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Suppose I have a left-stochastic matrix $\textbf{M}$, an initial probability vector $\textbf{π}$, and a desired probability state, b, which is the ith component of a probability vector $\textbf{b}$. (In other words, I don't care about all of the components of $\textbf{b}$, just one of them.)

I want to find the real power(s) of $\textbf{M}$ that gives me the $\textbf{b}$:

$\textbf{M}^{x}\textbf{π} = \textbf{b} \mid b_{i} = c$

I want to solve for x.

In general, how many solutions could there be? Is there an effective procedure for finding out (1) if a solution exists, (2) the actual solutions?

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Ok, to clarify my question, here's a simple example.

Is there a real number x that makes this equation true?

In general, will there always be such an x?

How can I find that x if it exists?

$\begin{bmatrix} .4 & .6 \\ .6 & .4 \end{bmatrix}^{x}\begin{bmatrix}1 \\ 0\end{bmatrix}=\begin{bmatrix}.5 \\ .5 \end{bmatrix}$

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    I don't think I understand the question. Are you just requiring that, for some fixed $i_0,\pi,b,M$ you have $(M^x \pi)_{i_0} = b_{i_0}$? – 2017-02-12
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    Now I understand the question. In this case you're probably going to want to try to diagonalize $M$, at which point you just have a system of coupled exponential equations to solve (which are still transcendental, but tractable). For example, your equation can be written as $\begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} (-0.2)^x & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0.5 \\ 0.5 \end{bmatrix}$. Now simplify the left side... – 2017-02-12
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    ...and you get $\begin{bmatrix} (-0.2)^x/2 + 0.5 \\ -(-0.2)^x/2 + 0.5 \end{bmatrix} = \begin{bmatrix} 0.5 \\ 0.5 \end{bmatrix}$, assuming I haven't made an error. Thus there won't be a solution, because your RHS is actually the stationary distribution, which is not reached in finite time unless it is also the initial distribution. – 2017-02-12

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