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This is a follow up to What is the difference between the forward and backward equations in a CTMC?

Let $Q^T$ be the transpose of $Q$.

Why $\pi Q =0$ is a steady state solution to the continuous time Markov chain, accounting for the forward case, but for the backward we get $ Q^T \pi^T =0^T$? Where does the transpose come from?

For the forward we have $\pi'(t)=\pi_t Q$, and the backward equation is: $\pi'^T(t)=Q \pi^T_t$.

In both cases, letting $t \rightarrow \infty$ we get $\pi Q =0$ and $Q \pi ^T=0^T$, respectively, no? Where is my mistake?

Daniel S.
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  • the backward equation is $P'(t) = Q P(t)$ and not $\pi'^T(t)=Q \pi^T_t$. The backward equation is a function of a matrix, not of a vector! – Daniel S. Mar 02 '21 at 03:30

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The backward equation is $P'(t) = Q P_t$.

Then, in steady state, $P'(t) = Q P_t = 0$. Indeed, this equation holds! It is a tautology. Although it is not helpful to find the steady state solution, as all its lines are linearly dependent, it is still true.

The forward equation, in contrast, leads to an alternative system of equations, $\pi Q = 0$, which is helpful to find the steady state solution to the system.

Daniel S.
  • 823