From Transition Intensities to Future-State Probabilities
A disability model describes movement among states over time. For example, a person who is active may become disabled, recover to active status, or die; a disabled person may recover or die. The transition intensities describe instantaneous rates of movement, while a transition probability answers a different question: what is the probability of occupying each state after a specified period?
This distinction matters in valuation. A benefit may depend on whether a person is disabled at a future date, not merely on whether disability began at some point during the period. Transition probabilities account for the entire sequence of possible state changes before that date, subject to the model assumptions.
Model and Matrix Setup
Assume a time-homogeneous continuous-time Markov model: transition intensities are constant over the period, and conditional on the current state, the future state development does not depend on the earlier path. Allow transitions from Active to Disabled at intensity \(\alpha\), from Active to Dead at intensity \(\mu_A\), from Disabled to Active at intensity \(\rho\), and from Disabled to Dead at intensity \(\mu_D\). Death is absorbing, so there are no transitions out of Dead.
An intensity is a rate per unit time, not a probability over a whole year. For instance, an intensity of 0.2 per year does not mean that the one-year transition probability is exactly 0.2. Other transitions can occur during the year, and the person may change state more than once.
With states ordered A, D, X, the generator matrix \(Q\) records the transition intensities off the diagonal. Each diagonal entry is the negative sum of the other entries in its row. This makes the row sums of \(Q\) equal to zero.
For a time-homogeneous model, the transition probability matrix is the matrix exponential \(P(t)=\exp(Qt)\). It can be characterized by the Kolmogorov differential equations. With the row convention used here, the forward equation is \(P'(t)=P(t)Q\), and the initial condition is \(P(0)=I\), the identity matrix. These equations describe how the probabilities change as the time horizon increases.
If the initial state is known to be \(i\), the required probabilities are row \(i\) of \(P(t)\). If the initial state is uncertain, let \(\boldsymbol{\pi}(0)\) be a row vector of initial-state probabilities. Then the future distribution is \(\boldsymbol{\pi}(t)=\boldsymbol{\pi}(0)P(t)\). This multiplication averages the transition probabilities over the possible starting states.
A Useful Closed Form in a Symmetric Case
A closed form makes the matrix exponential especially transparent when the disability and recovery intensities are equal and the two mortality intensities are equal. Let \(\alpha=\rho=b\) and \(\mu_A=\mu_D=m\), where \(b\) and \(m\) are constant rates. The transient part of the generator, for A and D, is then \(\begin{pmatrix}-(b+m)&b\\b&-(b+m)\end{pmatrix}\).
The vectors \((1,1)\) and \((1,-1)\) are eigenvectors of this two-state transient matrix. Their respective eigenvalues are \(-m\) and \(-(m+2b)\). Therefore the corresponding components of the matrix exponential are \(e^{-mt}\) and \(e^{-(m+2b)t}\). Combining the components gives the probabilities of being in A or D, before adding the absorbing death state.
The derivation also gives useful checks. The probability of remaining alive from either transient starting state is \(g\), because the first two entries of either transient row sum to \(g\). The death probability is consequently \(1-g\). Each transient row sums to \(g+(1-g)=1\), while the Dead row is \((0,0,1)\).
Worked Example: Future Probabilities From Active Status
A hypothetical disability model has constant annual intensities \(\alpha=\rho=0.2\) per year and \(\mu_A=\mu_D=0.3\) per year. A person is active at time 0. Find the probabilities of being active, disabled, or dead after one year. Use the symmetric-case formula and report probabilities to four decimal places.
Method. Here \(b=0.2\), \(m=0.3\), and \(t=1\) year. First calculate the two exponential terms:
Starting in Active, the probability of being Active after one year is \((g+h)/2=(0.7408182+0.4965853)/2=0.6187018\). The probability of being Disabled is \((g-h)/2=(0.7408182-0.4965853)/2=0.1221165\). The probability of being Dead is \(1-g=1-0.7408182=0.2591818\).
Thus the one-year Active row of the transition matrix, rounded to four decimal places, is \((0.6187, 0.1221, 0.2592)\). The probabilities sum to \(0.6187+0.1221+0.2592=1.0000\), as required.
Common mistake avoided: Multiplying the disability intensity by one year would give \(0.2\), but that is not the probability of being Disabled at the end of the year. The calculation above allows for competing death and for state changes during the year.
Working With an Initial Distribution
A starting population may contain people in more than one state. The correct calculation is to multiply the initial distribution by \(P(t)\), not to select a single row unless everyone starts in the same state. For a distribution \((\pi_A,\pi_D,0)\), the first future-state probability is \(\pi_Ap_{AA}(t)+\pi_Dp_{DA}(t)\). This is a weighted sum of conditional probabilities.
Worked Example: A Population Initially Split Across States
Consider the same hypothetical model as in the previous example: \(\alpha=\rho=0.2\) per year and \(\mu_A=\mu_D=0.3\) per year. At time 0, 80% of the group is Active and 20% is Disabled, with no one Dead. Find the state distribution after one year. Report probabilities to four decimal places.
The one-year transition matrix is obtained from the preceding calculation and the symmetry of this model:
The initial distribution is \(\boldsymbol{\pi}(0)=(0.8,0.2,0)\). Multiply it by \(P(1)\). The Active probability is \(0.8(0.6187018)+0.2(0.1221165)=0.4949614+0.0244233=0.5193847\). The Disabled probability is \(0.8(0.1221165)+0.2(0.6187018)=0.0976932+0.1237404=0.2214336\).
