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

Increasing and Decreasing Assurances and Annuities: Means and Variances

Updated 11 October 2026 · Fact-checked

Varying-benefit contracts pay an amount that rises or falls by a fixed step each year. The present value is a random variable Z. Find the mean as the sum of benefit × v^time × probability. Find the variance as E[Z²] − (E[Z])², squaring the benefit as well as discounting at v². Or split into level pieces.

Understand Assurances and Annuities with Varying Benefits

A level assurance pays the same sum whenever death occurs. A varying benefit assurance pays an amount that depends on when death occurs. In an increasing whole life assurance the benefit is 1 if death is in year 1, 2 if death is in year 2, and so on. In a decreasing term assurance over n years the benefit is n if death is in year 1, n − 1 in year 2, down to 1 in year n.

Every such contract has a present value random variable Z. Let K be the curtate future lifetime, so death is in policy year K + 1. For the increasing whole life assurance with the benefit paid at the end of the year of death, Z = (K + 1) v^(K+1). The mean is E[Z] = (IA)x = Σ (k + 1) v^(k+1) × k|qx, summed from k = 0. Here k|qx = kpx × qx+k is the probability of dying in year k + 1.

The same idea works for annuities. In an increasing annuity-due you receive 1 at time 0, 2 at time 1, 3 at time 2, and so on, while you are alive. The mean is (Iä)x = Σ (k + 1) v^k × kpx. You can read it as a sum of level deferred annuities, or as a sum of level annuities-due bought at each future age x + k, discounted and survived to that age. For a payment at the moment of death, the benefit can be the year-of-death number ⌈T⌉ or the exact time T. These give (IĀ)x and (ĪĀ)x, which are different contracts.

The variance needs one extra idea. Var(Z) = E[Z²] − (E[Z])². To get E[Z²], square the whole payment, benefit included, and discount at v². So the second moment of (IA)x is Σ (k + 1)² v^(2(k+1)) × k|qx. Squaring only v is the most common way to lose marks. For annuities, the neat level-annuity shortcut Var = (²A − A²) ÷ d² does not carry over to increasing benefits. You list the value of the annuity for each K, then take the probability-weighted average of the squares.

Key rules to remember

Increasing whole life assurance, end of year of death
(IA)x = Σ (k + 1) v^(k+1) × k|qx, for k = 0, 1, 2, …
Benefit is K + 1 at time K + 1. Here k|qx = kpx × qx+k.
Increasing term assurance
(IA)¹x:n = Σ (k + 1) v^(k+1) × k|qx, for k = 0 to n − 1
Same terms as the whole life case, but stopping at year n.
Decreasing term assurance
(DA)¹x:n = Σ (n − k) v^(k+1) × k|qx, for k = 0 to n − 1
Benefit is n in year 1 and 1 in year n. Death after n years pays nothing.
Increasing plus decreasing term
(IA)¹x:n + (DA)¹x:n = (n + 1) × A¹x:n
In each year the two benefits add to n + 1. Useful for finding one from the other.
Increasing assurance as a sum of level assurances
(IA)x = Σ v^k × kpx × Ax+k, for k = 0, 1, 2, …
Think of it as one level assurance bought at each age x + k, discounted and survived to that age.
Recursion for increasing assurance
(IA)x = v qx + v px [ (IA)x+1 + Ax+1 ]
If the life survives a year, the benefits become 2, 3, … which is 1 plus the sequence 1, 2, …
Increasing annuity-due
(Iä)x = Σ (k + 1) v^k × kpx, for k = 0, 1, 2, … Also (Iä)x = Σ v^k × kpx × äx+k
Payment k + 1 is made at time k if the life is alive then.
Increasing temporary annuity-due, certain payments
(Iä) n⌉ = (ä n⌉ − n v^n) ÷ d
Annuity-certain with payments 1, 2, …, n at times 0 to n − 1. Use for Y when K is fixed.
Variance of any varying-benefit present value
Var(Z) = E[Z²] − (E[Z])²
For (IA)x: E[Z²] = Σ (k + 1)² v^(2(k+1)) × k|qx. The benefit is squared and the discount is v², i.e. the rate j = (1 + i)² − 1.
Payment at moment of death, year-of-death benefit, under UDD
(IĀ)x = (i ÷ δ) × (IA)x
Benefit ⌈T⌉ paid at time T. Needs uniform distribution of deaths within each year. For the second moment use j = (1 + i)² − 1 and 2δ, so ²(IĀ)x = (j ÷ 2δ) × ²(IA)x.
Increasing assurance, benefit T at the moment of death
(ĪĀ)x = ∫ t v^t × tpx × μx+t dt, from 0 to ∞
Second moment: ∫ t² e^(−2δt) × tpx × μx+t dt. Square the t as well.

