Actuarial Mathematics for Modelling · Means and variances of assurance and annuity payments

Present Value Random Variables for Life Contingencies

Q: What is the difference between Tx and Kx?

Tx is the exact future lifetime of a life aged x and is continuous. Kx is the whole number of complete years lived, Kx = ⌊Tx⌋. Use Tx for benefits paid at the moment of death and Kx for benefits paid at year-end.

Q: Why is the death benefit exponent Kx + 1 and not Kx?

Kx counts complete years survived. If Kx = k, death occurs during year k+1 and the benefit is paid at the end of that year, at time k+1. So the discount exponent is Kx + 1.

Q: How do I write the present value variable for a term assurance?

Write the discounted sum assured when death occurs within the term, and 0 if the life survives the term. For end-of-year payment, Z = v^(Kx+1) if Kx < n and 0 otherwise.

Q: How is the variance of a present value random variable found?

Find E[Z] and E[Z²], then use Var(Z) = E[Z²] − (E[Z])². The second moment E[Z²] is computed like the mean but with the force of interest doubled to 2δ, for a unit sum assured.

Updated 11 October 2026 · Fact-checked

A present value random variable is the value today of a benefit whose amount or timing depends on how long a life survives. You write it as a function of the future lifetime Tx (or Kx), multiplied by the discount factor v raised to the payment time. Then you take its mean and variance.

Understand Present Value Random Variables for Life Contingencies

A life insurance benefit is paid only when something happens to a person, such as death or survival. So its present value is not a fixed number. It depends on how long the person lives. That makes it a random variable.

The starting point is the future lifetime of a life aged x. Tx is the complete future lifetime, a continuous random variable, so Tx ≥ 0. Kx = ⌊Tx⌋ is the curtate future lifetime, the whole number of complete years lived. It takes values 0, 1, 2, ... Death in year k+1 means Kx = k.

The discount factor is v = 1 ÷ (1 + i). A payment of 1 made at time t has present value v^t. With a force of interest δ, v^t = e^(−δt). The present value random variable is the benefit amount times v raised to the time of payment, where that time depends on Tx or Kx.

For an assurance, the payment time is the time of death (or the end of the year of death). For an annuity, payments are made while the life survives, so the number of payments depends on Kx. For an endowment, you combine a death benefit before the term ends and a survival benefit at the end.

The key skill is writing Z (assurance) or Y (annuity) correctly for each range of Tx or Kx. Once you have that, means and variances follow by summing or integrating over the distribution of the lifetime. This page is about defining the variable. The expected values come from it directly.

Key rules to remember

Discount factor: v = 1 ÷ (1 + i); v^t = e^(−δt); Use δ = ln(1 + i) to convert between the two forms.
Curtate future lifetime: Kx = ⌊Tx⌋; Kx = k means death between ages x+k and x+k+1.
Whole life assurance, paid at moment of death (sum assured 1): Z = v^Tx; Mean is Āx = E[v^Tx].
Whole life assurance, paid at end of year of death: Z = v^(Kx+1); Mean is Ax = E[v^(Kx+1)].
n-year term assurance, death-moment payment: Z = v^Tx if Tx < n; Z = 0 if Tx ≥ n; Written Ā¹x:n for the mean.
n-year term assurance, end-of-year payment: Z = v^(Kx+1) if Kx < n; Z = 0 if Kx ≥ n; Written A¹x:n for the mean, with the 1 above x only. The 1 above n (x:n¹) is reserved for the pure endowment.
n-year pure endowment: Z = 0 if Tx < n; Z = v^n if Tx ≥ n; Mean is v^n × npx.
n-year endowment assurance, death-moment payment: Z = v^Tx if Tx < n; Z = v^n if Tx ≥ n; Equivalent to Z = v^min(Tx, n).
n-year endowment assurance, end-of-year payment: Z = v^(Kx+1) if Kx < n; Z = v^n if Kx ≥ n; Equivalent to Z = v^min(Kx+1, n).
Whole life annuity-due (annual in advance, 1 per year): Y = ä(Kx+1) = (1 − v^(Kx+1)) ÷ d; Here d = iv = 1 − v. Payments at times 0, 1, ..., Kx.
Whole life annuity-immediate (annual in arrears): Y = a(Kx) = (1 − v^Kx) ÷ i; Payments at times 1, 2, ..., Kx. Equals 0 if Kx = 0. This is also equal to ä(Kx+1) − 1.
Continuous whole life annuity: Y = ā(Tx) = (1 − v^Tx) ÷ δ; Payment at rate 1 per year while alive.
n-year temporary annuity-due: Y = ä(Kx+1) if Kx < n; Y = ä(n) if Kx ≥ n; Equivalent to Y = ä(min(Kx+1, n)).

How to solve Present Value Random Variables for Life Contingencies questions

Use this method for any question that asks you to define or write down a present value random variable.

1Identify the benefit: death benefit, survival benefit, or regular payments. Note the sum assured or payment amount.
2Decide the payment timing: at the moment of death (use Tx), at the end of the year of death (use Kx + 1), or in advance or arrears for annuities.
3Write the payment time as a function of Tx or Kx for the full lifetime range.
4Split the range at the policy term n, and at any deferment period. State the value in each range, including 0 where nothing is paid.
5Write the present value as amount × v^(payment time). For an annuity, use the annuity function ä or a at the number of payments made.
6Check the boundaries. Test a small and a large lifetime and confirm the variable gives sensible values.
7If asked for the mean or variance, take the expected value over the distribution of Tx or Kx. Use Var(Z) = E[Z²] − (E[Z])², where Z² has force 2δ.

