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Let $\{X_n\}$ be an ergodic sequence of random variables, $X_n:(\Omega,\mathcal{F})\to (S,\mathcal{S})$ where the target set $S$ is a matrix ring. My question is,

Can the following limit be found almost surely? $$\frac{\displaystyle\sum_{N\ge i_1>i_2>\cdots>i_k\ge 1}X_{i_1}X_{i_2}\cdots X_{i_k}}{\displaystyle\binom{N}{k}}$$

I think this limit could be found, had the target set been a commutative ring, by an application of Newton's identity, but because of the noncommutative nature here, I cannot apply that principle here. Is there any way to tackle this problem?

Please make reference to any material available, as I do not have any proper background on non-commutative ring theory (nor on ergodic theory applied to rings, for that matter), apart from the basics of rings. Thanks in advance.

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  • $\begingroup$ In order for the question to make sense, you would need a notion of limits in the ring; also a unique notion of division by an integer.Maybe you want a matrix ring? Also, which limit are you taking? $N\to\infty$? $N$ and $k$ jointly going to $\infty$ somehow? $\endgroup$
    – Anthony Quas
    Commented Feb 17, 2016 at 17:46
  • $\begingroup$ @AnthonyQuas, I am exactly interested in a matrix ring. And yes, I am taking $N\to \infty$ and keeping $k$ fixed. $\endgroup$
    – Samrat Mukhopadhyay
    Commented Feb 17, 2016 at 18:15
  • $\begingroup$ I believe the answer in the case of matrix rings is just $(\mathbb EX_0)^k$. I'll write down some details later. $\endgroup$
    – Anthony Quas
    Commented Feb 17, 2016 at 19:54
  • $\begingroup$ Thanks@AnthonyQuas. It would be great if I can find some steps regarding how to get around the noncommutativity. $\endgroup$
    – Samrat Mukhopadhyay
    Commented Feb 17, 2016 at 20:13

1 Answer 1

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Let me give an answer for real matrices in the case $k=2$. I believe that larger values of $k$ can be handled by induction.

Let $A=\mathbb E X_0$. I will write out three sums for $k=2$:

\begin{align*} S_1&=\Big(X_0(X_1+\ldots+X_{N-1})+X_1(X_2+\ldots+X_{N-1})+ \ldots+X_{N-2}(X_{N-1})\Big)\\ S_2&=\Big((N-1)X_0A+(N-2)X_1A+\ldots+1X_{N-2}A\Big)\\ &=\Big((N-1)X_0+\ldots+X_{N-2}\Big)A\\ &=\Big((X_0+\ldots+X_{N-2})+(X_0+\ldots+X_{N-3})+\ldots+(X_{0})\Big)A\\ S_3&=\Big((N-1)A+(N-2)A+\ldots+A\Big)A=\binom{N}2A^2. \end{align*} By the ergodic theorem, each bracketed term in the $S_1$ differ from the corresponding bracketed term in the $S_2$ sum by $o(N)$, so that $S_1-S_2=o(N^2)$. Similarly, each bracketed terms in the final version of the $S_2$ sum differs from the corresponding bracketed terms in the $S_3$ sum by $o(N)$, so that $S_3-S_2=o(N^2)$ also. Hence $S_1-S_3=o(N^2)$ and your expression converges to $A^2$.

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  • $\begingroup$ Thanks a lot for the the answer. I think I understood the technique. Can you though kindly clarify a few points for me? The points may be trivial but because of my lack of proper background in Ergodic theory, I am asking them. I understand that $\sum_{i=10}^nX_i/n$ will be almost surely within an $\epsilon$ ball around $A$ for large $n$, from which your claim for the $o(N)$'s comes, right? But I do not understand how I can write $X_{N-2}$ as $o(N)+A$ $\endgroup$
    – Samrat Mukhopadhyay
    Commented Feb 18, 2016 at 7:00
  • $\begingroup$ Also, is the ergodic theorem that you applied to compare $S_2$ and $S_3$ is applied to the sequence $\{Y_n\}$ where $Y_0=(X_n+\cdots+X_0),\ Y_1=(X_{n-1}+\cdots+X_0),\cdots,\ Y_{n}=X_0$? $\endgroup$
    – Samrat Mukhopadhyay
    Commented Feb 18, 2016 at 7:03
  • $\begingroup$ The estimates in the comparison between $S_1$ and $S_2$ become weaker and weaker as you move from left to right. In the last one, $X_{N-1}$ is compared to $A$, so that if the $X_i$'s are bounded, the correct estimate should be $O(1)$. Nevertheless, the $o(N)$ bound is the right order for most of the sequence, and is too high for the large terms. The answer to your second question is yes. $\endgroup$
    – Anthony Quas
    Commented Feb 18, 2016 at 17:25
  • $\begingroup$ Ok, thanks a lot for the answers. It was really helpful. $\endgroup$
    – Samrat Mukhopadhyay
    Commented Feb 18, 2016 at 17:43

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