Mean and Variance of Assurance Benefits Explained

Understand Mean and Variance of Assurance Benefits

Key rules to remember

Whole life assurance, present value

Z = v^(K+1) (end of year of death); Z = v^T (at moment of death)

K is curtate future lifetime, T is complete future lifetime.

Mean of whole life assurance

E[Z] = Ax (discrete); E[Z] = Āx (continuous)

Ax = Σ v^(k+1) × kpx × q(x+k).

Second moment

E[Z²] = ²Ax (or ²Āx), the assurance value at force of interest 2δ

Use v² in place of v. Equivalent rate is 2i + i².

Variance of whole life assurance

Var(Z) = ²Ax − (Ax)²

Continuous case: ²Āx − (Āx)². Multiply by S² for sum assured S.

Term assurance, n years

Z = v^(K+1) if K+1 ≤ n, else 0. E[Z] = A¹x:n̅|. Var(Z) = ²A¹x:n̅| − (A¹x:n̅|)²

Z is zero if the life survives the term. Doubled-rate version works because Z² is also zero in that case.

Pure endowment, n years

Z = vⁿ if K ≥ n, else 0. E[Z] = Ax:n̅|¹ = vⁿ × npx. Var(Z) = v²ⁿ × npx × nqx

Equivalently ²Ax:n̅|¹ − (Ax:n̅|¹)² with ²Ax:n̅|¹ = v²ⁿ npx. The 1 sits over the n̅| for a pure endowment, and before the x for a term assurance.

Endowment assurance, n years

Z = v^(min(K+1, n)). E[Z] = Ax:n̅|. Var(Z) = ²Ax:n̅| − (Ax:n̅|)²

Z² = v^(2 min(K+1, n)), so doubled rate applies to the whole endowment.

Endowment as sum of parts

Ax:n̅| = A¹x:n̅| (term) + Ax:n̅|¹ (pure endowment)

Term and pure endowment cannot both be paid, so their product Z1×Z2 = 0. Then Var(Z1+Z2) = Var(Z1) + Var(Z2) − 2E[Z1]E[Z2].

Scaling

Var(S × Z) = S² × Var(Z)

Mean scales by S, variance by S².

Whole life split

Ax = A¹x:n̅| + vⁿ npx × A(x+n)

Useful when a term value is not given directly.

How to solve Mean and Variance of Assurance Benefits questions

1. 1Write down the benefit: sum assured, timing of payment (end of year of death or moment of death), term and any deferment.
2. 2Define Z as a function of K+1 or T, and state when Z is zero.
3. 3Find E[Z] as the standard assurance value at the given rate. Use tables, given values or a direct sum.
4. 4Find E[Z²]. Square Z, then recognise it as the same benefit with discount v². Use the doubled-rate value ²A, or compute directly using v² or force 2δ.
5. 5For an endowment built from parts, check the cross-term: Z1 × Z2 = 0 when the two events are mutually exclusive.
6. 6Compute Var(Z) = E[Z²] − (E[Z])². Multiply by S² if the sum assured is S.
7. 7Take the square root if a standard deviation is asked. State units in rupees where relevant, and check the variance is not negative.

Quickest way: Doubled-rate shortcut

Common mistakes in Mean and Variance of Assurance Benefits

• Writing Var(Z) = Ax² − (Ax)² or treating ²Ax as (Ax)².

  The superscript 2 looks like a square.

  Fix: Remember ²Ax is the assurance value at the doubled force of interest. It is a different number from (Ax)².

• Doubling the interest rate i instead of the force of interest.

  Doubling δ is the rule, so students double i by habit.

  Fix: Use v² = (1+i)^(−2). The equivalent rate is i′ = 2i + i², not 2i.

• Forgetting to multiply the variance by S² for a sum assured of S.

  Tables are per unit sum assured, and the mean scales by S, so students scale the variance by S too.

  Fix: Variance scales by S². Standard deviation scales by S.

• Using Var(Z1+Z2) = Var(Z1) + Var(Z2) for endowment assurance.

  The parts look independent.

  Fix: The parts are mutually exclusive, so they are strongly dependent. Include the −2E[Z1]E[Z2] term, or use ²Ax:n̅| − (Ax:n̅|)² directly.

• Assuming the term assurance variance is zero beyond the term or ignoring the zero outcome.

  Students forget Z = 0 when the life survives n years.

  Fix: Z is zero if death is after n years. The zero outcome still counts in E[Z] and E[Z²] through the probabilities.

• Mixing discrete and continuous assurances, such as using Ax with ²Āx.

  Notation is similar and tables are side by side.

  Fix: Match both moments to the same payment timing, both with bars or both without.

Worked examples