Age-Dependent Transition Intensities

Why Attained Age Matters

A transition rate may change as a person gets older. For example, a hypothetical insurance model might assign a different rate of transition from active to disabled at age 65 than at age 45. To represent this, the model must distinguish a person’s attained age from the time elapsed since the model began.

Suppose a person is age \(x\) at time \(0\), and let \(t\) be elapsed time in years. If age increases continuously at one year per year, attained age at time \(t\) is \(x+t\). An age-dependent transition intensity is therefore evaluated at \(x+t\), not at \(t\) alone. For a fixed starting age \(x\), the resulting intensity matrix can be written as a function of elapsed time, but its underlying specification is in terms of attained age.

Definition: An age-dependent transition intensity from state \(i\) to state \(j\) is a rate specified as a function of attained age \(y\), denoted \(\mu_{ij}(y)\). For a person starting at age \(x\), the intensity at elapsed time \(t\) is \(\mu_{ij}(x+t)\). Intensities are nonnegative rates, usually expressed per year. The diagonal entry is the negative of the sum of the off-diagonal intensities in that row.

This specification assumes that the transition rates depend on attained age as described and do not also depend on calendar date or on an unrecorded feature such as time already spent in the current state. If rates also depend on calendar time, that dependence must be represented separately. For this tutorial, assume age advances deterministically and all ages and elapsed times use the same time unit.

From an Age Function to an Intensity Matrix

Use the row convention: row \(i\) describes transitions out of state \(i\). At attained age \(y\), the off-diagonal entry in row \(i\), column \(j\) is \(\mu_{ij}(y)\). The diagonal entry is \(-\sum_{j\ne i}\mu_{ij}(y)\), so each row sums to zero. Forbidden transitions have intensity zero, and an absorbing state has no outgoing intensities.

For a person who starts at age \(x\), substitute \(y=x+t\) into every age-based intensity. This gives a matrix \(Q(x+t)\) at elapsed time \(t\). A model with rates fixed across ages instead uses the same intensity values at every attained age; such a model is age-homogeneous, even though its states may still change over time.

Q(y)=\bigl(q_{ij}(y)\bigr), \qquad q_{ij}(y)=\mu_{ij}(y)\ (i\ne j), \qquad q_{ii}(y)=-\sum_{j\ne i}\mu_{ij}(y).

The distinction is about the rate specification, not about whether a person’s age changes. In an age-dependent model, the same person can have different rates at different attained ages. In a model with rates fixed across ages, the rate assigned to a particular transition is constant as age changes. A rate can be constant with age for one transition and age-dependent for another.

Why the Diagonal Changes With Age

The diagonal entry is determined by the outgoing rates at that age. Over a short interval of length \(h\), conditional on being in state \(i\) at elapsed time \(t\), the probability of one particular transition to \(j\ne i\) is \(\mu_{ij}(x+t)h+o(h)\). The probability of leaving state \(i\) by any allowed route is therefore the sum of these first-order terms. The probability of remaining in \(i\) is one minus that sum, to first order.

\begin{aligned} P(\text{leave }i\text{ during }[t,t+h)\mid X(t)=i) &=\sum_{j\ne i}\mu_{ij}(x+t)h+o(h),\\ P(X(t+h)=i\mid X(t)=i) &=1-\left(\sum_{j\ne i}\mu_{ij}(x+t)\right)h+o(h). \end{aligned}

Thus the diagonal intensity is the negative total outgoing intensity evaluated at the same attained age as the off-diagonal rates. This is a local justification of the row-sum rule; it does not say that an intensity is itself a finite-interval transition probability.

Worked Example: Building Matrices at Two Attained Ages

Consider a hypothetical contract with states Active, Disabled, and Dead. Assume a person starts at age 60, age increases by one year per year, and all intensities below are annual rates. The only allowed transitions are Active to Disabled, Active to Dead, Disabled to Active, and Disabled to Dead. Dead is absorbing. Specify \(\mu_{AD}(y)=0.006+0.001(y-60)\), \(\mu_{A,\text{Dead}}(y)=0.004+0.0005(y-60)\), \(\mu_{DA}(y)=0.020\), and \(\mu_{D,\text{Dead}}(y)=0.010+0.001(y-60)\). Find the intensity matrices at elapsed times \(0\) and \(4\).

Method. First find attained age \(y=60+t\). Then evaluate each allowed off-diagonal rate at that age and take the negative row totals for the diagonal entries.

