Why Piecewise Constant Forces Matter for Lifetime Moments
A life table gives survival probabilities at integer ages, while a complete future lifetime can end at any point between those ages. Under a piecewise constant force assumption, survival decreases exponentially within each age interval, but its rate can change at an integer boundary. To calculate a mean or variance, the survival curve must therefore be integrated across all relevant intervals, including the final tail.
The key is to keep track of two quantities on every interval: the force of mortality that applies there, and the probability of having survived to the start of that interval. Omitting the second quantity, or continuing the wrong force beyond a boundary, changes the lifetime moments.
For a life-table year beginning at age \(x+j\), the annual death probability is \(q_{x+j}\), and the probability of surviving the year is \(p_{x+j}=1-q_{x+j}\). Under a constant force within that year, the force is \(\lambda_j=-\ln p_{x+j}=-\ln(1-q_{x+j})\). This conversion matches the model survival over one full year to the life-table survival probability.
Moment Formulas from the Survival Function
For a nonnegative lifetime, the first two raw moments can be obtained from the survival function. The first raw moment is the mean; the second raw moment is then used to calculate the variance. These formulas require the relevant integrals to be finite.
A short justification explains the factor of 2 in the second-moment formula. For each nonnegative value of \(T_x\), \(T_x^2=\int_0^{T_x}2t\,dt\). Taking expectations and reversing the order of integration gives \(E[T_x^2]=\int_0^\infty 2t\,P(T_x>t)\,dt\), which is the stated result.
Suppose an interval starts at future time \(a\), has length \(h\), force \(\lambda>0\), and survival probability \(C={}_ap_x\) at its start. Within the interval, \({}_{a+s}p_x=C e^{-\lambda s}\). Integrating this expression gives the contributions of the interval to the first and second raw moments.
Here \(M_1(a,h)\) is the interval’s contribution to \(E[T_x]\), and \(M_2(a,h)\) is its contribution to \(E[T_x^2]\). The factor \(a+s\) in the second-moment integral is important: the lifetime is measured from time 0, not from the beginning of the current interval.
If the final interval begins at \(a\) and continues indefinitely with force \(\lambda>0\), its contributions are instead \(C/\lambda\) to the mean and \(2C(a/\lambda+1/\lambda^2)\) to the second raw moment. These follow by integrating the same survival curve from \(a\) to infinity.
For each year, calculate \(p=1-q\) and \(\lambda=-\ln p\). Record where each force applies.
Multiply the survival probability at the previous boundary by \(e^{-\lambda h}\) for the interval just completed.
Add the interval contributions to \(E[T_x]\) and \(E[T_x^2]\), using the correct starting time \(a\) and starting survival \(C\).
Subtract the squared mean from the second raw moment. Confirm that the variance is nonnegative and that survival is continuous at each boundary.
Worked Calculations
The examples use a life aged \(x\), with \(T_x\) measured in years from age \(x\). A force stated “thereafter” applies continuously from the indicated boundary onward. The method is the same when the forces are obtained from life-table \(q\)-values: convert each annual survival probability first, then integrate using the resulting force.
Worked Example: A High First-Year Force Followed by a Lower Force
A hypothetical life table has \(q_x=0.75\) for the first year and \(q_{x+1}=0.50\) for every year thereafter. Assume a constant force within each year. Calculate the mean and variance of \(T_x\).
Method. Convert the annual survival probabilities to forces, calculate the moment contributions over the first year, and then integrate the remaining tail. This avoids applying the first-year force beyond age \(x+1\).
Let \(L=\ln 2\approx0.6931472\). The first-year survival probability is \(0.25\), so its force is \(-\ln(0.25)=\ln4=2L\). In every later year the survival probability is \(0.50\), so the force is \(L\). Thus survival is \(e^{-2Lt}\) for \(0\le t<1\), and \(0.25e^{-L(t-1)}\) for \(t\ge1\). The survival probability at the boundary is \(0.25\) under both expressions.
The first-year contribution to the mean is \((1-e^{-2L})/(2L)=0.75/(2L)=0.5410106\). The tail contribution is \(0.25/L=0.3606738\). Therefore \(E[T_x]=0.5410106+0.3606738=0.9016844\) years.
The first-year contribution to the second raw moment is \(2[1-e^{-2L}(1+2L)]/(2L)^2=0.4198396\). For the tail, use \(a=1\), \(C=0.25\), and \(\lambda=L\): \(2C(1/L+1/L^2)=0.5(1/L+1/L^2)\). Since \(1/L\approx1.4426950\) and \(1/L^2\approx2.08136898\), the tail contribution is \(1.7620320\) years squared.
Consequently, \(E[T_x^2]=0.4198396+1.7620320=2.1818716\) years squared. The variance is \(2.1818716-(0.9016844)^2\approx1.3688\) years squared. The nonnegative result and the matching boundary survival provide checks on the calculation.
Common mistake avoided: The tail integral uses \(t=1+s\), not just \(s\). In the second moment, the elapsed lifetime is \(1+s\), which is why the tail contribution includes both \(1/L\) and \(1/L^2\) terms.
