Poisson parameter. – queenbee Apr 1 '14 at 15:21