Infinitesimal Generator Entries and Small-Time Probabilities

Why Small-Time Probabilities Matter

In a continuous-time multiple-state model, transition intensities describe how quickly a process can leave one state for another. For a short interval, an intensity converts directly into an approximate transition probability. This link is essential when interpreting an intensity matrix and when setting up calculations that use it.

The approximation is local: it describes what happens over an interval of length \(h\) as \(h\) becomes small. It does not generally give the exact transition probabilities over a longer interval. In particular, the chance of a transition from state \(i\) to state \(j\) is approximately the corresponding intensity multiplied by \(h\), while the chance of no state change is approximately one minus the total rate of leaving state \(i\), multiplied by \(h\).

Definition: For a continuous-time Markov chain with states indexed by \(i,j\), the intensity \(q_{ij}(t)\), for \(i\ne j\), is the instantaneous rate of transition from state \(i\) to state \(j\) at time \(t\). The intensity matrix \(Q(t)=(q_{ij}(t))\) has off-diagonal entries \(q_{ij}(t)\). Its diagonal entry \(q_{ii}(t)\) is the negative total intensity of leaving state \(i\). In a time-homogeneous model, the intensities do not depend on \(t\), so the matrix is written \(Q\).

Write \(P_{ij}(t,t+h)\) for the probability of being in state \(j\) at time \(t+h\), given that the process is in state \(i\) at time \(t\). For a finite-state continuous-time Markov chain with finite, locally well-behaved intensities, the small-time relations are

P_{ij}(t,t+h)=q_{ij}(t)h+o(h), \qquad i\ne j,

and

P_{ii}(t,t+h)=1+q_{ii}(t)h+o(h).

Here \(o(h)\) denotes a remainder whose ratio to \(h\) tends to zero as \(h\) tends to zero. These are limiting approximations, not identities for a general positive interval. If time is measured in years, then \(h\) must be in years when the intensity is expressed per year.

Why the Diagonal Entry Gives the No-Transition Probability

The diagonal entry is negative because it records the rate of leaving the current state, not a rate of making a transition to a different state. If the process is currently in state \(i\), it can leave for any of the other states \(j\). Over a sufficiently short interval, the probability of leaving for a particular destination \(j\) is \(q_{ij}(t)h+o(h)\). Adding the probabilities of the possible destinations gives the short-time probability of leaving state \(i\).

P(\text{leave }i\text{ during }[t,t+h]\mid X(t)=i) = \left(\sum_{j\ne i}q_{ij}(t)\right)h+o(h).

The probability of no state change is one minus this probability. Therefore,

P_{ii}(t,t+h) = 1-\left(\sum_{j\ne i}q_{ij}(t)\right)h+o(h).

Thus the diagonal intensity is \(q_{ii}(t)=-\sum_{j\ne i}q_{ij}(t)\), and substituting it gives \(P_{ii}(t,t+h)=1+q_{ii}(t)h+o(h)\). This argument explains the diagonal formula in terms of competing exits from the current state; it does not say that the short-time approximation is exact for a finite interval.

Interpretation: For \(i\ne j\), multiply the intensity \(q_{ij}(t)\) by the short interval \(h\) to obtain the leading-order probability of a transition from \(i\) to \(j\). For state \(i\) itself, use \(1+q_{ii}(t)h\), or equivalently one minus the total exit intensity times \(h\).

Worked Applications

Worked Example: Short-Time Probabilities in a Three-State Model

A hypothetical income-protection model has states Active (A), Disabled (D), and Dead (X). It is time-homogeneous, with intensities per year \(q_{AD}=0.08\), \(q_{AX}=0.02\), \(q_{DA}=0.15\), and \(q_{DX}=0.05\). Dead is absorbing. Use the small-time approximation for an interval of \(h=0.1\) year, starting from Active and then starting from Disabled.

Method. For each starting state, multiply each off-diagonal intensity by \(0.1\). Find the diagonal entry by taking the negative total exit intensity, then use \(1+q_{ii}h\) for the no-transition probability.

