Symmetric Three-State Chains by Eigenvectors

Symmetry Reduces the Matrix Exponential

When a continuous-time Markov chain has the same rate for every move between distinct states, its three states play interchangeable roles. That symmetry makes the generator especially simple: one direction describes total probability, while the other two describe deviations from the uniform distribution. Diagonalizing along those directions gives the transition probabilities at every time without solving three separate differential equations.

Let $(X_t)_{t\ge0}$ be a continuous-time Markov chain on the finite state space $S=\{1,2,3\}$, defined on a probability space $(\Omega,\mathcal F,\mathbb P)$, with its natural filtration $(\mathcal F_t)_{t\ge0}$. Suppose each transition from a state to either of the other two states has rate $a$, where $a\ge0$. The generator $Q=(q_{ij})$ has off-diagonal entries $q_{ij}=a$ for $i\ne j$; its diagonal entries are chosen so each row sums to zero.

Definition: For this equal-rate three-state chain, the generator and transition semigroup are $$ Q=\begin{bmatrix}-2a&a&a\\a&-2a&a\\a&a&-2a\end{bmatrix}, \qquad P(t)=e^{tQ},\quad t\ge0. $$ The row sums of $Q$ are $-2a+a+a=0$, so $Q$ is a valid generator. If the initial distribution is the row vector $\mu$, then the distribution at time $t$ is $\mu P(t)$.

For $a>0$, the chain can move between every pair of states. The case $a=0$ is also included in the formulas, but it is a different process: no transitions occur. We will keep that degenerate case visible when discussing limits.

Eigenvectors and the Transition Formula

Let $\mathbf{1}=(1,1,1)^{\mathsf T}$, let $J$ be the $3$-by-$3$ matrix of ones, and let $I$ be the identity matrix. Then $Q=a(J-3I)$. The direction $\mathbf{1}$ represents equal coordinates; vectors whose coordinates sum to zero represent deviations from equality.

Theorem: For $a\ge0$, the generator has eigenvalue $0$ on the span of $\mathbf{1}$, and eigenvalue $-3a$ on the two-dimensional subspace of vectors whose coordinates sum to zero. Consequently, $$ P(t)=\frac13J+e^{-3at}\left(I-\frac13J\right). $$

Proof. Since $J\mathbf{1}=3\mathbf{1}$, we have $Q\mathbf{1}=a(3\mathbf{1}-3\mathbf{1})=0$. Now take any column vector $x$ whose coordinates sum to zero. Each coordinate of $Jx$ is that sum, so $Jx=0$, and therefore $Qx=a(0-3x)=-3ax$. The vector $\mathbf{1}$ and the zero-sum subspace have dimensions one and two, respectively, and their intersection contains only the zero vector. Together they span $\mathbb R^3$, so these eigenvectors account for the whole space, including when $a=0$.

Define $\Pi=\tfrac13J$. Direct multiplication gives $J^2=3J$, hence $\Pi^2=\Pi$. Also $I-\Pi$ is the projection onto the zero-sum subspace: $\Pi x$ has all coordinates equal to one third of the sum of the coordinates of $x$, and $(I-\Pi)x$ has coordinate sum zero. Thus $Q=0\Pi-3a(I-\Pi)$. On the range of $\Pi$, the matrix exponential multiplies by $e^{0t}=1$; on the range of $I-\Pi$, it multiplies by $e^{-3at}$. Combining these actions gives $e^{tQ}=\Pi+e^{-3at}(I-\Pi)$, as claimed. $\square$

Expanding the projections gives a particularly useful entrywise formula. Write $r=e^{-3at}$. Then $P(t)$ has diagonal entries $\tfrac13+\tfrac23r$ and off-diagonal entries $\tfrac13-\tfrac13r$. For $a,t\ge0$, $0<r\le1$, so the diagonal entries are nonnegative and the off-diagonal entries are nonnegative. Each row sums to $\tfrac13+\tfrac23r+2(\tfrac13-\tfrac13r)=1$.

P(t)= \begin{bmatrix} \frac{1+2e^{-3at}}{3}&\frac{1-e^{-3at}}{3}&\frac{1-e^{-3at}}{3}\\ \frac{1-e^{-3at}}{3}&\frac{1+2e^{-3at}}{3}&\frac{1-e^{-3at}}{3}\\ \frac{1-e^{-3at}}{3}&\frac{1-e^{-3at}}{3}&\frac{1+2e^{-3at}}{3} \end{bmatrix}.

