Why an Observation Changes the Calculation
A transition matrix describes how a state model evolves, but a calculation can change when new information arrives partway through the period. If the state at that time is observed exactly, the uncertainty about the current state disappears. If the observation is incomplete or imperfect, some uncertainty remains and must be updated before calculating what may happen next.
For example, a hypothetical long-term care model might track whether a person is independent, receiving care, or deceased. A report made at an intermediate date could identify the state exactly, indicate only that the person is alive, or be an imperfect screening result. These are different conditioning events. The central task is to represent the information at the observation time correctly, then use the transition probabilities for the interval after that time.
Suppose the initial distribution is \(\boldsymbol{\pi}(0)\), and an observation is made at time \(t\). First calculate the distribution immediately before using that observation: \(\boldsymbol{\pi}(t)=\boldsymbol{\pi}(0)P(0,t)\). Then condition this distribution on what was observed. Finally, use \(P(t,u)\) to obtain probabilities at a later time \(u\). For a time-homogeneous model, the transition matrix depends only on the interval length; for a time-inhomogeneous model, use the matrix for the actual start and end times.
Conditioning on Different Kinds of Observation
If the state is observed exactly to be \(j\), then the conditional distribution at time \(t\) puts probability one on state \(j\) and zero on every other state. In vector form this is the row vector \(\mathbf{e}_j\). The later distribution is therefore \(\mathbf{e}_jP(t,u)\), which selects row \(j\) of the transition matrix.
If the observation only establishes that the state belongs to a set \(A\), the updated probability of state \(i\) is its probability before observation divided by the total probability of the states in \(A\), provided \(i\in A\). States outside \(A\) receive updated probability zero.
An imperfect observation requires Bayes' rule. Let \(Y=y\) be the signal observed at time \(t\), and let \(L_i=P(Y=y\mid X_t=i)\) be the probability of that signal in state \(i\). If \(q_i=P(X_t=i)\) immediately before observing the signal, the posterior probability of state \(i\) is proportional to \(q_iL_i\). The normalizing denominator is the probability of observing the signal under the model.
The observation model matters: this formula assumes that the signal likelihoods \(L_i\) are specified for each current state and that the signal provides no additional information about future transitions once the current state is known. This is the relevant Markov assumption for using the updated distribution with \(P(t,u)\).
Worked Example: Exact Observation at the Intermediate Time
Worked Example: Exact Observation at the Intermediate Time
A hypothetical two-state model has states 1 and 2. At time 0, the distribution is \(\boldsymbol{\pi}(0)=(0.6,0.4)\). The transition matrix from time 0 to time 2 years is \(P(0,2)=\begin{pmatrix}0.75&0.25\\0.25&0.75\end{pmatrix}\). At time 2, the state is observed to be state 2 exactly. For the next year, use \(P(2,3)=\begin{pmatrix}0.8&0.2\\0.4&0.6\end{pmatrix}\). Find the conditional distribution at time 3.
Method. Propagate the initial distribution to time 2, condition on the exact observation, and then apply the one-year transition matrix.
Step 1: Propagate to the observation time.
Thus the probability of observing state 2 is \(0.45\). Because the observation identifies the state exactly, the distribution conditional on the observation is \((0,1)\), not the pre-observation distribution \((0.55,0.45)\).
Step 2: Propagate from the observed state.
Conditional on being in state 2 at time 2, the probabilities at time 3 are \(0.4\) for state 1 and \(0.6\) for state 2. The check is \(0.4+0.6=1\).
Common mistake avoided: Applying \(P(2,3)\) directly to \((0.55,0.45)\) would calculate the unconditional distribution at time 3, not the distribution given the observation. The exact observation replaces the uncertain intermediate distribution with \((0,1)\).
Why the Update-and-Propagate Formula Holds
The calculation follows from conditioning on the intermediate state. For an observation \(Y=y\), define \(\alpha_i=P(X_t=i\mid Y=y)\). To find the conditional probability of state \(j\) at time \(u\), partition the event by the possible states \(i\) at time \(t\). By the law of total probability and the Markov property,
The Markov property gives \(P(X_u=j\mid X_t=i,Y=y)=p_{ij}(t,u)\): conditional on the state at time \(t\), the earlier path and its observation do not further change the transition probabilities. In row-vector form, the result is \(\boldsymbol{\alpha}P(t,u)\). This reasoning also shows why the observation must be incorporated before the later transition matrix is applied.
Calculate \(\boldsymbol{\pi}(t)=\boldsymbol{\pi}(0)P(0,t)\), or use the appropriate distribution already given at time \(t\).
For an exact state, use a unit vector. For a state set, renormalize the probabilities in that set. For an imperfect signal, multiply each state probability by its signal likelihood and normalize.
Multiply the updated row vector by \(P(t,u)\), using the transition matrix for the correct interval.
The updated probabilities and the final probabilities should each be nonnegative and sum to one, subject to rounding.
Worked Example: Updating for an Imperfect Signal
Worked Example: Updating for an Imperfect Signal
A hypothetical model has three states. At time 0 the process is known to be in state 1, and the transition matrix over the first two years is \(P(0,2)=\begin{pmatrix}0.4&0.4&0.2\\0.2&0.6&0.2\\0.1&0.3&0.6\end{pmatrix}\). At time 2, a screening signal \(Y=+\) is observed. The probabilities of a positive signal in states 1, 2, and 3 are respectively \(0.2,0.8,0.4\). For the following year, \(P(2,3)=\begin{pmatrix}0.8&0.1&0.1\\0.2&0.5&0.3\\0.1&0.2&0.7\end{pmatrix}\). Find the probabilities of the three states at time 3 given the positive signal.
