As I was self studying, I came accross this exercise that made me think a lot about what a steady-state vector is. Wouldn't the probabilities change at each time increment with the following transition matrix?
$$P=\begin{bmatrix} 0 & 1/2 & 1/2 \\ 1/3 & 2/3 & 0 \\ 1 & 0 & 0 \\ \end{bmatrix} \ $$
Does anyone understand the question being asked below? Do we just sit down and observe the process for like infinity, hence just computing the steady state vector?
There are three spotlights flickering in a parking lot. Observations of the lights are made in regular increments. Given that one of the lights is on, there is a 50–50 chance either two or three lights will be on in the next time increment. Given two lights are on, there is a 2 in 3 chance two lights will remain on in the next time increment, otherwise one light will be on. If there are three lights on, in the next time increment there is always one light on. Find the probability that one, two or three lights are on in any given time increment.
I think that's a reasonable interpretation. Alternatively, maybe you could express your answer in terms of some initial probability vector $v$. In which case, the probabilities would be $vP^n$, for the $n$-th time increment.