From Force to Lifetime Functions
A force of mortality describes an instantaneous rate, not a probability over a finite interval. To obtain probabilities from a duration-dependent force, first integrate it over time. That integral determines survival; the CDF and density then follow from survival. Carrying out these steps in order helps prevent a common error: treating a force value as though it were itself a death probability.
Let \(T\) be a nonnegative, absolutely continuous lifetime measured from time 0. Assume the specified force \(\mu(t)\) is nonnegative and integrable over every finite interval, and that \(P(T>0)=1\). These assumptions give a continuous lifetime distribution with no point mass at the starting time. The time variable and the force must use reciprocal units; for example, if \(t\) is in years, \(\mu(t)\) is measured per year.
Cumulative force is dimensionless: it is a rate integrated over a duration. The force itself has units of inverse time, while \(S_T(t)\), \(F_T(t)\), and probabilities are dimensionless. A proper lifetime model in which death occurs eventually with probability 1 must also satisfy \(H(t)\to\infty\) as \(t\to\infty\). If \(H(t)\) instead has a finite limit, the survival function has a positive limit, so the model assigns positive probability to surviving indefinitely.
Why Integration Gives Survival
The force is the instantaneous rate of decrease in survival, relative to the probability of still being alive. In differential form, this is \(S_T'(t)=-\mu(t)S_T(t)\). This equation also explains why integrating the force produces an exponential.
For a differentiable survival function with positive values, divide \(S_T'(t)=-\mu(t)S_T(t)\) by \(S_T(t)\) to obtain \(S_T'(t)/S_T(t)=-\mu(t)\).
The left side integrates to \(\ln S_T(t)-\ln S_T(0)\), and the right side integrates to \(-\int_0^t\mu(s)\,ds\).
Since \(P(T>0)=1\), \(S_T(0)=1\) and \(\ln S_T(0)=0\). Thus \(\ln S_T(t)=-H(t)\), so \(S_T(t)=e^{-H(t)}\).
Because \(F_T(t)=1-S_T(t)\), differentiating gives \(f_T(t)=F_T'(t)=-S_T'(t)=\mu(t)S_T(t)\).
The last identity is an important consistency check: the density must equal the force multiplied by the probability of surviving to that duration. A nonnegative force therefore gives a nonnegative density. Also, \(F_T(0)=0\), \(S_T(0)=1\), and \(F_T(t)+S_T(t)=1\).
Worked Examples
Worked Example: A Linearly Increasing Force
A hypothetical device has a failure time \(T\), measured in years, with force \(\mu(t)=0.02+0.01t\) per year for \(t\ge0\). Under the absolutely continuous lifetime model, find the cumulative force, survival, CDF, and density at year 4. Also find the probability of failure in \((2,4]\). Use exact expressions until the final step, then round probabilities and density to four decimal places.
Method. Integrate the force from 0 to the duration of interest, then substitute the cumulative force into the survival formula. The force is nonnegative for \(t\ge0\), as required.
1. Cumulative force. The duration is in years, so integrate the per-year force with respect to years:
At \(t=4\), \(H(4)=0.02(4)+0.005(4^2)=0.08+0.08=0.16\). At \(t=2\), \(H(2)=0.02(2)+0.005(2^2)=0.04+0.02=0.06\).
2. Survival and CDF at year 4. The survival probability is \(S_T(4)=e^{-0.16}=0.8521438\), and the CDF is \(F_T(4)=1-e^{-0.16}=0.1478562\). Rounded to four decimal places, these are \(0.8521\) and \(0.1479\). They sum to 1 before rounding.
3. Density at year 4. The force at year 4 is \(\mu(4)=0.02+0.01(4)=0.06\) per year. Therefore \(f_T(4)=\mu(4)S_T(4)=0.06(0.8521438)=0.0511286\) per year, or \(0.0511\) per year to four decimal places.
4. Probability of failure in \((2,4]\). The event requires survival through year 2 and failure by year 4, so subtract the survival at year 4 from the survival at year 2:
The probability is \(0.0896\) to four decimal places. This method avoids treating the force as a finite-interval probability.
Worked Example: Checking the Density Against the CDF
A hypothetical biological lifetime, measured in years, has force \(\mu(t)=0.03+0.02t\) per year for \(t\ge0\). Find the cumulative force, survival, CDF, and density at year 3. Then calculate the probability that the lifetime is in \((1,3]\). Round probabilities and density to four decimal places.
Method. Derive all functions from the same cumulative force, then verify at year 3 that differentiating the CDF gives the stated density.
1. Integrate the force.
Thus \(H(3)=0.03(3)+0.01(3^2)=0.09+0.09=0.18\), while \(H(1)=0.03+0.01=0.04\).
2. Calculate the distribution functions at year 3. The survival is \(S_T(3)=e^{-0.18}=0.8352702\). Hence \(F_T(3)=1-e^{-0.18}=0.1647298\). Rounded to four decimal places, these are \(0.8353\) and \(0.1647\).
3. Calculate and check the density. At year 3, \(\mu(3)=0.03+0.02(3)=0.09\) per year. Therefore \(f_T(3)=0.09e^{-0.18}=0.0751743\) per year, or \(0.0752\) per year to four decimal places. Direct differentiation confirms the result:
4. Calculate the interval probability. Since \(S_T(1)=e^{-0.04}=0.9607894\),
The interval probability is \(0.1255\) to four decimal places. The method avoids calculating an interval probability from a force value at just one duration.
