I have an experiment where I throw a biased coin were with confidence interval [0.38, 0.42] and confidence level of 0.95 I will get heads. If I get heads then I throw a biased 6-sided die which gives me 1 with probability given in the interval [0.48, 0.52] and confidence level 0.90.

What is the probability of getting 1 and what is my confidence level for that result. If no intervals and no confidence levels the problem is trivial and could be done with simple multiplication, but I am pretty sure simple multiplication will not be a solution in interval case.

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    $\begingroup$ Hi Andre, I simplified the example in order to present it in "text book" style but at this point I have no idea whether this can be solved using off the shelve methods. $\endgroup$ – Bart Nov 5 '13 at 17:45

A confidence interval is a "post-data" interval estimate that is supposed to bracket the true parameter in %C of the samples. What you appear to be trying to do is to predict a future event. For this, you need a little more structure on your problem. In particular, a confidence interval is insufficient. What you really need is a distribution of possible head probabilities and probabilities of rolling 1. Then, you need to calculate the probability as such:

Let $C$ be the outcome of the coin toss and $X$ be the results of the die roll, $f_H(p)$ isthe density function on the probability of heads (i.e., $p$) and $f_{1|C=H}(q)$ is the density function for the probability of rolling a 1 (i.e, $q$) given that you got a head. Therefore,

$P(X=1)=E[1_{C=H}1_{X=1|C=H}]=E[1_{C=H}]E[1_{X=1|C=H}]$ where $1_{C=H}$ and $1_{X=1|C=H}$ are indicator functions that take value 1 when the event in subscript happens, and 0 otherwise. Conditional independence between the coin toss and die roll allow you to multiply expected values.

Now, $E[1_{C=H}]E[1_{X=1|C=H}]=\int\limits_0^1 \int\limits_0^1pf_H(p)qf_{1|C=H}(q)dqdp$. In other words, you multiply the expected values of the distributions on P(heads) and P(X=1|Heads). So, you will need more info to solve your problem as formulated.

IF you have the underlying data that produced each CI, then you can use methods from Bayesian prediction or predictive likelihood

  • $\begingroup$ In fact I don't have the underlying data and I am only given the CI + CL. I know that the underlying distributions will follow Normal Distribution. I assume that if I have CI I can compute compute minimal probability of getting 1 (by taking lower bounds) but I am not sure what happens to CL of my result. $\endgroup$ – Bart Nov 5 '13 at 19:39
  • $\begingroup$ Ok, if you don't have the data, then ignore the last sentence. You cannot get a lower bound by using the lower limits of the CIs, as they are not 100% CIs. The underlying distributions cannot be normal, as P(Heads) and P(X=1) are bounded between 0 and 1. Do you know the most likely P(Heads) and P(X=1)? If not, then your problem is underdefined. $\endgroup$ – user31668 Nov 5 '13 at 19:46
  • $\begingroup$ The only approach I can see for using only the intervals would be to rely on conditional independence of the coin and die and interpret the CI's and CL as probability intervals. In that case, you would multiply the interval endpoints and probabilities; i.e, $P((p\in [0.38, 0.42])\cap (q \in [0.48, 0.52])) = P(p\in [0.38, 0.42])P(q \in [0.48, 0.52])=P(P(X=1)\in [0.38\times.48,0.42\times0.52])\\$ $\endgroup$ – user31668 Nov 5 '13 at 20:18
  • $\begingroup$ P(X=1) cannot be calculated exactly without prior distributions on P(heads) and P(X=1|Heads) $\endgroup$ – user31668 Nov 5 '13 at 20:21

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