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A state vector X for a four-state Markov chain is such that the system is four times as likely to be in state 3 as in 4, is not in state 2, and is in state 1 with probability 0.2. Find the state vector X.
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The probability of all states add up to \(1\). Let \(a\) be the probability of state 4. \[ 0.2 + 0 + 4a + a =1 \]If you solve for \(a\), then you can use: \[ X = \begin{bmatrix}0.2 \\ 0 \\ 4a \\ a\end{bmatrix} \]
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hmmm
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