Worked Example: A Lower First-Year Force Followed by a Higher Force
For a different hypothetical life table, \(q_x=0.50\) in the first year and \(q_{x+j}=0.75\) for every \(j\ge1\). Assume constant force within each year. Calculate the mean and variance of \(T_x\).
Method. Separate the first-year contribution from the tail and use the survival probability at time 1 as the tail’s starting level. The common error is to integrate the later force from time 0 instead of from time 1.
With \(L=\ln2\), the first-year force is \(L\), and the force after time 1 is \(\ln4=2L\). Survival is \(e^{-Lt}\) for \(0\le t<1\), then \(0.5e^{-2L(t-1)}\) for \(t\ge1\). The boundary survival is \(0.5\).
The mean contribution from the first year is \((1-e^{-L})/L=0.5/L=0.7213475\). The tail contribution is \(0.5/(2L)=0.3606738\). Hence \(E[T_x]=1.0820213\) years.
The first-year contribution to \(E[T_x^2]\) is \(2[1-e^{-L}(1+L)]/L^2=(1-L)/L^2\approx0.6387\). With \(a=1\), \(C=0.5\), and \(\lambda=2L\), the tail contribution is \(2(0.5)[1/(2L)+1/(2L)^2]=1/(2L)+1/(4L^2)\approx1.2417\) years squared. Thus \(E[T_x^2]\approx1.8804\) years squared.
Finally, \(\operatorname{Var}(T_x)=1.8804-(1.0820213)^2\approx0.7096\) years squared. The tail is included, and the variance is nonnegative.
Common mistake avoided: The second-moment tail contribution cannot be calculated using only time since the boundary. The factor \(t\) in \(2\int t\,{}_tp_x\,dt\) is the full future lifetime, including the first year already elapsed.
Worked Example: Two Years at One Force, Then a Different Tail
A hypothetical life table has annual death probability \(0.50\) at ages \(x\) and \(x+1\), and annual death probability \(0.75\) from age \(x+2\) onward. Under constant force within each year, calculate the mean and variance of \(T_x\).
Method. Combine the first two years into one interval because their forces agree, but retain the boundary at time 2 for the tail. This makes the accumulated survival level at that boundary explicit.
Set \(L=\ln2\). The force is \(L\) for \(0\le t<2\), and \(2L\) thereafter. Survival at time 2 is \(e^{-2L}=0.25\), so the survival function is \(e^{-Lt}\) before time 2 and \(0.25e^{-2L(t-2)}\) afterward. Both pieces give \(0.25\) at the boundary.
The first two years contribute \((1-e^{-2L})/L=0.75/L=1.0820213\) years to the mean. The tail contributes \(0.25/(2L)=0.1803369\) years. Therefore \(E[T_x]=1.2623582\) years.
For the second raw moment, the first interval contributes \(2[1-e^{-2L}(1+2L)]/L^2\approx1.6794\) years squared. The tail, with \(a=2\), \(C=0.25\), and \(\lambda=2L\), contributes \(2(0.25)[2/(2L)+1/(2L)^2]\approx0.9815\) years squared. Thus \(E[T_x^2]\approx2.6609\) years squared.
The variance is \(2.6609-(1.2623582)^2\approx1.0673\) years squared. As a check, the mean exceeds one year, as expected given that \(0.25\) of the cohort survives to time 2 and then has a positive remaining expected lifetime.
Common mistake avoided: The tail starts with survival \(0.25\), not 1. The tail’s contribution must be weighted by the probability of reaching time 2.
Checks That Prevent Boundary and Tail Errors
The formulas are straightforward once each interval’s starting point and survival level are correct. Before accepting an answer, check that the pieces of the survival function agree at every boundary. At time \(a+h\), survival from the interval just completed is \(C e^{-\lambda h}\); this must equal the value used as \(C\) at the next interval.
Also distinguish the mean from the second raw moment. The mean integrates survival alone. The second raw moment weights survival by the full time \(t\), so later intervals receive a larger time factor even if their survival levels are small. A force change affects the slope of log survival, not the accumulated survival probability at the boundary.
Check Your Understanding
For each problem, assume constant force within every stated interval and a positive constant force throughout the final infinite tail. Round numerical answers to four decimal places.
- A life has force \(\ln2\) for the first year and force \(3\ln2\) thereafter. Find \(E[T_x]\) and \(\operatorname{Var}(T_x)\). (Answer: mean \(0.9618\) years; variance \(0.4258\) years squared.)
- A life-table model has \(q_x=0.75\) for the first year and \(q_{x+j}=0.50\) for every \(j\ge1\). Find the mean and variance of complete future lifetime. (Answer: mean \(0.9017\) years; variance \(1.3688\) years squared.)
- A life-table model has annual \(q=0.50\) for the first two years and annual \(q=0.75\) from year 3 onward. Find the mean and variance. (Answer: mean \(1.2624\) years; variance \(1.0673\) years squared.)
- A life has force \(0.4\) for the first two years and force \(0.8\) thereafter. Find the mean and variance, carrying survival at the two-year boundary into the tail. (Answer: mean \(1.9383\) years; variance \(2.2837\) years squared.)
Calculating complete lifetime moments from life-table intervals is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. The essential exam discipline is to make the interval boundaries explicit, preserve survival across them, and integrate the tail using time measured from the original starting age.