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Regular markov chains, a type of markov chain with transition matrices that have only positive entries after a certain power. Regular markov chains ensure that all products or entities will eventually obtain some market share or success in the long-term. Examples and calculations using matrices related to politicians and a pet snake. Students can use this document as study notes, summaries, or schemes and mind maps to understand the concept of regular markov chains.
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© Arizona State University, Department of Mathematics and Statistics 1 of 3
There are some Markov chains with very interesting properties. The first we will look at are called regular. Regularity comes from the properties of matrices. Recall that when you multiply matrices we can have non-zero matrices create zero products, or perhaps one the original matrices even though we never used the identity or other freaky occurrences.
Regular Markov Chains have transition matrices P such that for some n , has only positive entries (no zeros
n
allowed anywhere).
Thinking of what this means, it is always possible to move from one state to another, even if the first step wouldn’t allow it.
Just to be clear about how we know where we are during a stage of transition, recall that to find the probability of passing from state I to state j we need only look at the ijth^ entry_._ In the matrix to the right, our probability in moving from s2 to s1 is zero since the 21 th entry (p 21 ) is zero. If you started in state two, you would be stuck there forever. We’ll come back to that.
Look again at the transition matrix to the right. Clearly, it could not be regular. Taking the second and then successive powers of the transition matrix, that zero value is stuck forever in the 21 th^ entry (p 21 ).
Now look at the 3 by 3 matrix to the right. Enter it into your calculator and take it to successfully higher powers. You’ll see that the p 31 entry is non-zero after just the second step. It is regular at n = 2. Even so, should you begin in the initial state of C , you have no chance of moving to A immediately. If this matrix described product satisfaction since, this could mean that if you bought C first you would never settle for less.
However, if we seek the fixed probability vector , t , we find that the matrix stabilizes at
chain. The long-term behavior is exactly that. It does not describe what will happen at the first step from a given state.
As suggested, a regular Markov chain could demonstrate a level of satisfaction over some number of steps or years. The impact of regularity is that... given a list of products, say A, B, C,... , ALL of the products will eventually obtain some market share when the model is regular.
Example: Suppose the transition matrix above related to politicians A, B, C. Politician A is a scoundrel. He has been caught with his hand in the cash box too many times. Let’s analyze what the Markov model tells us about him.
From t , we would know that eventually 20 out of 261 voters will vote for that scoundrel A every time. This is true even if we started with politician C in office. We establish this by setting the initial probability distribution
However , the gratifying part is that scoundrel A will probably never win the office two times. Surprisingly, while C is the preferred candidate when A is initially in office, the eventual long-term candidate of choice is B. We would characterize his eventual success as a “landslide.”
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Example: Suppose we had the situation where we had the floor plan shown. We’ve let our pet snake get out of its cage again! We discovered from previous mistakes of this sort that the snake moves quickly around the house. It changes rooms every ten minutes. The snake has no particular pattern to the room changes. It just picks a door and slithers through randomly. It never stays in the room. As nearly as anyone can recall, the snake was in its cage an hour ago and the cage was found in room 3.
The transition matrix assumes the snake is equally likely to exit through any door. So if the snake starts in room one, it has two ways into room 2, one way into room 3 and no way to go directly to room 4. So it looks like this:
The next step is creating the initial probability distribution. Since the snake
Finally, one hour after escape is 6 ten-minute stages. So the final result is obtained using v^ (6)^ = v (0)^ P^6.
Using my calculator, I found this result: 0,0.5999871399,0.4000128601,0. From it I decide that I should search room 2 first then room 3. I can discard the notion that the snake is in rooms 1 or 4.
Definitely not. Applying successively higher powers, the matrix stabilizes at about step 15 to the one to the right. Lot’s of zero entries. Every larger power shows this same result.
There is also some mathematical nonsense happening. The results suggest that the snake moves instantaneously from room 1 to 4! There is no connection! Hmm!
With this new information, we need to create a new initial probability distribution. Considering the
snake’s penchant for equally likely doors, I would use and find v^ (0)^ = 0.20, 0.70, 0, 0.
v^ (3)^ = v (0)^ P^3 since half an hour is three stages.
1 2 3 4 0 2 1 0 1 3 3 2 2 0 0 1 3 3 (^3 1 0 ) 2 2 (^4 1 ) 0 2 2 0
1 2 3 4 1 .6 0 0. 2 0 .6 .4 0 3 0 .6 .4 0 4 .6 0 0.