Two successive binomial experiments performed by two different groups of individuals

Question

Imagine two experiments:

1) A group of $n=100$ individuals are under condition $A$, and they perform some test. Among these individuals, $y=5$ individuals succeed in the test.

2) A group of $m=1000$ individuals are under condition $\Delta$ and they perform some test with two possible issues: the first possible is that an individual falls under the above condition $A$, the second possible issue is another condition $B$ (obviously $B = \neg A$). Among these individuals, $x=2$ individuals fall under condition $A$.

Now consider an individual under condition $\Delta$ who perform experiment 2) and then experiment 1) in case when he falls under condition $A$. We are interested in the probability for this individual to succeed in the second test.

Of course this probability is naturally estimated by $x/m \times y/n$. But how to assess the uncertainty about this estimate ?

I have in mind a valuable Bayesian solution: using a (possibly noninformative) prior distribution on the first probability of success $\theta_1$ and another (possibly noninformative) prior distribution on the second probability of success $\theta_2$ then $(x,m)$ yields a posterior distribution for $\theta_1$ and $(y,n)$ yields a posterior distribution on $\theta_2$, and finally we get a posterior distribution on the probability of interest $\theta_1\theta_2$ (assuming independent posterior distributions of $\theta_1$ and $\theta_2$).

Do you know/imagine another solution ?

EDIT : Maybe the following frequentist solution is valuable : estimating the asymptotic variance of $\hat\theta_1$ and the asymptotic variance of $\hat\theta_2$ then we get an asymptotic variance of $\hat\theta_1 \hat\theta_2$ by multiplying the two variances. But, in fact, my real problem is a bit more complicated : there is a third experiment and I am interested in the estimation of the probability of two possible issues at the last stage.

EDIT2: I was tired when writing the previous "EDIT". More correctly I was having in mind the Delta-method to derive the asymptotic behaviour of $\hat\theta_1 \hat\theta_2 - \theta_1\theta_2$ and then an asymptotic confidence interval.

Answer 1

Your solution is right.

Let the probability of falling under condition A be $P$, and the probability of succeeding in the first test conditional on falling under condition A be $Q$. You want to find $PQ$.

Given $n$ and $y$, we have a beta-distributed belief over $P$ with parameters $y$ and $n-y$. Similarly, $Q$ has a beta-distributed belief with parameters $x$ and $m-x$. Let these distributions have density $f$ and $g$ respectively.

So, the likelihood on PQ is \begin{align} h(z) &= \int_0^1 f(p)g(z/p) \; dp \\ &\propto \int_0^1 p^{y-m}(1-p)^{n-y}\; z^x(p-z)^{m-x} dp. \end{align}