Two States, Two Switching Probabilities
A transition matrix for two states contains only two independent choices: the probability of switching away from each state. Naming those probabilities directly makes the matrix easier to construct and provides a compact way to compute its eigenvalues. The parameters also have a concrete interpretation: each describes one particular one-step transition, while the remaining probability in that row describes staying put.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, and let \((X_n)_{n\ge0}\) be a discrete-time Markov chain with state space \(S=\{0,1\}\) and transition matrix \(P=(p_{ij})\). Use the row-stochastic convention: \(p_{ij}=\mathbb{P}(X_{n+1}=j\mid X_n=i)\), and each row of \(P\) sums to \(1\). The transition probabilities do not depend on \(n\). For a chain with these transition probabilities, the first parameter is the probability of moving from state \(0\) to state \(1\); the second is the probability of moving from state \(1\) to state \(0\).
The bounds on \(a\) and \(b\) are exactly the probability constraints. The first row assigns probability \(1-a\) to staying at \(0\) and probability \(a\) to switching to \(1\); the second assigns probability \(b\) to switching to \(0\) and probability \(1-b\) to staying at \(1\). In particular, the row sums are \((1-a)+a=1\) and \(b+(1-b)=1\). The entries are all nonnegative precisely when \(a,b\in[0,1]\).
Why This Parameterization Is Complete
Proof. Suppose \(P=(p_{ij})\) is row-stochastic. Define \(a=p_{01}\) and \(b=p_{10}\). Nonnegativity and the row-sum conditions give \(0\le p_{01}\le1\) and \(0\le p_{10}\le1\). The first row sums to \(1\), so \(p_{00}=1-p_{01}=1-a\). The second row sums to \(1\), so \(p_{11}=1-p_{10}=1-b\). This gives the displayed form and also shows that \(a\) and \(b\) are uniquely determined by the off-diagonal entries. Conversely, if \(a,b\in[0,1]\), all four entries \(1-a,a,b,1-b\) are nonnegative, and the two row sums are \(1-a+a=1\) and \(b+1-b=1\). The matrix is therefore row-stochastic. \(\square\)
This proposition gives a useful modeling check. If a proposed two-state transition matrix does not have nonnegative entries and row sums equal to \(1\), it cannot describe a Markov chain under the row-stochastic convention. When it does, reading \(a\) from the top-right entry and \(b\) from the bottom-left entry immediately recovers the two switching probabilities.
The Eigenvalues in Terms of \(a\) and \(b\)
The characteristic polynomial gives the eigenvalues directly. For a two-by-two matrix, it is the determinant of \(P-\lambda I\), where \(I\) is the identity matrix and \(\lambda\) is the eigenvalue variable. The result below includes repeated eigenvalues: roots are counted with their algebraic multiplicities.
Proof. The characteristic polynomial is $$ \begin{aligned} \det(P-\lambda I) &=(1-a-\lambda)(1-b-\lambda)-ab\\ &=\lambda^2-(2-a-b)\lambda+1-a-b\\ &=(\lambda-1)\bigl(\lambda-(1-a-b)\bigr). \end{aligned} $$ The roots are therefore \(1\) and \(1-a-b\), counted with multiplicity. If \(a+b=0\), the two roots are both \(1\), so the factorization gives a repeated root. This establishes the eigenvalue statement in all cases. \(\square\)
The eigenvalue \(1\) also follows directly from the row sums: multiplying \(P\) by the column vector of ones leaves that vector unchanged. The other eigenvalue records how different the two rows are. In fact, subtracting the second row of \(P\) from the first gives $$ (1-a-b,\ a+b-1)=(1-a-b)(1,-1). $$ Thus the difference between the rows is scaled by \(1-a-b\). This is an algebraic description of the contrast between the two possible current states; it is not itself a formula for the chain's later transition probabilities.
Because \(a,b\in[0,1]\), their sum lies in \([0,2]\), so \(1-a-b\in[-1,1]\). A value close to \(1\) corresponds to small switching probabilities; a value close to \(-1\) corresponds to switching probabilities close to \(1\). If \(a+b=1\), the second eigenvalue is \(0\). These statements follow from the parameter range and eigenvalue expression, without requiring any assumption about the chain's initial distribution.
Worked Example: Unequal Switching Probabilities
Suppose \(a=\tfrac14\) and \(b=\tfrac12\). Construct the transition matrix, check its rows, identify the one-step interpretation, and calculate the eigenvalues.
From state \(0\), the switching probability is \(a=\tfrac14\), so the probability of staying at \(0\) is \(1-a=1-\tfrac14=\tfrac34\). From state \(1\), the switching probability is \(b=\tfrac12\), so the probability of staying at \(1\) is \(1-b=1-\tfrac12=\tfrac12\). Therefore $$ P=\begin{bmatrix}\tfrac34&\tfrac14\\[2pt]\tfrac12&\tfrac12\end{bmatrix}. $$ The first row sums to \(\tfrac34+\tfrac14=1\), and the second row sums to \(\tfrac12+\tfrac12=1\). All entries are nonnegative, so this is a valid transition matrix.
