Why the Time Scale Matters
A future-lifetime calculation can look almost correct while using the wrong time scale. The curtate future lifetime counts completed years; the complete future lifetime measures elapsed time, including the fraction of a year lived. Confusing those quantities can shift an event by a year, mishandle an interval endpoint, or use a life-table value from the wrong age.
Let \(T_x\) denote the complete future lifetime of a life aged \(x\), and let \(K_x\) denote the curtate future lifetime. Both are measured from age \(x\), but they describe different random variables. The notation \({}_tp_x=P(T_x>t)\) denotes the probability of surviving at least \(t\) years in the continuous-time survival model; \({}_tq_x=1-{}_tp_x\).
The floor relation is the key translation rule. For example, \(K_x\geq 2\) means \(T_x\geq 2\): the life survives for two complete years. In a continuous model this has the same probability as \(T_x>2\), namely \({}_2p_x\). By contrast, \(K_x>2\) means \(K_x\geq 3\), so its probability is \({}_3p_x\). The distinction between \(>\) and \(\geq\) matters for the integer-valued variable \(K_x\), even when integer-time endpoints have probability zero for \(T_x\).
Translate the Event Before Calculating
Before using a life table or a survival probability, rewrite the event in the variable actually being calculated. The following relations apply when \(T_x\) has no probability mass at integer times:
| Event | Equivalent complete-lifetime event | Probability |
|---|---|---|
| \(K_x=k\) | \(k\leq T_x<k+1\) | \({}_kp_x-{}_{k+1}p_x={}_kp_xq_{x+k}\) |
| \(K_x\geq k\) | \(T_x\geq k\) | \({}_kp_x\) |
| \(K_x>k\) | \(T_x\geq k+1\) | \({}_{k+1}p_x\) |
| \(K_x\leq k\) | \(T_x<k+1\) | \(1-{}_{k+1}p_x\) |
The identity \({}_kp_xq_{x+k}\) assumes the standard life-table notation: \({}_kp_x\) is survival from age \(x\) to age \(x+k\), and \(q_{x+k}\) is the probability of dying during the following year, conditional on being alive at age \(x+k\). If the survival model has a point mass at an integer time, endpoint conventions must be handled explicitly rather than discarded.
Worked Examples: Integer-Year Probabilities and Endpoints
Worked Example: A Curtate Value Is a Whole-Year Interval
For a hypothetical life aged \(x\), suppose \(q_x=0.10\), \(q_{x+1}=0.20\), and \(q_{x+2}=0.25\). Assume these are the applicable one-year mortality probabilities and that death times have no point masses at integer times. Calculate \(P(K_x=2)\), \(P(K_x\leq2)\), and \(P(K_x>1)\). Also calculate \(P(K_x=2\mid K_x\geq1)\).
Method. Translate each event into survival through the required integer years. The common mistake to avoid is treating \(K_x=2\) as death exactly two years after age \(x\); it means death during the third year of follow-up.
First, \(p_x=1-q_x=0.90\), \(p_{x+1}=1-q_{x+1}=0.80\), and \(p_{x+2}=1-q_{x+2}=0.75\).
To have \(K_x=2\), the life survives the first two years and dies in the third:
The event \(K_x\leq2\) means death before completing three years. Survival through all three years is \({}_3p_x=(0.90)(0.80)(0.75)=0.54\), so \(P(K_x\leq2)=1-0.54=0.46\).
The strict inequality \(K_x>1\) means \(K_x\geq2\), or survival for two complete years: \(P(K_x>1)={}_2p_x=(0.90)(0.80)=0.72\). Finally, \(K_x\geq1\) means survival through the first year, with probability \(0.90\). Hence \(P(K_x=2\mid K_x\geq1)=0.18/0.90=0.20\).
Check. The probabilities for \(K_x=0,1,2\) are \(0.10\), \((0.90)(0.20)=0.18\), and \(0.18\). Their sum is \(0.46\), agreeing with \(P(K_x\leq2)\). The conditional probability \(0.20\) also equals the probability of surviving year 2 and dying in year 3, given survival to age \(x+1\): \((0.80)(0.25)=0.20\).
Fractional Ages: Interpolate First, Then Condition
When the starting age is fractional, the life-table exposure at that age must be established before forming a conditional probability. Under the uniform distribution of deaths (UDD) assumption, the life-table survivor function is linear between successive integer ages. If \(0\leq s\leq1\), then \({}_s p_x=1-sq_x\), and the interpolated survivor count is \(\ell_{x+s}=(1-s)\ell_x+s\ell_{x+1}\). For a fractional-age life, conditional survival from age \(x+s\) to age \(x+s+t\) is the ratio of the corresponding survivor counts. UDD is an assumption about the distribution of deaths within each year; a different fractional-age model, such as constant force, can give a different answer.
Worked Example: Curtate Lifetime From a Fractional Starting Age
A hypothetical life table gives \(\ell_x=1000\), \(\ell_{x+1}=800\), and \(\ell_{x+2}=600\). Assume UDD in each year. For a life now aged \(x+0.5\), calculate \(P(K_{x+0.5}=0)\), \(P(K_{x+0.5}=1)\), and \(P(K_{x+0.5}\geq2)\).
Method. Interpolate the survivor count at each needed age, then divide by the count at the starting age. The common mistake to avoid is using \(\ell_x\) as the conditional denominator even though the life is already aged \(x+0.5\).
