Short-Interval Transition Probabilities From Intensities

From an Intensity Matrix to a Short-Interval Probability

A transition intensity describes a local rate, not a probability over a finite period. For a very short interval of length $h$, an intensity of $0.08$ per year suggests a transition probability close to $0.08h$. The approximation is useful only when the time unit is consistent and the interval is short enough for multiple transitions to be comparatively unlikely.

This tutorial develops the first-order approximation and explains what it leaves out. In particular, a direct transition can have probability of order $h$, while a transition requiring two jumps may have probability of order $h^2$. A zero first-order term therefore does not always mean the transition probability is zero.

Definition: For a continuous-time Markov model with states $1,\ldots,n$, let $P_{ij}(t,t+h)$ be the probability of being in state $j$ at time $t+h$, given state $i$ at time $t$. For $i\ne j$, the transition intensity is $$ q_{ij}(t)=\lim_{h\downarrow 0}\frac{P_{ij}(t,t+h)}{h}. $$ The diagonal intensity is $q_{ii}(t)=-\sum_{j\ne i}q_{ij}(t)$. The matrix $Q(t)=(q_{ij}(t))$ is the intensity matrix: its off-diagonal entries are nonnegative and every row sums to zero.

We use the convention that rows identify the state occupied at the start of the interval and columns identify the state at the end. Assume a finite-state continuous-time Markov model with locally bounded transition intensities. For the first-order approximation, assume the intensities are continuous at $t$. Then, as $h$ decreases to zero, the transition matrix satisfies $P(t,t+h)=I+hQ(t)+o(h)$, entry by entry.

P_{ij}(t,t+h)= \begin{cases} q_{ij}(t)h+o(h), & i\ne j,\\[4pt] 1+q_{ii}(t)h+o(h), & i=j. \end{cases}

Here $o(h)$ denotes a term that becomes negligible compared with $h$ as $h$ tends to zero. The formula gives a local approximation, not an exact finite-interval calculation. If rates are measured per year, $h$ must be in years; if rates are per month, $h$ must be in months.

Why the Diagonal Has a First-Order Decrease

For a process starting in state $i$, each possible direct move to a different state $j$ contributes approximately $q_{ij}(t)h$. The chance of still being in state $i$ must decrease to account for those exits. Since each row of an exact transition matrix sums to one, its first-order diagonal term is forced by the off-diagonal terms.

List the direct exits.
From state $i$, the first-order probability of a direct transition to $j\ne i$ is $q_{ij}(t)h$.

Add the exit probabilities.
The total first-order probability of leaving $i$ is $h\sum_{j\ne i}q_{ij}(t)$.

Use the row sum.
The diagonal must be one minus that total to first order, giving $1-h\sum_{j\ne i}q_{ij}(t)=1+q_{ii}(t)h$.

This argument explains why the diagonal entry is not simply one, even when the interval is short. A path that leaves $i$ and returns to $i$ requires at least two jumps, so under bounded intensities its probability is of order $h^2$. Such a return does not change the diagonal's first-order term.

Likewise, if a transition from $i$ to $j$ is impossible in one jump throughout the interval and requires at least two jumps, its probability is at most of order $h^2$ under bounded intensities. The first-order matrix records zero in that position. This is an approximation statement, not a claim that the transition is impossible. If an intensity is zero only at the instant $t$ but becomes positive immediately afterward, the probability is $o(h)$; it need not be of order $h^2$ without further regularity assumptions.

Worked Example: Building a First-Order Matrix for Three States

Consider a hypothetical income-protection model with states Active (A), Disabled (D), and Dead (X). At the interval's starting time, the intensities per year are A to D $=0.12$, A to X $=0.03$, D to A $=0.08$, and D to X $=0.20$. State X is absorbing. The process starts in A, and the interval is $h=0.02$ years. Find the first-order probabilities from A and D. No interest or benefit assumptions are needed for this transition calculation.

Method. Form each off-diagonal entry as its intensity times $h$. For each diagonal, subtract the sum of the state's outgoing intensities, multiplied by $h$.

