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- First identify which dimension of the matrix represents the current state. - Use the row for the current state \(B\). - Then use the column for the next state \(A\).

Use the entry in row \(B\), column \(A\). That entry is \(0.35\).

\(0.35\)

- Recall the two requirements for a row-stochastic matrix. - Check the sign of every entry. - Check each row sum separately.

Every entry is nonnegative. The first row sums to \(0.60+0.40=1\), and the second row sums to \(0.25+0.75=1\). Therefore, \(P\) is row-stochastic.

Yes. \(P\) is row-stochastic.

- Look for a row with a \(1\) in its own-state column and \(0\) elsewhere. - That row represents a state that cannot be left.

1. A state is absorbing if, once entered, the process stays there with probability \(1\). 2. The first row is \((1,0)\), so a process in state \(A\) remains in \(A\) with probability \(1\). 3. Therefore, state \(A\) is absorbing.

State \(A\).

- Under the row-stochastic convention, multiply the row state vector on the left by the transition matrix. - The resulting probabilities should add to \(1\).

1. For the row-stochastic convention, compute \(\mathbf{p}_1=\mathbf{p}_0T\). 2. Thus, \(\mathbf{p}_1=\begin{pmatrix}0.75&0.25\end{pmatrix}\begin{pmatrix}0.80&0.20\\0.40&0.60\end{pmatrix}=\begin{pmatrix}0.70&0.30\end{pmatrix}\).

\(\mathbf{p}_1=\begin{pmatrix}0.70&0.30\end{pmatrix}\).

- 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.

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.

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

- Square the one-step transition matrix to obtain two-step probabilities. - Use the sunny row and rainy column for the requested entry. - The matrix product adds the probabilities of the two possible intermediate-state paths.

1. The two-step transition matrix is \(T^2\). 2. Its sunny-to-rainy entry is \(0.70\cdot0.30+0.30\cdot0.60=0.21+0.18=0.39\). 3. Therefore the probability of rain two days from now, given sunshine today, is \(0.39\).

The sunny-to-rainy entry of \(T^2\) is \(0.39\), so the probability is \(0.39\).

- Use the defining row-sum property of a row-stochastic matrix. - Only one row contains the unknown entry. - Set that row's total equal to \(1\).

The entries in the third row must sum to \(1\): \(0.25+x+0.50=1\). Therefore, \(x=0.25\).

\(x=0.25\)

- Match the state-vector orientation to the row-stochastic convention. - Each next-state total combines contributions from both current states. - Check that the total number of customers is preserved.

Use the current row vector \(\begin{pmatrix}60&40\end{pmatrix}\): \(\begin{pmatrix}60&40\end{pmatrix}T =\begin{pmatrix}60\cdot0.70+40\cdot0.20&60\cdot0.30+40\cdot0.80\end{pmatrix} =\begin{pmatrix}50&50\end{pmatrix}\).

\(50\) customers in plan \(A\) and \(50\) customers in plan \(B\)

- Fix the stated state order before placing any probabilities. - Build one row at a time from the transitions out of that current state. - Check each completed row against the row-stochastic condition.

Each row records the probabilities from one current state to the three next states. Therefore, \(T=\begin{pmatrix}0.70&0.20&0.10\\0.15&0.75&0.10\\0.30&0&0.70\end{pmatrix}\). Each row sums to \(1\), as required.

\(\begin{pmatrix}0.70&0.20&0.10\\0.15&0.75&0.10\\0.30&0&0.70\end{pmatrix}\)

- Match the state-vector orientation to the stated transition convention. - In the row-stochastic form, multiply the row state vector on the left. - In the column-stochastic form, multiply the column state vector on the right of the matrix.

1. With the row-stochastic convention, \(\begin{pmatrix}70&30\end{pmatrix}R=\begin{pmatrix}68&32\end{pmatrix}\). 2. Since \(C=R^T=\begin{pmatrix}0.8&0.4\\0.2&0.6\end{pmatrix}\), the column-stochastic update is \(C\begin{pmatrix}70\\30\end{pmatrix}=\begin{pmatrix}68\\32\end{pmatrix}\). 3. Transposing the transition matrix and switching the state vector from row orientation to column orientation describes the same transition probabilities, so both conventions produce the same physical counts.

a) \(\begin{pmatrix}68&32\end{pmatrix}\) b) \(\begin{pmatrix}68\\32\end{pmatrix}\) c) The conventions are transposes of one another and use matching state-vector orientations, so they represent the same process.

- 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.

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: \(\mathbf{v}_2=M\mathbf{v}_1\) \(=\begin{pmatrix}0.8\cdot8800+0.1\cdot5800+0.05\cdot5400\\0.1\cdot8800+0.7\cdot5800+0.15\cdot5400\\0.1\cdot8800+0.2\cdot5800+0.8\cdot5400\end{pmatrix}\) \(=\begin{pmatrix}7890\\5750\\6360\end{pmatrix}\).

