Combining two pieces of evidence expressed as probabilities I have a hidden binary random variable Z that can have a value of either 1 or 0. There is some true probability P(Z=1) = z that I do not know.
I also have two separate pieces of "evidence" that give some information about the state of Z: 


*

*P(Z=1) = x

*P(Z=1) = y


I'm interested in estimating z by combining the evidence contained in x and y. In particular:


*

*Is there a simple formula I can use to express a good estimate for z in terms of x and y?

*If I "trust" one of the two pieces of evidence more than the other, is there a way of weighting the combination?


Both pieces of evidence can be viewed as imperfect estimates of the true probability z, from alternative sources that have access to some (but not all) information that might be useful for predicting the random variable Z (e.g they can observe some variables that are correlated with Z). E.g. you have two bookmakers who are independently giving you odds of a specific event occurring, but neither has perfect information so the odds they give you are different guesses at the true probability. 
So the motivation for this question is whether, knowing two separate predictions, you can somehow combine them to make a single better prediction of z.
 A: One way to think about this is to let each piece of "evidence" be represented by a Beta distribution, $B(a_i+1,b_i+1)$, where $a_i,b_i \ge 0$. Let $P_i(Z=1) = \int_0^1 x\,B(x|a_i+1,b_i+1)\,dx = (a_i+1)/(a_i+b_i+2)$. Associated with each piece of evidence is its precision, which is the inverse of its variance. The variance is $(a_i+1)(b_i+1)/[(a_i+b_i+2)^2 (a_i+b_i+3)]$. 
Combining a number of pieces of evidence produces another Beta distribution, $B(c+1,d+1)$ where $c=\sum a_i$ and $d=\sum b_i$. (Assuming the pieces of evidence are independent, the probability density function (PDF) for the combined evidence equals the product of the individual PDFs, normalized to integrate to one.) Thus the combined evidence produces $P_{\text{comb}}(Z=1) = (c+1)/(c+d+2)$. The individual precisions are taken into account automatically: more precise evidence gets more weight. Finally, the precision associated with the combined result is the inverse of the variance of its Beta distribution, $(c+1)(d+1)/[(c+d+2)^2(c+d+3)]$.
