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Landmarking and Dynamic Prediction · Tutorial 483 of 699

Worked Illness-Death Predictions at a Landmark

Calculate horizon-specific illness-death state probabilities from a landmark, and distinguish being ill at the horizon from having experienced illness before it.

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What You'll Learn

  • Define the healthy, ill, and dead states and condition predictions on a patient’s state at the landmark.
  • Calculate state probabilities under a constant-intensity illness-death model.
  • Multiply estimated transition matrices to obtain predictions at successive horizons.
  • Handle the equal-rate case without dividing by zero.
  • Check that predictions respect the allowed transitions and sum to one.
  • Distinguish current illness-state occupation from having ever become ill.

What an Illness-Death Prediction Answers

At a landmark time \(s\), a clinician may want to know the probability that a patient who is healthy then will be healthy, ill, or dead at a later time \(s+t\). For a patient already ill at \(s\), the corresponding prediction is the probability of still being ill or having died. These are conditional state-occupation probabilities: they describe the patient’s state at a specified future horizon, given the state at the landmark.

The state at the landmark matters. A patient healthy at \(s\) and a patient ill at \(s\) have different possible paths and different conditional predictions. The horizon \(t\) is elapsed time after the landmark; it is not the original time since study entry. The landmark state and the prediction horizon must therefore be stated together when reporting a probability.

Definition: In the illness-death model, state \(0\) denotes healthy, state \(1\) denotes ill, and state \(2\) denotes dead. The allowed transitions are \(0\to1\), \(0\to2\), and \(1\to2\); death is absorbing, and recovery from illness is not included. Write \(p_{ij}(t)\) for the probability of being in state \(j\) after \(t\) time units, conditional on being in state \(i\) at the landmark. Thus \(p_{ij}(t)=P\{X(s+t)=j\mid X(s)=i\}\) under the specified prediction model.

For a patient healthy at the landmark, the three horizon probabilities are \(p_{00}(t)\), \(p_{01}(t)\), and \(p_{02}(t)\). For a patient ill at the landmark, \(p_{11}(t)\) and \(p_{12}(t)\) are the possible state probabilities; \(p_{10}(t)=0\). Because the states are mutually exclusive and exhaustive, each applicable set of probabilities must sum to one.

Constant Transition Intensities

A useful model for hand calculation assumes a time-homogeneous continuous-time Markov illness-death process. In this model, the transition intensities are constant: \(\lambda_{01}=a\) per unit time for becoming ill while healthy, \(\lambda_{02}=b\) for dying while healthy, and \(\lambda_{12}=c\) for dying while ill. The Markov assumption means that, conditional on the current state, the future transition behaviour does not additionally depend on the patient’s earlier path or duration in that state. Constant intensities make the probabilities depend on elapsed time \(t\) only.

These assumptions are substantive. They are suitable for the calculations below, not automatic facts about illness-death data. In applications, a constant intensity may be a rough approximation over a particular period. If estimated transition probabilities vary over time, a time-varying approach such as multiplying Aalen–Johansen transition matrices is more appropriate. Both approaches still require a valid state definition and conditioning on the state at the landmark.

While healthy, the patient leaves state 0 at total rate \(q=a+b\). Consequently, the probability of remaining healthy for \(t\) time units is \(e^{-qt}\). While ill, the only possible transition is to death at rate \(c\), so the probability of remaining ill is \(e^{-ct}\).

$$ p_{00}(t)=e^{-(a+b)t},\qquad p_{11}(t)=e^{-ct},\qquad p_{12}(t)=1-e^{-ct}. $$

For a patient starting healthy, the probability of being ill and alive at the horizon requires two transitions in the right order: first \(0\to1\), then no death from state 1 before \(t\). If the first transition occurs at elapsed time \(u\), the probability contribution has density \(e^{-(a+b)u}a\), followed by survival in state 1 for \(t-u\), with probability \(e^{-c(t-u)}\).

$$ \begin{aligned} p_{01}(t) &=\int_0^t e^{-(a+b)u}\,a\,e^{-c(t-u)}\,du\\ &=a e^{-ct}\int_0^t e^{-(a+b-c)u}\,du\\ &=\frac{a\{e^{-ct}-e^{-(a+b)t}\}}{a+b-c}, \qquad a+b\ne c. \end{aligned} $$

The integrand is nonnegative, so the integral gives a nonnegative probability even when the displayed numerator and denominator are both negative. When \(a+b=c\), the integrand simplifies to \(a e^{-ct}\), giving \(p_{01}(t)=at e^{-ct}\). Since the only possible states at the horizon are healthy, ill, and dead, the remaining probability is assigned to death.

$$ p_{02}(t)=1-p_{00}(t)-p_{01}(t). $$

Worked Example: Predicting Two-Year State Probabilities

Suppose, hypothetically, that a constant-intensity model has \(a=0.20\) per year, \(b=0.10\) per year, and \(c=0.40\) per year. Calculate the two-year probabilities for patients healthy and ill at the landmark. The assumed illness-death structure permits only the transitions defined above, and the rates are taken to be constant over these two years.

