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Given a transition matrix, is the biggest component of the stationary distribution the one that correspond to the column whose sum of entries is the biggest among all columns?

(By "correspond" I mean $ \text{The } i\text{th component of the vector} \leftrightarrow \text{the }i\text{th column of the matrix}$)

It seems plausible since in the transition matrix the column with the biggest sum of the entries is the most "absorbing" state. My probability teacher told me he often asks his colleagues about this fact, never hearing it is a known fact.

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2 Answers 2

up vote 2 down vote accepted

Here's another counterexample that may throw some light on why this isn't the case:

$$P=\pmatrix{1-2\epsilon&\epsilon&\epsilon&0\\0&0&0&1\\0&0&0&1\\1&0&0&0},\quad\pi\approx\pmatrix{1-4\epsilon,\epsilon,\epsilon,2\epsilon}\;.$$

The last column has sum $2$, the first one $2-2\epsilon$, but the stationary distribution is concentrated almost entirely in the first state. This shows that it's not just the probabilities of transitions to a state that determine its stationary probability, but also the probabilities of transitions away from that state, and the stationary probabilities of the states that lead to it. In the example, though the first and last state have very similar probabilities for transitions to them, the first state has a much lower probability for a transition away from it.

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Thanks, now it's very intuitive why is not the case that such a thing had to hold. –  Emanuele Natale Oct 3 '11 at 21:01

Here is a counterexample:

$$P=\pmatrix{3/4&1/4&0\cr1/8&2/3&5/24\cr0&1/6&5/6},\quad \pi=\left({2\over11},{4\over11},{5\over11}\right).$$

The second column has the largest sum, but the second state has less probability than the third.

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Yes, $\pi$ is the unique stationary distribution. –  Byron Schmuland Oct 3 '11 at 20:45
1  
This is an irreducible, aperiodic Markov chain. –  Byron Schmuland Oct 3 '11 at 20:46
    
yep, I relized that and deleted my comment before the page was updated with your answer, thanks. –  Emanuele Natale Oct 3 '11 at 20:48

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