Relationship Between Assurance and Annuity Values

Understand Relationships Between Assurance and Annuity Values

Key rules to remember

Whole life identity (discrete)

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

Assurance paid at end of year of death; annuity-due. Here d = i ÷ (1 + i) = 1 − v.

Endowment assurance identity

A_{x:n̅|} = 1 − d·ä_{x:n̅|}

Endowment assurance pays at end of year of death or at the end of n years if the life survives. The annuity is the n-year temporary annuity-due.

Continuous identity

Ā_x = 1 − δ·ā_x

Benefit paid at the moment of death; annuity paid continuously. δ is the force of interest.

Annuity-due and immediate

ä_x = 1 + a_x

Use this to move between annuity-due and annuity-immediate before applying the identity.

Annuity recursion

ä_x = 1 + v·p_x·ä_{x+1}

First payment is certain. The rest is paid only if the life survives one year.

Assurance recursion

A_x = v·q_x + v·p_x·A_{x+1}

Pay 1 at the end of the year if death occurs; otherwise continue with the value at age x+1.

Term split of annuity

ä_x = ä_{x:n̅|} + nE_x·ä_{x+n}, where nE_x = vⁿ·ₙp_x

Splits a whole life annuity into an n-year temporary annuity and a deferred annuity.

Term assurance from endowment

A¹_{x:n̅|} = A_{x:n̅|} − nE_x

Removes the survival benefit from the endowment assurance.

Level premium link

P_x = A_x ÷ ä_x = 1 ÷ ä_x − d

Annual net premium for a whole life assurance of 1, payable annually in advance for life.

Variance of annuity from assurance

Var(ä_{K+1}) = (²A_x − (A_x)²) ÷ d²

Holds for whole life because the annuity present value is (1 − Z) ÷ d, where Z = v^(K+1). ²A_x is A_x at the rate of interest (1+i)² − 1.

How to solve Relationships Between Assurance and Annuity Values questions

1. 1Write down what is given and what is asked, in full actuarial notation. Note the age, the term and the interest rate.
2. 2Check the timing of each benefit. Is the assurance paid at end of year of death or at the moment of death? Is the annuity due, immediate or continuous? Match the identity to the timing.
3. 3Convert the annuity if needed. Use ä_x = 1 + a_x to move between due and immediate forms.
4. 4Calculate d = i ÷ (1 + i) (or δ = ln(1 + i) for continuous) before you substitute. Keep full decimal places.
5. 5Apply the identity to get the missing value. For a premium, then use P = A ÷ ä.
6. 6For a recursion question, split the first year: one branch for death (probability q_x) and one for survival (probability p_x). Discount each by v and add the value at x+1 where the contract continues.
7. 7For a proof, start from the right side or from the cash flows, and state which identity you use at each line.
8. 8Check that the answer is sensible. An assurance value for a whole life of 1 lies between 0 and 1, and A_x should rise with age.

Quickest way: Identity-first shortcut

Common mistakes in Relationships Between Assurance and Annuity Values

• Using a_x instead of ä_x in A_x = 1 − d·ä_x.

  The two annuity symbols look alike, and the question may give the immediate form.

  Fix: Convert first with ä_x = 1 + a_x. Then apply the identity. The identity uses the annuity-due only.

• Using i instead of d in the identity.

  Students remember the shape of the formula but not which rate multiplies the annuity.

  Fix: Compute d = i ÷ (1 + i) and write it at the top of your working. The identity is 1 = d·ä_x + A_x, interest paid in advance.

• Applying the discrete identity to a benefit paid at the moment of death.

  The benefit timing in the question is easy to skim past.

  Fix: If death benefit is paid immediately, use Ā_x = 1 − δ·ā_x. Do not mix Ā with ä, or A with ā.

• Writing the annuity recursion as ä_x = v·p_x·ä_{x+1} without the leading 1.

  Students confuse the due form with the immediate form a_x = v·p_x·(1 + a_{x+1}).

  Fix: The first due payment is at time 0 and is certain, so add 1. Check by setting p_x = 0: ä_x must equal 1.

• Leaving out the survival probability in the recursion, or using q_x in place of p_x.

  The two branches are mixed up in the rush.

  Fix: Draw a one-year tree. Death branch gets q_x and the payment. Survival branch gets p_x and the value at x+1. Check that the probabilities sum to 1.

• Forgetting that the variance formula needs ²A_x at the doubled force of interest.

  Students use A_x squared twice, or use A_x at the original rate for the first term.

  Fix: ²A_x is A_x calculated at rate (1 + i)² − 1. Then Var = (²A_x − (A_x)²) ÷ d².

Worked examples