Why an Intensity Is Not a Finite-Interval Probability

Why the Distinction Matters

In a continuous-time state model, an intensity describes how quickly a transition is occurring locally, conditional on the individual still being in the relevant state. A probability, by contrast, describes the chance of an event over a stated interval. Those quantities are related, but they are not the same. Treating an intensity as a finite-interval probability can produce an impossible answer, especially when the intensity is large or the interval is long.

This distinction is important whenever model inputs are given as rates per unit time but a question asks for a probability over days, months, or years. The interval length must be used, and the exact conversion depends on the model assumptions. This tutorial focuses on the simplest useful case: one allowed transition out of the starting state, with a constant intensity.

Definition: Suppose an individual is currently in state $i$, and a transition to state $j$ is allowed. The transition intensity $\lambda$ is the limiting conditional rate $$ \lambda=\lim_{h\downarrow 0}\frac{P(\text{transition from }i\text{ to }j\text{ during }[t,t+h)\mid\text{in state }i\text{ at }t)}{h}, $$ when the limit exists. Its units are inverse time, such as transitions per year. It is not itself a probability over a nonzero interval.

For a small interval of length $h$, the probability of the transition is approximately $\lambda h$. More precisely, under the regularity conditions implicit in the intensity definition, it is $\lambda h+o(h)$, where $o(h)$ divided by $h$ tends to zero as $h$ tends to zero. The approximation is useful for sufficiently short intervals; it does not make $\lambda h$ an exact probability for an arbitrary interval.

A probability must lie between zero and one. A rate has no such upper bound: for example, a rate of 2.4 transitions per year is meaningful. Multiplying that rate by half a year gives 1.2, but 1.2 cannot be a probability. The calculation has not shown that the model is invalid; it has shown that the small-interval approximation is being used outside its appropriate range.

From a Constant Intensity to a Finite-Interval Probability

Assume there are only two relevant states for the calculation: the individual starts in state $i$, and the only possible transition is from $i$ to state $j$. Once the transition occurs, the individual leaves $i$. Assume also that the transition intensity $\lambda$ is constant while the individual remains in $i$. Under these assumptions, the probability of remaining in $i$ for at least $t$ time units is $e^{-\lambda t}$. Therefore, the probability of making the transition within $t$ time units is $1-e^{-\lambda t}$.

P(\text{transition within }t)=1-e^{-\lambda t}.

The time unit in $t$ must match the unit used for $\lambda$. If $\lambda$ is per year, use $t$ in years. The formula is for the probability of leaving the starting state by the end of the interval in this one-exit, constant-intensity model. It is not a universal formula for any transition probability in a multi-state model.

Why the Exponential Formula Holds

Let $R(t)$ be the probability of still being in state $i$ after $t$ time units, given that the individual starts in $i$. Over a short additional interval of length $h$, the conditional probability of leaving $i$, given that the individual is still in $i$, is $\lambda h+o(h)$. Thus the conditional probability of remaining in $i$ over that short interval is $1-\lambda h+o(h)$. By the constant-intensity assumption, the same relation applies after any elapsed time $t$.

Write the short-interval relation.
$R(t+h)=R(t)[1-\lambda h+o(h)]$, because remaining in the starting state through $t+h$ requires both remaining through $t$ and then not transitioning in the next short interval.

Take the derivative.
Subtract $R(t)$, divide by $h$, and let $h$ decrease to zero. The resulting differential equation is $R'(t)=-\lambda R(t)$.

Apply the initial condition.
Since the individual starts in $i$, $R(0)=1$. Solving $R'(t)=-\lambda R(t)$ gives $R(t)=e^{-\lambda t}$.

Take the complement.
The individual either remains in $i$ through time $t$ or leaves $i$ by time $t$. Hence $P(\text{transition within }t)=1-R(t)=1-e^{-\lambda t}$.

The result is consistent with the small-interval approximation: for small $t$, $1-e^{-\lambda t}$ is approximately $\lambda t$. But as $t$ grows, the exact probability approaches one and never exceeds it. The rate $\lambda$ itself does not have to approach or remain below one.

Worked Example: Converting a Quarterly Probability

A hypothetical disability model has one allowed transition out of the active state, with a constant intensity of 0.8 per year. An individual is active at the start of a three-month interval. Calculate the probability of the transition during the interval, and compare it with the small-interval approximation. State the method and identify the mistake the exact calculation avoids.

Method. Use the constant-intensity, one-exit formula. First convert three months to years: $t=3/12=0.25$ years. The units now match the intensity’s unit of per year.

The exact probability is $$ 1-e^{-\lambda t}=1-e^{-(0.8)(0.25)}=1-e^{-0.2}. $$

Since $e^{-0.2}\approx0.8187308$, the transition probability is $1-0.8187308=0.1812692$, or 0.1813 to four decimal places.

The small-interval approximation is $\lambda t=(0.8)(0.25)=0.2000$. It overstates the exact probability by approximately $0.2000-0.1812692=0.0187308$. The approximation is not the exact quarterly probability.

Common mistake avoided: using $0.8$ as the probability, or reporting $0.2000$ as the exact answer. The first ignores the interval length; the second mistakes the small-interval approximation for the finite-interval probability. The calculated exact value is between zero and one, as a probability must be.

