# Beta Bernoulli - distribution of p conditional on successes

I have a Beta-Bernoulli process $$\textbf{X}=\{X_1, X_2...X_n\}$$, $$\textbf{p}=\{p_1, p_2...p_n\}$$ where: $$X_i \sim Bernoulli(p_i)$$ $$p_i \sim Beta(\alpha,\beta)$$

where $$\alpha$$ and $$\beta$$ are known.

Is there an analytical solution for distribution of $$p'$$ for the subset of trials that were successful: $$f_{p'|X_i=1}$$

It's easier to illustrate in a simulation setting. That is, given sequence $$\textbf{X}$$, let $$\textbf{X'}$$ be the subset of $$\textbf{X'}$$ containing successes only: $$\textbf{X}'=\{X_i|X_i = 1\} = \{1,1,1...1\}$$

What is the distribution of $$p$$ that generated $$\textbf{X'}$$?

For example, below is the result from a simulation. The blue line is the distribution of $$p$$ from all trials ($$p\sim Beta(2,2)$$). Orange line is the conditional distribution of $$p$$ from successes only. Green line is the distribution of $$p$$ from failures only. So, the overall distribution of $$p$$ is a mixture of the two.

I suppose this is supposed to be a Bayesian question

If the prior for $$p$$ is $$\operatorname{Beta}(\alpha, \beta)$$ with density proportional to $$p^{\alpha-1}(1-p)^{\beta-1}$$ and you then observe $$k$$ successes and $$n-k$$ failures in $$\mathbf{X}$$ with likelihood proportional to $$p^k(1-p)^{n-k}$$

then the posterior density for $$p$$ given $$\mathbf{X}$$ is proportional to $$p^{\alpha-1+ k}(1-p)^{\beta-1+ n-k}$$ so the posterior distribution for $$p$$ is $$\operatorname{Beta}(\alpha+k, \beta+n-k)$$

If $$\mathbf{X}$$ is a single observation and is a success then this mean the posterior distribution for $$p$$ is $$\operatorname{Beta}(\alpha+1, \beta)$$

I now believe you think you are asking a different question, where different values of $$p$$ are drawn from $$\operatorname{Beta}(\alpha, \beta)$$ giving $$p_1, p_2, \ldots, p_n$$, and then each $$X_i$$ is drawn from $$\operatorname{Bernoulli}(p_i)$$; you want to know the distribution of those $$p_i$$ corresponding to $$X_i=1$$.

This is equivalent to the final line of my initial answer on a single successful observation: the distribution in question is $$\operatorname{Beta}(\alpha+1, \beta)$$

To try and convince you, here are a couple of simulations in R of $$10^6$$ initial draws, first with $$\alpha=2, \beta=2$$ and then with $$\alpha=2, \beta=8$$, drawing the simulated densities for the successful cases (black) and comparing this with the theoretical densities for $$\operatorname{Beta}(\alpha+1, \beta)$$. They are a very close match

set.seed(2021)

positives <- function(samplesize, alpha, beta){
p <- rbeta(samplesize, alpha, beta)
X <- rbinom(samplesize, 1, p)
success <- (X == 1)
psuccess <- p[success]
plot(density(psuccess, from=0, to=1))
curve(dbeta(x, alpha+1, beta), col="red", from=0, to=1, add=TRUE)
c(mean(success))
}

positives(10^6, alpha=1, beta=2)


and the chart above, starting with $$p\sim \operatorname{Beta}(1,2)$$, shows a $$\operatorname{Beta}(2,2)$$ distribution for the $$p_i$$ corresponding to $$X_i=1$$.

and the chart above, starting with $$p\sim \operatorname{Beta}(2,2)$$, shows a $$\operatorname{Beta}(3,2)$$ distribution for the $$p_i$$ corresponding to $$X_i=1$$. Then

positives(10^6, alpha=2, beta=8)


and the chart above, starting with $$p\sim \operatorname{Beta}(2,8)$$, shows a $$\operatorname{Beta}(3,8)$$ distribution for the $$p_i$$ corresponding to $$X_i=1$$.

• Sorry if the question was unclear. I'm not trying to estimate the distribution of $p$ (the parameters $\alpha$ and $\beta$ are already known). I'm trying to estimate the distribution of $p'$ that generated all the successful trials. I updated the question for clarification. May 28 at 22:43
• @parasu You said "$X_i \sim Bernoulli(p)$" which suggested to me that $p$ was the parameter for all the trials May 29 at 0:35
• I understand the confusion now. For every $X_i$ there's a $p_i$ from which $X_i$ is sampled. I updated the post to clarify this. May 29 at 1:19
• @parasu That is in effect the same question worded a different way, but with the same answer. I have now illustrated this May 29 at 1:31
• you're absolutely right. I just checked this as well. It makes sense that it can be reduced to a single draw. Thank you! May 29 at 1:48