How to solve Assurances and Annuities with Varying Benefits questions

Use this method for any question on varying benefit assurances or annuities, whether it asks for a mean, a variance or a relationship.

  1. 1Write the contract as a payment table: for each possible K (or T), list the benefit amount, the time it is paid, and whether it is a death benefit or a survival payment.
  2. 2Decide the timing exactly. Is the benefit paid at the end of the year of death or at the moment of death? Does an annuity start at time 0 (due) or time 1 (immediate)? Does the first benefit equal 1 or something else?
  3. 3Find the probability of each outcome. For assurances use k|qx = kpx × qx+k. For annuities use kpx. Take these from the life table or the given q values.
  4. 4Compute the mean: sum of (benefit × v^time × probability). If the contract is a mix of level pieces, use the decomposition or relationship formulas to save work.
  5. 5For the variance, build E[Z²] by squaring each present value (benefit and discounting together), weighting by the same probabilities. Check that the squared discount factor is v^(2 × time).
  6. 6Compute Var(Z) = E[Z²] − (E[Z])². The result must be positive. If it is negative, there is an arithmetic error.
  7. 7For payments at the moment of death, state your assumption (UDD or constant force) and apply the correct conversion factor to both the mean and the second moment. Remember the second moment uses j and 2δ.
  8. 8Sanity check: the mean must lie between the smallest and largest possible present values. For term assurances, (IA)¹x:n + (DA)¹x:n must equal (n + 1)A¹x:n.

Quickest way: Column table for short terms

When to use it: Use this for a term of 2 to 5 years, or whenever probabilities are given as individual q values. It is the fastest and least error-prone method in the exam.

  1. Make columns: k, probability (k|qx or kpx), benefit or annuity value, present value Z.
  2. Add a column for Z² and fill it straight from the Z column.
  3. Multiply the probability column by Z to get the mean, and by Z² to get the second moment. Sum each.
  4. Subtract the mean squared from the second moment.
  5. Keep at least six decimal places in intermediate steps. Variances come from subtracting two close numbers.

Common mistakes in Assurances and Annuities with Varying Benefits

  • Squaring only the discount factor when finding the variance of an increasing assurance

    With level assurances the second moment is just the first moment at the rate j, so students assume the same shortcut works here.

    Fix: Square the whole present value. For (IA)x use (k + 1)² v^(2(k+1)). Writing Z² next to Z in a table removes the risk.

  • Mixing up the timing of the first benefit, for example using k v^(k+1) instead of (k + 1) v^(k+1)

    Death in year k + 1 pays k + 1, but the formula index starts at k = 0, so the offset is easy to lose.

    Fix: Write out the first two terms by hand. Year 1 pays 1 at time 1. Year 2 pays 2 at time 2. Then match the formula.

  • Using n − k + 1 or n − k − 1 as the benefit for a decreasing term assurance

    The benefit depends on whether the sum is indexed from k = 0 or k = 1.

    Fix: Check the end points. With k = 0 for year 1 the benefit is n − k, so year 1 pays n and year n pays 1.

  • Using the level annuity variance formula (²A − A²) ÷ d² for an increasing annuity

    That formula relies on ä = (1 − A) ÷ d, which only holds for a level annuity with benefit 1.

    Fix: List the annuity value Y for each K, with probabilities. Compute E[Y²] directly, then subtract the mean squared.

  • Treating (IĀ)x and (ĪĀ)x as the same thing

    The notation differs only by a bar, and both pay at the moment of death.

    Fix: (IĀ)x pays the whole-year number ⌈T⌉. (ĪĀ)x pays the exact time T. The first steps up each year. The second increases continuously.

  • Applying the UDD factor i ÷ δ to the second moment

    The same factor is reused automatically.

    Fix: For the second moment, use the rate j = (1 + i)² − 1 and force 2δ, giving the factor j ÷ 2δ.

Worked examples

Example 1

A 3-year decreasing term assurance is issued to a life aged x. The benefit is ₹3 lakh if death is in year 1, ₹2 lakh in year 2 and ₹1 lakh in year 3, paid at the end of the year of death. Take the unit as ₹1 lakh. Mortality: qx = 0.01, qx+1 = 0.02, qx+2 = 0.03. Interest is 5% per year. Find the mean and the variance of the present value, in units of ₹1 lakh and its square.

