MHB Markov Chains with No Limit: Proving Convergence for $A_n$

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The discussion focuses on the properties of the matrix P, defined as P = [[0, 1], [1, 0]], and its powers. It is established that P^n does not converge to a limit, alternating between the identity matrix for even n and P for odd n. However, the sequence A_n, defined as A_n = (1/(n+1))(I + P + P^2 + ... + P^n), does converge to a limit. The convergence of A_n is demonstrated through the averaging of the oscillating behavior of P. The key takeaway is that while P itself does not stabilize, the average of its powers does converge.
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Consider:

$P=\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$

Show that $P^n$ has no limit, but that: $A_n=\frac{1}{n+1}(I+P+P^2+\ldots+P^n)$ has a limit.

I can see that $P^{EVEN}=\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)$ and $P^{ODD}=\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$, so a steady state is never reached, but I can't figure out the second part.

Any suggestions?
 
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Jason said:
Consider:
$P=\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$
Show that $P^n$ has no limit, but that: $A_n=\frac{1}{n+1}(I+P+P^2+\ldots+P^n)$ has a limit.
$\sum\limits_{k = 0}^N {{P^k}} = \left\{ {\begin{array}{rl}{\tfrac{1}{2}\left[ {\begin{array}{rl}{N + 2}&N\\N&{N + 2}\end{array}} \right]}&{,N\text{ even}}\\{\tfrac{{N + 1}}{2}\left[ {\begin{array}{rl}1&1\\1&1\end{array}} \right]}&{,N\text{ odd}}\end{array}} \right.$
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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