Why Boundary Conditions Matter
A formula can look smooth and plausible yet fail to describe any lifetime. For example, a proposed survival curve that rises with time assigns a larger chance of survival to someone who has already survived longer. A proposed density whose total area is less than one leaves some probability unaccounted for. Boundary and shape checks catch these problems before a function is used in a model.
Let \(T_x\) denote the future lifetime of a life aged \(x\), measured from the present time. In this tutorial, assume \(T_x\) is a continuous, nonnegative random variable and is finite with probability one. Write its distribution function, survival function, and density as \(F_x(t)\), \(S_x(t)\), and \(f_x(t)\), respectively. The time variable \(t\) is measured in the same units as the lifetime.
Boundary Conditions and Shape
Because future lifetime cannot be negative, \(F_x(t)=0\) for \(t<0\). Because the lifetime is continuous, there is no probability of death at the single instant \(t=0\), so \(F_x(0)=0\). Consequently \(S_x(0)=1\): at the starting time, the life has survived zero elapsed time. These are starting-value conditions, not assumptions that the functions must be strictly increasing or strictly decreasing.
As \(t\) increases, the event \(\{T_x\leq t\}\) can only gain outcomes, so \(F_x(t)\) cannot decrease. Equivalently, the event \(\{T_x>t\}\) can only lose outcomes, so \(S_x(t)\) cannot increase. A function may have flat sections: a CDF can remain constant where the density is zero, and the corresponding survival function can remain constant there too.
Under the stated assumption that \(T_x\) is finite with probability one, the lifetime eventually ends with probability one. Thus \(F_x(t)\) tends to one and \(S_x(t)\) tends to zero as \(t\) tends to infinity. A candidate with a positive limiting survival value would not satisfy this finite-lifetime assumption.
| Function | Starting condition | Direction as time increases | Limit as \(t\to\infty\) |
|---|---|---|---|
| CDF \(F_x(t)\) | \(F_x(0)=0\) | Nondecreasing | 1 |
| Survival \(S_x(t)\) | \(S_x(0)=1\) | Nonincreasing | 0 |
| Density \(f_x(t)\) | No required value at \(t=0\) | Nonnegative almost everywhere | Its integral over the support is 1 |
A density is not itself a probability at a particular time. Its height can exceed one, and its value at a single point does not determine the probability of an event. The relevant normalization condition concerns area over the full support:
For a lifetime with a bounded maximum possible duration, the density is zero beyond that duration, so integrating over the bounded support gives the same total of one.
Why the Density Must Integrate to One
The relationship between the density and CDF explains the normalization check. For a continuous lifetime, the probability accumulated from time zero through time \(t\) is the area under the density up to \(t\). As \(t\) grows without bound, that accumulated probability approaches the probability that the lifetime is finite. Under our assumption, that probability is one.
This argument uses the ordinary density-CDF relationship for a continuous lifetime. A density must be nonnegative, since a negative area would make accumulated probability decrease. Conversely, nonnegativity alone is not enough: a nonnegative function with total area \(0.7\), for example, does not account for all possible lifetime outcomes.
The CDF and survival checks are linked by \(S_x(t)=1-F_x(t)\). Subtracting a nondecreasing function from one produces a nonincreasing function, and the starting values and limits correspond as well. It is still useful to check both when a problem supplies both formulas: an algebra or transcription error may mean they are not complements.
Worked Checks
Worked Example: Checking a Proposed CDF
A hypothetical benefit model proposes the future-lifetime CDF \(F_x(t)=0\) for \(t<0\), \(F_x(t)=t^2/16\) for \(0\leq t\leq4\), and \(F_x(t)=1\) for \(t>4\), with time measured in years. Assume the lifetime is continuous and finite with probability one. Check whether this is a valid CDF, find its density, and verify the density total.
Method. Check the starting value, joins, direction, and limiting value of the CDF. Then differentiate on the interval where it changes and integrate the resulting density.
At \(t=0\), \(F_x(0)=0^2/16=0\), as required. At \(t=4\), the middle expression gives \(4^2/16=16/16=1\), matching the value after four years. On \(0<t<4\), the derivative is \(t/8\), which is nonnegative. Thus the function is nondecreasing on that interval; it is constant before zero and after four years. Its limiting value is one.
Values assigned to the density at the endpoints do not affect its integral. All checks pass, so the proposed function is a valid CDF for a continuous lifetime. The density is nonnegative and has total area one.
Common mistake avoided: Checking only that the CDF starts at zero is insufficient. The join at \(t=4\), nondecreasing behavior, and limiting value must also be checked.
Worked Example: Validating a Piecewise Survival Function
A hypothetical model proposes the following survival function, with time in years: \(S_x(t)=1\) for \(t<0\), \(S_x(t)=1-t/4\) for \(0\leq t\leq1\), \(S_x(t)=3(3-t)/8\) for \(1<t\leq3\), and \(S_x(t)=0\) for \(t>3\). Check the boundary values and monotonicity, then derive the density and verify its total.
Method. Evaluate both expressions where the definition changes, check that survival never rises, and use \(f_x(t)=-S_x'(t)\) wherever the derivative exists.
