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Let $X_n$ be a sequence of identically distributed (e.g., binomial $B(1,1/2)$) random variables which are not independent (say, for any $n$ and $m$, $corr(X_n,X_m)=c$).

What can be said about the limit of their mean $\sum_n X_n/N$?

Motivation: a reason price variations assumed to be Gaussian is that they are composed by allegedly independent actions of many traders. However, in reality the traders are not independent, because, e.g., they all read the same news. So, $X_n$ is best viewed as a "set", not an "ordered sequence".

I am not sure if this question belongs here or on cross-validated.

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Limit in which sense? –  Davide Giraudo Nov 7 '12 at 20:43
any limit is fine –  sds Nov 8 '12 at 16:13
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2 Answers

What can be said about the limit of their mean $\sum_n X_n/N$?

Not much. Here is an example to show that the limit can be non deterministic.

Let $V$ and $(W_n)_n$ denote i.i.d. symmetric $-1/1$ Bernouli random variables (weights $\frac12$ on $+1$ and $-1$). Let $(U_n)_n$ denote independent i.i.d. $0/1$ Bernoulli random variables (weights $u$ on $1$ and $1-u$ on $0$). Let $X_n=U_nV+(1-U_n)W_n$. In words, each $X_n$ is $V$ or $W_n$, with probabilities $u$ and $1-u$ respectively. Then $\mathrm{Corr}(X_n,X_k)=u^2$ for every $n\ne k$ and $\frac1N\sum\limits_{n=1}^NX_n\to uV$ almost surely.

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This webpage may be helpful. The law of large numbers for dependent random variables is discussed near the end of the page. However, the relevant result there seems to require that $X_i$ and $X_j$ are asymptotically independent. In your example, the correlation between $X_i$ and $X_j$ remains constant (can you give a concrete example of such a sequence of $X_i$s?), so the result in that webpage may not apply.

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see the newly added "motivation" –  sds Nov 8 '12 at 16:28
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