Comparing Lifetime Models by Survival Curves

Why Compare Survival Curves?

A lifetime model describes how the probability of surviving changes as duration increases. Comparing two models at a specified duration answers a focused question: which model assigns the greater probability to a lifetime longer than that duration? The answer can be useful when checking alternative assumptions or explaining how different mortality patterns affect projected survival.

A comparison at one duration is deliberately limited. It does not, by itself, show that one model has higher survival at every duration, nor does it identify which model is more appropriate. Curves can cross: one model may assign higher survival at shorter durations and lower survival at longer ones.

Definition: For a future lifetime $T$ measured from a common starting point, its survival function is $S_T(t)=P(T>t)$, for $t\geq 0$. At duration $t$, a larger value of $S_T(t)$ means a greater modeled probability of surviving beyond $t$. For a life aged $x$, the corresponding probability is ${}_tp_x$.

For a continuous lifetime, the probability of dying at exactly one specified instant is zero, so using $P(T>t)$ or $P(T\geq t)$ gives the same survival probability. All comparisons below use the same time origin and the same duration units within each example. If durations are measured in years, a force of mortality specified per year must be combined with durations in years.

How to Evaluate and Compare Curves

A survival curve can be given directly as a function, or derived from a stated mortality model. Under a constant force of mortality $\mu$, the lifetime has an exponential distribution and its survival function is $S_T(t)=e^{-\mu t}$. Under a piecewise-constant force, the accumulated force over each part of the interval contributes to the survival probability.

Result: Suppose a continuous lifetime has force $\mu(s)$ at time $s$, and the force is nonnegative and integrable. Its survival probability to duration $t$ is $$ S_T(t)=\exp\left(-\int_0^t \mu(s)\,ds\right). $$ For a force that is constant on successive intervals, the integral is the sum of each interval's force multiplied by the time spent in that interval.

This formula follows from the force-survival differential equation $S_T'(t)=-\mu(t)S_T(t)$, with $S_T(0)=1$. Dividing by $S_T(t)$ and integrating from 0 to $t$ gives

\int_0^t \frac{S_T'(s)}{S_T(s)}\,ds =-\int_0^t\mu(s)\,ds, \qquad \log S_T(t)-\log S_T(0) =-\int_0^t\mu(s)\,ds.

Since $S_T(0)=1$, its logarithm is zero. Exponentiating gives the stated survival formula. The argument assumes the survival function is positive over the interval; the formula is applied here to continuous lifetime models with finite accumulated force.

For a direct comparison at duration $t$, calculate both survival probabilities using the same duration, then subtract if the size of the difference is useful. If $S_A(t)-S_B(t)=0.04$, model A assigns a probability four percentage points higher than model B of surviving beyond that duration. This is a difference between model probabilities, not a probability that one particular person survives under A but not B.

Worked Example: Comparing Two Constant-Force Models

A hypothetical analyst compares two continuous lifetime models, both starting at the same time. Model A has constant force $0.06$ per year; Model B has constant force $0.08$ per year. Calculate and compare their survival probabilities at 5 and 10 years. Give probabilities to four decimal places.

Method and assumptions. Each force is constant, measured per year, and applies from time zero onward. Thus each model has exponential survival $S(t)=e^{-\mu t}$. The durations are in years.

At 5 years, Model A gives $$ S_A(5)=e^{-0.06(5)}=e^{-0.30}=0.7408. $$

At the same duration, Model B gives $$ S_B(5)=e^{-0.08(5)}=e^{-0.40}=0.6703. $$

The difference is $0.7408-0.6703=0.0705$, so Model A assigns a probability 0.0705, or 7.05 percentage points, higher than Model B of surviving beyond 5 years.

At 10 years, the calculations are $$ S_A(10)=e^{-0.06(10)}=e^{-0.60}=0.5488, \qquad S_B(10)=e^{-0.08(10)}=e^{-0.80}=0.4493. $$

The difference is $0.5488-0.4493=0.0995$. Model A has higher survival at both selected durations. As a check, both models' survival probabilities have decreased between 5 and 10 years, as survival probabilities must. This comparison says nothing by itself about which assumptions are appropriate for a real population.

