Suppose we have an extremely large collection of red balls and green balls. If we let $R$ and $G$ be the events of drawing a red ball and drawing a green ball, respectively, and if we let $\Pr(R)=p$ then $\Pr(G)=1-p$. We also assume that the vast majority of the balls in our collection are green, implying $p<<1-p$.

We would like to estimate $p$ using the following experiment: sample from the collection 3 times by selecting 20 balls each time. Then starting with the uninformed prior distribution of $p$ as Beta$(1,1)$ we update the prior by Bayesian updating after each sample and compute the expected value of $p$ after completing the 3 samples.

So suppose we sample and get 20 green balls for each of the 3 samples. The final update of the prior distribution of $p$ is Beta$(1,61)$ and the expected value of $p$ is $E(p)=\frac{1}{62}\approx 0.01613$.

This seems to be a very high estimate and I think the problem is that I am starting the updating with an uninformed prior despite the fact that I know $p<< 1-p$. Is there a way to justify a more informed prior in order to get a stronger conclusion? For example, if I believe that $p\le 0.01$ start with the informed prior Beta$(2,100)$, my update for $p$ will be distributed as Beta$(2,160)$ and $E(p) = \frac{1}{81}>0.01$! That makes no sense to me. Any help would be appreciated.


1 Answer 1


Seeing 0/60 isn't strong evidence for $p\leq 0.01$, since seeing 0/60 is pretty much what you expect for $p\leq 1/60$. This means your posterior won't have high weight on $p\leq 0.01$ unless your prior does

If you are confident a priori that $p\leq 0.01$ you want the prior probability of that to be high. A Beta(2,100) prior still only has 26% probability of $p\leq 0.01$; the Beta(2,160) prior has about 48% probability, which seems plausible. Note that the prior and posterior are very skewed, so although the posterior median is close to 0.01, the posterior mean is higher.

If you had a Beta(2,200) prior, the prior mean would be 0.01 and the prior median a bit lower. The posterior mean would be about .0075, the posterior median about 0.0065, and the posterior probability of $p\leq 0.01$ about 75%

So, there's two things going on: the data don't provide much evidence that $p$ is very small rather that just small, and your prior is much weaker than what you describe as your actual beliefs.

  • $\begingroup$ Thank you! Your comments make perfect sense. If I solve $(1-p)^{60}=0.5$ for $p$, then that tells me I have a $50\%$ chance that $p\approx 0.011486$ based on the test results, so starting with a prior distribution of Beta$(2,160)$ is reasonable. Then the posterior is distributed as Beta$(2,220)$ so that $E(p)=\frac{1}{111}$. $\endgroup$ Commented Oct 24, 2020 at 5:19

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