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

  • Thread starter Thread starter Jason4
  • Start date Start date
  • Tags Tags
    Limit
AI Thread Summary
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.
Jason4
Messages
27
Reaction score
0
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?
 
Physics news on Phys.org
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.
$\sum\limits_{k = 0}^N {{P^k}} = \left\{ {\begin{array}{rl}{\tfrac{1}{2}\left[ {\begin{array}{rl}{N + 2}&N\\N&{N + 2}\end{array}} \right]}&{,N\text{ even}}\\{\tfrac{{N + 1}}{2}\left[ {\begin{array}{rl}1&1\\1&1\end{array}} \right]}&{,N\text{ odd}}\end{array}} \right.$
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...
Back
Top