# What is “symmetric property” for stationary distribution

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?

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 $D_2$.)
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.