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Feb 21, 2019 at 3:28 comment added Taylor Regarding the comment above, if $\pi$ is the stationary/invariant distribution, then $\pi \otimes \pi$ is the stationary distribution for the product chain. This is because $$ (\pi \otimes \pi)(T \otimes T) = (\pi T) \otimes (\pi T) = \pi \otimes \pi. $$ The first equality is just the "mixed-product" property of Kronecker products.
Oct 20, 2017 at 5:36 vote accept dineshdileep
Dec 1, 2016 at 6:26 comment added dineshdileep can you make any comment on the relation between the steady state probability vectors of both the chains?
Nov 29, 2016 at 10:39 history answered user83457 CC BY-SA 3.0