The eigenvalues are \(1\) and \(1-a-b\). Substituting the parameters gives $$ 1-\tfrac14-\tfrac12=1-\tfrac34=\tfrac14, $$ so the two eigenvalues are \(1\) and \(\tfrac14\). As a check using the characteristic polynomial, the trace is \(\tfrac34+\tfrac12=\tfrac54\), and the determinant is \(\tfrac34\cdot\tfrac12-\tfrac14\cdot\tfrac12=\tfrac38-\tfrac18=\tfrac14\). The two proposed eigenvalues have sum \(1+\tfrac14=\tfrac54\) and product \(1\cdot\tfrac14=\tfrac14\), matching the trace and determinant.
Worked Example: Equal Rows and a Zero Eigenvalue
Let \(a=\tfrac13\) and \(b=\tfrac23\). Find the transition matrix and eigenvalues, and explain what the rows show about the one-step transitions.
The probability of staying at \(0\) is \(1-a=1-\tfrac13=\tfrac23\), and the probability of staying at \(1\) is \(1-b=1-\tfrac23=\tfrac13\). Hence $$ P=\begin{bmatrix}\tfrac23&\tfrac13\\[2pt]\tfrac23&\tfrac13\end{bmatrix}. $$ Both row sums are \(\tfrac23+\tfrac13=1\), and every entry is nonnegative. The two rows are equal: from either starting state, the probabilities for the next state are \(\tfrac23\) for state \(0\) and \(\tfrac13\) for state \(1\).
The second eigenvalue is $$ 1-a-b=1-\tfrac13-\tfrac23=1-1=0. $$ The eigenvalues are \(1\) and \(0\). To check, the trace is \(\tfrac23+\tfrac13=1\), and the determinant is \(\tfrac23\cdot\tfrac13-\tfrac13\cdot\tfrac23=\tfrac29-\tfrac29=0\). The proposed eigenvalues have sum \(1+0=1\) and product \(1\cdot0=0\), as required.
Worked Example: Certain Switching in Both Directions
Take \(a=1\) and \(b=1\). Determine the matrix, verify that it is stochastic, and interpret its eigenvalues.
Since \(1-a=1-1=0\) and \(1-b=1-1=0\), $$ P=\begin{bmatrix}0&1\\1&0\end{bmatrix}. $$ The first row sums to \(0+1=1\), and the second row sums to \(1+0=1\); the entries are nonnegative. From state \(0\), the chain switches to state \(1\) with probability \(1\), and from state \(1\), it switches to state \(0\) with probability \(1\).
The eigenvalues are \(1\) and $$ 1-a-b=1-1-1=-1. $$ For a direct check, \(P-\lambda I=\begin{bmatrix}-\lambda&1\\1&-\lambda\end{bmatrix}\), whose determinant is \(\lambda^2-1=(\lambda-1)(\lambda+1)\). Its roots are \(1\) and \(-1\), agreeing with the parameter calculation.
Reading the Parameters Without Confusing the Rows
The first parameter always belongs to the first row's switch, and the second parameter always belongs to the second row's switch. Keeping that order fixed prevents a common error: \(a\) is not the probability of switching in both directions, and \(b\) is not the probability of staying at state \(1\). The stay probabilities are the complements \(1-a\) and \(1-b\).
Common mistake. For the matrix in the first worked example, a tempting but incorrect calculation is to use \(1-a+b\) as the second eigenvalue. Substitution gives \(1-\tfrac14+\tfrac12=\tfrac54\). This cannot be correct: the exact characteristic polynomial has roots \(1\) and \(1-a-b\), and the trace is \(\tfrac54\), so the eigenvalues must sum to \(\tfrac54\). The correct second eigenvalue is \(1-\tfrac14-\tfrac12=\tfrac14\); then \(1+\tfrac14=\tfrac54\), consistent with the trace. The error comes from adding \(b\) where the characteristic polynomial requires subtracting it.
Boundary values are legitimate, not exceptions to the parameterization. If \(a=0\), the transition from state \(0\) stays at \(0\) with probability \(1\); if \(a=1\), it switches to state \(1\) with probability \(1\). The same interpretation applies to \(b\) for transitions out of state \(1\). In the particular case \(a=b=0\), the matrix is the identity and both eigenvalues equal \(1\), as the characteristic polynomial shows. In the case \(a=b=1\), the chain switches with certainty and the eigenvalues are \(1\) and \(-1\), as computed above. The parameter formula covers both endpoints without changing its form.
These calculations are useful whenever a two-state model is specified verbally or its parameters are to be compared. The parameterization turns a description of switching into a matrix with checked row sums; the characteristic polynomial then gives its eigenvalues with only the two switching probabilities. The eigenvalue calculation is a matrix-level fact. It does not by itself supply the \(n\)-step transition probabilities, which require additional work.
Check Your Understanding
For each question, keep the row-stochastic convention and distinguish switching probabilities from staying probabilities.
- Write the two-state transition matrix when \(a=\tfrac25\) and \(b=\tfrac35\). Check both row sums and find its eigenvalues.
- A row-stochastic matrix has first row \((\tfrac34,\tfrac14)\) and second row \((\tfrac25,\tfrac35)\). Identify \(a\) and \(b\), then calculate the second eigenvalue.
- For what parameter values is the second eigenvalue \(0\)? Explain why the two rows are then equal.
- Can \(1-a-b\) be less than \(-1\) for valid switching probabilities? Justify your answer from the parameter constraints.
- For \(a=b=0\), write the matrix and state its eigenvalues, including their multiplicities.