From a Survival Curve to a Lifetime Distribution
A survival curve gives the probability that a lifetime exceeds each elapsed time. From it, we can recover the probability of having died by a particular time, the density of death times when a density exists, and the instantaneous force of mortality. The key is to check what the curve says at the boundaries and where each derived function is defined.
Let \(T\) be a nonnegative lifetime, measured in a specified time unit, and assume there is no probability of death at time zero. Write \(S_0(t)=P(T>t)\) for its survival function and \(F_T(t)=P(T\le t)\) for its cumulative distribution function. The subscript in \(S_0\) indicates that time is measured from the starting point of the model.
These formulas do not all have the same domain. The CDF is defined at every real time; for a nonnegative lifetime it is zero at negative times. A density formula requires an absolutely continuous distribution and applies where the derivative exists. The force additionally requires positive survival: once \(S_0(t)=0\), its ratio definition is not available. At a corner in a piecewise curve, a derivative-based density or force may not be defined at the corner itself.
For a valid survival curve with no mass at time zero, \(S_0(0)=1\), its values lie between zero and one, and it is nonincreasing. For a proper lifetime that eventually ends with probability one, \(S_0(t)\) tends to zero as \(t\) tends to infinity. A downward jump in a survival curve represents probability concentrated at one exact time; in that case, a density alone does not describe the whole distribution.
Why Differentiation Gives the Density
The link between survival and density follows from the complement rule and the definition of a density. It is not a separate assumption about mortality.
For \(t\ge 0\), the events \(T\le t\) and \(T>t\) are complements. Thus \(F_T(t)=1-S_0(t)\).
If the distribution is absolutely continuous, its density is the derivative of its CDF. Therefore \(f_T(t)=F_T'(t)=-S_0'(t)\) wherever that derivative exists.
When \(S_0(t)>0\), divide the density by the probability of still being alive: \(\mu_t=f_T(t)/S_0(t)=-S_0'(t)/S_0(t)\). This is the force, not the density itself.
An interval death probability can also be read directly from the survival curve. For \(0\le a<b\), the event \(a<T\le b\) has probability \(F_T(b)-F_T(a)\), which equals \(S_0(a)-S_0(b)\). The order matters: survival decreases as time increases, so this difference is nonnegative.
Worked Examples
Worked Example: Exponential Survival and an Interval Probability
A hypothetical device has lifetime \(T\), measured in years, with survival function \(S_0(t)=e^{-0.2t}\) for \(t\ge 0\), and \(S_0(t)=1\) for \(t<0\). Assume this is a continuous lifetime model. Find its CDF, density, force of mortality, and the probability that failure occurs after year 2 but by year 7.
Method and assumptions. Use the complement to obtain the CDF, then differentiate it. Because this survival function is smooth and positive for every finite \(t\ge 0\), the density and force formulas apply throughout that domain.
CDF. For \(t\ge 0\), \(F_T(t)=1-S_0(t)=1-e^{-0.2t}\). For \(t<0\), \(F_T(t)=0\).
Density. Differentiating the CDF gives \(f_T(t)=0.2e^{-0.2t}\) for \(t\ge 0\). Its value at zero from the right is \(0.2\), consistent with the derivative of the CDF there from the right.
Force. Since \(S_0(t)>0\), \(\mu_t=f_T(t)/S_0(t)=(0.2e^{-0.2t})/e^{-0.2t}=0.2\) per year.
Interval probability. Apply \(P(2<T\le7)=S_0(2)-S_0(7)\):
The result is \(0.4237\), or 42.37% to two decimal places as a percentage. Check: \(S_0(2)>S_0(7)\), so the interval probability is positive and less than one. The common mistake is reversing the survival values, which would give a negative probability.
Worked Example: A Finite Support and a Singular Endpoint
A hypothetical component has lifetime \(T\), measured in years, with \(S_0(t)=(1-t/12)^3\) for \(0\le t\le12\), \(S_0(t)=0\) for \(t>12\), and \(S_0(t)=1\) for \(t<0\). Assume the lifetime is continuous. Derive its CDF, density, and force on the appropriate domains, then find \(P(4<T\le9)\).
Method and assumptions. The curve decreases from one to zero over the interval \([0,12]\), so the lifetime is confined to that interval. Differentiate only on the interior where the survival curve is smooth.
CDF. For \(0\le t\le12\), \(F_T(t)=1-(1-t/12)^3\). For \(t<0\), it is zero; for \(t>12\), it is one. The expressions agree at both endpoints: \(F_T(0)=0\) and \(F_T(12)=1\).
Density. On \(0<t<12\),
The density is zero outside the support. The formula gives a right-hand density limit at zero and a left-hand limit of zero at 12; values assigned at individual endpoints do not change probabilities.
Force. For \(0\le t<12\), divide by the positive survival probability:
At \(t=12\), survival is zero, so the force ratio is undefined; the expression tends to infinity as \(t\) approaches 12 from below. It is also undefined after 12, when no lives remain under this model.
