Question about Bayesian Inference, Posterior Distribution

thehairygorilla

New member
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 $$\displaystyle P(E)$$?

Klaas van Aarsen

MHB Seeker
Staff member
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

New member
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.

Last edited: