Why Same-State Probabilities Can Fall Then Rise

Why “Still in the Same State” Does Not Always Mean “No Change”

A person can begin in a state, leave it, and later return. Consequently, the probability of occupying the starting state at time \(t\) includes paths with transitions as well as paths with no transitions. It need not behave like a holding-time survival probability, which counts only the chance of making no transition during the interval.

This distinction matters whenever a model allows transitions in more than one direction. A same-state probability can initially decrease as probability leaves the starting state, then rise as probability returns. The three-state example below makes this effect explicit using a hypothetical model with constant transition intensities.

A Three-State Cycle

Consider three states, \(A\), \(B\), and \(C\), with transitions only in the directed cycle \(A\to B\to C\to A\). Suppose each transition has constant intensity \(\lambda>0\) per year. The model is time-homogeneous: its transition intensities do not vary with calendar time or elapsed time. Each time the process enters a state, its holding time there is exponential with rate \(\lambda\), and the next state is determined by the stated cycle.

Using rows for starting states and columns for ending states, the intensity matrix is

The diagonal entry in each row is the negative of the total rate of leaving that state. The transition matrix after elapsed time \(t\) is \(P(t)=e^{Qt}\). We focus on its \(A,A\) entry, \(p_{AA}(t)\).

Why the Formula Has This Form

For this equal-rate cycle, the same-state probability can be written in a compact form:

One justification uses the matrix exponential and a scalar differential equation. The characteristic polynomial of \(Q\) is \(r(r^2+3\lambda r+3\lambda^2)\), so \(Q\) satisfies \(Q^3+3\lambda Q^2+3\lambda^2Q=0\). Since \(P(t)=e^{Qt}\), differentiating \(P\) gives \(P'=PQ\). Applying the polynomial identity to \(P\) shows that \(f(t)=p_{AA}(t)\) satisfies

The initial values follow from \(P(0)=I\), \(P'(0)=Q\), and \(P''(0)=Q^2\). In particular, \(Q_{AA}=-\lambda\) and \((Q^2)_{AA}=\lambda^2\), so \(f(0)=1\), \(f'(0)=-\lambda\), and \(f''(0)=\lambda^2\). The characteristic roots of the differential equation are \(0\) and \(-3\lambda/2\pm i\sqrt{3}\lambda/2\). Therefore its real-valued solution has the form

Substitution of \(f(0)=1\) and the two derivative conditions gives \(c=1/3\), \(C=2/3\), and \(D=0\). This yields the stated expression. The constant \(1/3\) is also the long-run probability of each state in this symmetric cycle; the exponential factor makes the oscillations around that value decay.

What Determines Whether the Probability Rises?

Differentiating the formula gives

At \(t=0\), this derivative is \(-\lambda\), so the probability initially decreases. It reaches its first minimum when the phase is \(2\pi/3\). For phases strictly between \(2\pi/3\) and \(\pi\), the bracket is positive, so the probability rises over that interval. Later it can fall and rise again in smaller oscillations, because the oscillating term is damped by \(e^{-3\lambda t/2}\).

The same conclusion can be understood as a balance of probability leaving and returning. For a time-homogeneous model, \(P'(t)=P(t)Q\). The \(A,A\) entry of that equation gives

The first term is the rate at which probability mass currently in \(A\) leaves \(A\). The second is the rate at which mass in \(C\) moves into \(A\). Once return flow is large enough relative to outflow, the net derivative can be positive. At the initial instant, no mass is in \(C\), so there is outflow but no return flow.

Common Mistake and Practical Interpretation

A frequent error is to equate \(p_{AA}(t)\) with the probability of no transition. In the cycle, the no-transition probability from \(A\) is \(e^{-\lambda t}\), but this counts only paths that stayed in \(A\) continuously. It omits paths that left \(A\) and returned. For example, at the second time in the first worked example, the no-transition probability is \(e^{-5.4414}\approx0.0043\), while the same-state probability is about \(0.3304\). The much larger endpoint probability includes return paths.

The distinction is useful in actuarial models where a state can be entered again, such as hypothetical models of temporary disability, recovery, or care status. A same-state transition probability describes status at the endpoint; it does not say whether the person remained in that state at every time in between. If the product or question depends on whether a transition occurred at all, an endpoint transition probability alone is not enough.