Worked Example: A Force That Changes Formula
A hypothetical sensor has a failure time in years and a force given by \(\mu(t)=0.02\) per year for \(0\le t\le2\), and \(\mu(t)=0.02+0.01(t-2)\) per year for \(t>2\). Find the cumulative force, survival, CDF, and density at year 4, and find the probability of failure in \((1,4]\). Round probabilities and density to four decimal places.
Method. Integrate each force formula over the interval where it applies. For durations past year 2, keep the accumulated force from the first interval; do not restart the integral at the change point.
1. Write the cumulative force. For \(0\le t\le2\),
For \(t>2\), the first interval contributes \(0.02(2)=0.04\). Integrating the force after year 2 gives
At year 4, \(H(4)=0.04+0.02(2)+0.005(2^2)=0.04+0.04+0.02=0.10\). At year 1, \(H(1)=0.02\).
2. Find survival and the CDF at year 4. \(S_T(4)=e^{-0.10}=0.9048374\), so \(F_T(4)=1-e^{-0.10}=0.0951626\). Rounded to four decimal places, these are \(0.9048\) and \(0.0952\).
3. Find the density at year 4. The applicable force is \(\mu(4)=0.02+0.01(4-2)=0.04\) per year. Thus \(f_T(4)=0.04e^{-0.10}=0.0361935\) per year, or \(0.0362\) per year to four decimal places.
4. Find the interval probability. Since \(S_T(1)=e^{-0.02}=0.9801987\),
This is \(0.0754\) to four decimal places. The cumulative force is continuous at year 2: both formulas give \(H(2)=0.04\). The interval calculation uses that accumulated value rather than discarding the force experienced before the change point.
Common Errors and Consistency Checks
A force function, its integral, and the resulting probabilities are related but are not interchangeable. In particular, \(\mu(t)\) is an instantaneous rate, \(H(t)\) is cumulative force, and \(F_T(t)\) is a probability by duration \(t\). Use the stated force over the entire interval from 0 to \(t\), including any changes in its formula.
- Using the force as the CDF. Setting \(F_T(t)=\mu(t)\) is incorrect. The force has units of inverse time and may even exceed 1; a CDF is a probability between 0 and 1.
- Forgetting to integrate. The exponent in \(e^{-H(t)}\) uses \(H(t)=\int_0^t\mu(s)\,ds\), not the force evaluated at the endpoint.
- Using the wrong interval order. For \(a<b\), \(P(a<T\le b)=S_T(a)-S_T(b)=F_T(b)-F_T(a)\). Reversing the survival values would produce a negative result.
- Dropping earlier accumulation in a piecewise model. When the force changes at a time \(c\), the cumulative force after \(c\) includes \(\int_0^c\mu(s)\,ds\) as well as the integral from \(c\) to the duration.
A compact check is to confirm \(H(0)=0\), \(S_T(0)=1\), and \(F_T(t)=1-S_T(t)\). Where the force is differentiable, also check \(F_T'(t)=f_T(t)=\mu(t)S_T(t)\). These identities must describe one and the same lifetime model.
Check Your Understanding
For each problem, derive the cumulative force before calculating the requested lifetime quantities. Times are in years; round probabilities and densities to four decimal places.
- A hypothetical component has force \(\mu(t)=0.01+0.01t\) per year. Find \(H(2)\), \(S_T(2)\), \(F_T(2)\), and \(f_T(2)\). (Answer: \(H(2)=0.0400\), \(S_T(2)=0.9608\), \(F_T(2)=0.0392\), and \(f_T(2)=0.0288\) per year.)
- A hypothetical lifetime has force \(\mu(t)=0.02+0.01t\) per year. Find \(P(1<T\le3)\) and \(f_T(3)\). (Answer: \(P(1<T\le3)=0.0750\) and \(f_T(3)=0.0450\) per year.)
- A hypothetical device has force \(\mu(t)=0.006t^2\) per year. Find \(H(5)\), \(S_T(5)\), \(F_T(5)\), \(f_T(5)\), and \(P(2<T\le5)\). (Answer: \(H(5)=0.2500\), \(S_T(5)=0.7788\), \(F_T(5)=0.2212\), \(f_T(5)=0.1168\) per year, and \(P(2<T\le5)=0.2053\).)
- A hypothetical sensor has force \(\mu(t)=0.03\) per year for \(0\le t\le1\), and \(\mu(t)=0.03+0.02(t-1)\) per year for \(t>1\). Find \(H(3)\), \(S_T(3)\), \(F_T(3)\), and \(f_T(3)\). (Answer: \(H(3)=0.1300\), \(S_T(3)=0.8781\), \(F_T(3)=0.1219\), and \(f_T(3)=0.0615\) per year.)
- A hypothetical lifetime has force \(\mu(t)=0.01(1+t)^2\) per year. Find \(H(2)\), \(S_T(2)\), \(F_T(2)\), and \(f_T(2)\). (Answer: \(H(2)=0.0867\), \(S_T(2)=0.9170\), \(F_T(2)=0.0830\), and \(f_T(2)=0.0825\) per year.)