At \(t=0\), \(S_x(0)=1\). At the join \(t=1\), the first expression gives \(1-1/4=3/4\), while the second expression approaches \(3(3-1)/8=6/8=3/4\). At \(t=3\), the second expression gives zero, matching the value after three years. The slope is \(-1/4\) on \(0<t<1\), \(-3/8\) on \(1<t<3\), and zero after time three. Each slope is nonpositive, so survival is nonincreasing; it tends to zero.
The density on the two intervals is the negative of the corresponding survival slope:
The survival function starts at one, is continuous at its joins, is nonincreasing, and tends to zero. Its implied density is nonnegative and integrates to one, so the candidate passes the checks.
Common mistake avoided: The second interval has width \(3-1=2\), not width three. Using the wrong width would give an incorrect density total.
Worked Example: Normalizing a Proposed Density
A hypothetical underwriting model specifies an unnormalized density shape \(g(t)=t\) for \(0\leq t\leq2\), \(g(t)=4-t\) for \(2<t\leq4\), and zero elsewhere. Find the constant \(c\) that makes \(f_x(t)=c\,g(t)\) a valid density, then derive the CDF and check its endpoints.
Method. Integrate the proposed shape over its full support to find its total area. The normalizing constant must make the resulting area one.
The area on the first piece is \(\int_0^2 t\,dt=[t^2/2]_0^2=2\). The area on the second piece is \(\int_2^4(4-t)\,dt=[4t-t^2/2]_2^4=8-(8-2)=2\). The total area under \(g\) is \(2+2=4\), so \(c=1/4\). The normalized density is therefore \(t/4\) on the first piece and \((4-t)/4\) on the second, and zero outside.
For \(0\leq t\leq2\), integrate the first piece: \(F_x(t)=\int_0^t s/4\,ds=t^2/8\). At \(t=2\), this gives \(4/8=1/2\). For \(2<t\leq4\), add the probability accumulated by time two to the area from two to \(t\):
At \(t=2\), the second expression gives \(1-4/8=1/2\), agreeing with the first piece. At \(t=4\), it gives \(1\). The density is nonnegative on its support, and its total is \((1/4)(2+2)=1\). The CDF starts at zero, is nondecreasing because its density is nonnegative, and reaches one at the end of the support.
Common mistake avoided: The raw shape \(g\) is not a density just because it is nonnegative. Its area is four, so it must be divided by four before it can represent a probability distribution.
Common Boundary-Condition Traps
- Reversing the direction. A CDF cannot decrease, and a survival function cannot increase. Check slopes on each smooth piece and also check the values at joins.
- Checking the value at zero but not the limit. A function can start at the right value and still fail to approach one as a CDF or zero as a survival function.
- Confusing a density's height with total probability. The density need not be at most one. Its integral, rather than its maximum value, must equal one.
- Ignoring the support. A future lifetime has no probability below zero. A formula that assigns positive CDF probability to negative times contradicts the model.
- Assuming strict monotonicity. Flat sections are allowed. The required conditions are nondecreasing for a CDF and nonincreasing for survival, not strictly changing at every time.
A reliable validation order is to state the support and lifetime assumptions, check starting values and joins, check monotonicity, check the limiting values, and then verify density nonnegativity and total area when a density is supplied. This sequence is useful whenever a problem gives a candidate function, asks whether it can model a lifetime, or requires a missing normalizing constant.
Check Your Understanding
For each problem, apply the support, boundary, monotonicity, and total-probability checks as appropriate.
- A proposed CDF is zero for \(t<0\), \(F_x(t)=t/6\) for \(0\leq t\leq6\), and one for \(t>6\). Is it valid for a continuous lifetime? State its density on the interior of its support. (Answer: Yes; \(f_x(t)=1/6\) for \(0<t<6\), with total integral 1.)
- A proposed survival function is \(S_x(t)=1+t/5\) for \(0\leq t\leq1\), and \(S_x(t)=0\) for \(t>1\). Identify the first failed check. (Answer: It increases from \(S_x(0)=1\) to \(S_x(1)=6/5\), so it is not a valid survival function.)
- An unnormalized density shape is \(g(t)=2\) for \(0\leq t\leq1\), \(g(t)=1\) for \(1<t\leq3\), and zero elsewhere. Find the normalizing constant \(c\) for \(f_x(t)=c\,g(t)\). (Answer: The area is \(2(1)+1(2)=4\), so \(c=1/4\).)
- A candidate CDF is zero for \(t<0\), \(F_x(t)=t/4\) for \(0\leq t\leq2\), and \(F_x(t)=3/4\) for \(t>2\). It is nondecreasing and starts at zero. Is it valid under the assumption that the lifetime is finite with probability one? (Answer: No; its limit is \(3/4\), not 1.)
- A density is \(f_x(t)=t/8\) for \(0\leq t\leq2\) and zero elsewhere. Find its total integral and determine whether it is normalized. (Answer: \(\int_0^2 t/8\,dt=4/16=1/4\); it is not normalized.)