Basically a modelling exercise, 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. $p_n,q_n$ and the correlations are all deterministic.

I am interested in what happens when $N$, the number of Bernoulli trials, tends to infinity. Intuitively, the information in the system should increase since we have more attempts to observe the true values, and our prediction should become better and better. However, I have recently been told that this isn't quite that easy, see

Convergent Statistic for Bernoulli Random Variable

Now, to make things feasible, let us make some further restrictions. Let $\Sigma_N$ be the covariance matrix for the $N$ first trials, then we assume that if $N'>N$ then $\Sigma_N$ is a block sub-matrix of $\Sigma_{N'}$. I.e. increasing $N$ does not change anything about the already existing random variables. Furtheremore, we assume that there are a fixed number $M$ of random variables $y_m$ such that every $v_n$ can be written $v_n=\chi_n(y_1,...,y_M)$ where $\chi_n$ is a deterministic indicator function. This way, we ensure that as $N$ increases the amount of randomness in the system remains fixed.

My question now is, what happens to the random quantities $E[v_n|\mathcal{S}_N]$ and $E|v_n-E[v_n|\mathcal{S}_N]|$ as $N$ increases, say by a finite amount? What happens to them if $N\to \infty$? Here $\mathcal{S}_N$ is the sigma-field generated by all the $s_n$, $\mathcal{S}_N=\sigma(s_1,...,s_N)$ so that $E[v_n|\mathcal{S}_N]$ is an $\mathcal{S}_N$ measurable random variable.

Remark: I am not sure what to assume about the $y_m$. If it is easier, I am willing to assume that they have some particular nice and easy continuous distribution, like a Gaussian.

To see why this is not a duplicate, note that when we do not impose the restriction of a finite amount of underlying randomness, the question has no real fruitful answer, except that there isn't really one. This was explained to me in the other question which prompted this one.

  • $\begingroup$ Possible duplicate of Convergent Statistic for Bernoulli Random Variable $\endgroup$ – Xi'an Oct 22 '16 at 13:35
  • 1
    $\begingroup$ @Xi'an I am the author of both questions and no, they are not the same question. The other question was answered in the sense that the phrasing was to general for there to be a solution, whereas this question possibly has a solution. $\endgroup$ – Winston Oct 22 '16 at 13:37
  • $\begingroup$ It would make more sense to rephrase the previous question, instead of creating another and almost similar question, as the previous one has been commented and answered. $\endgroup$ – Xi'an Oct 22 '16 at 13:43
  • 2
    $\begingroup$ OK, I actually asked in that previous question whether I should create another one, and the answer was affirmative. $\endgroup$ – Winston Oct 22 '16 at 14:26

Your Answer

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

Browse other questions tagged or ask your own question.