From A, the total exit intensity is $0.12+0.03=0.15$ per year. Thus the first-order probabilities are

P_{AA}\approx 1-(0.15)(0.02)=0.997,\qquad P_{AD}\approx(0.12)(0.02)=0.0024,\qquad P_{AX}\approx(0.03)(0.02)=0.0006.

Check the row: $0.997+0.0024+0.0006=1.0000$. From D, the total exit intensity is $0.08+0.20=0.28$ per year, so

P_{DA}\approx(0.08)(0.02)=0.0016,\qquad P_{DD}\approx1-(0.28)(0.02)=0.9944,\qquad P_{DX}\approx(0.20)(0.02)=0.0040.

Check the row: $0.0016+0.9944+0.0040=1.0000$. Since X is absorbing, its row is $(0,0,1)$. The resulting first-order matrix is

I+hQ\approx \begin{pmatrix} 0.9970 & 0.0024 & 0.0006\\ 0.0016 & 0.9944 & 0.0040\\ 0 & 0 & 1 \end{pmatrix}.

Each entry is dimensionless because a rate per year was multiplied by years. The row sums check the arithmetic, but they do not make this matrix the exact transition matrix for an interval of $0.02$ years. The approximation omits terms beyond first order. This method avoids the common mistake of treating a transition intensity such as $0.12$ as a probability of $12\%$ over any chosen interval.

Transitions That Require More Than One Jump

The first-order formula concerns direct transitions. To see why a transition can be absent at order $h$ but present at order $h^2$, consider a route that requires first moving from $i$ to an intermediate state and then from that state to $j$. Each jump must occur within the short interval. The possible time for the second jump depends on when the first jump occurred, producing a second factor of the interval length.

Worked Example: A Two-Jump Transition with Zero Direct Intensity

Consider a hypothetical three-state process with states A, B, and C. The only possible moves are A to C at constant intensity $0.40$ per year and C to B at constant intensity $0.50$ per year; B is absorbing. The process starts in A, and $h=0.10$ years. Find the first-order approximation to the probability of reaching B, and determine the leading two-jump contribution. There is no separate mortality, interest, or payment assumption in this state-transition model.

The direct intensity from A to B is zero. Therefore, the first-order approximation is $P_{AB}(h)\approx 0$. But reaching B is possible by the path A to C to B. For constant intensities, the leading two-jump contribution is obtained by integrating over the time of the first jump. If that jump occurs at time $s$, the remaining interval for the second jump has length $h-s$. To second order, the contribution is

(0.40)(0.50)\int_0^{0.10}(0.10-s)\,ds =(0.20)\left[\frac{(0.10)^2}{2}\right] =(0.20)(0.005) =0.001.

Thus the first-order approximation reports zero, while the leading probability contribution through the two-jump route is $0.001$, or $0.1\%$. Terms of higher order are not included in this calculation. The first-order approximation to A remaining in A is $1-(0.40)(0.10)=0.96$, and to A moving directly to C is $(0.40)(0.10)=0.04$; these sum to one at first order.

The distinction is the point: $P_{AB}$ is zero at order $h$, not necessarily zero altogether. The integral also shows why the leading contribution contains $h^2/2$, rather than just $h^2$. This calculation assumes constant intensities and includes only the leading two-jump term; it is not an exact finite-interval transition probability. It avoids the mistake of interpreting a zero entry in $I+hQ$ as proof that no multi-step route exists.

Using the Intensity at the Start of the Interval

When intensities vary with time, the first-order approximation uses $Q(t)$, the intensity matrix at the start of the interval. Replacing it by $Q(t+h)$, or by an interval average, can change the answer at second order. That difference vanishes relative to $h$ as the interval shrinks, but it matters if the task specifically asks for the first-order approximation.

Worked Example: Evaluating a Time-Varying Intensity Locally

Consider a hypothetical process with states A, B, and C. The only possible moves are A to C, with intensity $q_{AC}(t)=0.04+0.02t$ per year, and C to B, with constant intensity $0.08$ per year. State B is absorbing. Time $t$ is measured in years. The process starts in A at $t=3$, and the interval length is $h=0.25$ years. Find the first-order probabilities from A and C, and classify the A-to-B entry by order. No interest or mortality assumption is required.

