Two-State Switching Chains with Unequal Rates

Two Switching Rates, Four Transition Probabilities

A two-state continuous-time Markov chain can switch in either direction at its own rate. When those rates differ, the chain does not spend equal long-run time in the two states, and the transition probabilities from state $0$ need not match those from state $1$. Solving the forward equations makes both effects explicit: one exponential term controls how quickly the initial condition is forgotten, while the rates determine the long-run proportions.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(X_t)_{t\ge0}$ be a time-homogeneous continuous-time Markov chain with state space $S=\{0,1\}$, adapted to its natural filtration $\mathcal{F}_t=\sigma(X_s:0\le s\le t)$. Suppose the switching rate from $0$ to $1$ is $\alpha>0$, and the rate from $1$ to $0$ is $\beta>0$, with $\alpha\ne\beta$. Its generator is

Q= \begin{bmatrix} -\alpha & \alpha\\ \beta & -\beta \end{bmatrix}.

Each generator row sums to zero: $-\alpha+\alpha=0$ and $\beta-\beta=0$. Write $P(t)=(p_{ij}(t))_{i,j\in S}$, where $p_{ij}(t)=\mathbb{P}(X_t=j\mid X_0=i)$. The transition matrix is row-stochastic: its $i$th row gives the probabilities at time $t$ when the chain starts in state $i$. The forward Kolmogorov equation is $P'(t)=P(t)Q$, with $P(0)=I$.

Theorem: Set $r=\alpha+\beta$. The transition matrix of this chain is $$ P(t)= \begin{bmatrix} \frac{\beta}{r}+\frac{\alpha}{r}e^{-rt} & \frac{\alpha}{r}(1-e^{-rt})\\[4pt] \frac{\beta}{r}(1-e^{-rt}) & \frac{\alpha}{r}+\frac{\beta}{r}e^{-rt} \end{bmatrix}, \qquad t\ge0. $$ In particular, the probability of switching by time $t$ depends on the rate out of the starting state, and both rows converge to $(\beta/r,\alpha/r)$ as $t$ tends to infinity.

Solving the Forward Equations

The equation $P'=PQ$ evolves each row of $P(t)$. For the row starting at state $0$, let $x(t)=p_{01}(t)$. Since the row sums to one, $p_{00}(t)=1-x(t)$. The forward equation in column $1$ gives

x'(t)=\alpha p_{00}(t)-\beta p_{01}(t) =\alpha-(\alpha+\beta)x(t),\qquad x(0)=0.

Multiplying by $e^{rt}$ gives $\frac{d}{dt}(e^{rt}x(t))=\alpha e^{rt}$. Integrating from $0$ to $t$,

e^{rt}x(t)=\frac{\alpha}{r}(e^{rt}-1), \qquad p_{01}(t)=x(t)=\frac{\alpha}{r}(1-e^{-rt}).

Therefore $p_{00}(t)=1-p_{01}(t)=\frac{\beta}{r}+\frac{\alpha}{r}e^{-rt}$. For the row starting at state $1$, let $y(t)=p_{10}(t)$, so $p_{11}(t)=1-y(t)$. The forward equation in column $0$ yields

y'(t)=\beta p_{11}(t)-\alpha p_{10}(t) =\beta-(\alpha+\beta)y(t),\qquad y(0)=0.

Multiplication by $e^{rt}$ and integration give $e^{rt}y(t)=\frac{\beta}{r}(e^{rt}-1)$, and hence $p_{10}(t)=\frac{\beta}{r}(1-e^{-rt})$ and $p_{11}(t)=1-p_{10}(t)=\frac{\alpha}{r}+\frac{\beta}{r}e^{-rt}$.

Proof of the theorem. The preceding calculations solve the two scalar forward equations for the off-diagonal entries, using their initial values $p_{01}(0)=p_{10}(0)=0$. The row sums of a transition matrix are one, so the diagonal entries are the complements of those off-diagonal entries. This gives the displayed formula. To check the initial condition, set $t=0$: the entries are $p_{00}(0)=1$, $p_{01}(0)=0$, $p_{10}(0)=0$, and $p_{11}(0)=1$. To check that the formula solves the forward equation, its off-diagonal derivatives are $p'_{01}(t)=\alpha e^{-rt}$ and $p'_{10}(t)=\beta e^{-rt}$. The forward equation gives $\alpha p_{00}(t)-\beta p_{01}(t)=\alpha e^{-rt}$ and $\beta p_{11}(t)-\alpha p_{10}(t)=\beta e^{-rt}$, by substitution. The diagonal derivatives are the negatives of the corresponding off-diagonal derivatives, as required by the row sums of $Q$. Finally, the scalar equation $z'=c-rz$ with a fixed initial value has at most one solution: the difference $w$ of any two solutions satisfies $w'=-rw$, so $(e^{rt}w(t))'=0$, and $w(0)=0$ implies $w(t)=0$ for every $t$. Thus the entries obtained are the unique solution of the forward equations. $\square$

