Time-Inhomogeneous Intensity Matrices

Why Let Intensities Vary With Time?

In a continuous-time state model, the chance of a transition over a very short interval is described by an intensity. A model is time-inhomogeneous when those intensities may change with time. For example, a hypothetical benefit contract might have an increasing rate of transition from active to disabled status as the model’s time coordinate advances. A single fixed matrix cannot represent that change; instead, the model uses an intensity matrix whose entries are functions of time.

The matrix at a given time still follows the same structural rules as a time-homogeneous intensity matrix: off-diagonal entries are nonnegative transition intensities, and each diagonal entry makes its row sum to zero. What changes is that an entry can have a different value at different times. This tutorial focuses on representing and evaluating that time-varying matrix, not on calculating finite-interval transition probabilities.

Definition: For a continuous-time Markov model with states \(0,1,\ldots,m\), let \(q_{ij}(t)\) be the transition intensity from state \(i\) to state \(j\) at time \(t\), for \(i\ne j\). The time-inhomogeneous intensity matrix is \(Q(t)=[q_{ij}(t)]\), with diagonal entries defined by \(q_{ii}(t)=-\sum_{j\ne i}q_{ij}(t)\). Thus, every entry of \(Q(t)\), including each diagonal entry, is a function of time.

Meaning and Structure of the Matrix

For \(i\ne j\), the intensity is defined locally by

q_{ij}(t)=\lim_{h\downarrow 0}\frac{P(\text{state at }t+h\text{ is }j\mid\text{state at }t\text{ is }i)}{h},

when this limit exists. Its units are transitions per unit of time. The matrix uses a fixed ordering of states: row \(i\) describes the possible exits from state \(i\), and column \(j\) identifies the destination. The functions are understood to be defined on the time range of the model, with nonnegative off-diagonal values there.

The model is Markov when, conditional on the current state and time, the future state behavior does not additionally depend on the earlier path. Allowing the intensities to vary with time does not change that requirement. It changes the rates used at different times.

The diagonal is not an additional transition rate to another state. It is the negative total exit intensity from its row. Consequently, a state with no possible exits has a zero row, including a zero diagonal; such a state is absorbing.

Why the Diagonal Must Also Vary

The row-sum rule follows from the fact that the probabilities of all possible states after a short interval must sum to one. Suppose the process is in state \(i\) at time \(t\). For each other state \(j\), the probability of moving from \(i\) to \(j\) over a short interval of length \(h\) is \(q_{ij}(t)h+o(h)\). The probability of remaining in \(i\) is therefore one minus the sum of these outgoing probabilities, up to terms smaller than order \(h\). Dividing the resulting first-order change by \(h\) gives the diagonal rule.

\begin{aligned} 1 &=P(X_{t+h}=i\mid X_t=i)+\sum_{j\ne i}P(X_{t+h}=j\mid X_t=i)\\ &=1+q_{ii}(t)h+\sum_{j\ne i}q_{ij}(t)h+o(h),\\ 0&=q_{ii}(t)+\sum_{j\ne i}q_{ij}(t),\\ q_{ii}(t)&=-\sum_{j\ne i}q_{ij}(t). \end{aligned}

Here \(o(h)\) denotes terms whose ratio to \(h\) tends to zero as \(h\) tends to zero. This is a local argument; it does not say that \(q_{ij}(t)h\) is an exact transition probability over an arbitrary interval. The diagonal has to be recalculated whenever any outgoing intensity changes with time.

Worked Example: Evaluating a Three-State Matrix at a Given Time

A hypothetical income-protection model has states \(W\) (working), \(D\) (disabled), and \(X\) (dead), in that order. Time \(t\) is measured in years, and the intensities, in transitions per year, are \(q_{WD}(t)=0.04+0.01t\), \(q_{WX}(t)=0.01+0.002t\), \(q_{DW}(t)=0.20+0.02t\), and \(q_{DX}(t)=0.03+0.005t\). State \(X\) is absorbing. Construct and evaluate the intensity matrix at \(t=2\).

Method. Evaluate the stated off-diagonal functions first, place them in the appropriate row and column, and then obtain each diagonal from the negative sum of its row’s outgoing intensities.

At \(t=2\), the working-to-disabled intensity is \(0.04+0.01(2)=0.060\) per year, and the working-to-dead intensity is \(0.01+0.002(2)=0.014\) per year. The total exit intensity from \(W\) is \(0.060+0.014=0.074\) per year, so \(q_{WW}(2)=-0.074\) per year.

