A Piecewise-Intensity Mortality and Disability Example

Why Piecewise Intensities Need Careful Cohort Updates

Mortality and disability risks can change over time. A hypothetical disability-income portfolio, for example, might have one pair of cause-specific intensities during an initial period and a different pair later. To calculate exits and exposure over the full period, use the intensities that apply in each interval and carry the surviving active cohort forward before starting the next calculation.

The key distinction is that an intensity is a rate per unit of time, not an interval probability. Under constant competing intensities, the total intensity determines the probability of remaining active, while the relative sizes of the cause-specific intensities determine how exits are allocated between causes. Expected exposure is calculated separately for each interval and then added.

Definition: In a two-cause model, let \(\lambda_m\) be the mortality intensity and \(\lambda_d\) the disability intensity while a member is active. In an interval of length \(h\), assume both intensities are constant, exits are absorbing for the active state, and there are no other causes of exit or entries into the cohort. The total intensity is \(\lambda=\lambda_m+\lambda_d\). For a cohort of \(N\) active members at the start, the expected number remaining active is \(Ne^{-\lambda h}\). If \(\lambda>0\), expected exits are \(N\frac{\lambda_m}{\lambda}(1-e^{-\lambda h})\) by mortality and \(N\frac{\lambda_d}{\lambda}(1-e^{-\lambda h})\) by disability. Expected active-state exposure, in person-years when time is measured in years, is \(N\frac{1-e^{-\lambda h}}{\lambda}\). If \(\lambda=0\), there are no exits and exposure is \(Nh\).

Calculating One Interval at a Time

For a piecewise-constant model, apply the one-interval formulas using the intensities and duration for that interval. The starting cohort for the next interval is the expected number remaining at the end of the current interval. Cohort counts in these calculations may be fractional: they are expected counts, not claims that a fractional person is observed.

Set the interval inputs.
Record its starting active cohort \(N\), length \(h\), and cause-specific intensities. Confirm that rates and durations use compatible time units.

Find total intensity and survival.
Set \(\lambda=\lambda_m+\lambda_d\). The fraction remaining active is \(e^{-\lambda h}\), so the ending cohort is \(Ne^{-\lambda h}\).

Allocate exits and calculate exposure.
Multiply total exits by each cause’s share of total intensity. Calculate exposure for the interval and add it to exposure accumulated in earlier intervals.

Update and reconcile.
Use the ending cohort as the next interval’s starting cohort. Check that the interval’s cause-specific exits sum to its total exits, and that starting cohort less total exits equals ending cohort.

Why the Interval Formulas Hold

For a member active at the start of an interval, the probability of no exit by time \(t\) is \(e^{-\lambda t}\), because the total exit intensity is the constant \(\lambda\). Thus the probability of being active at time \(t\) is \(e^{-\lambda t}\), and the expected active cohort at that time is \(Ne^{-\lambda t}\).

The instantaneous probability of a mortality exit in a short time \(dt\), conditional on still being active, is approximately \(\lambda_m dt\). Therefore the unconditional probability of a mortality exit in \([t,t+dt]\) is approximately \(e^{-\lambda t}\lambda_mdt\). Integrating this rate over the interval gives the cause-specific exit probability:

\int_0^h e^{-\lambda t}\lambda_m\,dt =\frac{\lambda_m}{\lambda}(1-e^{-\lambda h}), \qquad \lambda>0.

Multiplying by \(N\) gives the expected number of mortality exits. Replacing \(\lambda_m\) by \(\lambda_d\) gives disability exits. For exposure, add the expected active cohort over time: its integral is \(N\int_0^h e^{-\lambda t}\,dt=N(1-e^{-\lambda h})/\lambda\). When \(\lambda=0\), the cohort stays active throughout, so its exposure is \(Nh\).

Worked Example: Two Intervals With Both Causes Active

A hypothetical cohort has 500 active members. During the first 0.5 years, mortality and disability intensities are 0.2 and 0.2 per year. During the following 1 year, they are 0.1 and 0.3 per year. Find the expected exits by cause in each interval, the final active cohort, and total exposure. Use the piecewise-constant competing-intensity method.

