Generate state matrices by recursion

State matrices by recursion

A state matrix is a column listing the count in each state, and each new state is found by multiplying the transition matrix on the left. The initial state S₀ is before any applications, while a direct jump uses a power of the transition matrix

1. Use a state column List the state counts vertically in the declared state order \(\displaystyle S_n=\begin{bmatrix}u_n\\v_n\end{bmatrix}\)
2. Multiply on the left The transition matrix acts on the current state column from the left \(\displaystyle S_{n+1}=TS_n\)
3. Count applications S₀ is before any application, so the subscript n counts applications of the transition matrix \(\displaystyle S_0\text{ is initial},\quad S_2=T^2S_0\)
4. Use a power for a jump To reach a distant index directly, raise T to the index and apply the power to S₀ \(\displaystyle S_n=T^nS_0\)

Worked example

Use the transition matrix to calculate S₁, S₂ and S₃ from the initial state

1. \(\displaystyle S_0=\begin{bmatrix}100\\50\end{bmatrix}\)The initial state is the column before any transition-matrix application
2. \(\displaystyle S_1=TS_0=\begin{bmatrix}0.7&0.2\\0.3&0.8\end{bmatrix}\begin{bmatrix}100\\50\end{bmatrix}=\begin{bmatrix}70+10\\30+40\end{bmatrix}=\begin{bmatrix}80\\70\end{bmatrix}\)Multiply T by the initial column to obtain the state after one application
3. \(\displaystyle S_2=TS_1=\begin{bmatrix}0.7&0.2\\0.3&0.8\end{bmatrix}\begin{bmatrix}80\\70\end{bmatrix}=\begin{bmatrix}56+14\\24+56\end{bmatrix}=\begin{bmatrix}70\\80\end{bmatrix}\)Use the state at index 1 as the column for the second application
4. \(\displaystyle S_3=TS_2=\begin{bmatrix}0.7&0.2\\0.3&0.8\end{bmatrix}\begin{bmatrix}70\\80\end{bmatrix}=\begin{bmatrix}49+16\\21+64\end{bmatrix}=\begin{bmatrix}65\\85\end{bmatrix}\)A third left multiplication gives the state at index 3.

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