Interval probability. Evaluate the survival values at the two endpoints:
The probability is \(0.2807\) to four decimal places. Check: the CDF difference gives the same result, since \(F_T(9)-F_T(4)=S_0(4)-S_0(9)\). A common mistake is to report a finite force at \(t=12\); the density-to-survival ratio cannot be evaluated when survival is zero.
Worked Example: Recovering Functions Across a Join
A hypothetical population has lifetime \(T\), measured in years, with the following survival curve:
Assume there is no probability concentrated at the join \(t=5\), and that the model is continuous on each side. Obtain the CDF, density, and force on each interval, then calculate \(P(3<T\le8)\).
Method and assumptions. First check the join: the first branch gives \(1-0.08(5)=0.6\), and the second gives \(0.6e^0=0.6\). The survival curve is continuous there, but its slopes differ, so derivatives at the join require care.
CDF. For \(t<0\), \(F_T(t)=0\). For \(0\le t\le5\), \(F_T(t)=0.08t\). For \(t>5\), \(F_T(t)=1-0.6e^{-0.1(t-5)}\). Both branches give \(F_T(5)=0.4\).
Density. Differentiating on the two open intervals gives \(f_T(t)=0.08\) for \(0<t<5\), and \(f_T(t)=0.06e^{-0.1(t-5)}\) for \(t>5\). There is no point mass at \(t=5\), because the CDF is continuous there. The one-sided density values at the join differ, so the ordinary derivative at exactly \(t=5\) is not defined.
Force. For \(0\le t<5\), \(\mu_t=0.08/(1-0.08t)\). For \(t>5\), \(\mu_t=0.1\), because the factor \(0.6e^{-0.1(t-5)}\) cancels in the density-to-survival ratio. The force has different one-sided limits at \(t=5\); its derivative-based value there is not defined.
Interval probability. At \(t=3\), \(S_0(3)=1-0.08(3)=0.76\). At \(t=8\), \(S_0(8)=0.6e^{-0.1(3)}=0.6e^{-0.3}\). Therefore,
The answer is \(0.3155\) to four decimal places. Check: the probability is positive because survival has decreased from \(0.76\) to approximately \(0.4445\). Do not infer an atom at the join merely because the slope changes; an atom would require a jump in the survival curve.
Domain Checks and Common Traps
A reliable calculation starts by locating the intervals on which the given survival formula applies. Then check whether it is differentiable there and whether it is positive before forming a force. For an interval probability, use the CDF difference or its equivalent survival difference.
- Reversing the interval difference. The incorrect calculation \(P(a<T\le b)=S_0(b)-S_0(a)\) is nonpositive for a decreasing curve. For the first example, this wrong order gives \(e^{-1.4}-e^{-0.4}=-0.4237\). The correction is \(S_0(a)-S_0(b)\), giving \(0.4237\).
- Calling the density the force. The density is \(f_T(t)=-S_0'(t)\); the force is \(f_T(t)/S_0(t)\). They are generally different functions and have different interpretations.
- Ignoring the support endpoint. A formula derived where survival is positive does not automatically apply at or beyond an endpoint where survival reaches zero.
- Differentiating across a corner. A piecewise curve may be continuous but not differentiable at its join. Use branch-specific derivatives and do not assign a derivative-based force there unless it exists.
Check Your Understanding
For each problem, state the relevant domain as well as the requested quantity.
- A hypothetical lifetime in years has \(S_0(t)=e^{-0.25t}\) for \(t\ge0\). Find \(F_T(2)\), \(f_T(2)\), and \(\mu_2\). (Answer: \(F_T(2)=0.3935,\ f_T(2)=0.1516,\ \mu_2=0.25\) per year.)
- A continuous lifetime has \(S_0(t)=(1-t/8)^2\) for \(0\le t\le8\), and zero for \(t>8\). Find \(\mu_4\) and \(P(2<T\le6)\). (Answer: \(\mu_4=0.5\) per year; \(P(2<T\le6)=0.5000\).)
- A lifetime has \(S_0(t)=1-0.05t\) for \(0\le t\le4\), and \(S_0(t)=0.8e^{-0.2(t-4)}\) for \(t>4\). Find \(F_T(7)\), \(f_T(7)\), and \(\mu_7\). (Answer: \(F_T(7)=0.5610,\ f_T(7)=0.0878,\ \mu_7=0.2\) per year.)
- A continuous lifetime has \(S_0(t)=(1-t/6)^2\) for \(0\le t\le6\), and zero after 6. Find \(P(1<T\le4)\) and \(\mu_3\), and state whether the force is defined at \(t=6\). (Answer: \(P(1<T\le4)=7/12=0.5833,\ \mu_3=2/3\) per unit time; the force is undefined at \(t=6\) because survival is zero.)