The death probability is \(0.8(0.2591818)+0.2(0.2591818)=0.2591818\). Therefore the distribution is \((0.5194,0.2214,0.2592)\), to four decimal places. The unrounded probabilities sum to 1; the rounded probabilities also sum to 1.0000.
Method and check. Weight each starting-state row by its initial proportion, then check that the resulting probabilities sum to one. A common error is to use only the Active row because Active is the larger starting group; doing so ignores the 20% who start Disabled.
Transition Over Multiple Periods
For a time-homogeneous model, transition matrices satisfy the Chapman–Kolmogorov relationship \(P(s+t)=P(s)P(t)\). To see why, condition on the state occupied at time \(s\). The probability of moving from \(i\) to \(j\) over \(s+t\) is the sum, over each possible intermediate state \(k\), of the probability of reaching \(k\) at time \(s\) multiplied by the probability of reaching \(j\) from \(k\) over the next \(t\) units. Matrix multiplication performs exactly this sum.
This identity allows a long-period calculation to be checked by splitting the period into shorter intervals. It does not mean that the state at the intermediate date can be ignored: every possible intermediate state contributes to the product.
Worked Example: Two-Year Probabilities From One-Year Transitions
Use the same hypothetical model, with \(\alpha=\rho=0.2\) per year and \(\mu_A=\mu_D=0.3\) per year. A person is Active at time 0. Calculate the two-year probability of being Disabled, and the full two-year state distribution. Report probabilities to four decimal places.
Method. First use the one-year transition probabilities \(p=0.6187018\), \(q=0.1221165\), and \(r=0.2591818\), where \(p\) is either transient-state diagonal entry, \(q\) is either transient-state off-diagonal entry, and \(r\) is the one-year death probability from either transient state. By the Chapman–Kolmogorov relationship, \(P(2)=P(1)P(1)\).
For a person starting Active, the two-year Active probability includes the paths A-to-A-to-A and A-to-D-to-A. It is \(p^2+q^2\). Using the closed form gives \((e^{-0.6}+e^{-1.4})/2=(0.5488116+0.2465970)/2=0.3977043\).
The two-year Disabled probability includes the paths A-to-A-to-D and A-to-D-to-D. It is \(pq+qp=2pq\), or \((e^{-0.6}-e^{-1.4})/2=(0.5488116-0.2465970)/2=0.1511073\). The probability of being Dead by year two is \(1-e^{-0.6}=1-0.5488116=0.4511884\).
The resulting two-year distribution is \((0.3977,0.1511,0.4512)\). Its entries sum to 1.0000 after rounding. In particular, \(0.1511\) is the probability of being Disabled at year two, not the probability of ever having become Disabled during the two-year period.
Common mistake avoided: Using only the direct one-year Active-to-Disabled probability, or multiplying that probability by two, misses the intermediate-state paths and is not a valid two-year transition calculation. Matrix multiplication includes all possible intermediate states.
Interpreting and Checking the Results
A transition probability \(p_{AD}(t)\) concerns the state at the end of the period. It includes paths that may involve multiple transitions, such as Active-to-Disabled-to-Active-to-Disabled, when those paths are permitted by the model. It is therefore different from a probability defined by a particular transition history, such as the probability of a first disability onset. State occupancy and the history of how that state was reached are not interchangeable.
For a three-state disability model, useful checks include:
- Every entry of \(P(t)\) must be between 0 and 1, inclusive.
- Each row must sum to 1, because the model includes all possible states at time \(t\).
- The Dead row must remain \((0,0,1)\), because death is absorbing.
- At \(t=0\), \(P(0)=I\): everyone is in their starting state with probability 1.
- For a specified initial distribution, \(\boldsymbol{\pi}(0)P(t)\) must also have nonnegative entries summing to 1.
These checks can expose errors in row and column orientation, omitted death probabilities, or incorrect use of intensities as finite-period probabilities. In applications, the resulting state probabilities can be combined with state-contingent amounts and payment timing to evaluate expected cash flows. The probability calculation itself should come first; payment amounts and discounting are separate valuation steps.
Check Your Understanding
Assume in each question that the model is time-homogeneous, uses the states Active, Disabled, and Dead in that order, and has death as an absorbing state. Report probabilities to four decimal places.
- A hypothetical model has equal Active-to-Disabled and Disabled-to-Active intensities of 0.1 per year, and both mortality intensities are 0.2 per year. Starting Active, find the state distribution after one year. (Answer: Active 0.7445, Disabled 0.0742, Dead 0.1813.)
- In a different hypothetical model, both disability and recovery intensities are 0.15 per year, and both mortality intensities are 0.1 per year. Starting Disabled, find the state distribution after two years. (Answer: Active 0.1847, Disabled 0.6340, Dead 0.1813.)
- Use disability and recovery intensities of 0.2 per year and mortality intensities of 0.3 per year. If 60% start Active and 40% start Disabled, find the state distribution after one year. (Answer: Active 0.4201, Disabled 0.3208, Dead 0.2592.)
- In the model with disability and recovery intensities of 0.1 per year and mortality intensities of 0.2 per year, find the probability of being Disabled after two years given Active at time 0. Then find the probability of being Disabled conditional on being alive at year two. (Answer: Disabled and alive probability 0.1105; conditional probability 0.1648.)