By UDD, the starting survivor count is \(\ell_{x+0.5}=(1000+800)/2=900\). At age \(x+1.5\), it is \(\ell_{x+1.5}=(800+600)/2=700\).
The event \(K_{x+0.5}=0\) is death during the first year after the fractional starting age, before reaching age \(x+1.5\). Its probability is \(1-700/900=200/900=2/9\).
For \(K_{x+0.5}=1\), the life must survive to age \(x+1.5\) and die before age \(x+2.5\). Since the table provides \(\ell_{x+2}=600\), and the calculation instead needs age \(x+2.5\), interpolate between ages \(x+2\) and \(x+3\). The necessary value \(\ell_{x+3}\) has not been supplied, so \(P(K_{x+0.5}=1)\) cannot be determined from the given data alone.
For the same reason, \(P(K_{x+0.5}\geq2)\) requires the survivor count at age \(x+2.5\), which also cannot be determined from the supplied table. The available value \(\ell_{x+2}=600\) instead gives the different probability \(P(T_{x+0.5}>1.5)=600/900=2/3\), corresponding to survival to age \(x+2\), not to age \(x+2.5\).
Check. This example is intentionally a data-sufficiency trap: the first requested probability is determined, but the other two are not. Interpolating \(\ell_{x+2.5}\) would require \(\ell_{x+3}\). A calculation that silently substitutes \(\ell_{x+2}\) would use a 1.5-year duration where a two-year duration is required.
Complete Lifetimes and Conditional Probabilities
A complete-lifetime event can end at a fractional duration, so its probability generally requires survival probabilities at fractional times. Under a constant force of mortality \(\mu\) over the relevant ages, the survival function is \({}_tp_x=e^{-\mu t}\). At a fractional starting age within the same constant-force segment, conditional survival for another \(t\) years is also \(e^{-\mu t}\). The starting age in the condition matters: a probability conditional on being alive at \(x+s\) is computed from that age, not from birth or from age \(x\).
Worked Example: Do Not Round a Complete Lifetime to an Integer Year
Assume a hypothetical life has constant force of mortality \(\mu=\ln 2\) per year over the ages considered. The life is known to be alive at age \(x+0.5\). Calculate the probability of surviving another 1.5 years, \(P(T_{x+0.5}>1.5)\), and compare it with \(P(K_{x+0.5}>1)\).
Method. Use the continuous survival model for the fractional duration, then translate the curtate event using the floor relation. The common mistake to avoid is replacing a 1.5-year duration by either one or two years without first identifying the event.
For constant force, conditional survival for \(t\) years is \(e^{-\mu t}\). Therefore
For the curtate event, \(K_{x+0.5}>1\) is equivalent to \(K_{x+0.5}\geq2\), which requires survival for at least two complete years after age \(x+0.5\). Thus
The probabilities differ because the first event requires survival for 1.5 years and the second for two years. As expected, survival for two years is less likely than survival for 1.5 years. Probabilities are rounded to four decimal places.
Common Exam Traps
- Using the wrong integer boundary. \(K_x>k\) means \(K_x\geq k+1\), not \(K_x\geq k\). Translate the inequality before substituting a survival probability.
- Using a curtate event for a complete-time interval. The probability \(P(K_x=k)\) covers the whole interval from \(k\) to \(k+1\); it is not the probability that \(T_x=k\).
- Forgetting the conditioning age. Given survival to age \(x+s\), divide later survivor counts by \(\ell_{x+s}\), not by \(\ell_x\). Under a specified continuous model, use conditional survival from the attained age.
- Interpolating without the required model or data. UDD and constant force are distinct assumptions. Also check that the life-table ages needed for the requested duration are actually available.
- Assuming endpoint equivalence without checking the model. For a continuous lifetime distribution, \(P(T_x=k)=0\). If the model includes a point mass at an integer time, strict and inclusive endpoints can have different probabilities.
Check Your Understanding
For each problem, identify the time scale and conditioning event before calculating.
- A hypothetical life has \(q_x=0.15\) and \(q_{x+1}=0.20\). Assuming no point masses at integer times, find \(P(K_x=1)\), \(P(K_x\geq2)\), and \(P(K_x\leq1)\). (Answer: \(0.17,\ 0.68,\ 0.32\), respectively.)
- A life table has \(\ell_x=1200\), \(\ell_{x+1}=900\), and \(\ell_{x+2}=600\), with UDD within each year. For a life aged \(x+0.25\), find the probability of surviving another \(0.75\) years. (Answer: \(900/1125=0.8000\).)
- Under a constant force with annual survival probability \(p_x=0.64\), calculate \(P(T_x>1.5)\) and \(P(K_x>1)\). (Answer: \(0.5120\) and \(0.4096\), respectively.)
- Under UDD, a table gives \(\ell_x=1000\) and \(\ell_{x+1}=700\). For a life aged \(x+0.6\), calculate the probability of dying before reaching age \(x+1\). (Answer: \((820-700)/820=6/41\approx0.1463\).)
- A hypothetical life has \(q_x=0.12\), \(q_{x+1}=0.25\), and \(q_{x+2}=0.40\). Find \(P(K_x=2)\), \(P(K_x\geq2)\), and \(P(K_x=2\mid K_x\geq2)\). (Answer: \(0.2640,\ 0.6600,\ 0.4000\), respectively.)