For $t\ge0$, $0<e^{-rt}\le1$, and $\alpha/r,\beta/r$ are positive and sum to one. Each displayed entry is nonnegative. The row sums can also be checked directly: the first is $\frac{\beta}{r}+\frac{\alpha}{r}e^{-rt}+\frac{\alpha}{r}(1-e^{-rt})=1$, and the second is $\frac{\beta}{r}(1-e^{-rt})+\frac{\alpha}{r}+\frac{\beta}{r}e^{-rt}=1$. Thus the formula gives a stochastic matrix at every time.

Examples and Checks

Worked Example: A Faster Switch Out of State 0

Take $\alpha=2$, $\beta=1$, and $t=\frac{\ln 2}{3}$. The generator is $Q=\begin{bmatrix}-2&2\\1&-1\end{bmatrix}$; its row sums are $-2+2=0$ and $1-1=0$. Here $r=2+1=3$ and $e^{-rt}=e^{-3(\ln 2/3)}=e^{-\ln 2}=\frac12$.

Substituting into the formula gives

P(t)= \begin{bmatrix} \frac13+\frac23\cdot\frac12 & \frac23(1-\frac12)\\[3pt] \frac13(1-\frac12) & \frac23+\frac13\cdot\frac12 \end{bmatrix} = \begin{bmatrix} \frac23&\frac13\\[2pt] \frac16&\frac56 \end{bmatrix}.

The row sums are $\frac23+\frac13=1$ and $\frac16+\frac56=1$. Starting from state $0$, the probability of being in state $1$ at this time is $\frac13$; starting from state $1$, the probability of being in state $0$ is $\frac16$. These are different because the rates out of the two starting states differ.

Check: At $t=0$, the formula gives the identity. Also, the derivative of $p_{01}(t)=\frac23(1-e^{-3t})$ at zero is $2$, exactly the rate from $0$ to $1$.

Worked Example: An Unequal Initial Distribution

Take $\alpha=1$, $\beta=3$, and $t=\frac{\ln 4}{4}$. The generator is $Q=\begin{bmatrix}-1&1\\3&-3\end{bmatrix}$; its row sums are $-1+1=0$ and $3-3=0$. Now $r=1+3=4$, and $e^{-rt}=e^{-\ln 4}=\frac14$. Substitution gives

P(t)= \begin{bmatrix} \frac34+\frac14\cdot\frac14 & \frac14(1-\frac14)\\[3pt] \frac34(1-\frac14) & \frac14+\frac34\cdot\frac14 \end{bmatrix} = \begin{bmatrix} \frac{13}{16}&\frac3{16}\\[2pt] \frac9{16}&\frac7{16} \end{bmatrix}.

The row sums are $\frac{13}{16}+\frac3{16}=1$ and $\frac9{16}+\frac7{16}=1$. If the initial distribution is $\mu_0=(\frac12,\frac12)$, its distribution at time $t$ is the row vector $\mu_0P(t)$: the probability of state $0$ is $\frac12\cdot\frac{13}{16}+\frac12\cdot\frac9{16}=\frac{11}{16}$, and the probability of state $1$ is $\frac12\cdot\frac3{16}+\frac12\cdot\frac7{16}=\frac5{16}$. These values sum to one.

Check by the scalar equation: If $z(t)=\mathbb{P}(X_t=1)$, then $z'=\alpha-rz=1-4z$, with $z(0)=\frac12$. Its solution is $z(t)=\frac14+(\frac12-\frac14)e^{-4t}$. At the specified time this is $\frac14+\frac14\cdot\frac14=\frac5{16}$, agreeing with the matrix calculation.

Worked Example: Checking the Semigroup Property

For the chain with $\alpha=2$, $\beta=1$, take $s=t=\frac{\ln 2}{3}$. As checked above, each generator row sums to zero, and the transition matrix at either time is $\begin{bmatrix}\frac23&\frac13\\\frac16&\frac56\end{bmatrix}$, whose row sums are both one. The Chapman–Kolmogorov relation requires $P(s)P(t)=P(s+t)$. Multiplying the two matrices gives

P(s)P(t)= \begin{bmatrix} \frac49+\frac1{18}&\frac29+\frac5{18}\\[2pt] \frac19+\frac5{36}&\frac1{18}+\frac{25}{36} \end{bmatrix} = \begin{bmatrix} \frac12&\frac12\\[2pt] \frac14&\frac34 \end{bmatrix}.