First interval. The total intensity is \(\lambda_1=0.2+0.2=0.4\) per year and the duration is \(h_1=0.5\) years. The cumulative intensity is \(0.4(0.5)=0.2\), so the remaining cohort is \(500e^{-0.2}=409.3654\). Total exits are \(500-409.3654=90.6346\). The cause shares are \(0.2/0.4=0.5\) each, giving mortality exits \(90.6346(0.5)=45.3173\) and disability exits \(45.3173\). Exposure is \(500(1-e^{-0.2})/0.4=226.5866\) person-years.

Second interval. Start with the unrounded remaining cohort \(500e^{-0.2}\). Here \(\lambda_2=0.1+0.3=0.4\) per year and \(h_2=1\), so the ending cohort is \(500e^{-0.2}e^{-0.4}=500e^{-0.6}=274.4058\). Total exits are \(500(e^{-0.2}-e^{-0.6})=134.9596\). Mortality’s share is \(0.1/0.4=0.25\), giving \(33.7399\) mortality exits; disability’s share is \(0.3/0.4=0.75\), giving \(101.2197\) disability exits. Exposure is \(500e^{-0.2}(1-e^{-0.4})/0.4=337.3989\) person-years.

Result and check. Total expected mortality exits are \(45.3173+33.7399=79.0572\), and total expected disability exits are \(45.3173+101.2197=146.5370\). Total exposure is \(226.5866+337.3989=563.9855\) person-years. The exits sum to \(225.5942\), and \(500-225.5942=274.4058\), matching the final active cohort. Rounded to two decimals, the results are 79.06 mortality exits, 146.54 disability exits, 274.41 active members, and 563.99 person-years.

Common mistake avoided: The second interval starts with about 409.37 members, not the original 500. Applying its intensities to the original cohort would count members who already exited during the first interval.

Worked Example: Mortality Followed by Disability

A hypothetical cohort of 300 starts a 0.75-year interval with mortality intensity 0.2 per year and disability intensity zero. It is followed by a 1.25-year interval with mortality intensity zero and disability intensity 0.3 per year. Calculate the exits by cause, final active cohort, and total exposure. Keep the cohort update explicit.

First interval. The total intensity is 0.2 per year and the cumulative intensity is \(0.2(0.75)=0.15\). The remaining cohort is \(300e^{-0.15}=258.2124\). There are \(300-258.2124=41.7876\) mortality exits and no disability exits. Exposure is \(300(1-e^{-0.15})/0.2=208.9380\) person-years.

Second interval. The starting cohort is 258.2124, and only disability operates. The cumulative intensity is \(0.3(1.25)=0.375\). The final cohort is \(300e^{-0.15}e^{-0.375}=300e^{-0.525}=177.4666\). Thus disability exits are \(258.2124-177.4666=80.7458\), with no mortality exits. Exposure is \(300e^{-0.15}(1-e^{-0.375})/0.3=269.1526\) person-years.

Result and check. Total mortality exits are 41.7876, total disability exits are 80.7458, and total exposure is \(208.9380+269.1526=478.0906\) person-years. The cohort check is \(300-41.7876-80.7458=177.4666\). Rounded to two decimals, the results are 41.79 mortality exits, 80.75 disability exits, 177.47 active members, and 478.09 person-years.

Common mistake avoided: Zero intensity for a cause in an interval means that cause has no exits in that interval; it does not mean the cohort has no exits if the other cause has positive intensity.

Worked Example: Three Intervals, Including a Zero-Cause Interval

A hypothetical cohort of 1,200 active members passes through three intervals: 0.25 years with mortality intensity 0.4 and disability intensity 0.2; 0.5 years with mortality intensity zero and disability intensity 0.4; and 0.75 years with mortality intensity 0.2 and disability intensity 0.1. Find the exits by cause in each interval, final active cohort, and total exposure. Calculate each interval using its actual starting cohort.

Interval 1. The total intensity is 0.6 and the cumulative intensity is \(0.6(0.25)=0.15\). The ending cohort is \(1{,}200e^{-0.15}=1{,}032.8496\). Total exits are 167.1504. The mortality and disability shares are \(0.4/0.6=2/3\) and \(0.2/0.6=1/3\); exits are 111.4336 and 55.7168, respectively. Exposure is \(1{,}200(1-e^{-0.15})/0.6=278.5841\) person-years.

