Means and variances of assurance and annuity payments: formula sheet

Key formulas

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)).

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.

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.

Assurance–annuity identity (annual)

A_x = 1 − d·ä_x, so ä_x = (1 − A_x) ÷ d

d = i ÷ (1 + i) = i·v. Holds for whole life. Not valid with a_x (annuity-immediate) unless you convert first.

Continuous identity

Ā_x = 1 − δ·ā_x, so ā_x = (1 − Ā_x) ÷ δ

δ = ln(1 + i). Both the assurance and the annuity are continuous.

Endowment version

A_{x:n|} = 1 − d·ä_{x:n|}

Uses the n-year endowment assurance and the n-year temporary annuity-due. The whole life form is the special case n → ∞.

Annuity-immediate and annuity-due

ä_x = 1 + a_x

The first payment is at time 0 in the due form and at time 1 in the immediate form.

Assurance recursion

A_x = v·q_x + v·p_x·A_{x+1}

One-year-ahead argument. For term assurance use A¹_{x:n|} = v·q_x + v·p_x·A¹_{x+1:n−1|}.

Annuity-due recursion

ä_x = 1 + v·p_x·ä_{x+1}

For the annuity-immediate: a_x = v·p_x·(1 + a_{x+1}).

UDD conversion for death benefits

Ā_x = (i ÷ δ)·A_x; Ā¹_{x:n|} = (i ÷ δ)·A¹_{x:n|}

Applies to whole life and term assurances paid at death. A pure endowment is not changed because it pays only on survival.

Variance of assurance

Var(Z) = ²A_x − (A_x)²

²A_x is the same assurance valued at force of interest 2δ, which is interest rate 2i + i².

Variance of annuity-due

Var(Y) = (²A_x − (A_x)²) ÷ d²

From Y = (1 − Z)/d. Divide by d², not d. Continuous: Var(ā_T) = (²Ā_x − (Ā_x)²) ÷ δ².

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.

Quick revision

• Z is the present value random variable. E[Z] is its mean, and Var(Z) = E[Z²] − (E[Z])².
• For a whole life assurance paid at the end of the year of death, Z = v^(K+1). With payment at the moment of death, Z = v^T.
• E[Z²] is the mean of the same benefit at force of interest 2δ, so that v becomes v².
• For an annuity-due, Y = ä_(K+1) with K the curtate future lifetime.
• ä_x = (1 − A_x) ÷ d, where d = i ÷ (1 + i) = 1 − v. This holds for whole life with the same mortality basis.
• Var(Y) = Var(Z) ÷ d² for the whole life annuity-due, where Var(Z) = ²A_x − (A_x)².
• Term and endowment assurances: split Z into a death part and a survival part, and be careful with the cross-term.
• Endowment assurance: A_(x:n) = A¹_(x:n) + A_(x:n)¹, where the second term is the pure endowment.
• Recursion: A_x = v q_x + v p_x A_(x+1), and ä_x = 1 + v p_x ä_(x+1).
• Varying benefits: write an increasing benefit as a sum of level benefits, or use the standard (IA) notation.
• Always state your interest rate and mortality table at the start of an answer.
• Check the sign and size of your answer: A_x lies between 0 and 1 for a unit benefit.

Common mistakes

• Using v^Kx instead of v^(Kx+1) for end-of-year death benefits. 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. 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.
• Writing Var(Z) = Ax² − (Ax)² or treating ²Ax as (Ax)². 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. Fix: Use v² = (1+i)^(−2). The equivalent rate is i′ = 2i + i², not 2i.
• Using ²A_x = (A_x)² in the variance formula. 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. Fix: Adding or subtracting a constant never changes a variance. Var(a_x) = Var(ä_x).
• Using i instead of d in A = 1 − d·ä. Fix: Compute d = i/(1 + i) first and write it down. With a_x (annuity-immediate) the link is A_x = 1 − d(1 + a_x), which still uses d.
• Calculating ²A_x at the same interest rate as A_x. Fix: ²A_x is A_x evaluated with v² (interest rate 2i + i²). It is not the square of A_x. Take the variance as ²A_x − (A_x)².
• Squaring only the discount factor when finding the variance of an increasing assurance 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) 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.

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.
• Always write Z explicitly and say when it is zero. Examiners give marks for the definition.
• State the doubled rate: v² or 2δ, or i′ = 2i + i². This is a common mark in written answers.
• For endowment questions, show either the direct doubled-rate method or the cross-term method, and say why the cross-term is used.