Common mistake avoided: Comparing the force values alone does not give the requested probability difference. The force identifies the model, but the duration must also be included in $e^{-\mu t}$.

Piecewise Models and Crossing Curves

With a piecewise force, first identify how much time falls in each interval. Do not apply the force from the final interval to the entire duration. The accumulated force is the sum of force-times-duration contributions, and survival is the exponential of the negative of that sum.

Worked Example: A Change in Which Model Has Higher Survival

Consider two hypothetical continuous lifetime models. Model A has constant force $0.06$ per year. Model B has force $0.03$ per year for the first 4 years, then $0.12$ per year thereafter. Compare survival at 2, 4, 6, and 8 years, and determine whether the ranking changes.

Method and assumptions. Both models begin at time zero. The forces are annual and piecewise constant as stated. For Model B, the accumulated force is $0.03t$ when $0\leq t\leq4$, and $0.03(4)+0.12(t-4)$ when $t>4$.

At 2 years, the accumulated forces are $0.06(2)=0.12$ for A and $0.03(2)=0.06$ for B. Therefore, $$ S_A(2)=e^{-0.12}=0.8869, \qquad S_B(2)=e^{-0.06}=0.9418. $$ Model B is higher by $0.9418-0.8869=0.0549$.

At 4 years, the accumulated forces are $0.06(4)=0.24$ for A and $0.03(4)=0.12$ for B. Thus $$ S_A(4)=e^{-0.24}=0.7866, \qquad S_B(4)=e^{-0.12}=0.8869. $$ Model B is higher by $0.8869-0.7866=0.1003$.

At 6 years, A accumulates $0.06(6)=0.36$. B accumulates $0.03(4)+0.12(6-4)=0.12+0.24=0.36$. Hence $$ S_A(6)=S_B(6)=e^{-0.36}=0.6977. $$

At 8 years, A accumulates $0.06(8)=0.48$. B accumulates $0.03(4)+0.12(8-4)=0.12+0.48=0.60$. Therefore, $$ S_A(8)=e^{-0.48}=0.6188, \qquad S_B(8)=e^{-0.60}=0.5488. $$ Now A is higher by $0.6188-0.5488=0.0700$.

The ranking changes: B has higher survival before 6 years, both models agree at 6 years, and A has higher survival at 8 years. The equality can also be checked for any $t\geq4$: their accumulated forces are $0.06t$ and $0.12t-0.36$. Equating them gives $0.06t=0.12t-0.36$, so $t=6$. The computed values are consistent with that crossing.

Common mistake avoided: At 8 years, applying B's later force $0.12$ to all 8 years would give the wrong accumulated force. The first 4 years must be counted at $0.03$, and only the remaining 4 years at $0.12$.

Directly Specified Survival Functions

A force model is not necessary when the survival functions themselves are given. Evaluate each function at the requested durations, check that the inputs use the stated time scale, and interpret each result as a probability of exceeding that duration. A valid survival curve starts at 1, is nonincreasing, and has values between 0 and 1.

Worked Example: A Close Comparison at Different Durations

Two hypothetical models for a future lifetime in years have survival functions $$ S_C(t)=\frac{1}{1+0.10t}, \qquad S_D(t)=e^{-0.07t}, \qquad t\geq0. $$ Compare them at 5, 10, and 12 years. Give probabilities to four decimal places and identify whether the same model is higher at all three durations.

Method and assumptions. Treat the supplied functions as the full survival models for nonnegative durations in years. Both equal 1 at $t=0$, remain positive, decrease as $t$ increases, and tend to zero. Substitute each duration directly.

At 5 years, $$ S_C(5)=\frac{1}{1+0.10(5)}=\frac{1}{1.5}=0.6667, \qquad S_D(5)=e^{-0.35}=0.7047. $$ Model D is higher by $0.7047-0.6667=0.0380$.

At 10 years, $$ S_C(10)=\frac{1}{2}=0.5000, \qquad S_D(10)=e^{-0.70}=0.4966. $$ Model C is higher by $0.5000-0.4966=0.0034$.

