Steady states and regularity
In a steady state, the counts do not change from one step to the next, although transfers can continue. With a regular transition matrix, the state approaches its unique steady state for a fixed total. For a two-state model, balance the two off-diagonal flows and conserve the total
- Balance the flows At a steady state, the total leaving each state equals the total entering it. In a two-state model, this balances the two flows \(\displaystyle aL_1=bL_2\)
- Use the off-diagonal proportions For a regular two-state model, if a is the proportion leaving state 1 and b is the proportion leaving state 2, the long-run ratio is b to a. \(\displaystyle a=0.3,\quad b=0.6\Rightarrow L_1:L_2=0.6:0.3=2:1\)
- Conserve the total The entries of every state matrix have the same total as the initial state \(\displaystyle L_1+L_2=S_{0,1}+S_{0,2}\)
- Allow movement to continue A steady state means the totals stay fixed, not that no one moves between states \(\displaystyle 0.3 \times 40=0.6 \times 20=12\)
- Find a strictly positive power A transition matrix is regular if some positive integer power has every entry strictly greater than zero. If T itself has all positive entries, the power 1 is sufficient \(\displaystyle (T^k)_{ij}>0\quad\text{for all }i,j\text{ and some integer }k\ge 1\)
- Do not reject a zero entry too early A zero entry in T does not rule out regularity. Several transitions may provide a route that is unavailable in a single step
- Distinguish a steady state from settling A steady state satisfies TL=L. With a regular transition matrix, the state approaches its unique steady state for a fixed total, regardless of the starting split. A non-regular matrix need not settle from every starting split \(\displaystyle TL=L\)
Worked example
The columns of T describe the current state and the rows describe the next state. The total is 120. Find the long-run state and the number moving each way
- \(\displaystyle a=T_{21}=0.25,\quad b=T_{12}=0.5\)The off-diagonal entries give the proportions leaving state 1 and state 2.
- \(\displaystyle L_1:L_2=b:a=0.5:0.25=2:1\)The long-run counts have the ratio of the second off-diagonal proportion to the first
- \(\displaystyle L_1+L_2=120,\quad L_1=80,\quad L_2=40\)The ratio has three parts, so each part is 120 divided by 3, or 40.
- \(\displaystyle 0.25 \times 80=20,\quad 0.5 \times 40=20\)The equal flows confirm the long-run state while movement continues
Common mistakes
Difficulty levels
Questions about this topic
What does steady state mean in a transition model?+
The state counts remain unchanged from one application of the transition matrix to the next. People or items can still move between states when the flows balance
Why is the long-run ratio b to a?+
In a regular two-state model, if a leaves state 1 and b leaves state 2, balance requires a times the state 1 count to equal b times the state 2 count. Rearranging gives the ratio b to a. With more states, balance the total inflow and outflow for each state; individual pairs need not balance
Why does the total stay constant?+
Every column of the transition matrix totals 1, so all of the current population is distributed among the states at the next step. No count is created or lost
Does a zero entry mean a transition matrix is not regular?+
No. Calculate powers: if some positive integer power has every entry strictly positive, the matrix is regular. Finding zeros in only a few powers is not a proof that every power has zeros
Does finding a steady state prove that every starting split settles?+
No. A matrix that swaps two states has a steady equal split but makes unequal splits alternate. For a regular transition matrix, the same fixed-total steady state is approached from every starting split
Can I use a calculator?+
A calculator is allowed unless a question is marked no calculator
Which year level and chapter is this?+
Year 11, VCE Units 1 & 2, General Mathematics, chapter 4: Matrices