Why the Floor Relation Matters
Curtate future lifetime counts the number of whole years a life aged \(x\) completes after age \(x\). Complete future lifetime measures the actual time lived after age \(x\), including any fractional year. They describe the same future lifetime on different scales: one is integer-valued, while the other can take fractional values. The floor relation connects them and lets us translate an event stated with one variable into an event stated with the other.
This translation is especially useful when a problem mixes conditions such as “has completed two years” and “dies after two and a half years.” The endpoint at an integer age matters: completing at least \(n\) years means reaching time \(n\), whereas having curtate lifetime greater than \(n\) means reaching time \(n+1\).
Definitions and the Exact Relation
For example, if \(T_x=3.6\), then \(K_x=3\): the life completed three full years but not four. If \(T_x=4\) exactly, then \(K_x=4\). This is a pointwise relation, so it does not require a mortality distribution assumption. In the usual continuous lifetime models, the probability of death at any particular exact integer time is zero, but the floor definition also specifies what happens at such a time.
The left endpoint is included because a life that reaches exactly time \(k\) has completed \(k\) years. The right endpoint is excluded because a life that reaches exactly time \(k+1\) has completed \(k+1\) years, not \(k\).
Deriving the Event Translations
The basic translation follows directly from the definition of the floor. For an integer \(n\geq 0\), \(K_x\geq n\) means that the greatest integer not exceeding \(T_x\) is at least \(n\). That occurs exactly when \(T_x\geq n\). Replacing a non-strict inequality on \(K_x\) with a strict one changes the corresponding time threshold by a full year.
\(K_x=\lfloor T_x\rfloor\), so \(K_x\) is the integer \(k\) precisely over the interval \(k\leq T_x<k+1\).
\(\lfloor T_x\rfloor\geq n\) holds if and only if \(T_x\geq n\). If \(T_x<n\), its floor is at most \(n-1\); if \(T_x\geq n\), its floor is at least \(n\).
\(\lfloor T_x\rfloor>n\) holds if and only if \(T_x\geq n+1\). Also, \(\lfloor T_x\rfloor\leq n\) holds if and only if \(T_x<n+1\).
Together, these give a compact set of event translations. Intersections can be translated one condition at a time, then reduced to the interval satisfying both conditions.
The last equivalence applies to integers \(0\leq a\leq b\). It shows why \(a\leq K_x\leq b\) is one continuous-time interval, not a collection of separated intervals: \(T_x\) may take any value from \(a\) up to, but not including, \(b+1\).
Using Survival Probabilities
Let \({}_tp_x=P(T_x\geq t)\) denote the probability that a life aged \(x\) survives at least \(t\) years. With this convention, the event \(K_x=k\) is \(k\leq T_x<k+1\). Subtracting the probability of reaching time \(k+1\) from the probability of reaching time \(k\) leaves exactly the probability of being in that interval. This subtraction is valid whether or not the model assigns probability to exact times, provided survival uses the stated “at least” convention.
The survival subtraction is the useful probability calculation after the event has been translated. For a mixed event, first intersect the time intervals. For instance, \(K_x=2\) and \(T_x>2.4\) together require \(2.4<T_x<3\). A difference of survival probabilities can then evaluate that interval; if the lifetime distribution is continuous, endpoint distinctions at fixed times do not change the probability.
Worked Example: A Curtate-Lifetime Range From a Survival Table
A hypothetical life table gives \({}_2p_x=0.86\), \({}_3p_x=0.78\), \({}_4p_x=0.69\), and \({}_5p_x=0.62\). Find the probability that \(K_x\) is either 2 or 4. Use the survival-interval method and identify the endpoint issue to avoid.
Method. The two possible values are separate events, so calculate \(P(K_x=2)\) and \(P(K_x=4)\) from adjacent survival probabilities, then add them. The supplied values are treated as probabilities of surviving at least the stated number of years.
For \(K_x=2\), the event is \(2\leq T_x<3\). Therefore,
For \(K_x=4\), the event is \(4\leq T_x<5\). Therefore,
The events \(K_x=2\) and \(K_x=4\) are disjoint, so
Check. Both component probabilities are nonnegative, and their sum is \(0.15\), which is no greater than 1. The endpoint error to avoid is assigning deaths at time 3 to \(K_x=2\); the floor relation places \(T_x=3\) in \(K_x=3\), not \(K_x=2\).
