Why Define an Intensity Through a Limit?
In a continuous-time state model, a transition can occur at any instant, rather than only at fixed observation dates. A probability over a fixed period answers how likely a change is during that period. A transition intensity instead describes the local rate of transition from one state to another, conditional on occupying the starting state at a specified time. The shrinking-interval limit makes that distinction precise.
For example, a hypothetical long-term care model might distinguish an active state, a care state, and a dead state. The chance of being in care one year from now can include several possible paths. The intensity from active to care concerns the local transition from active to care, not the probability of ending the year in care by any route.
The condition \(i\ne j\) matters: this definition concerns a change to a different state. The conditioning is on occupying \(i\) at time \(t\), not merely on having started in \(i\) at time zero. In a Markov model, the current state supplies the relevant state information for the future; the earlier material on the Markov assumption gives that modeling framework.
The limit may depend on \(t\). A model need not assume the same transition behavior at all times. To use the definition, the conditional probability must be specified under the model, and the ratio must have a finite limit as \(h\) decreases to zero. A probability over a large interval by itself does not specify that limit.
Endpoint Probabilities and Direct Transitions
The definition uses the state at the end of the interval. That is important because an endpoint event can result from more than one transition. Starting in \(i\), the path \(i\to k\to j\) can end in \(j\) even though the first transition was not to \(j\). Thus, in general, it is not correct to equate the endpoint probability \(p_{ij}(t,h)\) exactly with the probability that the first transition is directly from \(i\) to \(j\).
Under regular conditions, however, paths involving two or more transitions do not alter the limiting rate. For example, suppose the model has finitely many states and transition rates are bounded in a neighborhood of \(t\). Then the chance of two or more transitions in an interval of length \(h\) is of order \(h^2\), and so divided by \(h\) it tends to zero. The endpoint probability and the probability of a single direct transition therefore have the same first-order limit. This is a statement about the limit, not an assertion that their probabilities are exactly equal for positive \(h\).
Start with \(X(t)=i\); do not substitute an unconditional population probability for this condition.
Use the probability of \(X(t+h)=j\), allowing every path that can end in \(j\).
Evaluate \(\lim_{h\downarrow0}p_{ij}(t,h)/h\), if it exists.
Worked Calculations
Worked Example: A Conditional Probability Given as a Polynomial
A hypothetical rehabilitation model has distinct states \(A\) (active) and \(R\) (receiving rehabilitation). At a fixed time \(t\), the model specifies that, conditional on being active at \(t\), the probability of being in rehabilitation at \(t+h\) is \(0.024h-0.004h^2\), for \(0<h\leq 1\), where \(h\) is measured in years. Other endpoint outcomes have the remaining probability. Find the intensity from \(A\) to \(R\) at \(t\). No interest or mortality assumption is needed for this conditional state-transition calculation.
Method. Use the defining conditional endpoint probability and its limit. The stated probability is nonnegative on the specified interval: \(0.024h-0.004h^2=h(0.024-0.004h)>0\) for \(0<h\leq1\).
Divide by \(h\): \(p_{AR}(t,h)/h=(0.024h-0.004h^2)/h=0.024-0.004h\).
Now let \(h\) decrease to zero: \(\alpha_{AR}(t)=\lim_{h\downarrow0}(0.024-0.004h)=0.024\) per year.
The interval length was measured in years, so the rate is per year. The quadratic term affects probabilities over positive intervals, but vanishes after division by \(h\) and taking the limit. A common mistake would be to keep the quadratic term as part of the limiting rate; it contributes zero to this limit.
Worked Example: A Probability From an Exponential Waiting Time
In a hypothetical two-state model, a person currently in state \(W\) waits for a transition to state \(C\). The waiting time is exponentially distributed with parameter \(0.06\) per year, and \(C\) is absorbing. Find the intensity from \(W\) to \(C\), using the probability of being in \(C\) after \(h\) years given \(W\) now.
Assumptions and distribution. The waiting time to the only possible transition has survival probability \(e^{-0.06h}\); once the transition occurs, the person remains in \(C\). There are no other states or transitions in this model. Therefore, being in \(C\) at \(t+h\) is equivalent to having made the transition by then.
The conditional endpoint probability is \(p_{WC}(t,h)=1-e^{-0.06h}\). Hence \(\alpha_{WC}(t)=\lim_{h\downarrow0}(1-e^{-0.06h})/h\).
Using the derivative of \(e^{-0.06h}\) at \(h=0\), or the standard limit \(\lim_{h\to0}(1-e^{-ch})/h=c\) for \(c>0\), gives \(\alpha_{WC}(t)=0.06\) per year. The rate is constant in \(t\) under the stated model.
Check. For any \(h>0\), \(1-e^{-0.06h}\) is a probability, while \(0.06\) is the limiting probability per year. They are not the same quantity for a fixed interval. The calculation avoids the mistake of reporting a one-year transition probability as the intensity.
Worked Example: An Indirect Path Can Also End in the Target State
Consider a hypothetical model with states \(i\), \(k\), and \(j\). While in \(i\), the first transition has constant total rate \(\lambda=0.08\) per year; its destination is \(j\) at rate \(\alpha=0.03\) per year or \(k\) at rate \(\beta=0.05\) per year. In \(k\), transition to \(j\) has rate \(\gamma=0.40\) per year. State \(j\) is absorbing. Starting in \(i\) at time \(t\), calculate the intensity to \(j\), taking account of all paths that can end in \(j\).
