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.


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}

  • $\begingroup$ Thanks, but I know my solution is right ! $\endgroup$ – Stéphane Laurent Mar 12 '12 at 12:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.