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Suppose $X_1,...,X_n$ is a sequence of stationary correlated random variables in $\{-1,+1\}$ such that : $\mathbb{P}[X_i = +1] = p$, $\mathbb{P}[X_i = -1] = 1-p$ with $p\in (0,1)$, and with a correlation $\rho_{j}$ is defined by: $$ \rho_{j}:= \frac{\mathbb{E}[X_iX_{i+j}] - \mathbb{E}[X_i]\mathbb{E}[X_{i+j}]}{\mathbb{E}[X_i]\mathbb{E}[X_{i+j}]}\text{ (independent of }i\text{ by stationarity)}$$

My question concerns the expectation of the product: $$P_n:=\mathbb{E}\left[\prod_{i=1}^n X_i\right]$$

Clearly, if the $X_i$ are independent, then $P_n = (2p-1)^n$ .

My question is the following:

If $\rho_j =O(e^{-\kappa j})$ when $j\to\infty$ for some $\kappa>0$, what can we say about $P_n$? Can we compute it explicitly ? Can we obtain an asymptotic formula in the limit $n\to \infty$?

Thanks for your help !


A more specific question (proof or counter-example):

Is it true that one can find a sequence $c_n = O(e^{-\lambda n})$ for some $\lambda>0$ such that $$P_n = \prod_{i=1}^n (2p-1+c_i)$$

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I'm not sure there's enough information here to determine the answer. In particular, for Bernoulli- and Bernoulli-like-distributed random variables, the correlations are not enough to determine the joint distribution of more than two of them. For example, if $$ X_i = \left.\begin{cases} 1 & \text{with probability }1/2 \\ 0 & \text{with probability }1/2 \end{cases}\right\}\text{ for }i=1,2 $$ and these two are independent, and $X_3$ is the mod-2 sum of $X_1$ and $X_2$, then all three of the correlations are 0, and they're pairwise independent, but obviously they're not independent. – Michael Hardy Jul 14 '11 at 0:03
Note that the usual definition of correlation divides the covariance by the product of the standard deviations, not the expectations. One case where your $P_n$ can be computed explicitly is that of a Markov chain. – Robert Israel Jul 14 '11 at 1:35
@Robert : could you explain how you compute it explicitely in the case of a Markov chain? @Michael : Yes you are right, the correlation does not determine the joint distribution. However, my intuition is that the 'information' about $X_i$ for $i$ close to $n$ 'contained' in the product $\prod_{i=1}^{n-1}X_i$ is very 'weak' for large $n$, so that I expect something like $\mathbb{E}[X_n \prod_{i=1}^{n-1}X_i] \approx \mathbb{E}[\prod_{i=1}^{n-1}X_i] (\mathbb{E}[X_n] + c_n)$ with $c_n \to 0$ very fast (exponentially?). But I do not know if this is correct, and how to prove it. – user16215 Jul 14 '11 at 22:31

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