Interval 2. Mortality intensity is zero, so all exits are due to disability. The cumulative intensity is \(0.4(0.5)=0.2\). The ending cohort is \(1{,}200e^{-0.35}=845.6257\), and exits are \(1{,}032.8496-845.6257=187.2239\), all by disability. Exposure is \(1{,}200e^{-0.15}(1-e^{-0.2})/0.4=468.0597\) person-years.

Interval 3. The total intensity is 0.3 and the cumulative intensity is \(0.3(0.75)=0.225\). The ending cohort is \(1{,}200e^{-0.575}=675.2458\), and total exits are \(845.6257-675.2458=170.3799\). The mortality and disability shares are \(0.2/0.3=2/3\) and \(0.1/0.3=1/3\); exits are 113.5866 and 56.7933. Exposure is \(1{,}200e^{-0.35}(1-e^{-0.225})/0.3=567.9329\) person-years.

Result and check. Total mortality exits are \(111.4336+113.5866=225.0202\); total disability exits are \(55.7168+187.2239+56.7933=299.7340\). Total exposure is \(278.5841+468.0597+567.9329=1{,}314.5767\) person-years. The exits sum to 524.7542, and \(1{,}200-524.7542=675.2458\). Rounded to two decimals, the results are 225.02 mortality exits, 299.73 disability exits, 675.25 active members, and 1,314.58 person-years.

Common mistake avoided: The zero mortality intensity in interval 2 eliminates mortality exits only for that interval. It does not erase mortality exits from interval 1 or prevent mortality exits when that cause resumes in interval 3.

Common Exam Traps and Checks

A frequent error is to treat an intensity as an interval probability. For example, with a mortality intensity of 0.3 per year and a disability intensity of 0.1 per year over half a year, the total intensity is 0.4 and the total exit probability is \(1-e^{-0.4(0.5)}=1-e^{-0.2}\), not 0.4. The cause shares are 0.75 and 0.25, so the corresponding cause probabilities are \(0.75(1-e^{-0.2})\) and \(0.25(1-e^{-0.2})\). Using 0.3 and 0.1 as the cause probabilities ignores both the interval length and the competing risk.

For any piecewise calculation, use unrounded intermediate cohort values when carrying the calculation forward. Rounding each interval’s ending cohort before the next update can create small discrepancies in exits and exposure. At the end, reconcile total exits against the initial and final cohort, and compare summed interval exposure with the sum of the interval-specific exposure calculations. These checks are especially useful in problems aligned with the SOA Exam ALTAM (Advanced Long-Term Actuarial Mathematics) syllabus.

Key takeaway: On every interval, calculate survival from the sum of active cause-specific intensities, allocate total exits in proportion to those intensities, integrate the active cohort for exposure, and carry the remaining cohort into the next interval.

Check Your Understanding

For each problem, treat the stated intensities as constant within each interval, assume exits are absorbing, and round final numerical answers to two decimals.

A cohort of 100 has mortality intensity 0.12 and disability intensity 0.08 for 0.5 years. Find expected exits by cause, remaining active members, and exposure. (Answer: mortality exits 5.71; disability exits 3.81; active 90.48; exposure 47.58 person-years.)
A cohort of 200 has mortality intensity 0.1 and disability intensity 0.3 for 1 year, followed by 0.5 years with mortality intensity zero and disability intensity 0.2. Find total exits by cause, final active members, and exposure. (Answer: mortality exits 16.48; disability exits 62.21; active 121.31; exposure 228.63 person-years.)
A cohort of 400 has mortality and disability intensities both equal to 0.4 for 0.25 years. Find exits by cause, remaining active members, and exposure. (Answer: mortality exits 36.25; disability exits 36.25; active 327.49; exposure 90.63 person-years.)
A cohort of 150 has mortality intensity 0.2 and disability intensity 0.1 for 1 year, then mortality intensity 0.05 and disability intensity 0.15 for 2 years. Find total exits by cause, final active members, and total exposure. (Answer: mortality exits 35.08; disability exits 40.44; active 74.49; exposure 312.77 person-years.)
A cohort of 80 has zero mortality and disability intensities for 1 year, followed by 0.5 years with mortality and disability intensities both equal to 0.2. Find exits by cause, final active members, and total exposure. (Answer: mortality exits 7.25; disability exits 7.25; active 65.50; exposure 116.25 person-years.)

What You'll Learn