One-Way Transitions With Constant Intensities

Introduction

A transition intensity describes the instantaneous rate of moving between states; it is not itself the probability of making that transition over a year or another specified interval. To construct transition probabilities, the rate must be combined with a model for how long the individual remains in the starting state. When the intensity is constant and there is only one possible destination, the holding time has an exponential distribution, giving a particularly direct calculation.

This tutorial develops the transition probability matrix for a model with one transient state and one absorbing destination. The model is deliberately simple: it isolates how a constant intensity determines the probability of remaining in the starting state and the probability of having made the one-way transition by time \(t\). The calculations are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

The One-Way Constant-Intensity Model

Definition: Consider two states, labelled 0 and 1. An individual in state 0 moves to state 1 at constant intensity \(\lambda\), where \(\lambda \geq 0\). State 1 has no exits. Thus, state 0 is transient and state 1 is an absorbing destination. The holding time \(T\) in state 0 is exponentially distributed with rate \(\lambda\) when \(\lambda>0\).

The labels are just labels; their meaning depends on the hypothetical situation being modelled. For example, state 0 might represent being covered before a specified event, and state 1 might represent that the event has occurred. The assumption that state 1 is absorbing means the model records no transition back from state 1. It is appropriate only when that simplification fits the question being modelled.

With a constant intensity \(\lambda\), the holding time in state 0 has exponential survival function \(P(T>t)=e^{-\lambda t}\). Therefore, by time \(t\), the probability of still being in state 0 is \(e^{-\lambda t}\). Since the only alternative is that the transition has occurred and state 1 has been reached, that probability is \(1-e^{-\lambda t}\).

Use the row-source, column-destination convention: entry \((i,j)\) is the probability of being in state \(j\) after time \(t\), given a starting state \(i\). The transition probability matrix is therefore

\mathbf{P}(t)= \begin{pmatrix} p_{00}(t) & p_{01}(t)\\ p_{10}(t) & p_{11}(t) \end{pmatrix} = \begin{pmatrix} e^{-\lambda t} & 1-e^{-\lambda t}\\ 0 & 1 \end{pmatrix}, \qquad t\geq 0.

The second row follows from the assumption that state 1 has no exits: starting in state 1, the individual remains there with probability 1. The time \(t\) and intensity \(\lambda\) must use reciprocal units. For example, if \(t\) is measured in years, \(\lambda\) must be expressed per year.

Why the Matrix Entries Have This Form

The derivation uses the holding-time distribution and the fact that there are only two possible states. It does not require a general matrix-exponential calculation.

Write the no-transition probability.
Starting in state 0, the individual remains there at time \(t\) exactly when the holding time exceeds \(t\). Thus \(p_{00}(t)=P(T>t)=e^{-\lambda t}\).

Use the only alternative destination.
Starting in state 0, the individual must be in either state 0 or state 1 at time \(t\). These events are mutually exclusive and exhaustive, so \(p_{01}(t)=1-p_{00}(t)=1-e^{-\lambda t}\).

Apply the no-exit assumption for state 1.
Starting in state 1, the individual stays there, so \(p_{10}(t)=0\) and \(p_{11}(t)=1\). These four entries give the stated matrix.

This also checks the key probability conditions. Each entry is between 0 and 1 for \(t\geq0\), and each row sums to 1:

e^{-\lambda t}+(1-e^{-\lambda t})=1, \qquad 0+1=1.

At \(t=0\), \(e^0=1\), so the matrix is the identity matrix: there has been no time for a transition. If \(\lambda=0\), the transition never occurs and the matrix is also the identity for every \(t\). For \(\lambda>0\), the probability of being in state 0 decreases as \(t\) increases, while the probability of having reached state 1 increases toward 1.

Worked Example: A Two-Year Event Probability

Method: Use the exponential holding-time survival probability, then use the two-state row sum to obtain the transition probability. Suppose a hypothetical covered population is in state 0 until a specified event occurs, after which it is recorded in state 1 permanently. Assume a constant intensity of \(\lambda=0.25\) per year, and assume each individual begins in state 0. Find the transition matrix after 2 years.

The rate is per year and the interval is 2 years, so the exponent is \(\lambda t=(0.25)(2)=0.50\). The probability of remaining in state 0 is

p_{00}(2)=e^{-0.50}=0.6065307\ldots

The probability of reaching state 1 is the complement:

p_{01}(2)=1-e^{-0.50}=1-0.6065307\ldots=0.3934693\ldots

Starting in state 1, the individual remains in state 1, so \(p_{10}(2)=0\) and \(p_{11}(2)=1\). Rounding probabilities to four decimal places gives

\mathbf{P}(2)= \begin{pmatrix} 0.6065 & 0.3935\\ 0 & 1 \end{pmatrix}.

Check: The first row sums to \(0.6065+0.3935=1.0000\), and the second row sums to \(0+1=1\). A common mistake is to report \(\lambda t=0.50\) as the transition probability. It is the exponent in the survival calculation, not the probability; the correct transition probability is \(1-e^{-0.50}=0.3935\).

Worked Example: Converting Months to the Rate's Time Unit

Method: Put the rate and elapsed time in consistent units before applying the exponential formula. In a hypothetical benefit model, state 0 means the covered person has not yet entered a specified benefit state, and state 1 means entry has occurred and is treated as permanent. Assume \(T\sim\operatorname{Exponential}(\lambda)\) with \(\lambda=0.08\) per month. Find the transition matrix after 15 months for an individual who begins in state 0.

Both the rate and time are already expressed using months. The exponent is

\lambda t=(0.08\text{ per month})(15\text{ months})=1.20.