Since $s+t=\frac{2\ln 2}{3}$, $e^{-3(s+t)}=e^{-2\ln 2}=\frac14$. Substituting this value into the formula for $P(s+t)$ gives $p_{00}=\frac13+\frac23\cdot\frac14=\frac12$, $p_{01}=\frac23(1-\frac14)=\frac12$, $p_{10}=\frac13(1-\frac14)=\frac14$, and $p_{11}=\frac23+\frac13\cdot\frac14=\frac34$. Thus the product agrees entry by entry; its row sums are $\frac12+\frac12=1$ and $\frac14+\frac34=1$.

Stationary Proportions and Long-Time Behavior

Proposition: For $\alpha,\beta>0$, the row vector $\pi=(\frac{\beta}{\alpha+\beta},\frac{\alpha}{\alpha+\beta})$ is stationary, meaning $\pi P(t)=\pi$ for every $t\ge0$. For every initial distribution $\mu_0$, $\mu_0P(t)$ converges to $\pi$ as $t$ tends to infinity.

Proof. Write $r=\alpha+\beta$ and $u=e^{-rt}$. The vector $\pi$ has nonnegative entries summing to $\frac{\beta}{r}+\frac{\alpha}{r}=1$. Its first coordinate after multiplication by $P(t)$ is $\frac{\beta}{r}(\frac{\beta}{r}+\frac{\alpha}{r}u)+\frac{\alpha}{r}\frac{\beta}{r}(1-u)=\frac{\beta}{r}$. Its second coordinate is $\frac{\beta}{r}\frac{\alpha}{r}(1-u)+\frac{\alpha}{r}(\frac{\alpha}{r}+\frac{\beta}{r}u)=\frac{\alpha}{r}$. Hence $\pi P(t)=\pi$. For an arbitrary initial distribution $\mu_0=(m_0,m_1)$, the state-1 probability at time $t$ is $m_0\frac{\alpha}{r}(1-u)+m_1(\frac{\alpha}{r}+\frac{\beta}{r}u)=\frac{\alpha}{r}+(m_1-\frac{\alpha}{r})u$. Because $r>0$, $u=e^{-rt}$ tends to zero as $t$ tends to infinity. The state-1 probability therefore tends to $\frac{\alpha}{r}$, and the state-0 probability, its complement, tends to $\frac{\beta}{r}$. This proves the asserted convergence. $\square$

The positive-rate assumption matters for this conclusion. If both rates are zero, then $Q$ is the zero matrix and $P(t)=I$ at every time: starting in state $0$ stays at $0$, while starting in state $1$ stays at $1$. There is no single limiting distribution shared by all initial distributions. In the positive-rate case, the stationary proportions also have a direct interpretation: the chain spends a fraction $\beta/r$ of its long-run time in state $0$ and $\alpha/r$ in state $1$. The state with the smaller rate of switching out receives the larger stationary proportion.

Common mistake: reversing the switching rates. In the first example, assigning the rate $\beta=1$ to the probability of moving from $0$ to $1$ would give the incorrect expression $\frac{\beta}{r}(1-e^{-rt})$. At $t=\frac{\ln2}{3}$, this wrongly gives $\frac13(1-\frac12)=\frac16$, instead of the correct $\frac23(1-\frac12)=\frac13$. The correction follows from the forward equation at time zero: $p'_{01}(0)=\alpha p_{00}(0)-\beta p_{01}(0)=\alpha=2$. The probability of a $0$-to-$1$ transition initially increases at rate $\alpha$, not $\beta$.

Check Your Understanding

Use the forward equations and the transition formula to answer these questions; check probabilities by confirming that each row sums to one.

For $\alpha=3$ and $\beta=1$, write the generator and compute $P(t)$ at $t=\frac{\ln2}{4}$.
For general positive $\alpha\ne\beta$, what are $p'_{01}(0)$ and $p'_{10}(0)$, and what do they represent?
If the initial distribution is $(m_0,m_1)$, derive the probability of state $1$ at time $t$ from $\mu_0P(t)$.
Verify that $(\frac{\beta}{\alpha+\beta},\frac{\alpha}{\alpha+\beta})$ is a probability vector and explain why it remains unchanged under $P(t)$.
Why does setting both switching rates to zero prevent convergence to one common limiting distribution for every initial state?

Two-State Switching Chains with Unequal Rates

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