First evaluate the A-to-C intensity at the start time:

q_{AC}(3)=0.04+(0.02)(3)=0.10\text{ per year}.

The total exit intensity from A at $t=3$ is $0.10$ per year. Therefore,

P_{AC}(3,3.25)\approx(0.10)(0.25)=0.025,\qquad P_{AA}(3,3.25)\approx1-(0.10)(0.25)=0.975.

The A-to-B direct intensity is zero, so its first-order entry is zero. Reaching B from A requires the two jumps A to C and C to B. As both intensities are bounded on this interval, that route contributes at order $h^2$, not at order $h$. From C, the only exit intensity is $0.08$ per year, giving

P_{CB}(3,3.25)\approx(0.08)(0.25)=0.020,\qquad P_{CC}(3,3.25)\approx1-(0.08)(0.25)=0.980.

Both row sums are one to first order. Using the endpoint value $q_{AC}(3.25)=0.105$ would give $0.105(0.25)=0.02625$, not the requested start-of-interval first-order approximation $0.025$. The two values differ by $0.00125$, a term of order $h^2$ here because the intensity changes linearly. This method avoids mixing an endpoint rate with a start-time approximation.

Common Mistakes and Practical Checks

A first-order matrix is a local calculation. Before using it, identify the interval's time unit, read each row as the starting state, and use the intensity at the correct time. Then check that the off-diagonal probabilities are nonnegative and that each approximate row sums to one. For a state with total outgoing intensity $\lambda_i(t)$, the diagonal approximation is $1-\lambda_i(t)h$; for a sufficiently short interval this is nonnegative.

Using a rate as a probability. An intensity of $0.10$ per year gives a first-order transition probability of $0.10h$, not $0.10$ for every interval length.
Forgetting the diagonal exit term. The diagonal is $1-$ the sum of outgoing intensities times $h$, not one.
Ignoring the time unit. A one-month interval is $1/12$ year when rates are per year, not one year.
Reading a zero first-order entry as impossible. A zero direct intensity can coexist with a positive probability through a multi-jump route.
Treating the approximation as exact. Row sums can equal one in $I+hQ(t)$, even though higher-order transition effects are omitted.

This approximation is the basic local translation from intensities to probabilities in continuous-time state models. It is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus because exam problems require careful interpretation of intensity matrices, time units, and transition paths. The central check is whether the proposed move can happen directly within one jump or requires additional jumps within the interval.

Check Your Understanding

For each problem, use the first-order approximation unless the question asks for a higher-order classification.

A two-state model has intensity $0.07$ per year from state 1 to state 2 and $0.03$ per year from state 2 to state 1. Starting in state 1, find the two first-order probabilities over $h=0.10$ years. (Answer: $P_{12}\approx0.007$, $P_{11}\approx0.993$.)
A state has two direct exits with intensities $0.05$ and $0.15$ per year. Find the first-order exit probabilities and the probability of remaining in the state over $h=0.02$ years. (Answer: exit probabilities $0.001$ and $0.003$; remaining probability $0.996$.)
From A, the only move is to C at intensity $0.30$ per year; from C, the only move is to B at intensity $0.20$ per year. Both intensities are constant. Starting in A, what is the first-order A-to-B entry, and what is the leading two-jump contribution for $h=0.05$ years? (Answer: first-order entry $0$; leading contribution $(0.30)(0.20)(0.05)^2/2=0.000075$.)
At time $t=4$ years, a process has $q_{12}(t)=0.02+0.01t$ per year and $q_{13}(t)=0.06$ per year, with no other direct exits from state 1. Find the first-order row for state 1 over $h=0.10$ years. (Answer: $P_{12}\approx0.006$, $P_{13}\approx0.006$, $P_{11}\approx0.988$.)
In a process, A can move directly to B at intensity $0.04$ per year, or to C at intensity $0.10$ per year. C can move to B at intensity $0.08$ per year. Find the first-order A-to-B probability over $h=0.05$ years, and state the order of the additional contribution through C. (Answer: direct first-order probability $0.002$; the route through C contributes at order $h^2$.)

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