Thus the probability of remaining in state 0 is

p_{00}(15)=e^{-1.20}=0.3011942\ldots

The probability of reaching state 1 is

p_{01}(15)=1-0.3011942\ldots=0.6988058\ldots

The row for an individual starting in state 1 is \((0,1)\). Rounding probabilities to four decimal places, the full matrix is

\mathbf{P}(15)= \begin{pmatrix} 0.3012 & 0.6988\\ 0 & 1 \end{pmatrix}.

Check: The first row sums to \(0.3012+0.6988=1.0000\), and the second row sums to 1. The calculation would be wrong if 15 months were treated as 15 years while retaining a rate per month: the product in the exponential must be dimensionless, with the rate and elapsed time expressed in matching units.

Worked Example: Recovering an Intensity From a Transition Probability

Method: Rearrange the no-transition probability to find the rate, then use that rate at a second duration. A hypothetical model has one transient state and one absorbing destination, and its constant intensity is unknown. The model gives a probability of 0.80 of remaining in state 0 after 4 years. Find the implied intensity per year and the transition matrix after 10 years.

For a constant intensity, \(p_{00}(4)=e^{-4\lambda}\). Set this equal to 0.80 and take natural logarithms:

0.80=e^{-4\lambda}, \qquad \ln(0.80)=-4\lambda, \qquad \lambda=-\frac{\ln(0.80)}{4} =\frac{0.2231436\ldots}{4} =0.0557859\ldots

The implied intensity is \(0.05579\) per year to four significant figures. At 10 years, the probability of remaining in state 0 is

p_{00}(10)=e^{-10(0.0557859\ldots)} =e^{-0.5578589\ldots} =0.5724334\ldots

The probability of reaching state 1 is \(p_{01}(10)=1-0.5724334\ldots=0.4275666\ldots\). Therefore, to four decimal places,

\mathbf{P}(10)= \begin{pmatrix} 0.5724 & 0.4276\\ 0 & 1 \end{pmatrix}.

Check: The first row sums to \(0.5724+0.4276=1.0000\), and the second row sums to 1. Also, the probability of remaining in state 0 has decreased from 0.80 at 4 years to 0.5724 at 10 years, as expected for a positive constant intensity. A common mistake is to solve for \(\lambda\) without the factor of 4; the known probability applies to a 4-year interval, so its exponent is \(-4\lambda\), not \(-\lambda\).

Interpretation and Limits of the Model

A transition probability is conditional on the starting state. In particular, the first row describes probabilities for someone starting in state 0, while the second row describes someone starting in state 1. If the starting state is known to be 0, the relevant row is the first row; the probability of being in state 1 after time \(t\) is \(1-e^{-\lambda t}\). The full matrix includes both possible starting states so that it describes the model consistently.

The exponential formula depends on the assumptions, not just on the number \(\lambda\). It applies here because the transition intensity from state 0 is constant over time and state 1 has no exits. If the intensity varies with time, a single constant-rate exponential expression will generally not give the correct survival probability. If there is more than one possible destination from state 0, the complement of the no-transition probability gives the probability of leaving state 0, but does not by itself specify the probability of reaching each destination.

The model also assumes that state membership is completely described by the current state for purposes of this transition calculation. In particular, no extra duration-dependent rate is being used. If the rate depends on how long the individual has already been in state 0, the simple constant-intensity expression may not apply without representing that duration dependence appropriately.

Common Exam Traps

Treating an intensity as an interval probability. The intensity \(\lambda\) is a rate. Over time \(t\), the no-transition probability is \(e^{-\lambda t}\), and the transition probability is \(1-e^{-\lambda t}\).
Using incompatible time units. A rate per month must be multiplied by time in months, or converted before use with time in years.
Reversing the matrix convention. With rows as starting states and columns as destination states, the probability of moving from state 0 to state 1 is in row 0, column 1.
Leaving out the absorbing-state row. A transition matrix includes probabilities for each possible starting state. Starting in state 1 gives the row \((0,1)\), not a blank row.
Assuming the model fits without checking. The formula relies on a constant intensity and one possible destination from state 0, with no exits from state 1.

Key takeaway: For a constant intensity \(\lambda\) from transient state 0 to absorbing state 1, \(p_{00}(t)=e^{-\lambda t}\) and \(p_{01}(t)=1-e^{-\lambda t}\). Include the no-exit row \((0,1)\), use consistent time units, and verify that each row sums to 1.

Check Your Understanding

For each question, assume the stated one-way model has a constant intensity, one transient state 0, and one absorbing state 1. Round probabilities to four decimal places unless another form is requested.

A model has intensity \(0.40\) per year. Find its transition matrix after 2 years. (Answer: \(\begin{pmatrix}0.4493&0.5507\\0&1\end{pmatrix}\))
A hypothetical transition has intensity \(0.06\) per month. Find the probabilities of remaining in state 0 and reaching state 1 after 10 months. (Answer: \(p_{00}=0.5488,\ p_{01}=0.4512\))
The probability of remaining in state 0 after 3 years is 0.75. Find the implied constant intensity per year and the transition probabilities after 8 years. (Answer: \(\lambda=0.09589\) per year; \(p_{00}(8)=0.4644,\ p_{01}(8)=0.5356\))
There are 100 individuals, all initially in state 0, and the intensity is \(0.15\) per year. Use the model to find the expected number in each state after 2 years. (Answer: state 0: \(100e^{-0.30}=74.08\); state 1: \(100(1-e^{-0.30})=25.92\))
A population starts with 30% in state 0 and 70% in state 1. The intensity from state 0 to state 1 is \(0.20\) per year. Find the population proportions in both states after 4 years. (Answer: state 0: \(0.30e^{-0.80}=0.1348\); state 1: \(1-0.1348=0.8652\))

What You'll Learn