For a patient healthy at the landmark, \(a+b=0.30\) per year. The probability of still being healthy after two years is

$$ p_{00}(2)=e^{-0.30(2)}=e^{-0.6}=0.5488. $$

The illness-state probability is

$$ \begin{aligned} p_{01}(2) &=\frac{0.20\{e^{-0.40(2)}-e^{-0.30(2)}\}}{0.30-0.40}\\ &=\frac{0.20(0.4493-0.5488)}{-0.10}\\ &=0.1990. \end{aligned} $$

Thus the probability of death by the two-year horizon, meaning being in state 2 at that time, is

$$ p_{02}(2)=1-0.5488-0.1990=0.2522. $$

The values sum to \(0.5488+0.1990+0.2522=1.0000\). Under this model, a patient healthy at the landmark has a 54.88% probability of being healthy, a 19.90% probability of being ill and alive, and a 25.22% probability of being dead two years later. For a patient already ill at the landmark,

$$ p_{11}(2)=e^{-0.40(2)}=0.4493,\qquad p_{12}(2)=1-0.4493=0.5507. $$

These are conditional predictions for different landmark states, not a comparison of two randomly assigned groups.

Predictions From Estimated Transition Matrices

The constant-rate formulas are not the only way to calculate predictions. With estimates of transition probabilities over successive intervals, the state distribution at the end of an interval is multiplied by the interval’s transition matrix. This approach allows transition probabilities to differ from one interval to the next. The matrix must use the same state ordering throughout, and its rows must correspond to the starting state while its columns correspond to the destination state.

The following hypothetical matrices illustrate this calculation. They are assumed to be valid estimated transition matrices over the two one-year intervals following the landmark. Their zero entries reflect the illness-death model’s prohibited transitions; each row sums to one.

$$ P_1= \begin{pmatrix} 0.70&0.20&0.10\\ 0&0.80&0.20\\ 0&0&1 \end{pmatrix}, \qquad P_2= \begin{pmatrix} 0.75&0.15&0.10\\ 0&0.70&0.30\\ 0&0&1 \end{pmatrix}. $$

These matrices could be obtained from Aalen–Johansen transition estimates under appropriate conditions, including independent censoring for estimation and a Markov interpretation when chaining the conditional transition matrices. The example assumes those conditions and adequate data support for the transitions. It does not assume that the two intervals have identical transition probabilities.

Worked Example: Chaining Two Estimated Intervals

Using the hypothetical matrices above, calculate the two-year state probabilities for patients healthy and ill at the landmark. A healthy patient has initial state distribution \((1,0,0)\). After the first interval this becomes \((0.70,0.20,0.10)\). Multiplication by \(P_2\) gives

$$ \begin{aligned} (0.70,0.20,0.10)P_2 &=\big( 0.70(0.75),\, 0.70(0.15)+0.20(0.70),\\ &\hspace{2em}0.70(0.10)+0.20(0.30)+0.10(1) \big)\\ &=(0.525,0.245,0.230). \end{aligned} $$

The two-year probabilities are therefore 0.525 of being healthy, 0.245 of being ill and alive, and 0.230 of being dead. They add to one. Starting instead in state 1, the first-interval distribution is \((0,0.80,0.20)\), so

$$ (0,0.80,0.20)P_2 =\big(0,\;0.80(0.70),\;0.80(0.30)+0.20(1)\big) =(0,0.560,0.440). $$

For a patient ill at the landmark, the estimated probabilities are 0.560 of still being ill and 0.440 of being dead at two years. The calculation carries forward the full state distribution after the first interval; it does not apply the second matrix to the original landmark state.

Edge Cases and Interpretation

Worked Example: When the Healthy Exit Rate Equals the Ill Death Rate

Suppose, hypothetically, \(a=0.25\), \(b=0.15\), and \(c=0.40\) per year, and consider a two-year horizon. Here \(a+b=c=0.40\), so the formula with denominator \(a+b-c\) cannot be used directly. The equal-rate result applies:

$$ \begin{aligned} p_{00}(2)&=e^{-0.40(2)}=0.4493,\\ p_{01}(2)&=0.25(2)e^{-0.40(2)}=0.2247,\\ p_{02}(2)&=1-0.4493-0.2247=0.3260. \end{aligned} $$

The probabilities are nonnegative and sum to one. From the ill state, \(p_{11}(2)=0.4493\) and \(p_{12}(2)=0.5507\). This case is a useful numerical check: a zero denominator signals a need for the limiting formula, not a probability of zero.