The disabled-to-working intensity is \(0.20+0.02(2)=0.240\) per year, and the disabled-to-dead intensity is \(0.03+0.005(2)=0.040\) per year. The total exit intensity from \(D\) is \(0.240+0.040=0.280\) per year, so \(q_{DD}(2)=-0.280\) per year. Since \(X\) is absorbing, all entries in its row are zero.

Q(2)= \begin{pmatrix} -0.074 & 0.060 & 0.014\\ 0.240 & -0.280 & 0.040\\ 0 & 0 & 0 \end{pmatrix} \quad\text{per year}.

Check. The row sums are \(-0.074+0.060+0.014=0\), \(-0.280+0.240+0.040=0\), and \(0+0+0=0\). A common mistake is to leave the diagonal at a value from another time even though the outgoing rates have changed. Recomputing it from the evaluated rates avoids that error.

Each Matrix Entry Can Depend on Time

The functional notation \(Q(t)\) is more than a label: it means that the entries are evaluated at the same time argument. If the off-diagonal intensities are given as functions, the diagonal functions follow by adding the outgoing functions in each row and taking the negative. It is not necessary for every off-diagonal function to change; some may be constant while others vary.

For example, if a row has two outgoing intensities \(a(t)\) and \(b(t)\), its diagonal function is \(-a(t)-b(t)\), not a constant chosen once. A matrix can be valid at one time but invalid at another if an off-diagonal entry becomes negative or if a diagonal no longer equals the negative row total. Rates must be specified on a domain where the model is intended to operate.

Worked Example: Constructing a Matrix With a Reversible Transition

Consider a hypothetical contract with states \(A\) (active), \(L\) (lapsed), and \(D\) (dead), in that order. The rates, in transitions per year, are \(q_{AL}(t)=0.02+0.003t\), \(q_{AD}(t)=0.008+0.001t\), \(q_{LA}(t)=0.04\), and \(q_{LD}(t)=0.02+0.002t\); death is absorbing. Find \(Q(5)\).

Method. Evaluate every outgoing rate at year 5, including the constant rate, and compute the diagonal separately for each transient-state row.

From state \(A\), \(q_{AL}(5)=0.02+0.003(5)=0.035\) per year and \(q_{AD}(5)=0.008+0.001(5)=0.013\) per year. Thus, \(q_{AA}(5)=-(0.035+0.013)=-0.048\) per year.

From state \(L\), \(q_{LA}(5)=0.040\) per year and \(q_{LD}(5)=0.02+0.002(5)=0.030\) per year. Thus, \(q_{LL}(5)=-(0.040+0.030)=-0.070\) per year. The row for \(D\) is zero because death is absorbing.

Q(5)= \begin{pmatrix} -0.048 & 0.035 & 0.013\\ 0.040 & -0.070 & 0.030\\ 0 & 0 & 0 \end{pmatrix} \quad\text{per year}.

Check. The first row sums to \(-0.048+0.035+0.013=0\); the second sums to \(-0.070+0.040+0.030=0\); and the death row sums to zero. The possibility of a transition from \(L\) back to \(A\) is represented by the off-diagonal entry \(q_{LA}(t)\), not by changing the meaning of the diagonal.

Worked Example: Building the Full Matrix From Rate Functions

A hypothetical rehabilitation model has states \(H\) (home), \(R\) (receiving rehabilitation), and \(D\) (dead), in that order. For \(0\le t\le4\) years, its intensities per year are \(q_{HR}(t)=0.05+0.01t\), \(q_{HD}(t)=0.02\), \(q_{RH}(t)=0.12-0.01t\), and \(q_{RD}(t)=0.04+0.005t\); \(D\) is absorbing. Write \(Q(t)\) and evaluate it at \(t=3\).

Method. First place the four given functions according to the stated state order. Then express each transient-state diagonal as the negative sum of the two intensities leaving that state.

For the \(H\) row, \(q_{HH}(t)=-[(0.05+0.01t)+0.02]=-0.07-0.01t\). For the \(R\) row, \(q_{RR}(t)=-[(0.12-0.01t)+(0.04+0.005t)]=-0.16+0.005t\). The death row is zero. Therefore,

Q(t)= \begin{pmatrix} -0.07-0.01t & 0.05+0.01t & 0.02\\ 0.12-0.01t & -0.16+0.005t & 0.04+0.005t\\ 0 & 0 & 0 \end{pmatrix} \quad\text{per year}.