Worked Example: A Transition Matrix at an Exact Time

Suppose the common rate is $a=2$, and find $P(t)$ at $t=\tfrac{\ln 2}{6}$. First, the generator is $$ Q=\begin{bmatrix}-4&2&2\\2&-4&2\\2&2&-4\end{bmatrix}. $$ Each row sums to $-4+2+2=0$, so the generator has the required row sums. The exponent in the transition formula is $-3at=-3\cdot2\cdot\tfrac{\ln2}{6}=-\ln2$, and hence $r=e^{-\ln2}=\tfrac12$. The diagonal entries are $\tfrac13+\tfrac23\cdot\tfrac12=\tfrac23$; each off-diagonal entry is $\tfrac13-\tfrac13\cdot\tfrac12=\tfrac16$. Therefore $$ P(t)=\begin{bmatrix}\frac23&\frac16&\frac16\\\frac16&\frac23&\frac16\\\frac16&\frac16&\frac23\end{bmatrix}. $$ Check: Each row sums to $\tfrac23+\tfrac16+\tfrac16=1$, and the matrix is symmetric, as the equal rates suggest.

Applying the Formula to an Initial Distribution

The spectral form also makes the distribution at time $t$ transparent. For any initial row distribution $\mu=(\mu_1,\mu_2,\mu_3)$, where $\mu_i\ge0$ and $\mu_1+\mu_2+\mu_3=1$, multiplication by $\Pi$ gives the uniform row vector $(\tfrac13,\tfrac13,\tfrac13)$. Consequently, $$ \mu P(t)=\left(\frac13,\frac13,\frac13\right)+e^{-3at}\left(\mu-\left(\frac13,\frac13,\frac13\right)\right). $$ Each initial deviation from uniformity is reduced by the same factor $e^{-3at}$.

Worked Example: Evolving a Nonuniform Initial Distribution

Take $a=1$, initial distribution $\mu=(\tfrac12,\tfrac13,\tfrac16)$, and time $t=\tfrac{\ln2}{3}$. The generator is $$ Q=\begin{bmatrix}-2&1&1\\1&-2&1\\1&1&-2\end{bmatrix}, $$ and each row sums to $-2+1+1=0$. Here $e^{-3at}=e^{-3\cdot1\cdot(\ln2/3)}=\tfrac12$. Thus the first component of $\mu P(t)$ is $\tfrac13+\tfrac12(\tfrac12-\tfrac13)=\tfrac13+\tfrac1{12}=\tfrac5{12}$. The second is $\tfrac13+\tfrac12(\tfrac13-\tfrac13)=\tfrac13$. The third is $\tfrac13+\tfrac12(\tfrac16-\tfrac13)=\tfrac13-\tfrac1{12}=\tfrac14$. Hence $$ \mu P(t)=\left(\frac5{12},\frac13,\frac14\right). $$ Check: The components sum to $\tfrac5{12}+\tfrac4{12}+\tfrac3{12}=1$. Using the matrix formula, the first component is $\tfrac12\cdot\tfrac23+\tfrac13\cdot\tfrac16+\tfrac16\cdot\tfrac16=\tfrac13+\tfrac1{18}+\tfrac1{36}=\tfrac5{12}$, agreeing with the spectral calculation.

Stationarity and Limiting Behavior

Theorem: If $a>0$, the unique stationary distribution is $\pi=(\tfrac13,\tfrac13,\tfrac13)$, and for every initial distribution $\mu$, $\mu P(t)$ converges to $\pi$ as $t\to\infty$. If $a=0$, every initial distribution is stationary and the distribution remains equal to its initial value, so there is no convergence to a common distribution independent of $\mu$.

Proof. First suppose $a>0$. A stationary distribution is a probability row vector satisfying $\pi P(t)=\pi$ for all $t\ge0$. The uniform vector satisfies this because $(\tfrac13,\tfrac13,\tfrac13)\Pi=(\tfrac13,\tfrac13,\tfrac13)$ and $(\tfrac13,\tfrac13,\tfrac13)(I-\Pi)=0$. To show uniqueness, if $\pi$ is stationary, differentiating $\pi P(t)=\pi$ at $t=0$, using $P'(0)=Q$, gives $\pi Q=0$. Writing $\pi=(\pi_1,\pi_2,\pi_3)$, the first coordinate of $\pi Q=0$ says $-2a\pi_1+a\pi_2+a\pi_3=0$. Since $a>0$ and $\pi_1+\pi_2+\pi_3=1$, this implies $3\pi_1=1$. The second and third coordinate equations give $3\pi_2=1$ and $3\pi_3=1$, respectively. Thus $\pi$ must be uniform.