Assumptions. The waiting time to leave \(i\) is exponential with parameter \(\lambda=\alpha+\beta=0.08\), and the destination probabilities at exit are \(\alpha/\lambda\) for \(j\) and \(\beta/\lambda\) for \(k\). The waiting time in \(k\) to move to \(j\) is exponential with parameter \(\gamma\). Once in \(j\), the state persists. These assumptions allow the endpoint event to include both a direct transition \(i\to j\) and the two-transition path \(i\to k\to j\).
The probability of a direct transition to \(j\) by time \(h\) is \((\alpha/\lambda)(1-e^{-\lambda h})\). The probability of the indirect path is the integral over the time \(s\) of the first transition to \(k\): \(\int_0^h \beta e^{-\lambda s}(1-e^{-\gamma(h-s)})\,ds\). Thus the endpoint probability is the sum of these two terms; it is not just the direct-transition term.
Divide the direct term by \(h\). Since \(\lim_{h\downarrow0}(1-e^{-\lambda h})/h=\lambda\), its limit is \((\alpha/\lambda)\lambda=\alpha=0.03\).
For the indirect term, use \(1-e^{-x}\leq x\) for \(x\geq0\), and \(e^{-\lambda s}\leq1\). Its probability is bounded above by \(\int_0^h\beta\gamma(h-s)\,ds=\beta\gamma h^2/2=(0.05)(0.40)h^2/2=0.01h^2\). After division by \(h\), this upper bound is \(0.01h\), which tends to zero. The indirect path therefore contributes zero to the limiting rate, even though it does contribute to the endpoint probability for positive \(h\).
Combining the two limits gives \(\alpha_{ij}(t)=0.03\) per year. The common mistake is to claim that the endpoint event is exactly the event of a first transition to \(j\). The correction is to include the indirect path in the endpoint probability and show that its contribution divided by \(h\) vanishes.
Justification of the Limiting Definition
The limit identifies local transition behavior because, over a very short interval, a direct transition contributes at order \(h\), while paths requiring two transitions contribute at a smaller order when rates are bounded. The following argument states the required comparison without assuming that endpoint and first-transition events are identical.
Starting in \(i\), a path ending in \(j\ne i\) must have at least one transition. It may have exactly one transition \(i\to j\), or it may involve two or more transitions, such as \(i\to k\to j\).
For a finite-state model whose transition rates are bounded near \(t\), the chance of a first transition during a length-\(h\) interval is bounded by a constant times \(h\). Conditional on that transition, the chance of another transition before the interval ends is also bounded by a constant times \(h\). Thus the chance of at least two transitions is bounded by a constant times \(h^2\).
The contribution of paths with at least two transitions, divided by \(h\), is bounded by a constant times \(h\), which tends to zero. Consequently those paths do not change the limiting probability per unit time.
This justification explains why the endpoint-probability definition agrees with the local rate of a direct transition under these regularity conditions. It does not say that the endpoint event and the direct-transition event have equal probabilities for any fixed positive \(h\). If rates are not bounded or the required limit does not exist, the conclusion must not be assumed without further analysis.
Common Exam Traps
- Dropping the conditioning. The quantity is conditional on \(X(t)=i\). A probability calculated for a mixed population in several states is not the transition intensity from \(i\).
- Using an initial-state condition at the wrong time. The definition conditions on the state occupied at time \(t\), not automatically on the state at time zero.
- Ignoring indirect paths as though they were impossible. An endpoint in \(j\) can follow an intermediate state. Include such paths when writing the endpoint probability; under bounded rates, show that their contribution is negligible in the limit.
- Stopping before taking the limit. For an exponential waiting time with parameter \(0.06\) per year, the one-year probability is \(1-e^{-0.06}\), approximately \(0.0582\). Calling \(0.0582\) the intensity is wrong: the intensity is \(\lim_{h\downarrow0}(1-e^{-0.06h})/h=0.06\) per year.
- Forgetting the time unit. If \(h\) is measured in months, the resulting intensity is per month; if it is measured in years, it is per year.
Check Your Understanding
For each problem, use the stated conditioning and time unit. Assume the specified conditional probabilities are valid for sufficiently small positive \(h\).
- In a hypothetical model, \(P(X(t+h)=b\mid X(t)=a)=0.015h+0.002h^2\), with \(h\) in years. Find the intensity from \(a\) to \(b\). (Answer: \(0.015\) per year)
- A two-state model has an exponential waiting time from \(r\) to absorbing state \(s\), with parameter \(0.025\) per month. Find the intensity from \(r\) to \(s\) using the conditional endpoint limit. (Answer: \(0.025\) per month)
- At time \(t\), a model specifies \(p_{uv}(t,h)=0.04h/(1+0.04h)\), with \(h\) measured in years. Calculate \(\lim_{h\downarrow0}p_{uv}(t,h)/h\). (Answer: \(0.04\) per year)
- In a model with states \(i,k,j\), rates from \(i\) to \(j\) and \(i\) to \(k\) are \(0.02\) and \(0.03\) per year, respectively; the rate from \(k\) to absorbing \(j\) is \(0.50\) per year. What is the intensity from \(i\) to \(j\), accounting for the indirect route? (Answer: \(0.02\) per year)
- For a time-dependent model, \(p_{ab}(t,h)=(0.01+0.002t)h+0.003h^2\), where \(t\) and \(h\) are measured in years. Find the intensity at \(t=3\) years. (Answer: \(0.016\) per year)