Step 1: Find the distribution at time 2. The initial distribution is \((1,0,0)\), so it selects the first row of \(P(0,2)\):
Step 2: Apply Bayes' rule. Multiply each state probability by the likelihood of a positive signal in that state. The unnormalized values are \(0.4(0.2)=0.08\), \(0.4(0.8)=0.32\), and \(0.2(0.4)=0.08\). Their sum, the probability of a positive signal, is \(0.08+0.32+0.08=0.48\). Hence
Step 3: Propagate the updated distribution.
The final distribution is \((17/60,23/60,20/60)\), or \((0.2833,0.3833,0.3333)\) to four decimal places. The fractions sum to \(60/60=1\), confirming the calculation.
Common mistake avoided: The signal likelihoods are not the posterior probabilities. For example, \(0.8\) is the probability of a positive signal given state 2; the posterior probability of state 2 is \(0.32/0.48=2/3\), after accounting for the prior state distribution and normalizing.
Worked Example: Observation of a State Set
Worked Example: Observation of a State Set
At an intermediate time \(t\), a hypothetical model has state distribution \((0.2,0.5,0.3)\). An audit establishes that the process is in either state 1 or state 2, but cannot distinguish between them. For the following interval, the transition matrix is \(P(t,u)=\begin{pmatrix}0.7&0.2&0.1\\0.1&0.6&0.3\\0.2&0.2&0.6\end{pmatrix}\). Find the conditional distribution at time \(u\).
Step 1: Condition on the reported set. The probability of being in the reported set is \(0.2+0.5=0.7\). The conditional probabilities of states 1 and 2 are \(0.2/0.7=2/7\) and \(0.5/0.7=5/7\); state 3 has conditional probability zero. Thus the updated distribution is \((2/7,5/7,0)\).
Step 2: Apply the later transition matrix.
The result is \((19/70,34/70,17/70)\). The numerator check is \(19+34+17=70\). Treating the set observation as if it identified state 2 would incorrectly use only the second row of the matrix. The correct calculation retains the relative probabilities of states 1 and 2 after conditioning.
Checks and Common Traps
A conditional distribution is a probability distribution under the information actually observed. Before multiplying by a later transition matrix, ask whether the intermediate state is known exactly, known only to lie in a set, or inferred from a signal. The corresponding updates are different.
- Skipping the update: Propagating the unconditional distribution answers a different question from one conditioned on an observation.
- Using likelihoods as posterior probabilities: A signal's probability given a state must be combined with the probability of that state before the signal.
- Discarding uncertainty after a partial observation: If an observation identifies a set of states, retain all states in the set and renormalize their probabilities.
- Using the wrong interval matrix: The transition after observation is \(P(t,u)\), not the matrix for the interval before the observation. In a time-inhomogeneous model, the actual start and end times matter.
- Reversing multiplication order: With row distributions, multiply on the right: \(\boldsymbol{\alpha}P(t,u)\). Check dimensions and that the resulting entries sum to one.
These conditional calculations are useful whenever information arrives between valuation dates or model intervals. They keep the state probabilities consistent with the information available at each time, a basic requirement for applying multiple-state transition models in actuarial work. The methods here are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, update for the stated observation before applying the later transition matrix.
- A process starts in state 1. Over the first interval its transition matrix is \(\begin{pmatrix}0.7&0.3\\0.2&0.8\end{pmatrix}\). At the intermediate time, state 2 is observed exactly. The next-interval matrix is \(\begin{pmatrix}0.9&0.1\\0.4&0.6\end{pmatrix}\). What is the conditional probability of state 1 at the end? (Answer: \(0.4000\))
- At the observation time, the distribution is \((0.3,0.7)\). An observation establishes that the state is in the set \(\{1\}\). The next transition matrix is \(\begin{pmatrix}0.6&0.4\\0.1&0.9\end{pmatrix}\). Find the conditional distribution one interval later. (Answer: \((0.6000,0.4000)\))
- At an intermediate time, the state distribution is \((0.5,0.5)\). A positive signal has likelihoods \(0.2\) in state 1 and \(0.8\) in state 2. The next-interval matrix is \(\begin{pmatrix}0.8&0.2\\0.3&0.7\end{pmatrix}\). Find the probability of state 2 at the end given the positive signal. (Answer: \(0.6167\))
- A process starts in state 1, with first-interval matrix \(\begin{pmatrix}0.5&0.3&0.2\\0.2&0.5&0.3\\0.1&0.2&0.7\end{pmatrix}\). At the intermediate time, a report says the state is either 2 or 3. For the next interval use \(\begin{pmatrix}0.7&0.2&0.1\\0.2&0.5&0.3\\0.1&0.3&0.6\end{pmatrix}\). Find the conditional probability of state 3 at the end. (Answer: \(0.4714\))
- At an intermediate time the state distribution is \((0.2,0.5,0.3)\). A signal has state-specific likelihoods \((0.5,0.2,0.8)\), and the following transition matrix is \(\begin{pmatrix}0.6&0.3&0.1\\0.1&0.7&0.2\\0.2&0.2&0.6\end{pmatrix}\). Find the conditional distribution at the later time given the signal. (Answer: \((0.2444,0.4333,0.3222)\))