Working with a Piecewise Survival Function
A lifetime model may use one formula during an initial period and another formula later. This can represent a change in the modeled mortality pattern without requiring a new probability calculation for every time interval. The key is to use the correct formula on each part of the timeline, and to check how the formulas connect.
Let \(T\) be a nonnegative lifetime, measured in a consistent time unit. Write \(S_0(t)=P(T>t)\) for its survival function. A piecewise definition gives a separate expression for \(S_0(t)\) on each stated interval. Once the survival values at the relevant endpoints are known, interval probabilities follow directly. Where the survival function is differentiable, its derivative also gives the density and the force of mortality.
A valid survival function is nonincreasing, takes values between zero and one, and begins at \(S_0(0)=1\) when the lifetime is strictly positive. If the lifetime is proper, \(S_0(t)\) tends to zero as \(t\) increases without bound. For a piecewise model that describes an absolutely continuous lifetime, the formulas should join continuously at their shared boundaries. Their slopes need not match: the density, and hence the force, may have different one-sided values at a boundary.
Interval Probabilities from Endpoint Values
The probability of a lifetime falling in an interval can be found without integrating a density. For \(0\leq a<b\), the event \(a<T\leq b\) consists of lifetimes that have survived past \(a\), but not past \(b\). Therefore,
This identity applies whether or not the interval crosses a formula boundary. If \(a\) and \(b\) lie in different pieces, evaluate \(S_0(a)\) and \(S_0(b)\) using their respective formulas. For an absolutely continuous model, the same probability can also be found by integrating the density across the interval. The endpoint formula is usually simpler and helps avoid unnecessary integration.
Deriving Density and Force on Each Piece
On an interval where a survival formula is differentiable, differentiate that formula and change the sign to obtain the density. Then divide by the survival value to obtain the force. This last step is important: the force is not generally the density itself.
The density measures probability per unit of time near \(t\); the force measures the instantaneous rate of failure among those who have survived to \(t\). The relationship can be justified from the conditional probability of failure in a short interval. For small \(h>0\),
If \(S_0\) is differentiable at \(t\), then \(S_0(t+h)=S_0(t)+S_0'(t)h+o(h)\), where \(o(h)/h\) tends to zero as \(h\) tends to zero. Substituting gives
Thus the conditional probability of failure over a short interval is approximately \(\mu_t h\), with \(\mu_t=-S_0'(t)/S_0(t)\). At a boundary where the slope changes, the derivative may not exist at the boundary itself. The density and force can then have distinct left- and right-hand limits. A single point has probability zero in an absolutely continuous model, so this does not change interval probabilities.
Identify which formula applies at \(a\) and which applies at \(b\). Check the boundary convention in the definition.
Calculate \(S_0(a)\) and \(S_0(b)\) separately, using the matching piece for each time.
Use \(S_0(a)-S_0(b)\). If needed, check that the result is between zero and one.
For a density or force at a particular time, differentiate only the formula that applies on that interval.
Divide the density by survival. Both density and force are rates per unit of the time scale used for \(t\).
Worked Calculations
Worked Example: An Interval That Crosses a Boundary
A hypothetical lifetime, measured in years, has survival function
Find \(P(0.5<T\leq 3.5)\), and find the density and force at \(t=1\) and \(t=3.5\). Give the probability to four decimals and rates to four significant figures.
Method and boundary check. The first formula gives \(S_0(2)=(1-2/4)^2=0.25\), and the second gives \(0.25e^0=0.25\) at the join. The formulas are continuous there, so we can use the endpoint probability formula across the boundary.
At \(t=0.5\), the first piece applies:
At \(t=3.5\), the second piece applies:
Therefore,
On the first interval, differentiation gives
Thus \(f_T(1)=\frac12(0.75)=0.375\) per year and \(\mu_1=0.5/0.75=0.6667\) per year. On the second interval, \(f_T(t)=0.2S_0(t)\), so \(\mu_t=0.2\) per year. In particular, \(f_T(3.5)=0.2(0.1852045552\ldots)=0.0370409110\ldots\) per year and \(\mu_{3.5}=0.2000\) per year. The interval probability is positive and less than one, as required.
Common mistake avoided: Using the first formula at \(3.5\) would extend it beyond its stated interval. The endpoint survival values must each come from the piece that covers that endpoint.
Worked Example: A Polynomial Piece Followed by Exponential Survival
A hypothetical lifetime in years has
Find \(P(1<T\leq 4)\), and find the density and force at \(t=1\). Round the probability to four decimals and the rates to four decimals.
Method and boundary check. At \(t=2\), the first piece gives \((1-2/5)^2=0.6^2=0.36\), matching the second piece at \(t=2\). For the probability, use the survival values at 1 and 4, which lie in different pieces.
