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I am trying to figure out why the following holds true and would like to ask for your help.

Given the following Bayesian model: $$ y_i \sim \text{Poisson}(\mu_i \theta_i)\\\mu_i \sim \text{Gamma}(\alpha, \beta) \\\alpha \sim \text{Exponential}(a)\\\beta \sim \text{Gamma}(b,c) $$ for available observations $y_i$ with $i=1,...,n$ and parameters $\theta_i, a, b, c$ fixed the posterior distribution of the parameters $\mu_i, \alpha, \beta$ is:

$$ \pi(\{\mu_i\}, \alpha, \beta | \{y_i\},\{\theta_i\},a,b,c) \propto \prod_{i=1}^n [f(y_i | \mu_i, \theta_i\pi(\mu_i|\alpha, \beta)]\pi(\alpha|a)\pi(\beta|b,c) $$

How is this equation derived?

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1 Answer 1

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Bayes's Rule (ignoring fixed parameters) implies $$\pi(\{\mu_i\}, \alpha, \beta \mid \{y_i\}, \{\theta_i\}, a, b, c) \propto f(\{y_i\} \mid \{\mu_i\}, \{\theta_i\} , \alpha, \beta, a, b, c) \cdot \pi(\{\mu_i\}, \alpha, \beta\mid \{\theta_i\}, a, b, c). $$

The first term decomposes as $$f(\{y_i\} \mid \{\mu_i\}, \{\theta_i\} , \alpha, \beta, a, b, c) = f(\{y_i\} \mid \{\mu_i\}, \{\theta_i\}) = \prod_{i=1}^n f(y_i \mid \mu_i, \theta_i).$$

The second term decomposes as $$\pi(\{\mu_i\}, \alpha, \beta\mid \{\theta_i\}, a, b, c) = \underbrace{\pi(\{\mu_i\} \mid \alpha, \beta)}_{= \prod_{i=1}^n \pi(\mu_i \mid \alpha, \beta)} \pi(\alpha \mid a) \pi(\beta \mid b,c).$$

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  • $\begingroup$ Thank you very much! $\endgroup$ Aug 22, 2020 at 11:03

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