Quickest way: Write the variable by timeline

When to use it: Use this when you need the variable fast and the contract has a term, deferment or several benefits.

Draw a line from 0 with the term n marked.
Mark where payments occur if death happens before n, and if the life survives to n.
For each region, write the amount and v to the power of the payment time.
For annuities, count payments: annuity-due pays Kx + 1 times, annuity-immediate pays Kx times.
Combine the pieces using min(), for example v^min(Kx+1, n) for endowment assurance.

Common mistakes in Present Value Random Variables for Life Contingencies

Using v^Kx instead of v^(Kx+1) for end-of-year death benefits.
Kx counts complete years lived, so students forget the benefit is paid at the end of the following year.
Fix: If death is in year k+1, payment is at time k+1. Always write the exponent as Kx + 1.
Mixing up Tx and Kx.
Both measure future lifetime, so they look interchangeable.
Fix: Use Tx for payment at the moment of death or for continuous annuities. Use Kx when payment is at year-end or annuities are paid annually.
Leaving out the zero branch for term assurance and pure endowment.
Students focus on when a payment is made and forget the cases where none is made.
Fix: State the value for every range of Tx or Kx, including 0.
Writing the annuity-due variable as ä(Kx) and the immediate as a(Kx+1).
Payment counts are confused with the year of death.
Fix: Annuity-due pays at time 0 through Kx, so Kx + 1 payments: ä(Kx+1). Annuity-immediate pays Kx times: a(Kx).
Calculating Var(Z) by squaring v rather than using a second moment at force 2δ.
Students treat the variance as Var(v^T) = (v^T)².
Fix: Use E[Z²] = E[v^(2T)], computed at force 2δ, and subtract (E[Z])².
Treating the endowment assurance as the sum of two unrelated terms and double-counting the survival range.
Term assurance and pure endowment ranges are not kept separate.
Fix: Term part applies only when Tx < n. Pure endowment applies only when Tx ≥ n. Their sum is v^min(Tx, n).

Worked examples

Example 1

A life aged x buys a 10-year endowment assurance with sum assured ₹5,00,000. The death benefit is paid at the end of the year of death. The effective annual interest rate is 5%. Write the present value random variable in terms of Kx. Then find its value if the life dies in the 4th policy year, and if the life survives the term.

Show the solution

Payment is at the end of the year of death if Kx < 10, and at time 10 if Kx ≥ 10.
So Z = 5,00,000 × v^(Kx+1) for Kx = 0, 1, ..., 9, and Z = 5,00,000 × v^10 for Kx ≥ 10.
Equivalent form: Z = 5,00,000 × v^min(Kx+1, 10).
Death in the 4th policy year means Kx = 3, so payment is at time 4.
v = 1 ÷ 1.05, so v^4 = 1 ÷ 1.05^4 = 1 ÷ 1.21550625 = 0.822702.
Z = 5,00,000 × 0.822702 = ₹4,11,351 (approximately).
Survival to the end of the term gives payment at time 10. 1.05^10 = 1.628895, so v^10 = 0.613913.
Z = 5,00,000 × 0.613913 = ₹3,06,957 (approximately).

Answer: Z = 5,00,000 × v^min(Kx+1, 10). Death in year 4 gives about ₹4,11,351. Survival to the end of the term gives about ₹3,06,957.

Example 2

A life aged x buys a whole life annuity-due paying ₹1 per year in advance. The interest rate is 4% per year effective. (a) Define the present value random variable Y in terms of Kx. (b) Find Y if Kx = 2.

Show the solution

An annuity-due pays at times 0, 1, 2, ..., Kx. That is Kx + 1 payments.
So Y = ä(Kx+1), which equals (1 − v^(Kx+1)) ÷ d.
For Kx = 2, there are 3 payments, at times 0, 1 and 2.
v = 1 ÷ 1.04 = 0.961538. v² = 0.924556.
Y = 1 + 0.961538 + 0.924556 = 2.886094.
Check with the formula: d = 0.04 ÷ 1.04 = 0.038462. v³ = 0.888996. Y = (1 − 0.888996) ÷ 0.038462 = 0.111004 ÷ 0.038462 = 2.8861.
The two methods agree.

Answer: Y = ä(Kx+1) = (1 − v^(Kx+1)) ÷ d. For Kx = 2, Y ≈ 2.8861 (per ₹1 of annual payment).

Exam tips

Write the variable for every range of Tx or Kx. Examiners award marks for the zero branch and the boundary at n.
State your timing assumption (moment of death or end of year) before you write the formula.
Use notation exactly as in the IAI tables: Z for assurances, Y for annuities, Kx and Tx for lifetimes.
For endowment questions, use v^min(Kx+1, n) to show you understand the structure.
In Paper B, define the variable first on paper, then code it. Check one case by hand against your R or Excel output.

Practice questions from Means and variances of assurance and annuity payments