It follows that
At \(t=1\), the first formula applies. Its derivative is
Hence \(f_T(1)=0.4(0.8)=0.32\) per year. Dividing by survival gives \(\mu_1=0.32/0.64=0.5\) per year. These values are consistent with the density-force relationship \(f_T(1)=\mu_1S_0(1)=0.5(0.64)=0.32\).
Common mistake avoided: The force is not just the derivative’s magnitude. Here the density is \(0.32\), while the force is \(0.32/0.64=0.50\) per year.
Worked Example: Conditional Probability Across Two Pieces
A hypothetical lifetime, measured in years, has survival function
Find the probability that the lifetime lies between 1 and 4 years, given survival past year 1. Also find the density and force at \(t=4\). Give the conditional probability to four decimals and rates to four significant figures.
Method and continuity check. At \(t=2\), the first and second formulas both give \(e^{-0.2}\). At \(t=5\), the second gives \(e^{-0.2}e^{-0.9}=e^{-1.1}\), matching the third. The requested interval ends at \(4\), so the second piece applies at that endpoint.
The unconditional probability of \(1<T\leq 4\) is \(S_0(1)-S_0(4)\). To condition on survival past 1, divide by \(P(T>1)=S_0(1)\). Equivalently,
The endpoint values are \(S_0(1)=e^{-0.1}=0.9048374180\ldots\) and \(S_0(4)=e^{-0.2}e^{-0.3(4-2)}=e^{-0.8}=0.4493289641\ldots\). Thus the unconditional interval probability is \(0.9048374180\ldots-0.4493289641\ldots=0.4555084539\ldots\), and
At \(t=4\), the second piece has force \(0.3\) per year because its derivative is \(-0.3S_0(t)\). Therefore \(f_T(4)=0.3S_0(4)=0.3(0.4493289641\ldots)=0.1347986892\ldots\) per year, and \(\mu_4=0.3000\) per year. The conditional probability is below one, and the density again equals force times survival.
Common mistake avoided: The unconditional interval probability is not the requested conditional probability. Conditioning on \(T>1\) requires division by \(S_0(1)\).
Common Traps and Practical Checks
Piecewise formulas are manageable when endpoint evaluation, differentiation, and conditioning are kept separate. Before accepting a result, check the following:
- Check the interval labels. A formula is valid only on its stated range. An endpoint at a boundary can be evaluated from either side only if the formulas agree there.
- Check the direction of subtraction. Since \(S_0\) decreases, \(S_0(a)-S_0(b)\) must be nonnegative for \(a<b\).
- Check continuity and monotonicity. A jump or an increase in a purported survival function may signal a model inconsistency. If there is a jump, the distribution has an atom there, and the ordinary density alone does not describe all probability.
- Differentiate the local formula. The density on a piece is the negative derivative of that piece; the force is then the density divided by survival.
- Keep time units consistent. If \(t\) is measured in years, the density and force are per year. A rate specified per month cannot be substituted into a formula using years without conversion.
The central advantage of a piecewise survival model is that it supports direct calculations across several time regimes. Use endpoint survival values for interval probabilities, and use local derivatives for density and force. These checks are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, identify the applicable piece before calculating. Answers are rounded as indicated.
- A hypothetical lifetime in years has \(S_0(t)=1-0.08t\) for \(0\leq t\leq 5\), and \(S_0(t)=0.6e^{-0.2(t-5)}\) for \(t>5\). Find \(P(2<T\leq 7)\), to four decimals. (Answer: \(0.4378\))
- A lifetime has \(S_0(t)=(1-t/6)^3\) for \(0\leq t\leq 3\), followed by \(S_0(t)=0.125e^{-0.1(t-3)}\) for \(t>3\). Find \(f_T(2)\) and \(\mu_2\), to four decimals. (Answer: \(f_T(2)=0.2222\) per time unit; \(\mu_2=0.7500\) per time unit)
- A lifetime has \(S_0(t)=e^{-0.12t}\) for \(0\leq t\leq 3\), and \(S_0(t)=e^{-0.36}e^{-0.08(t-3)}\) for \(t>3\). Find \(P(2<T\leq 5)\), to four decimals. (Answer: \(0.1921\))
- A lifetime has \(S_0(t)=1-0.05t\) for \(0\leq t\leq 4\), and \(S_0(t)=0.8e^{-0.25(t-4)}\) for \(t>4\). Find \(P(3<T\leq 6)\) and \(\mu_2\), to four decimals. (Answer: \(P=0.3648\); \(\mu_2=0.0556\) per time unit)
- A lifetime has \(S_0(t)=e^{-0.1t}\) for \(0\leq t\leq 2\), and \(S_0(t)=e^{-0.2}e^{-0.2(t-2)}\) for \(t>2\). Find \(P(2<T\leq 5\mid T>2)\), to four decimals. (Answer: \(0.4512\))