Convergent Statistic for Bernoulli Random Variable I am currently working on my senior thesis and have encountered a small statistical problem which I need to solve but I lack the rigorous training to do so with sufficient precision myself.
The problem can be stated as follows: Suppose there are N bernoulli trials, $v_n$, with probability $p_n$ of heads. Suppose further there is for each bernoulli trial a bernoulli "signal" about the outcome of the trial. More precisely, for each $v_n$ there is a bernoulli random variable $s_n$ which has conditional distribution
$$[s_n |v_n =1] \sim Be(q_n)$$
$$[s_n |v_n =0] \sim Be(1-q_n)$$
i.e. it reveals the true type of $v_n$ with probability $q_n$ and further assume $q_n>0.5$. Suppose now further that all the $v_n$ are dependent in some sense. Say $|Cov(v_n,v_m)|>\rho_0$ for some $\rho_0 \in \mathbb{R}$ and for all pairs $n,m$. I.e. this inequality holds for all pairs but they do not necessarily have the same correlation coefficient.
My problem now is that, I want to show that if I observe $N \to \infty$ of the $s_n$, then I will with probability $1-\epsilon$ (or $1$?) know the true value of any $v_n$.
Intuitively, I think this should hold, but I lack the references to prove it. So I would very much appreciate if someone could point me in the right direction converning references which may be relevant (or even better, knows how to prove this).
My idea: Maybe we can show that $E[v_n|\mathcal{S}]$ is a consistent estimator? Here $\mathcal{S}$ is the sigma field generated by all the $s_n$. Not sure if this is the right approach or even correct concept though.
EDIT: I am willing to be quite lax in specifying dependence. If there is one concept which is more favorable for proving this than another, I am defenitely willing to give it a go.
Thanks in advance.
 A: 
"My problem now is that, I want to show that if I observe $N \to
 \infty$ of the $s_n$, then I will with probability $1-\epsilon$ (or
$1$?) know the true value of any $v_n$."

In the comments it has been clarified that the parameters of the model $p_n, q_n, \rho_{m,n}$ are a priori known. So you are not after estimating them, but rather you're after predicting/estimating perfectly (almost surely) the random variables $v_n$ per se.
Certainly, if the $v_n$'s were independent, the only relevant information for some $v_k$ would be $s_k$, and it is evident that with it alone, no perfect prediction is possible -by applying standard probability rules and expressions, one can obtain that
$$\text{Prob} (v_k =1 \mid s_k =1) = \frac {q_kp_k}{1+2q_kp_k-p_k-q_k} <1$$
So your case rests on the existence of dependence between the $v_n$'s.
Let's examine a case a bit more clear and certainly more favorable: assume that we actually observe the sequence of $\{v_n\}$'s, and not the sequence of imperfect signals $\{s_n\}$'s. Would that be enough to predict the next $v_n$ in line?
No: the existence of stochastic/statistical dependence between random variables does not make the one perfectly predictable given the others. It should be the case that the one under prediction is a deterministic function of the others to be able to obtain perfect prediction in principle. And even if we observe an "infinite" sequence of $\{v_n\}$, still, the next $v_n$ in line could not be predicted perfectly -an inherent "randomness/unpredictability" will always remain, exactly because $v_n$ is a non-degenerate random variable.
...and in your case, we are in a worse situation: we only observe the sequence of imperfect signals $\{s_n\}$ -an additional source of randomness/imperfection exists here.
Concepts such as asymptotic consistency usually refer to the ability, when information accrues beyond bound, of estimating perfectly an unknown constant, not a non-degenerate random variable. When the target is a random variable, then we talk about a "predictor", not an "estimator", and we can still use the concept of a "consistent predictor", although we would better make the clarification.
