MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 3 down vote accepted

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.

share|cite|improve this answer
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? – jonalm Apr 6 '10 at 18:52
Not really. Perhaps you'll need to tell us the full construction of your Bernoullis. – fedja Apr 6 '10 at 22:31
What exactly do I need to specify? I think any process will do, as long as it satisfy the autocorrelation and the symmetry. – jonalm Apr 7 '10 at 4:26
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. – fedja Apr 10 '10 at 5:31

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

share|cite|improve this answer
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" ( 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. – jonalm Apr 6 '10 at 12:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.