Worked Example: An Intensity Above One

A hypothetical care model allows a transition from independent living to a care state. While the individual remains independent, the transition intensity is constant at 2.4 per year. Find the probability of entering care within six months. Explain why multiplying the intensity by the interval is not a valid probability calculation.

Method. There is one allowed exit from the starting state, and its intensity is constant, so use $1-e^{-\lambda t}$. Six months is $t=6/12=0.5$ years.

The small-interval product is $$ \lambda t=(2.4)(0.5)=1.2. $$

Because 1.2 exceeds one, it cannot be a probability. It is only the linear approximation, and it is poor for this interval. The exact probability is $$ 1-e^{-(2.4)(0.5)}=1-e^{-1.2}. $$

Using $e^{-1.2}\approx0.3011942$, the result is $1-0.3011942=0.6988058$, or 0.6988 to four decimal places. The result is a valid probability.

Check. The probability of remaining in the independent state for six months is approximately 0.3012. Adding it to the probability of entering care gives $0.3012+0.6988=1.0000$, subject to rounding.

Common mistake avoided: concluding that an intensity of 2.4 per year is impossible because it is greater than one. The intensity is a rate, not a probability. The finite-interval probability is calculated from the rate and the interval under the stated model.

Conditional and Unconditional Interval Probabilities

An interval probability may be conditional on still occupying the starting state at the beginning of that interval, or it may be unconditional from the original starting time. These are different questions. Under a constant intensity, the probability of leaving during the next interval, conditional on still being in the state at that interval’s start, is calculated using the interval length alone. The unconditional probability also accounts for the chance of having left before that interval began.

Worked Example: The Second-Year Probability

A hypothetical long-term care model has one allowed transition from the active state to a care state, with a constant intensity of 1.5 per year. A person starts active. Calculate (a) the probability of entering care during the first year, (b) the probability of entering care during the second year given that the person is still active at the start of that year, and (c) the unconditional probability of entering care during the second year. Check the total probability of entering care within two years.

Method. Apply the constant-intensity formula to one-year intervals. For the conditional probability in part (b), the individual is known to be active at the interval’s start, so the same one-year calculation applies.

For either one-year interval, the probability of transitioning during that year conditional on being active at its start is $$ 1-e^{-1.5}\approx1-0.2231302=0.7768698. $$

Thus (a) the probability of entering care in the first year is 0.7769. Part (b), the conditional probability of entering care during the second year given that the person is still active at the start of year two, is also 0.7769.

For part (c), the person must first remain active through year one, with probability $e^{-1.5}\approx0.2231302$. Multiply that by the conditional second-year transition probability: $$ e^{-1.5}(1-e^{-1.5})=e^{-1.5}-e^{-3} \approx0.2231302-0.0497871=0.1733431. $$

The unconditional probability of entering care during the second year is therefore 0.1733. The probability of entering care within two years is $1-e^{-3}\approx1-0.0497871=0.9502129$, or 0.9502. The interval probabilities check: $0.7768698+0.1733431=0.9502129$.

Common mistake avoided: reporting 0.7769 as the unconditional probability of entering care during year two. It is conditional on remaining active until that year begins. The unconditional probability must also include the probability of surviving in the active state through the first year.

Exam Traps and Model Limits

The exact conversion $1-e^{-\lambda t}$ depends on the assumptions stated above. In particular, it describes departure from one starting state when the only relevant transition has a constant intensity while the individual remains there. If the intensity changes over time, if more than one transition is possible, or if the question asks for occupancy of a state after several possible state changes, this one-exit formula cannot automatically be applied as the probability of a particular transition. The model must be examined before selecting a formula.

A reliable calculation separates three steps: identify the quantity being asked for, match the time units, and select either an approximation or an exact formula suited to the model. The most common errors are mixing years and months, treating a rate as a probability, and confusing a conditional probability for an interval with the unconditional probability of an event during that interval.

Key takeaway: An intensity is an instantaneous conditional rate, measured in inverse time; it need not be less than one. For one possible exit with constant intensity $\lambda$, the probability of leaving within $t$ time units is $1-e^{-\lambda t}$, not $\lambda$ and, except as a short-interval approximation, not $\lambda t$.

Check Your Understanding

Use the model assumptions in each question carefully. Give probabilities to four decimal places.

A hypothetical model has one possible transition out of a state, with constant intensity 0.3 per year. Find the probability of transitioning within one month. (Answer: 0.0247)
A constant intensity of 1.2 per year applies to a single allowed exit. Find the exact probability of exiting within one year, and state why the intensity-times-time product is not the answer. (Answer: 0.6988; the product is 1.2, which is not a probability)
A single-exit model has a constant intensity of 0.6 per month. Find the probability of transition within two months. (Answer: 0.6988)
A hypothetical transition has constant intensity 2 per year. Find the probability of transition in an interval of 0.4 years, and compare it with the small-interval approximation. (Answer: exact probability 0.5507; approximation 0.8000)
A single-exit model has constant intensity 0.9 per year. Find the probability of transition within three years. (Answer: 0.9328)

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