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If you have a Markov chain, such that

$x_i A = x_{i+1}$,

is it true that:

$x_i = x_{i+1} A^{-1}$

Further, in the strange case where $A$ may be singular, is it appropriate to compute the Moore-Penrose pseudo-inverse to get an approximate answer?

Thanks!

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    $\begingroup$ Sure, just multiply both sides by $A^{-1}$. $\endgroup$ – Qiaochu Yuan Aug 10 '17 at 3:07
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In an earlier question, you asked about a Markov chain for which all rows of the transition matrix are the same. Clearly, that is a singular case. So in that particular singular case, the transition matrix would be of no use in approximating the state at the previous step.

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    $\begingroup$ Hi. Yes, but would the Moore-Penrose inverse not tell you anything? $\endgroup$ – Thomas Moore Aug 10 '17 at 3:39
  • $\begingroup$ In the case where the transition matrix has all rows the same, the steps are independent, so it seems to me nothing can help with approximation. Reworded my answer for clarity. $\endgroup$ – BruceET Aug 10 '17 at 3:53
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    $\begingroup$ @ThomasMoore: not much, just how one dimension of the space changed. Not anything about how all the 0-eigenvalued ones were. $\endgroup$ – mathreadler Aug 10 '17 at 12:40
  • $\begingroup$ Analogous: I roll a fair die 10 times independently. What info about the 9th roll can be gained from the 10th? After some thought, I guess there is one less interesting kind of transition matrix from a probability point of view: the $n \times n$ identity (1's on the diagonal); a chain with no movement at all. $\endgroup$ – BruceET Aug 10 '17 at 18:59
  • $\begingroup$ Hi @BruceET thank you for your very helpful comments and answers! Very enlightening. $\endgroup$ – Thomas Moore Aug 10 '17 at 21:33

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