At \(t=3\), the \(H\)-to-\(R\) rate is \(0.05+0.01(3)=0.080\), and the \(H\)-to-\(D\) rate is \(0.020\). Their sum is \(0.100\), giving \(q_{HH}(3)=-0.100\). The \(R\)-to-\(H\) rate is \(0.12-0.01(3)=0.090\), while the \(R\)-to-\(D\) rate is \(0.04+0.005(3)=0.055\). Their sum is \(0.145\), giving \(q_{RR}(3)=-0.145\). Hence,

Q(3)= \begin{pmatrix} -0.100 & 0.080 & 0.020\\ 0.090 & -0.145 & 0.055\\ 0 & 0 & 0 \end{pmatrix} \quad\text{per year}.

Check. The row sums are \(-0.100+0.080+0.020=0\), \(-0.145+0.090+0.055=0\), and zero for the absorbing state. The stated domain also matters: these functions are specified for \(0\le t\le4\), not as an unrestricted description for all times. A common error is to substitute \(t=3\) into only the off-diagonal entries and retain a diagonal from another time; the diagonal must agree with the same-time rates.

Common Mistakes and Practical Checks

A reliable way to read or construct \(Q(t)\) is to fix the state order, identify the allowed destinations in each row, evaluate the outgoing intensities at the requested time, and then calculate the diagonal. Check both the signs and the row sums. In particular, an off-diagonal entry must be nonnegative on the model’s stated time domain, while a diagonal is nonpositive because it is the negative of a sum of nonnegative rates.

Treating an intensity as a probability. An intensity is a rate per unit time, not the probability of a transition over a year or another finite interval. For a short interval of length \(h\), \(q_{ij}(t)h\) gives a first-order approximation, not an exact finite-interval probability.
Using inconsistent times within a matrix. When asked for \(Q(t_0)\), evaluate all time-varying entries and diagonals at \(t_0\).
Forgetting the diagonal after evaluating rates. A diagonal is determined by the other entries in its row; it is not an independent function to guess.
Misreading the state order. The position of an entry depends on the stated row and column order. A rate from \(R\) to \(D\) belongs in row \(R\), column \(D\).
Giving an absorbing state a negative diagonal. If the state has no outgoing transitions, its entire row is zero.

Time-inhomogeneous intensity matrices provide the appropriate local description when transition rates change with time. They preserve the familiar intensity-matrix structure while making that structure explicit at each time. Being able to construct and check \(Q(t)\) is a foundation for later work with evolving state models and is aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, use the stated state order and units. Answers are provided for checking.

A hypothetical two-state model has states \(C\) (covered) and \(D\) (dead), with \(q_{CD}(t)=0.015+0.002t\) per year and \(D\) absorbing. Find \(Q(4)\). (Answer: \(\begin{pmatrix}-0.023&0.023\\0&0\end{pmatrix}\) per year.)
A model has states \(U\) (under care), \(R\) (recovered), and \(D\) (dead). At \(t=2\), its outgoing rates are \(q_{UR}=0.03+0.005t\), \(q_{UD}=0.01\), \(q_{RU}=0.08\), and \(q_{RD}=0.02+0.004t\), in transitions per year; death is absorbing. Find the full matrix at that time. (Answer: \(\begin{pmatrix}-0.050&0.040&0.010\\0.080&-0.108&0.028\\0&0&0\end{pmatrix}\) per year.)
In state \(A\), the only exits are to \(B\) and \(D\), with \(q_{AB}(t)=0.06-0.01t\) and \(q_{AD}(t)=0.02\), in transitions per year. Find the \(A\)-row diagonal at \(t=3\). (Answer: \(q_{AA}(3)=-0.050\) per year.)
For states \(1,2,3\), a proposed matrix at a particular time has rows \((-0.05,0.03,0.02)\), \((0.04,-0.05,0.01)\), and \((0,0,0)\), with entries in transitions per year. Does it satisfy the intensity-matrix sign and row-sum conditions? (Answer: Yes; all off-diagonal entries are nonnegative and each row sums to zero.)
On \(0\le t\le6\), a proposed off-diagonal intensity is \(q_{12}(t)=0.02-0.005t\) per year. Is this function a valid transition intensity throughout the stated domain, including at \(t=5\)? (Answer: No; \(q_{12}(5)=0.02-0.005(5)=-0.005\) per year, which is negative.)

What You'll Learn