At \(t=0\), attained age is \(60+0=60\). The rates are \(\mu_{AD}(60)=0.006+0.001(0)=0.006\), \(\mu_{A,\text{Dead}}(60)=0.004+0.0005(0)=0.004\), \(\mu_{DA}(60)=0.020\), and \(\mu_{D,\text{Dead}}(60)=0.010+0.001(0)=0.010\). The Active row total is \(0.006+0.004=0.010\); the Disabled row total is \(0.020+0.010=0.030\). Hence

Q(60)= \begin{pmatrix} -0.010 & 0.006 & 0.004\\ 0.020 & -0.030 & 0.010\\ 0 & 0 & 0 \end{pmatrix}.

At \(t=4\), attained age is \(60+4=64\). The rates are \(\mu_{AD}(64)=0.006+0.001(4)=0.010\), \(\mu_{A,\text{Dead}}(64)=0.004+0.0005(4)=0.006\), \(\mu_{DA}(64)=0.020\), and \(\mu_{D,\text{Dead}}(64)=0.010+0.001(4)=0.014\). The row totals are \(0.010+0.006=0.016\) and \(0.020+0.014=0.034\). Therefore

Q(64)= \begin{pmatrix} -0.016 & 0.010 & 0.006\\ 0.020 & -0.034 & 0.014\\ 0 & 0 & 0 \end{pmatrix}.

Check. Each row sums to zero, and every off-diagonal entry is nonnegative. The Disabled-to-Active rate remains \(0.020\) because its specified function is constant in age; the other changing rates increase as attained age increases. The common mistake avoided here is using the starting age for every evaluation rather than updating to attained age.

Age-Dependent Versus Fixed-Age Rates

To compare specifications, consider one transition from Healthy to Dead with annual intensity \(\mu_{H,\text{Dead}}(y)=0.008+0.0015(y-50)\). This is a linear age-dependent rate: every one-year increase in attained age raises the intensity by \(0.0015\). By contrast, a model that specifies \(\mu_{H,\text{Dead}}(y)=0.008\) for all \(y\) has a rate fixed across ages. The latter does not increase simply because the insured becomes older.

For a fixed starting age \(x\), the age-dependent version can be written as a time function, \(\mu_{H,\text{Dead}}(x+t)=0.008+0.0015(x+t-50)\). This expression depends on elapsed time because age changes with time, but it remains an age-based model. If two people are both age 50 at their respective starting dates, this specification gives the same rate to both at that age, provided the model has no separate calendar-time effect.

Worked Example: Comparing a Rate at Different Ages

A hypothetical two-state model has Healthy and Dead states. Assume the only transition is Healthy to Dead, age is measured in years, and the annual intensity is \(\mu_{H,\text{Dead}}(y)=0.008+0.0015(y-50)\). A person starts at age 50. Find the intensity matrix at the start and after 4 years, then compare with a model whose intensity is fixed at \(0.008\) at all ages.

Method. Evaluate the age function at the attained age at each time. In a two-state model with Dead absorbing, the Healthy diagonal is the negative of the Healthy-to-Dead intensity.

At the start, attained age is 50, so the intensity is \(0.008+0.0015(50-50)=0.008\). Thus

Q(50)= \begin{pmatrix} -0.008 & 0.008\\ 0 & 0 \end{pmatrix}.

After 4 years, attained age is \(50+4=54\). The intensity is \(0.008+0.0015(54-50)=0.008+0.006=0.014\), giving

Q(54)= \begin{pmatrix} -0.014 & 0.014\\ 0 & 0 \end{pmatrix}.

Under the fixed-across-ages alternative, the intensity remains \(0.008\) at both ages, so its matrix is \(\begin{pmatrix}-0.008&0.008\\0&0\end{pmatrix}\) at ages 50 and 54. Check. The age-dependent rate has increased by \(0.014-0.008=0.006\), exactly four annual age increments times \(0.0015\). The common mistake avoided is treating the initial rate as constant when the stated function changes with age.

Age Thresholds and the Time-Unit Trap

Some specifications change at a particular attained age, such as a hypothetical disability transition rate that increases at age 65. The condition must be applied to attained age, including the boundary convention stated in the model. If the rate is defined differently for ages below 65 and ages 65 or older, attained age 65 uses the second branch. The elapsed time at which the change occurs depends on starting age: a person who starts at age 63 reaches age 65 after 2 years.

