MHB Question about Bayesian Inference, Posterior Distribution

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To find the probability of event E, given the posterior distribution of the probabilities \( p_i \), one must recognize that \( p_i \) is a random variable. The correct approach is to express the probability as \( P(E|p_1,...,p_i,...,p_{|I|})=\prod_{i \in I} p_i^{k_i}(1-p_i)^{1-k_i} \). The goal is to compute the marginal probability \( P(E) \) by integrating over the distribution of \( p_i \). This involves using the posterior distribution of \( p_i \) to evaluate the integral for \( P(E) \). Understanding the relationship between the posterior distribution and the event probabilities is crucial for accurate computation.
thehairygorilla
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I have a posterior probability of $$p_i $$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 $$P(E)$$?
 
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thehairygorilla said:
I have a posterior probability of $$p_i $$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 $$P(E)$$?

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?
 
I like Serena said:
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.
 
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First trick I learned this one a long time ago and have used it to entertain and amuse young kids. Ask your friend to write down a three-digit number without showing it to you. Then ask him or her to rearrange the digits to form a new three-digit number. After that, write whichever is the larger number above the other number, and then subtract the smaller from the larger, making sure that you don't see any of the numbers. Then ask the young "victim" to tell you any two of the digits of the...

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