I When are Markov Matrices also Martingales?

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Are there known conditions under which a Markov Chain is also a Martingale? I know only that the only Random Walk that is a Martingale is the symmetric one, i.e., p= 1-p =1/2.
 
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If you have an nxn Markov matrix the first thing you need to decide is what does expected value mean? You have n discrete observations, are they the numbers 1 through n or something else? If your Markov chain is a random walk on a graph of n vertices the question doesn't even obviously make sense
 
WWGD said:
TL;DR Summary: I know only that the only Random Walk that is a Martingale is the symmetric one, i.e., p= 1-p =1/2.
Risk neutral probabilities are martingales but <> 0.5

https://en.wikipedia.org/wiki/Risk-neutral_measure

Can arbitrarily create a martingale with any binomial process
 
Thank you both. Im thinking about income decile transition matrices. If the matrix M tends to stabilize at abouth the kth iteration, I assume the dominant eigenvalues of M would be close to the kth root of unity, right?
 
Related question: Let M be an ## n \times n ## constant matrix, with constant value ##c## and let ##k ## a positive Integer. Is this formular correct : ##M^{k}=n^{k-1} c^{k} J ##, where ##J## is the constant ## n \times n ## matrix with ##c ==1##?
 
WWGD said:
Related question: Let M be an ## n \times n ## constant matrix, with constant value ##c## and let ##k ## a positive Integer. Is this formular correct : ##M^{k}=n^{k-1} c^{k} J ##, where ##J## is the constant ## n \times n ## matrix with ##c ==1##?
The formula is correct.
 
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