My Math Forum Markov Chains

February 2nd, 2015, 10:26 AM   #1
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Markov Chains

Hi , I am trying to classify the subsets of this Markov chain

I am unsure if one of the subsets is {2, 3, 4} or it is {2, 4}.

So 2 and 4 obviously communicate but I have not seen an example where another number joins two communicating classes in this way (i.e. 3).

Could someone help me clear this up please?
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 February 2nd, 2015, 12:18 PM #2 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 You need to explain the diagram.
February 2nd, 2015, 02:50 PM   #3
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Hello, interestedinmaths!

Quote:
 I am trying to classify the subsets of this Markov chain I am unsure if one of the subsets is {2, 3, 4} or it is {2, 4}. $\color{blue}{\text{I don't understand what "subset" means.}}$

You seem to have this matrix:

$\quad \begin{pmatrix}0&1&0&0&0&0 \\ 0&0&\frac{1}{2}&\frac{1}{2}&0&0 \\ 0&0&0&1&0&0\\ 0&1&0&0&0&0 \\ 0 & \frac{1}{3} &0& \frac{1}{3} &0& \frac{1}{3} \\ \frac{1}{2} &0&0&0&\frac{1}{2}&0 \end{pmatrix}$

What exactly is the question?

February 2nd, 2015, 07:05 PM   #4
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I suspect that this is the sort of thing we are after
Quote:
 A set of states C is a communicating class if every pair of states in C communicates with each other, and no state in C communicates with any state not in C... A communicating class is closed if the probability of leaving the class is zero, namely that if i is in C but j is not, then j is not accessible from i.
By these definitions $\{2,3,4\}$ is a closed communicating class and \$\{5,6\} is a communicating class. There are no other communicating classes.

Subsidiary definitions:
Quote:
 A state j is said to be accessible from a state i (written i → j) if a system started in state i has a non-zero probability of transitioning into state j at some point... A state i is said to communicate with state j (written i ↔ j) if both i → j and j → i.
Source: Wikipedia

 February 3rd, 2015, 01:05 AM #5 Newbie   Joined: Dec 2012 Posts: 27 Thanks: 0 Thanks for all the replies. Apologies for not being more descriptive. I have the matrix that soroban mentions and I wanted to draw the corresponding Markov chain. When drawing it, I was not sure how to identify the communicating classes. I was stuck deciding whether it was {2,4} and {5,6} or {2, 3, 4} and {5, 6}. So could it be said as 4 communicates to 2, and 2 communicates to 3, 3 to 4, and 4 back to 2 they all communicate with each other?
 February 3rd, 2015, 03:29 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra Yes. 2 and 4 are not a communicating class because from 4 you can get to 3 and back again. (The same goes for 2). Thanks from interestedinmaths

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