For convergence, the formula already derived gives $\mu P(t)=\pi+e^{-3at}(\mu-\pi)$. Since $a>0$, $e^{-3at}\to0$, so each coordinate converges to the corresponding coordinate of $\pi$. If $a=0$, then $Q$ is the zero matrix and $P(t)=e^{tQ}=I$ for every $t$; consequently $\mu P(t)=\mu$. Every probability distribution is stationary, and distinct initial distributions remain distinct. This proves both cases. $\square$

Worked Example: Convergence from Two Different States

Let $a=\tfrac12$ and $t=\tfrac{\ln2}{3}$. The generator is $$ Q=\begin{bmatrix}-1&\frac12&\frac12\\\frac12&-1&\frac12\\\frac12&\frac12&-1\end{bmatrix}. $$ Each row sums to $-1+\tfrac12+\tfrac12=0$. The exponential factor is $e^{-3at}=e^{-3\cdot(1/2)\cdot(\ln2/3)}=\tfrac12$. The transition matrix therefore has diagonal entries $\tfrac13+\tfrac23\cdot\tfrac12=\tfrac23$ and off-diagonal entries $\tfrac13-\tfrac13\cdot\tfrac12=\tfrac16$: $$ P(t)=\begin{bmatrix}\frac23&\frac16&\frac16\\\frac16&\frac23&\frac16\\\frac16&\frac16&\frac23\end{bmatrix}. $$ If the chain starts in state $1$, its distribution at time $t$ is the first row; from state $2$, it is the second row. The total variation distance between these distributions is half the sum of the absolute coordinate differences: $$ \frac12\left(\left|\frac23-\frac16\right|+\left|\frac16-\frac23\right|+\left|\frac16-\frac16\right|\right)=\frac12. $$ Check: The two rows are probability distributions since each sums to $\tfrac23+\tfrac16+\tfrac16=1$. Their initial distributions, concentrated at different states, have total variation distance $1$; at the chosen time it is $\tfrac12$. The exact formula shows that the difference between any two initial distributions is multiplied by $e^{-3at}$, which here equals $\tfrac12$.

Reading the Decay Rate Correctly

The two nonconstant eigenvectors decay at the same rate, $e^{-3at}$. For example, $v_1=(1,-1,0)^{\mathsf T}$ and $v_2=(1,1,-2)^{\mathsf T}$ both have coordinate sum zero, so $Qv_1=-3av_1$ and $Qv_2=-3av_2$. Their coefficients in an initial distribution describe its deviation from uniformity. This is why the chain has one simple relaxation rate despite having two independent directions of deviation.

Common mistake. Using $e^{-at}$ as the decay factor because the rate on each individual edge is $a$ misses the combined effect of all three states. The relevant nonzero eigenvalue is $-3a$, not $-a$. For instance, with $a=1$ and $t=\tfrac{\ln2}{3}$, the correct factor is $e^{-3\cdot1\cdot(\ln2/3)}=\tfrac12$, giving diagonal probability $\tfrac13+\tfrac23\cdot\tfrac12=\tfrac23$. The incorrect factor $e^{-t}=2^{-1/3}$ would give a different diagonal probability and would not satisfy the matrix exponential generated by $Q$.

The transition formula is useful whenever one needs probabilities at a fixed time, the long-run distribution, or a quantitative description of how fast the initial condition is forgotten. Its simplicity depends on the exact symmetry of the rates. If the rates differ, the equal-coordinate and zero-sum decomposition may no longer diagonalize the generator in this form; one must instead analyze that generator’s own eigenstructure.

Check Your Understanding

Use the generator and transition formula in this tutorial to answer the following. In each numerical calculation, verify that the resulting distribution sums to one.

For common rate $a=3$, write out the generator and check every row sum.
At what time does the nonconstant eigenvector component have decayed to one quarter when $a=1$?
Starting from $(0,1,0)$, find the distribution at time $t=\tfrac{\ln2}{3a}$ for $a>0$.
Why does the uniform distribution remain unchanged under $P(t)$ for every $t\ge0$?
Describe the stationary distributions and limiting behavior when $a=0$, and explain why the positive-rate convergence conclusion does not apply.