Worked Example: Combining a Curtate Condition With a Fractional-Time Condition
In a hypothetical model, the force of mortality is constant at \(0.25\) per year for a life aged \(x\). Find \(P(K_x=2,\ T_x>2.4)\). Give the probability to four decimal places.
Model and method. Under a constant force of mortality \(\mu=0.25\) per year, the survival probability is \({}_tp_x=e^{-0.25t}\). First intersect the event intervals: \(K_x=2\) requires \(2\leq T_x<3\), and the additional condition requires \(T_x>2.4\). Together they require \(2.4<T_x<3\).
For this continuous model, the probability of that interval is the survival probability at 2.4 years minus the survival probability at 3 years:
Rounded to four decimal places, the probability is \(\boxed{0.0764}\). As a check, the probability is positive because survival to 2.4 years is more likely than survival to 3 years. It is also less than the probability of surviving to 2.4 years.
Common mistake avoided. The condition \(K_x=2\) does not mean \(T_x=2\), and it does not mean \(2<T_x\leq3\). Its exact interval is \(2\leq T_x<3\); applying the additional condition then raises the lower endpoint to 2.4.
Worked Example: A Conditional Probability With Two Curtate Conditions
For a hypothetical life aged \(x\), suppose \({}_1p_x=0.90\), \({}_2p_x=0.80\), and \({}_4p_x=0.60\). Find \(P(K_x\geq2\mid K_x<4)\). State the event represented by the denominator before calculating.
Method. Translate each curtate condition to a condition on \(T_x\). The numerator event, including the conditioning event, is \(K_x\geq2\) and \(K_x<4\), or equivalently \(2\leq T_x<4\). The conditioning event \(K_x<4\) is \(T_x<4\).
Thus the numerator probability is the probability of surviving to time 2 but not to time 4:
The denominator is the probability of not surviving to time 4:
Therefore,
Check. The numerator event is contained in the denominator event, and \(0.20/0.40\) is between 0 and 1. The common error is to use \({}_4p_x\) as the denominator; that is the probability of reaching time 4, whereas the condition \(K_x<4\) means \(T_x<4\).
Common Endpoint and Inequality Traps
The most reliable approach is to translate every condition before using a probability formula. The following comparisons show why the inequality symbol cannot be casually changed.
| Curtate event | Equivalent complete-lifetime event | Interpretation |
|---|---|---|
| \(K_x\geq n\) | \(T_x\geq n\) | At least \(n\) full years are completed |
| \(K_x>n\) | \(T_x\geq n+1\) | At least \(n+1\) full years are completed |
| \(K_x\leq n\) | \(T_x<n+1\) | The life does not reach \(n+1\) full years |
| \(K_x<n\) | \(T_x<n\) | The life does not reach \(n\) full years |
For example, \(K_x>2\) does not translate to \(T_x>2\). A life with \(2.7\) years of future lifetime has \(K_x=2\), so it fails the first condition. The correct translation is \(T_x\geq3\). In a continuous model, \(T_x\geq3\) and \(T_x>3\) have the same probability, but they are not the same event as definitions; keeping the exact translation prevents errors in models with point masses or when combining conditions.
A second trap is taking a complement incorrectly. The complement of \(K_x\geq n\) is \(K_x<n\), which translates to \(T_x<n\), not \(T_x\leq n\). Likewise, the complement of \(K_x\leq n\) is \(K_x>n\), equivalent to \(T_x\geq n+1\).
Check Your Understanding
Translate each event carefully before using the given survival information.
- A hypothetical life table gives \({}_1p_x=0.94\) and \({}_4p_x=0.70\). Find \(P(1\leq K_x\leq3)\). (Answer: \(0.24\))
- For a life with \({}_3p_x=0.76\), find \(P(K_x>2)\). (Answer: \(0.76\))
- A hypothetical life table gives \({}_2p_x=0.88\) and \({}_5p_x=0.55\). Find \(P(2\leq T_x<5)\), and express the event using \(K_x\). (Answer: \(0.33;\ 2\leq K_x\leq4\))
- Suppose \({}_1p_x=0.90\), \({}_2p_x=0.82\), and \({}_4p_x=0.58\). Find \(P(K_x\geq2\mid K_x<4)\). (Answer: \(4/7\), or \(0.5714\))
- Under a hypothetical constant force of mortality \(\mu=0.10\) per year, find \(P(K_x=1,\ T_x>1.5)\). Give the answer to four decimal places. (Answer: \(e^{-0.15}-e^{-0.20}\approx0.0420\))