From Annual Mortality to Curtate Lifetime Probabilities
The curtate future lifetime records the number of whole years a person aged \(x\) lives after age \(x\). Its possible values are nonnegative integers. To find the probability of one particular value, combine two events: the person must survive the required whole years, then die during the next year. This distinction matters because \(q_x\) is a conditional probability of death in one year, not the probability of surviving to a later year.
This tutorial develops the point probabilities from annual death probabilities and life-table counts. It also shows how to calculate tail and conditional probabilities, and how to check whether the results are consistent. The methods are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
The Point Probability Formula
For \(K_x=k\), the life must first survive \(k\) years and then die during the next year. The survival probability is \({}_k p_x\), and the probability of death in the following year, conditional on survival to age \(x+k\), is \(q_{x+k}\). Therefore:
For \(k=0\), no prior survival is required, so \({}_0p_x=1\) and \(P(K_x=0)=q_x\). For \(k=1\), the life must survive the first year and die during the second, so \(P(K_x=1)=p_xq_{x+1}\). Each later point probability follows the same pattern.
The annual death probabilities can be combined to obtain the survival probability:
Consequently, an equivalent expression for the point probability is:
The product of survival probabilities is empty when \(k=0\), and is defined as 1. The formula uses \(q_{x+k}\) because the death occurs in the year after the \(k\) completed years of survival.
Using Life-Table Counts
If \(\ell_y\) denotes the number of lives at exact age \(y\) in a life table, then the one-year survival and death probabilities are:
These ratios are conditional on being alive at age \(y\). In particular, the probability of \(K_x=k\) can be calculated directly from the table as:
To see why, the numerator counts the lives from the original age-\(x\) cohort who reach age \(x+k\) but not age \(x+k+1\); the denominator is the number alive at age \(x\). This direct count ratio agrees with multiplying the annual conditional probabilities.
Worked Example: Point Probabilities from Life-Table Counts
A hypothetical life table has \(\ell_x=1600\), \(\ell_{x+1}=1440\), \(\ell_{x+2}=1080\), \(\ell_{x+3}=810\), and \(\ell_{x+4}=486\). Find \(q_x\), \(q_{x+1}\), and \(q_{x+2}\), and then calculate \(P(K_x=0)\), \(P(K_x=1)\), \(P(K_x=2)\), and \(P(K_x=3)\). Also find \(P(K_x\ge4)\).
Method. First calculate each annual death probability from consecutive life-table counts. Then use survival probabilities for the completed years and the death probability for the next year.
At age \(x\):
At age \(x+1\):
At age \(x+2\):
The probability of dying in the first year is:
The probability of surviving the first year and dying in the second is:
For \(K_x=2\), survive the first two years and die in the third:
For \(K_x=3\), the table gives \(q_{x+3}=(810-486)/810=324/810=0.40\), so \(p_{x+3}=0.60\). Therefore:
Survival to age \(x+4\) has probability:
The answers are consistent: \(P(K_x=0)+P(K_x=1)+P(K_x=2)+P(K_x=3)+P(K_x\ge4)=0.10+0.225+0.16875+0.2025+0.30375=1\). A common mistake would be to multiply the three death probabilities \(q_xq_{x+1}q_{x+2}=(0.10)(0.25)(0.25)=0.00625\) and report that as \(P(K_x=2)\). That product is not the probability of surviving two years and dying in the third; the correct calculation uses \(p_xp_{x+1}q_{x+2}=0.16875\).
Point Probabilities, Tail Probabilities, and Conditional Probabilities
A point probability answers whether \(K_x\) equals one exact integer. A tail probability answers whether the life survives at least a specified number of complete years. By definition:
For example, \(K_x\ge3\) means the life reaches age \(x+3\). It does not specify what happens after that age. The probability of dying in one of a range of years is the sum of the corresponding point probabilities. Equivalently, the probability of dying after at least \(r\) but before completing \(s\) years, where \(s>r\), is:
The second equality follows because surviving \(r\) years either leads to death before completing \(s\) years or survival to age \(x+s\). These two outcomes partition the event \(K_x\ge r\).
Conditional calculations require care about the conditioning event. Given that the life has survived to age \(x+r\), the probability of dying after exactly \(m\) additional complete years is:
This is a curtate lifetime measured from age \(x+r\). It uses the age-specific probabilities from that age onward, rather than repeating the probabilities from age \(x\).
Worked Example: Combining Point and Tail Probabilities
For a hypothetical life aged \(x\), suppose \(q_x=0.08\), \(q_{x+1}=0.12\), \(q_{x+2}=0.20\), and \(q_{x+3}=0.25\). Calculate \(P(K_x=3)\), \(P(K_x\ge3)\), and \(P(1\le K_x\le3)\).