Show the solution
  1. Probabilities of death in each year: 0|qx = 0.01. 1|qx = 0.99 × 0.02 = 0.0198. 2|qx = 0.99 × 0.98 × 0.03 = 0.9702 × 0.03 = 0.029106.
  2. Discount factors: v = 0.952381, v² = 0.907029, v³ = 0.863838.
  3. Present values Z: year 1 gives 3v = 2.857143. Year 2 gives 2v² = 1.814059. Year 3 gives v³ = 0.863838.
  4. Mean: 0.01 × 2.857143 = 0.028571. 0.0198 × 1.814059 = 0.035918. 0.029106 × 0.863838 = 0.025143. Sum = 0.089632.
  5. Squares of Z: 2.857143² = 8.163265. 1.814059² = 3.290810. 0.863838² = 0.746216.
  6. Second moment: 0.01 × 8.163265 = 0.081633. 0.0198 × 3.290810 = 0.065158. 0.029106 × 0.746216 = 0.021719. Sum = 0.168510.
  7. Variance = 0.168510 − 0.089632² = 0.168510 − 0.008034 = 0.160476.

Answer: Mean ≈ 0.0896 (about ₹8,963 per ₹1 lakh unit), variance ≈ 0.1605 in squared units of ₹1 lakh.

Example 2

A life aged x buys a 3-year temporary increasing annuity-due. It pays 1 at time 0, 2 at time 1 and 3 at time 2, each only if the life is alive. Use the same mortality as before (qx = 0.01, qx+1 = 0.02, qx+2 = 0.03) and interest of 5% per year. Find the mean and variance of the present value Y.

Show the solution
  1. Survival probabilities: 0px = 1, 1px = 0.99, 2px = 0.99 × 0.98 = 0.9702.
  2. Mean = Σ (k + 1) v^k × kpx. Time 0: 1 × 1 × 1 = 1. Time 1: 2 × 0.952381 × 0.99 = 1.885714. Time 2: 3 × 0.907029 × 0.9702 = 2.640. Mean = 1 + 1.885714 + 2.640 = 5.5257.
  3. For the variance, list Y by outcome. Death in year 1 (K = 0): Y = 1, probability 0.01.
  4. Death in year 2 (K = 1): Y = 1 + 2v = 2.904762, probability 1px × q = 0.99 × 0.02 = 0.0198.
  5. Survive two years (K ≥ 2): Y = 1 + 2v + 3v² = 5.625850, probability 0.9702.
  6. Check the mean: 0.01 × 1 + 0.0198 × 2.904762 + 0.9702 × 5.625850 = 0.010000 + 0.057514 + 5.458 = 5.5257. This matches.
  7. Second moment: 0.01 × 1 = 0.010000. 0.0198 × 8.437643 = 0.167065. 0.9702 × 31.650188 = 30.707013. Sum = 30.884078.
  8. Variance = 30.884078 − 5.525711² = 30.884078 − 30.533482 = 0.350596.

Answer: Mean ≈ 5.526, variance ≈ 0.35 (using the unrounded mean 5.525711, variance ≈ 0.3506).

Exam tips

  • Always write the payment table first, even in a written question. Method marks are awarded for a correct setup, and it makes the variance easy.
  • Say clearly whether benefits are at the end of the year of death or at the moment of death. If the question says moment of death, state the assumption used (such as UDD) before converting.
  • Remember that (IA)¹x:n + (DA)¹x:n = (n + 1) A¹x:n. It often turns a long calculation into a short one.
  • In MCQs, a wrong variance often comes from squaring only the discount. Check that your answer uses the squared payment before selecting an option.
  • In the computer-based Paper B, build the table in Excel or R with columns for probability, Z and Z². Use SUMPRODUCT or sum(p * Z) and sum(p * Z^2), and show the formula in the working.

Practice questions from Means and variances of assurance and annuity payments

Assurances and Annuities with Varying Benefits: frequently asked questions

What is the difference between (IA)x and (Ia)x?

(IA)x is an increasing whole life assurance, which pays K + 1 at the end of the year of death. (Ia)x is an increasing life annuity-immediate, where payment k is made at time k if the life is alive. Look at the capital letter: A is assurance, a is annuity.

How do I find the variance of an increasing annuity?

List the present value Y for each possible curtate future lifetime K, with its probability. Compute E[Y] and E[Y²] from the table, then use Var(Y) = E[Y²] − (E[Y])². The level annuity shortcut using d does not apply.

How do I calculate the present value of a decreasing term assurance?

Sum (n − k) × v^(k+1) × k|qx for k = 0 to n − 1. Alternatively, find (IA)¹x:n and use (DA)¹x:n = (n + 1) A¹x:n − (IA)¹x:n.

When can I use the factor i ÷ δ for payments at the moment of death?

Use it when deaths are uniformly distributed within each year of age (UDD) and the benefit is a whole-number year-of-death amount ⌈T⌉. It converts (IA)x into (IĀ)x. For the second moment use j = (1 + i)² − 1 and 2δ instead.