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Consider a stochastic process $X_t$ , $t \in 1,2,3,..,N $.

$X_t$ is a Bernoulli variable and $\Pr(X_t=1) = p$ for all $t$. The Autocovariance function $\gamma(|s-t|)= E[(X_t - p)(X_s -p)]$ is given

$ \gamma(k) = \frac{1}{2} (|k-1|^{2H} - 2|k|^{2H} + |k+1|^{2H}). $

For a constant $H\in (0,1)$ This is the same autocovariance as for fractional gaussian noise (increments of the fractional brownian motion), and give a autocovariance which falls like a power law when $k$ goes to infinity.

Let X and Y be process with the given properties, I am interested in the following probability distribution:

$ \Pr\left(\sum_{i=0}^N X_i Y_i = k\right) $

That is the distribution of the overlap of two such processes. For $H=1/2$ the process is not correlated and I have the simple result that $\Pr(X_t Y_t)=p^2$, and that

$ \Pr\left(\sum_{i=0}^N X_i Y_i = k\right) = {N \choose k} p^{2k} (1-p^2)^{N-k}. $

But for $H\neq 1/2$, I do not know how to deal with the long range correlation. Is there a way to proceed on this problem? I regret i never took a class in Stochastic Analysis, and I really hope the question makes sense. Any help or input would be highly appreciated.

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2 Answers 2

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One thing you should understand is that Bernoulli is not Gaussian: the autocorrelation function does not determine the process uniquely. In particular, the fact that the Bernoulli variables are not correlated doesn't mean that they are independent. For instance, the 3 step process that takes the paths (0,0,0),(0,1,1),(1,1,0),(1,0,1) with probability $1/4$ each has no autocorrelations but $\sum_{i=0}^2 X_iY_i$ is never $3$ here. So, your formula fails for this process. We need to know much more than just autocorrelations to answer your question.

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  • $\begingroup$ I also have a symmetry condition. But I'm not sure how to state it properly. All path have a non-zero probability, and if a path have some probability $f(p)$, possibly depending on the marginal probability $p$, then the probability of the 'mirror' path (1->0 and 0->1), should have the probability $f(1−p)$. Will this condition suffice for the question to be well posed? $\endgroup$
    – jonalm
    Commented Apr 6, 2010 at 18:52
  • $\begingroup$ Not really. Perhaps you'll need to tell us the full construction of your Bernoullis. $\endgroup$
    – fedja
    Commented Apr 6, 2010 at 22:31
  • $\begingroup$ What exactly do I need to specify? I think any process will do, as long as it satisfy the autocorrelation and the symmetry. $\endgroup$
    – jonalm
    Commented Apr 7, 2010 at 4:26
  • $\begingroup$ Full definition. The symmetry adds almost nothing (take any process and mix it with its symmetric copy). The information is still insufficient for a unique answer. $\endgroup$
    – fedja
    Commented Apr 10, 2010 at 5:31
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I'm not an expert, but my impression is that constructing this process isn't trivial. Here are two papers more-or-less on the subject:

Strong approximation of fractional Brownian motion by moving averages of simple random walks, by SZABADOS Tamas

Fractional Brownian motion, random walks and binary market models, Tommi Sottinen

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  • $\begingroup$ Thank you for the input. I can generate a sequence like this by writing the N binary variates as $X_n=\mathbb{I}_{0}({Z_n})$, where $\mathbb{I}_{A}(b)=0$ if $b\in A$ and 1 else, and where $Z_n$ is a sum of a set of $N(N+1)/2$ Poisson variables. The algorithm is explained in detail in: "A simple method for Generating Correlated Binary Variates" (jstor.org.stable/2684925). Writing the whole expression in terms of Poisson variables is very messy, so I was hoping to utilize the structure of the covariance function directly to calculate the distribution I'm interested in. $\endgroup$
    – jonalm
    Commented Apr 6, 2010 at 12:26

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