From Active, the total exit intensity is \(0.08+0.02=0.10\) per year, so \(q_{AA}=-0.10\). The transition probabilities to first order are

P_{AD}(0.1)\approx 0.08(0.1)=0.008,\qquad P_{AX}(0.1)\approx 0.02(0.1)=0.002.

The probability of remaining Active is

P_{AA}(0.1)\approx 1+(-0.10)(0.1)=0.990.

From Disabled, the total exit intensity is \(0.15+0.05=0.20\) per year, so \(q_{DD}=-0.20\). The corresponding approximations are

P_{DA}(0.1)\approx 0.15(0.1)=0.015,\qquad P_{DX}(0.1)\approx 0.05(0.1)=0.005, $$ $$ P_{DD}(0.1)\approx 1+(-0.20)(0.1)=0.980.

As a check on the arithmetic, the Active-state approximations sum to \(0.008+0.002+0.990=1.000\), and the Disabled-state approximations sum to \(0.015+0.005+0.980=1.000\). These checks are consistent with the first-order calculation. The values remain approximations, with omitted \(o(h)\) terms.

Common mistake avoided: The Active-to-Dead probability uses \(q_{AX}=0.02\), not the total Active exit intensity \(0.10\). The total exit intensity is used for the diagonal probability, not for any one destination.

Worked Example: Time-Varying Rates and a Mixed Starting Distribution

A hypothetical rehabilitation model has states Healthy (H), Disabled (D), and Dead (X). At time \(t=5\) years, the intensities per year are \(q_{HD}(5)=0.03\), \(q_{HX}(5)=0.01\), \(q_{DH}(5)=0.12\), and \(q_{DX}(5)=0.04\); Dead is absorbing. At time 5, the state probabilities are \(0.60\) Healthy, \(0.40\) Disabled, and \(0\) Dead. Approximate the state probabilities \(0.25\) year later using the intensities at time 5.

Method. Use the local intensities at the start of the interval and multiply by \(h=0.25\). This gives a first-order approximation for each starting state; then weight each resulting row by the starting probabilities.

Starting Healthy, the exit intensity is \(0.03+0.01=0.04\) per year. Therefore,

P_{HD}(5,5.25)\approx 0.03(0.25)=0.0075,\qquad P_{HX}(5,5.25)\approx 0.01(0.25)=0.0025, $$ $$ P_{HH}(5,5.25)\approx 1-0.04(0.25)=0.9900.

Starting Disabled, the exit intensity is \(0.12+0.04=0.16\) per year. Thus,

P_{DH}(5,5.25)\approx 0.12(0.25)=0.0300,\qquad P_{DX}(5,5.25)\approx 0.04(0.25)=0.0100, $$ $$ P_{DD}(5,5.25)\approx 1-0.16(0.25)=0.9600.

Now combine the starting states. The probability of being Healthy at time \(5.25\) is approximately \(0.60(0.9900)+0.40(0.0300)=0.5940+0.0120=0.6060\). The Disabled probability is \(0.60(0.0075)+0.40(0.9600)=0.0045+0.3840=0.3885\). The Dead probability is \(0.60(0.0025)+0.40(0.0100)=0.0015+0.0040=0.0055\).

The resulting approximate distribution is \((0.6060,0.3885,0.0055)\), in the order H, D, X. Its entries sum to \(0.6060+0.3885+0.0055=1.0000\). Since the intensities may change during the interval, this calculation uses their values at time 5 and is not an exact finite-interval solution.

Common mistake avoided: Applying only the Healthy row would ignore the 40% who start Disabled. Each row must be weighted by the probability of starting in its corresponding state.

Worked Example: Monthly Probabilities From Annual Intensities

A hypothetical pension model has states Active (A), Disabled (D), Retired (R), and Dead (X). The time-homogeneous annual intensities are \(q_{AD}=0.02\), \(q_{AR}=0.10\), \(q_{AX}=0.005\), \(q_{DA}=0.20\), \(q_{DX}=0.03\), and \(q_{RX}=0.04\). Dead is absorbing. Starting in Active, approximate the probabilities of each state after one month, using \(h=1/12\) year. Round the final probabilities to six decimal places.

Method. Keep the time unit consistent: convert one month to \(1/12\) year, since all intensities are annual. The total Active exit intensity is \(0.02+0.10+0.005=0.125\) per year, so \(q_{AA}=-0.125\).

