Question about Bayesian Inference, Posterior Distribution

[thehairygorilla]

I have a posterior probability of \(\displaystyle p_i \)which is based on a Beta prior and some data from a binomial distribution:

I have another procedure:

$P(E)=\prod_{i \in I} p_i^{k_i}(1-p_i)^{1-k_i}$

which gives me the probability of a specific event of successes and failures for the set of $I$ in a model. Given the posterior distribution for $p_i$, how do I find \(\displaystyle P(E)\)?

 

[I like Serena]

I have a posterior probability of \(\displaystyle p_i \)which is based on a Beta prior and some data from a binomial distribution:

I have another procedure:

$P(E)=\prod_{i \in I} p_i^{k_i}(1-p_i)^{1-k_i}$

which gives me the probability of a specific event of successes and failures for the set of $I$ in a model. Given the posterior distribution for $p_i$, how do I find \(\displaystyle P(E)\)?

Hi thehairygorilla, welcome to MHB!

The event $E$ consists of a combination of $k_i$ for $i\in I$.
To find the probability $P(E)$ we would fill in those $k_i$ and the given $p_i$ in the formula, wouldn't we?

 

[thehairygorilla]

Hi thehairygorilla, welcome to MHB!

The event $E$ consists of a combination of $k_i$ for $i\in I$.
To find the probability $P(E)$ we would fill in those $k_i$ and the given $p_i$ in the formula, wouldn't we?

So not really. $p_i$ is a random variable. Better notation would be $P(E|p_1,...,...p_i,...,p_{|I|})=\prod_{i \in I} p_i^{k_i}(1-p_i)^{1-k_i}$ and I would be trying to find the marginal probability $P(E)$. Given the $p_i$s, $P(E|p_1,...,...p_i,...,p_{|I|})$ would be in terms of those random variables.