Mean and Variance of Annuity Payments from Assurance Variables

Understand Mean and Variance of Annuity Payments

Key rules to remember

Whole life annuity due: random variable

Y = ä_(K+1|) = (1 − v^(K+1)) ÷ d = (1 − Z) ÷ d

K is the curtate future lifetime. Z = v^(K+1) is the whole life assurance present value.

Mean of whole life annuity due

ä_x = (1 − A_x) ÷ d

Valid at the same interest rate for both A_x and d.

Variance of whole life annuity due

Var(Y) = (²A_x − (A_x)²) ÷ d²

²A_x is A_x calculated with v replaced by v² (force of interest 2δ). Do not square A_x to get ²A_x.

Annuity immediate

a_x = ä_x − 1 ; Var(a_(K|)) = Var(ä_(K+1|))

Subtracting a constant changes the mean only.

Continuous whole life annuity

ā_x = (1 − Ā_x) ÷ δ ; Var = (²Ā_x − (Ā_x)²) ÷ δ²

Uses T, the complete future lifetime, and δ instead of d.

Temporary annuity due

ä_(x:n|) = (1 − A_(x:n|)) ÷ d ; Var = (²A_(x:n|) − (A_(x:n|))²) ÷ d²

A_(x:n|) here is the endowment assurance, not the term assurance.

Deferred annuity due: mean

n|ä_x = (ₙEₓ − n|A_x) ÷ d, where ₙEₓ = v^n ₙpₓ

ₙEₓ = v^n ₙpₓ is the pure endowment value. The result equals ä_x − ä_(x:n|).

Deferred annuity due: variance

Var(Y) = E[Y²] − (E[Y])², with E[Y²] = [v^(2n) ₙpₓ − 2 v^n × n|A_x + ²(n|A_x)] ÷ d²

Write Y = (v^n − v^(K+1)) ÷ d if K ≥ n, and Y = 0 if K < n. Squaring and taking expectations over K ≥ n gives the E[Y²] above. E[Y] is the mean n|ä_x.

How to solve Mean and Variance of Annuity Payments questions

1. 1Write the annuity present value as a random variable in terms of K (or T). Decide whether payments are at the start (due), end (immediate) or continuous, and whether it is temporary or deferred.
2. 2Express it using the assurance variable: Y = (1 − Z) ÷ d for due, with Z = v^(K+1) (whole life) or the endowment variable (temporary).
3. 3Find the mean: E[Y] = (1 − E[Z]) ÷ d. For an immediate annuity, subtract 1 from the due mean.
4. 4Find the variance: Var(Y) = Var(Z) ÷ d². Compute Var(Z) = ²A − A². Use δ in place of d for continuous annuities.
5. 5Check that you have the correct ²A. It is at the rate of interest j = (1 + i)² − 1, using the same mortality, so v² replaces v.
6. 6For deferred annuities, write Y as zero if K < n and (v^n − v^(K+1)) ÷ d if K ≥ n. Compute E[Y] and E[Y²] directly, then Var = E[Y²] − (E[Y])².
7. 7Calculate d = i ÷ (1 + i) to full accuracy, then give mean and variance (or standard deviation) with units and a sense check: variance must be positive.

Quickest way: Assurance first, then divide by d²

Common mistakes in Mean and Variance of Annuity Payments

• Using ²A_x = (A_x)² in the variance formula.

  Students confuse the second moment with the square of the mean. This makes the variance zero.

  Fix: ²A_x is E[Z²] = E[v^(2(K+1))]. It is A_x calculated at the rate (1 + i)² − 1. Use it as given or compute it at the new rate.

• Subtracting 1 from the variance of the annuity due to get the variance of the immediate annuity.

  The mean changes by 1, so students assume the variance also shifts.

  Fix: Adding or subtracting a constant never changes a variance. Var(a_x) = Var(ä_x).

• Dividing by d instead of d² in the variance.

  Students copy the mean formula and forget that variance scales with the square of the constant.

  Fix: Var(cZ) = c² Var(Z). With c = 1 ÷ d, divide by d².

• Using the term assurance instead of the endowment assurance for a temporary annuity due.

  The annuity pays for at most n years, which looks like a term contract.

  Fix: Y = ä_(min(K+1,n)|) = (1 − v^(min(K+1,n))) ÷ d. The assurance variable is the endowment assurance.

• Using d = 1 − i or d = 1 ÷ (1 + i).

  Mixing up the discount rate with other interest functions.

  Fix: d = i ÷ (1 + i) = 1 − v. For i = 5%, d = 0.05 ÷ 1.05 = 0.047619.

• Assuming Var(deferred annuity) = Var(whole life) − Var(temporary).

  The relationship ä_x = ä_(x:n|) + n|ä_x holds for means, so students assume it holds for variances.

  Fix: Variances do not add because the two parts are not independent. Work from E[Y²] and the mean directly.

Worked examples