Worked Example: Applying an Age-Based Threshold

A hypothetical policyholder starts in Active at age 63. The annual transition intensity from Active to Disabled is \(0.004\) for attained ages below 65 and \(0.012\) at ages 65 and above. The annual Active-to-Dead intensity is \(0.003+0.001(y-63)\), where \(y\) is attained age. Disabled and Dead are not needed to evaluate the Active row. Find that row at elapsed times \(1.5\) and \(2\) years.

Method. Convert each elapsed time to attained age before choosing the threshold branch or evaluating the second intensity.

At \(t=1.5\), attained age is \(63+1.5=64.5\), which is below 65. Thus \(\mu_{A,D}(64.5)=0.004\), and \(\mu_{A,\text{Dead}}(64.5)=0.003+0.001(64.5-63)=0.003+0.0015=0.0045\). The total outgoing intensity is \(0.004+0.0045=0.0085\), so the Active row has diagonal \(-0.0085\).

At \(t=2\), attained age is \(63+2=65\). The boundary convention puts age 65 in the second branch, so \(\mu_{A,D}(65)=0.012\). Also, \(\mu_{A,\text{Dead}}(65)=0.003+0.001(65-63)=0.005\). The total is \(0.012+0.005=0.017\), and the diagonal is \(-0.017\).

Check and trap. The two Active rows are \((-0.0085,0.004,0.0045)\) and \((-0.017,0.012,0.005)\), with columns ordered Active, Disabled, Dead. A wrong calculation would compare elapsed time \(t=2\) directly with the age threshold 65 and incorrectly keep the first branch. The correction is to compare attained age \(63+2=65\) with the threshold.

Practical Checks Before Using an Age Function

Before evaluating an age-dependent intensity matrix, identify the age variable and the time unit used by each function. Then check the attained age at the evaluation time, apply any boundary conditions, and calculate the diagonal only after evaluating all outgoing rates. These steps help catch errors that can otherwise produce a valid-looking but incorrectly indexed matrix.

Record the starting age.
Write \(x\) and confirm whether age is measured in years or another unit.

Convert elapsed time to attained age.
For age measured in years, use \(y=x+t\), keeping \(t\) in years.

Evaluate each permitted transition rate.
Substitute attained age into the specified function and use zero for forbidden transitions.

Set each diagonal entry and check the matrix.
Negate each row’s total outgoing intensity and verify that every row sums to zero.

These representations are useful whenever a long-term model needs transition rates to change with age. The essential distinction is simple: a function of attained age is evaluated at \(x+t\); a rate fixed across ages stays the same as age changes. Confusing these specifications can alter every later calculation that uses the intensity matrix. The age-indexing skills here are aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Check Your Understanding

For each problem, identify attained age before evaluating any age-based rate. All intensities are annual rates, and age and time are measured in years.

A person starts at age 40. A transition intensity is \(\mu_{01}(y)=0.003+0.0004(y-40)\). Find the intensity at elapsed time 5 and the corresponding two-state intensity matrix, with state 1 absorbing. (Answer: attained age 45; intensity \(0.005\); matrix \(\begin{pmatrix}-0.005&0.005\\0&0\end{pmatrix}\).)
From state A, the only exits are to B and D, with \(\mu_{AB}(y)=0.004+0.001(y-50)\) and \(\mu_{AD}(y)=0.002+0.0005(y-50)\). Starting at age 52, find the A row at elapsed time 4. (Answer: attained age 56; rates \(0.010\) and \(0.005\); A row \((-0.015,0.010,0.005)\), with columns A, B, D.)
A transition intensity is \(0.020\) below attained age 75 and \(0.030\) at age 75 or older. A person starts at age 73. Find the rate at elapsed times 1.9 and 2. (Answer: attained ages 74.9 and 75; rates \(0.020\) and \(0.030\), respectively.)
For a two-state Healthy-Dead model, compare the rate at elapsed time 10 for a person starting at age 40 under \(\mu(y)=0.005+0.0002(y-40)\) with a model fixed at \(0.005\) across ages. State the attained age and each rate. (Answer: attained age 50; age-dependent rate \(0.007\); fixed-across-ages rate \(0.005\).)
At attained age \(y\), a state has three possible outgoing intensities: \(0.006+0.001(y-60)\), \(0.004\), and \(0.002+0.0005(y-60)\). A person starts at age 60. Find the three off-diagonal entries and the diagonal entry at elapsed time 2. (Answer: attained age 62; off-diagonal entries \(0.008\), \(0.004\), and \(0.003\); diagonal entry \(-0.015\).)

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