The off-diagonal probabilities from Active are

P_{AD}\approx 0.02\left(\frac{1}{12}\right)=0.0016667,\quad P_{AR}\approx 0.10\left(\frac{1}{12}\right)=0.0083333,\quad P_{AX}\approx 0.005\left(\frac{1}{12}\right)=0.0004167.

The probability of no state change is

P_{AA}\approx 1-0.125\left(\frac{1}{12}\right) =1-0.0104167=0.9895833.

To check the construction, the unrounded first-order values sum to \(0.0016667+0.0083333+0.0004167+0.9895833=1.0000000\). Rounded to six decimal places, the probabilities are \(P_{AD}=0.001667\), \(P_{AR}=0.008333\), \(P_{AX}=0.000417\), and \(P_{AA}=0.989583\); their rounded sum is also \(1.000000\).

Common mistake avoided: Using \(h=1\) because the interval is “one month” would treat the interval as one year. With annual intensities, the interval must be entered in years.

Common Errors and Practical Checks

Small-time calculations are brief, but the same errors recur. The following checks help keep the interpretation and arithmetic consistent.

Using the wrong sign on the diagonal. A diagonal intensity is negative. If the total exit intensity is \(r\), the no-transition approximation is \(1-rh\), not \(1+rh\).
Using the total exit rate for a specific destination. For a transition from \(i\) to \(j\), use \(q_{ij}h\). Add the destination-specific intensities only to find the total exit rate from \(i\).
Mixing time units. An annual intensity must be multiplied by an interval in years; a monthly intensity must be multiplied by an interval in months.
Treating an approximation as exact. The omitted \(o(h)\) terms matter for finite intervals. Do not use the small-time formula as an exact answer for a long period unless a separate result justifies it.
Forgetting the starting state. A transition probability is conditional on the state at the start of the interval. If the starting state is uncertain, combine the conditional probabilities using the starting-state distribution.

Key takeaway: For a short interval \(h\), an off-diagonal entry gives \(P_{ij}\approx q_{ij}h\), while the diagonal entry gives \(P_{ii}\approx 1+q_{ii}h\). Since \(q_{ii}\) is the negative total exit intensity, it determines the probability of no state change to first order.

Check Your Understanding

Use the first-order small-time approximation unless a question asks you to identify a limitation.

A model has states \(U,V,W\). From \(U\), the intensities are \(q_{UV}=0.06\) and \(q_{UW}=0.02\) per year. Find the approximate probabilities of moving to \(V\), moving to \(W\), and remaining in \(U\) over \(0.05\) year. (Answer: \(0.003,\ 0.001,\ 0.996\), respectively.)
From Disabled, a hypothetical care model has recovery intensity \(0.18\) per year and death intensity \(0.06\) per year. Find the probabilities of recovery, death, and no state change over \(0.1\) year. (Answer: \(0.018,\ 0.006,\ 0.976\), respectively.)
At time \(t=3\), the time-varying intensities from state \(M\) are \(q_{MN}(3)=0.07\) and \(q_{MX}(3)=0.03\) per year. Use the intensities at \(t=3\) to approximate the probabilities of moving to \(N\), moving to \(X\), and remaining in \(M\) over \(0.2\) year. State why these are not necessarily exact probabilities for the full interval. (Answer: \(0.014,\ 0.006,\ 0.980\); they are first-order approximations based on the rates at \(t=3\), which may change during the interval.)
A process starts with probabilities \(0.25\) in state \(A\) and \(0.75\) in state \(B\). Per-year rates are \(q_{AB}=0.04,\ q_{AX}=0.01,\ q_{BA}=0.10,\ q_{BX}=0.02\), with \(X\) absorbing. Approximate the distribution after \(0.1\) year, combining the conditional rows. (Answer: \(P(A)=0.25625,\ P(B)=0.74200,\ P(X)=0.00175\).)
An intensity is \(0.003\) per month. Approximate the probability of the corresponding transition over two months, and the probability of no transition if this is the only exit from the current state. (Answer: transition probability \(0.006\); no-transition probability \(0.994\).)

What You'll Learn