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

- A stationary distribution is unchanged by one transition. - Use the condition that the two probabilities sum to \(1\). - Substitute one probability in terms of the other into one component of the stationary equation.

From the first component of the stationary equation, \(p=0.80p+0.30q\). Since \(q=1-p\), \(p=0.80p+0.30(1-p)=0.50p+0.30\). Thus, \(0.50p=0.30\), so \(p=0.60\) and \(q=0.40\).

\(\begin{pmatrix}0.60&0.40\end{pmatrix}\)

- A stationary distribution is unchanged when the transition matrix is applied. - Use one state probability together with the fact that the two probabilities sum to \(1\). - After finding the probabilities, convert them to counts using the total membership.

1. A stationary distribution is unchanged by one transition, so \(\begin{pmatrix}p&1-p\end{pmatrix}T=\begin{pmatrix}p&1-p\end{pmatrix}\). 2. Using the active-state component gives \(0.85p+0.10(1-p)=p\). 3. Simplifying gives \(0.75p+0.10=p\), so \(0.25p=0.10\) and \(p=0.40\). 4. Therefore, the stationary distribution is \(\begin{pmatrix}0.40&0.60\end{pmatrix}\). 5. With \(5000\) members, the corresponding long-run counts are \(0.40\cdot5000=2000\) active and \(0.60\cdot5000=3000\) inactive.

a) \(\begin{pmatrix}0.40&0.60\end{pmatrix}\) b) About \(2000\) active members and \(3000\) inactive members in the long run.

- Compute the first states by row-vector multiplication. - Subtract the two component update equations to obtain a recurrence for their difference. - Combine the limiting difference with the fact that the two probabilities always sum to \(1\). - A stationary state is unchanged by multiplication by \(T\).

1. \(\mathbf p_1=(0.75,0.25)\), \(\mathbf p_2=(0.625,0.375)\), and \(\mathbf p_3=(0.5625,0.4375)\). 2. If \(\mathbf p_n=(a_n,b_n)\), then \(d_n=a_n-b_n\). The update gives \(d_{n+1}=(0.75a_n+0.25b_n)-(0.25a_n+0.75b_n)=0.5(a_n-b_n)=0.5d_n\). 3. Since \(d_0=1\), \(d_n=(0.5)^n\to0\). Also \(a_n+b_n=1\), so \(a_n=\frac{1+d_n}{2}\to\frac12\) and \(b_n=\frac{1-d_n}{2}\to\frac12\). 4. Finally, \((0.5,0.5)T=(0.5,0.5)\), so the limiting state is stationary.

a) \(\mathbf p_1=(0.75,0.25)\), \(\mathbf p_2=(0.625,0.375)\), \(\mathbf p_3=(0.5625,0.4375)\) b) \(d_{n+1}=0.5d_n\), so \(d_n=(0.5)^n\) c) \(\mathbf p_n\to(0.5,0.5)\) d) \((0.5,0.5)T=(0.5,0.5)\)

- 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.

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 \(0\le z\le0.5\).

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 \(0\le z\le0.5\).

- Apply the transition matrix repeatedly to the current row distribution. - An absorbing state has probability \(1\) of staying in itself once reached. - Think about what repeated factors such as \(0.6^n\) and \(0.7^n\) do as the number of steps grows.

1. \(\mathbf{p}_1=\mathbf{p}_0T=\begin{pmatrix}0.6&0.3&0.1\end{pmatrix}\). 2. \(\mathbf{p}_2=\mathbf{p}_1T=\begin{pmatrix}0.36&0.39&0.25\end{pmatrix}\). 3. \(\mathbf{p}_3=\mathbf{p}_2T=\begin{pmatrix}0.216&0.381&0.403\end{pmatrix}\). 4. State \(C\) is absorbing because a process in \(C\) stays in \(C\) with probability \(1\). 5. The probability of remaining in \(A\) for \(n\) consecutive steps is \(0.6^n\), which approaches \(0\). Once the process reaches \(B\), the probability of remaining there for another \(m\) consecutive steps is \(0.7^m\), which also approaches \(0\). 6. There is no transition out of \(C\). Thus probability continually leaves the nonabsorbing states and accumulates in \(C\), so the long-run distribution is \(\begin{pmatrix}0&0&1\end{pmatrix}\).

a) \(\mathbf{p}_1=\begin{pmatrix}0.6&0.3&0.1\end{pmatrix}\), \(\mathbf{p}_2=\begin{pmatrix}0.36&0.39&0.25\end{pmatrix}\), \(\mathbf{p}_3=\begin{pmatrix}0.216&0.381&0.403\end{pmatrix}\) b) State \(C\) c) Probability eventually leaves states \(A\) and \(B\), while state \(C\) cannot be left, so the distribution approaches \(\begin{pmatrix}0&0&1\end{pmatrix}\).