How to find the probability of an unobserved binary variable from repeated noisy observations? Let $Y \in \{0,1\}; P(Y=1)=\beta$. We have no observations of $Y$.
Instead, we observe a sample of $A$,$B$. We can assume that $P(A,B|Y)=P(A|Y)P(B|Y)$; $P(A=Y)=P(B=Y)=\alpha$; and that $P(A=Y|Y)=P(A=Y)$, P(B=Y|Y)=P(B=Y). In other words, $A$ and $B$ are binary variables, are conditionally independent given $Y$, and have the same error rate and their errors are independent of $Y$.
$\alpha$ is unknown, but I have convinced myself it is easy to estimate since it is determined by $P(A \ne B)$.
How can we estimate $\beta$?
Based on a comment on page 352 of the book "Measurement Error in Nonlinear Models" by Carroll et. al 2006, I believe that it is possible to consistently estimate $\beta$. But I'm having trouble working it out. I also recall doing a homework problem like this in the past, but I don't recall where.
I have worked out that $P(A\ne B) = 2\alpha (1-\alpha)$.
Carroll
et al. also suggest that with a sample of $A$,$B$,$C$ and $P(A,B,C|Y)=P(A|Y)P(B|Y)P(C|Y)$ we can estimate $\beta$ even if $P(A=Y) \ne P(B=Y) \ne P(C=Y)$.
EDIT: Prior versions of the question did not assume that error rates were independent of $Y$.  This assumption seems important to make the problem solvable, at least in the case of two replicates.

I think the answer you provided has a mistake. You substitute $P(A=1|Y=1)$ as $\alpha$. This is wrong according to your definition of $\alpha$ above. $ \alpha = P(A=Y) =  P(A=1, Y=1) +  P(A=0, Y=0)$, no?
I haven't found the solution, but I have developed some intuition that may be helpful.
Note that the joint probability distribution of two binary variables $A$ and $Y$ where $P(A=1) = \alpha$ and $P(Y=1) = \beta$ can be written as:





A=0
A=1




Y=0
$\bar\alpha\bar\beta + k$
$\alpha\bar\beta - k$


Y=1
$\bar\alpha\bar\beta - k$
$\alpha\beta + k$




Where $\bar\alpha = 1 - \alpha$, $\bar\beta =  1- \beta$, and $k$ controls the covariance between $A$ and $Y$:
$cov(A,Y) = E[AY] - E[A]E[Y] $
$cov(A,Y) =  \overbrace{\alpha\beta + k}^{P(A=1, Y=1)} - \alpha\beta = k$
We can now calculate the covariance between $A$ and $B$ in the setup you proposed (where $P(B=1) = \alpha$)  and find that:
$cov(A,B) = E[AB] - E[A]E[B]  $
$cov(A,B) = E[AB] - \alpha^2 $
$cov(A,B) = P(A=1, B=1 | Y = 0)  + P(A=1, B=1 | Y = 1) - \alpha^2 $
$cov(A,B) = P(A=1| Y = 0) P(B=1| Y = 0)  +P(A=1| Y = 1) P(B=1| Y = 1)  - \alpha^2 $
$cov(A,B) = P(A=1| Y = 0)^2  +P(A=1| Y = 1)^2  - \alpha^2 $
If we substitute this using the table above, we can find that:
$cov(A,B) = k^2 Var[Y]^{-1} = \frac{cov(A,Y)^2}{Var[Y]}$
