I have the one step transition matrix

$$\pmatrix{0 & \alpha & 0 & \beta \\ \alpha & 0 & \beta & 0 \\ 0 & \beta & 0 & \alpha \\ \beta & 0 & \alpha & 0 \\}$$

I want to work out the stationary distribuion. So I end up with

$$ \pi_1 = \alpha \pi_2 + \beta \pi_4$$ $$ \pi_2 = \alpha \pi_1 + \beta \pi_3$$ $$ \pi_3 = \beta \pi_2 + \alpha \pi_4$$ $$ \pi_4 = \beta \pi_1 + \alpha \pi_3$$

and it says that by noticing the symmetric property of all four states, we can deduce that all the $\pi$'s are equal to each other.

What is the symmetric property? Is it because we can see they are all in a similar form to $A = \alpha A + \beta B$ and $B = \alpha B + \beta A$ and so we get that they're symmetric?


Every node is equivalent to every other node.

To see this, draw the graph of this Markov chain:


Each of these four figures is a graph of the chain--all are equally good representations of it. The nodes are consistently represented by color and the transitions by edge style: the solid edges are, say, the $\alpha$ transitions and the dashed edges are therefore the $\beta$ transitions. (The graphs are related by graph automorphisms: the second and fourth are obtained by the geometric equivalent of a mirror image of the first and third, respectively, while the third is obtained from the first as a horizontal mirror image. The two reflections generate a group isomorphic to the Dihedral group $D_2$.)

Because the graph can be drawn in these four ways, all of which are identical except for the node coloring, it is apparent that the blue node is the same as the node in the upper left, upper right, lower left, and lower right corners of the first graph. That shows all nodes are equivalent. (Intuitively, each node "sees" the same environment within the graph. If the nodes were not labeled and you were standing on the graph at one of them, you could not tell which of them it happened to be.)

Assuming neither of $\alpha$ nor $\beta$ is zero, all nodes will be connected, whence there is a stationary state. Because all nodes are equivalent, they must have the same weights $\pi_i$ in this stationary state. (If they did not have the same weights, making additional transitions would result in new distributions that are weighted averages of the original distribution. The extreme--maximum and minimum--values among the $\pi_i$ would thereby be altered unless they already were equal.)

Of course, once you have seen this, it's simple to check by plugging in $1/4=\pi_i$ to the four simultaneous linear equations and noting they are satisfied. Note that if $\alpha=0$ or $\beta=0$ the conclusion is false (there are stationary states in which some of the $\pi_i$ differ and in fact there are initial states that reach no limiting state). It's worthwhile reviewing the argument to see at which points it breaks down in such cases.


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