When are Markov Matrices also Martingales?

[WWGD]

TL;DR

Two con

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.

 

[Office_Shredder]

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

 

[BWV]

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

 

[WWGD]

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?

 

[WWGD]

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 $c ==1$?

 

[Gavran]

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.