Aimathic – Transition matrices – Free Math Worksheets

Transition matrices

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Difficulty: 1 – 5

53071312

A dynamic system has states \(S_1, S_2, S_3, S_4\). The entry \(m_{ij}\) of its transition matrix \(M\) is the proportion moving from state \(S_j\) to state \(S_i\). Construct the \(4\times4\) matrix using these entries: \(m_{11}=0.75\), \(m_{21}=0.15\), \(m_{41}=0.10\) \(m_{12}=0.20\), \(m_{22}=0.80\) \(m_{33}=0.60\), \(m_{43}=0.40\) \(m_{14}=0.05\), \(m_{34}=0.05\), \(m_{44}=0.90\) All entries not listed are \(0\).

Hints

- Use the first index for the row and the second index for the column. - Fill unspecified entries with zero. - Check each column sum after constructing the matrix.

Solution

1. There are four states, so \(M\) has four rows and four columns. 2. The first index determines the row, and the second determines the column. 3. Place the given entries in their indexed positions and fill the remaining positions with zeros: \(M=\begin{pmatrix}0.75&0.20&0&0.05\\0.15&0.80&0&0\\0&0&0.60&0.05\\0.10&0&0.40&0.90\end{pmatrix}\). 4. Each column sums to \(1\), which is consistent with the stated transition convention.

Answer

\(M=\begin{pmatrix}0.75&0.20&0&0.05\\0.15&0.80&0&0\\0&0&0.60&0.05\\0.10&0&0.40&0.90\end{pmatrix}\)

53071912

Three competing car-sharing companies, \(A\), \(B\), and \(C\), share a city market. The entry \(m_{ij}\) gives the proportion of customers who move from company \(j\) to company \(i\), including customers who stay with the same company. The weekly transition matrix is \(M=\begin{pmatrix}0.8&0.1&0.05\\0.1&0.7&0.15\\0.1&0.2&0.8\end{pmatrix}\). At week \(0\), the \(20{,}000\) customers are distributed as follows: \(10{,}000\) with \(A\), \(6000\) with \(B\), and \(4000\) with \(C\). 1) Find the customer distribution after one week. 2) Find the customer distribution after two weeks.

Hints

- Write the current customer counts as a column state vector. - Multiply the transition matrix by the current state vector. - Use the week-1 result as the input for week 2. - Check that the total number of customers remains constant.

Solution

1. The initial state vector is \(\mathbf{v}_0=\begin{pmatrix}10{,}000\\6000\\4000\end{pmatrix}\). After one week, \(\mathbf{v}_1=M\mathbf{v}_0\) \(=\begin{pmatrix}0.8\cdot10{,}000+0.1\cdot6000+0.05\cdot4000\\0.1\cdot10{,}000+0.7\cdot6000+0.15\cdot4000\\0.1\cdot10{,}000+0.2\cdot6000+0.8\cdot4000\end{pmatrix}\) \(=\begin{pmatrix}8800\\5800\\5400\end{pmatrix}\). 2. Apply the transition matrix again: \(\mathbf{v}_2=M\mathbf{v}_1\) \(=\begin{pmatrix}0.8\cdot8800+0.1\cdot5800+0.05\cdot5400\\0.1\cdot8800+0.7\cdot5800+0.15\cdot5400\\0.1\cdot8800+0.2\cdot5800+0.8\cdot5400\end{pmatrix}\) \(=\begin{pmatrix}7890\\5750\\6360\end{pmatrix}\).

Answer

1) After one week: \(A: 8800\), \(B: 5800\), \(C: 5400\). 2) After two weeks: \(A: 7890\), \(B: 5750\), \(C: 6360\).

53075412

A row-stochastic transition matrix has nonnegative entries and each row sum equals \(1\). Let \(P=\begin{pmatrix}0.6&0.4\\0.2&0.8\end{pmatrix}\). a) Find all matrices \(Q=\begin{pmatrix}x&y\\z&w\end{pmatrix}\) that commute with \(P\), so that \(PQ=QP\). b) Among the matrices from part a, find all matrices \(Q\) that are also row-stochastic transition matrices.

Hints

- Multiply \(P\) and \(Q\) in both orders. - Equate corresponding entries to find relationships among the variables. - For part b), impose the row-sum condition. - Use nonnegativity to restrict the parameter interval.

Solution

1. Compute \(PQ=\begin{pmatrix}0.6x+0.4z&0.6y+0.4w\\0.2x+0.8z&0.2y+0.8w\end{pmatrix}\) and \(QP=\begin{pmatrix}0.6x+0.2y&0.4x+0.8y\\0.6z+0.2w&0.4z+0.8w\end{pmatrix}\). 2. Equating corresponding entries gives \(y=2z\) and \(w=x+z\). The remaining equations produce the same conditions. Thus, \(Q=\begin{pmatrix}x&2z\\z&x+z\end{pmatrix}\), where \(x, z\in\mathbb{R}\). 3. For \(Q\) to be row-stochastic, its first row must satisfy \(x+2z=1\), so \(x=1-2z\). The second row then also sums to \(1\). 4. Nonnegativity requires \(z\ge0\), \(1-2z\ge0\), and \(1-z\ge0\). Therefore, \(0\le z\le0.5\). Hence, \(Q=\begin{pmatrix}1-2z&2z\\z&1-z\end{pmatrix}\), where \(0\le z\le0.5\).

Answer

a) \(Q=\begin{pmatrix}x&2z\\z&x+z\end{pmatrix}\), where \(x, z\in\mathbb{R}\). b) \(Q=\begin{pmatrix}1-2z&2z\\z&1-z\end{pmatrix}\), where \(0\le z\le0.5\).

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