Method. Convert the annual death probabilities to survival probabilities: \(p_x=0.92\), \(p_{x+1}=0.88\), \(p_{x+2}=0.80\), and \(p_{x+3}=0.75\).
To have \(K_x=3\), the life survives the first three years and dies in the fourth:
To survive at least three complete years:
For the range \(1\le K_x\le3\), add the probabilities of \(K_x=1\), \(K_x=2\), and \(K_x=3\):
As a check, \(P(1\le K_x\le3)=P(K_x\ge1)-P(K_x\ge4)\). Here \(P(K_x\ge1)=0.92\) and \(P(K_x\ge4)=(0.64768)(0.75)=0.48576\), so the difference is \(0.92-0.48576=0.43424\), as required. The trap avoided is to interpret \(P(K_x\ge3)\) as the probability of dying during the third year; it is instead the probability of reaching age \(x+3\).
Worked Example: Conditioning on Survival to a Later Age
A hypothetical table has \(\ell_x=2500\), \(\ell_{x+1}=2250\), \(\ell_{x+2}=1800\), \(\ell_{x+3}=1260\), \(\ell_{x+4}=1008\), and \(\ell_{x+5}=756\). Given that a life aged \(x\) survives at least two years, calculate the probability that \(K_x=4\).
Method. The condition \(K_x\ge2\) means the life has reached age \(x+2\). From that age, \(K_x=4\) requires surviving two more years and then dying during the year from age \(x+4\) to \(x+5\).
The relevant annual survival probabilities are:
The death probability at age \(x+4\) is:
Therefore:
A direct check uses the unconditional probability and divides by the conditioning probability. The unconditional probability is:
The conditioning probability is \(P(K_x\ge2)=1800/2500=0.72\). Thus \(0.1008/0.72=0.14\), agreeing with the conditional calculation. The common error is to use \(q_{x+4}\) alone: that omits surviving from age \(x+2\) to age \(x+4\).
Checks and Common Exam Traps
A calculated probability distribution should respect the structure of the life table. Useful checks include:
- Point versus tail: \(P(K_x=k)\) includes death in the year after \(k\) completed years; \(P(K_x\ge k)\) includes survival to age \(x+k\).
- Correct annual factors: use \(p_{x+j}=1-q_{x+j}\) for each year the life must survive, then use \(q\) for the year of death.
- Consistent starting age: conditional on survival to age \(x+r\), use mortality rates beginning at \(x+r\).
- Life-table ratio check: \({}_kp_x=\ell_{x+k}/\ell_x\), and \(P(K_x=k)=(\ell_{x+k}-\ell_{x+k+1})/\ell_x\).
- Total probability: probabilities for all possible values of \(K_x\) sum to 1 when the table includes the ultimate age of death. If only a few years are listed, their sum is less than 1 by the probability of surviving beyond those years.
The key distinction is between conditional annual death probabilities and unconditional probabilities for a life starting at age \(x\). A \(q\) value applies only after reaching its associated age. The preceding survival factors convert that conditional annual probability into a point probability for \(K_x\).
Check Your Understanding
For each problem, use the stated annual mortality probabilities or life-table counts. Give probabilities to four decimal places where needed.
- Suppose \(q_x=0.15\) and \(q_{x+1}=0.20\). Calculate \(P(K_x=1)\). (Answer: \(0.1700\))
- A hypothetical table has \(\ell_x=1200\), \(\ell_{x+1}=1020\), \(\ell_{x+2}=765\), and \(\ell_{x+3}=612\). Calculate \(P(K_x=2)\). (Answer: \(0.1275\))
- Suppose \(q_x=0.10\), \(q_{x+1}=0.20\), and \(q_{x+2}=0.30\). Find \(P(K_x\in\{0,1,2\})\) and \(P(K_x\ge3)\). (Answer: \(0.4960\) and \(0.5040\))
- Given \(K_x\ge2\), suppose \(q_{x+2}=0.25\) and \(q_{x+3}=0.40\). Calculate the conditional probability that \(K_x=3\). (Answer: \(0.3000\))
- A hypothetical table has \(\ell_x=2000\), \(\ell_{x+1}=1700\), \(\ell_{x+2}=1190\), and \(\ell_{x+3}=833\). Calculate \(P(K_x=2)\) and \(P(K_x\ge3)\). (Answer: \(0.1785\) and \(0.4165\))
- Suppose \(q_x=0.10\), \(q_{x+1}=0.25\), and \(P(K_x=2)=0.16875\). Find \(q_{x+2}\), and state which survival factors are needed in the point-probability formula. (Answer: \(q_{x+2}=0.2500\); use \(p_xp_{x+1}q_{x+2}\))