There is an important distinction between being ill at the horizon and having become ill by the horizon. The probability \(p_{01}(t)\) counts patients who are in state 1 at time \(s+t\). It does not count those who entered state 1 and then died before the horizon. In the constant-rate model, the probability that a healthy patient has entered state 1 at least once by time \(t\) is \(a\{1-e^{-(a+b)t}\}/(a+b)\): the first exit from state 0 must be illness, and must occur by \(t\). With the rates in the first example, this probability by two years is \(0.20(1-e^{-0.6})/0.30=0.3008\), which exceeds the 0.1990 probability of occupying state 1 at two years. Use the estimand that matches the question.

Checks, Alternatives, and Limitations

Before reporting an illness-death prediction, check the state ordering, the conditioning state, the elapsed horizon, and the model assumptions. For a healthy starting state, \(p_{00}(t)\) should not increase with \(t\) under this model, because state 0 has no incoming transitions. The probability of being dead should not decrease, because death is absorbing. The probability of occupying state 1 need not increase monotonically: it can rise as patients become ill and later fall as they die. A claim that illness-state occupation must always rise confuses an entry probability with a state-occupation probability.

  • Check the matrix orientation. With row distributions, update by multiplying on the right. A mistaken transpose can send probability along the wrong transitions.
  • Check probability arithmetic. Every row of a transition matrix and every predicted state distribution must sum to one, apart from rounding.
  • Check model support. A fitted or supplied matrix with a positive probability for recovery contradicts the three-state illness-death model used here.
  • Check the time scale. Rates measured per year require \(t\) in years; probabilities over intervals must start at the landmark and cover the stated horizon.
  • Check the conditional population. A prediction for patients healthy at the landmark does not describe patients who were healthy at initial study entry but are ill by the landmark.

Constant-intensity formulas are compact and interpretable, but they impose a strong time-homogeneity assumption. Interval-specific Aalen–Johansen transition matrices can reflect changes over time without requiring one set of constant rates, provided the transition estimates are suitable for the target population and the Markov conditioning is appropriate. Either approach can fail when there is inadequate risk-set support, informative censoring, or important history dependence not represented by the current state. The earlier material on Markov state-occupation prediction and Aalen–Johansen state-occupation estimates after a landmark gives the general framework; this tutorial has focused on carrying out the illness-death calculations.

A frequent mistake is to report \(p_{01}(t)+p_{02}(t)\) as the probability of ever becoming ill. That sum is the probability of not being healthy at the horizon. It includes direct deaths from state 0, whether or not illness occurred first. The correction is to define the target event precisely and use a path-based probability for ever entering illness, rather than treating a horizon state probability as a history indicator.

Check Your Understanding

For each question, identify the conditioning state and horizon before calculating. Solutions are included below.

  1. In the constant-rate model with \(a=0.10\), \(b=0.10\), and \(c=0.30\) per year, calculate the one-year probability of remaining healthy from state 0.
  2. Using the same rates, calculate the one-year probability of remaining ill from state 1 and the probability of death by one year from state 1.
  3. A distribution after one interval is \((0.60,0.30,0.10)\), and the next transition matrix is \( \begin{pmatrix}0.80&0.10&0.10\\0&0.75&0.25\\0&0&1\end{pmatrix} \). Calculate the distribution after the second interval.
  4. Why is the probability of being ill at a horizon generally not the probability of having ever become ill by that horizon?

Solutions. (1) The healthy exit rate is \(a+b=0.20\) per year, so \(p_{00}(1)=e^{-0.20}=0.8187\). (2) From state 1, \(p_{11}(1)=e^{-0.30}=0.7408\), and \(p_{12}(1)=1-0.7408=0.2592\); these sum to one. (3) Multiply the row distribution by the matrix: the healthy probability is \(0.60(0.80)=0.48\); the ill probability is \(0.60(0.10)+0.30(0.75)=0.285\); the death probability is \(0.60(0.10)+0.30(0.25)+0.10(1)=0.235\). The result \((0.480,0.285,0.235)\) sums to one. (4) A patient may enter the ill state and then die before the horizon. Such a patient has ever become ill but is counted in state 2, not state 1, at the horizon.

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