Why a steady state question never gives you the starting distribution, and how to find the answer without one.
Bangkok commuters switch between the BTS and driving each week. Of BTS users, 80% stay with the BTS; of drivers, 30% switch to the BTS. Three districts start completely differently: all BTS, all driving, and half and half. Press play and watch the three lines.
All three districts have the same matrix. Only their starting week differs.
Write the state as a column, BTS on top. Each column of T holds the destinations for one starting state, so every column sums to 1.
| from BTS | from driving | |
|---|---|---|
| to BTS | 0.8 | 0.3 |
| to driving | 0.2 | 0.7 |
sn+1 = T sn, and so sn = Tn s0. To get the state in week 10, raise the matrix to the tenth power on your calculator and multiply, rather than stepping ten times by hand.
Check which way round your matrix is. Written as columns-from, the state is a column and T goes on the left. Written as rows-from, the state is a row and it goes on the left instead: sn+1 = sn T. Both appear in textbooks. The test that settles it in two seconds is which direction sums to 1.
The steady state is the vector the chain stops changing: Ts = s. Notice what is absent from that equation. There is no s0 in it, which is exactly why the three districts above agree.
Ts = s alone is not enough. The two rows give the same information, so they cannot pin down two unknowns. You must bring in a + b = 1. Candidates who omit it get a direction and no answer, and a scaled multiple of (0.6, 0.4) is not a probability distribution.
An absorbing state. If a state keeps all of its own probability, like a machine that once broken stays broken, then everything drains into it and the long-run answer is that state with certainty.
Strict alternation. A matrix that swaps the two states every step cycles for ever and never settles, so there is nothing for the powers of T to converge to. Convergence needs a chain that can reach every state and does not march in lockstep, which every examination question you meet will satisfy.
1. In this matrix, what proportion of drivers switch to the BTS each week? Give a decimal.
2. A district starts all driving, so s0 = (0, 1). Find the proportion on the BTS after one week.
3. In the long run, what proportion uses the BTS? Give a decimal.
4. A question asks only for the steady state and gives no starting distribution. This is because:
Setting up Ts = s and writing a + b = 1. The second equation is a mark in its own right and the answer is impossible without it.
Multiplying in the right order. T s0 and s0 T are different matrices, and only one of them matches how you wrote T.
Answering in context. “0.6” on its own is thin; “in the long run 60% of commuters use the BTS” is the sentence being marked.
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