At 12 years, $$ S_C(12)=\frac{1}{2.2}=0.4545, \qquad S_D(12)=e^{-0.84}=0.4317. $$ Model C is higher by $0.4545-0.4317=0.0228$. Thus the ranking at 5 years differs from the ranking at 10 and 12 years. The comparison at 10 years is particularly close, but the values are not equal at the displayed precision.

Common mistake avoided: Rounding too early can obscure a close comparison. Keep the expressions or extra calculator digits during subtraction, and round the reported probabilities and differences only at the final step.

Interpreting Results and Avoiding Traps

A survival probability always answers a duration-specific question. If $S_A(t)>S_B(t)$, the precise interpretation is that model A assigns a greater probability than model B to a lifetime exceeding $t$. It does not establish that every individual represented by A would outlive one represented by B, or that A is higher at other durations.

The corresponding death probability by duration $t$ is $1-S_T(t)$. Because this is the complement of survival, the model with higher survival has lower death probability by that same duration. For example, in the first worked example at 5 years, the death probabilities are $1-0.7408=0.2592$ for A and $1-0.6703=0.3297$ for B.

Exam trap: Do not reverse the conclusion by comparing death probabilities as if they were survival probabilities. The incorrect conclusion “B has higher survival because $0.3297>0.2592$” uses the death probabilities and assigns their ordering to survival. The correction is to compare survival directly: $0.7408>0.6703$, so A has higher survival at 5 years. Equivalently, B has the higher probability of death by 5 years.

A reliable comparison follows a short sequence:

Confirm the time scale.
Check that the duration and any force parameter use compatible units, such as years and force per year.

Evaluate both survival models.
Use the same duration for each model, and account for every interval if a force is piecewise constant.

State the comparison in context.
Name the model with the higher value and say that it assigns a greater probability of surviving beyond that duration.

Limit the conclusion.
If only selected durations were evaluated, do not claim a ranking at all durations unless it has been established separately.

Comparing survival curves is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus because it tests the interpretation and evaluation of lifetime probabilities. The central discipline is simple: compare like durations, preserve the model's time structure, and make only the conclusion supported by the values calculated.

Check Your Understanding

For each problem, compare the stated models at the specified duration or durations. Give probabilities to four decimal places.

Two continuous lifetime models have constant forces $0.05$ and $0.07$ per year. Compare their survival probabilities at 6 years and state which is higher. (Answer: $e^{-0.30}=0.7408$ versus $e^{-0.42}=0.6570$; the model with force $0.05$ is higher by $0.0838$.)
Model A has constant force $0.04$ per year. Model B has force $0.01$ per year for the first 5 years and $0.08$ per year thereafter. Compare survival at 5 and 10 years. (Answer: At 5 years, A is $e^{-0.20}=0.8187$ and B is $e^{-0.05}=0.9512$, so B is higher. At 10 years, A is $e^{-0.40}=0.6703$ and B is $e^{-0.45}=0.6376$, so A is higher.)
Two supplied survival functions are $S_U(t)=1/(1+0.08t)$ and $S_V(t)=e^{-0.075t}$, with $t$ in years. Compare at 4 and 8 years. (Answer: At 4 years, $S_U=0.7576$ and $S_V=0.7408$, so U is higher by $0.0168$. At 8 years, $S_U=0.6098$ and $S_V=0.5488$, so U is higher by $0.0610$.)
Model R has constant force $0.04$ per year. Model Q has force $0.02$ per year for the first 3 years and $0.08$ per year thereafter. Compare at 3 and 6 years. (Answer: At 3 years, R is $e^{-0.12}=0.8869$ and Q is $e^{-0.06}=0.9418$, so Q is higher. At 6 years, R is $e^{-0.24}=0.7866$ and Q is $e^{-(0.06+0.24)}=0.7408$, so R is higher.)
Model M has constant force $0.05$ per year. Model N has force $0.02$ per year for the first 4 years and $0.08$ per year thereafter. Find the duration after year 4 at which their survival probabilities are equal. (Answer: For $t\geq4$, the accumulated forces are $0.05t$ and $0.02(4)+0.08(t-4)=0.08t-0.24$. Equating gives $0.05t=0.08t-0.24$, so $t=8$ years.)

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