Present Value Random Variables and Actuarial Notation Explained

Understand Present Value Random Variables and Actuarial Notation

Key rules to remember

Whole life assurance, end of year of death

Z = v^(K_x+1); A_x = E[Z] = Σ v^(k+1) × k|q_x, summed over k = 0, 1, 2, ...

v = 1 ÷ (1 + i). The sum runs until the end of the mortality table.

Whole life assurance, payable at death

Z = v^(T_x); Ā_x = E[Z] = ∫ v^t × tp_x × μ_(x+t) dt, from 0 to ∞

The bar means payment at the moment of death. Continuous time uses the force of mortality μ.

Term assurance

Z = v^(K_x+1) if K_x < n, else 0; A¹_(x:n) = Σ v^(k+1) × k|q_x, summed over k = 0 to n−1

The superscript 1 over x marks the death benefit. The value is zero if the life survives n years.

Pure endowment

Z = v^n if T_x > n, else 0; A_(x:n)^1 = v^n × np_x

The superscript 1 over the term n marks the survival benefit. Z takes only two values.

Endowment assurance

Z = v^(min(K_x+1, n)); A_(x:n) = A¹_(x:n) + A_(x:n)^1

The endowment is a term assurance plus a pure endowment.

Second moment of an assurance

E[Z²] = ²A_x = A_x calculated at v² (force of interest 2δ)

Use the same method with v replaced by v². Do not square A_x to get this.

Variance of an assurance

Var(Z) = ²A_x − (A_x)²

For a benefit of S, the variance is S² × Var(Z).

Whole life annuity-due variable

Y = ä_(K_x+1) = (1 − v^(K_x+1)) ÷ d; d = i ÷ (1 + i) = 1 − v

Y is a linear function of Z.

Mean of annuity-due from assurance

ä_x = E[Y] = (1 − A_x) ÷ d

The continuous version is ā_x = (1 − Ā_x) ÷ δ.

Variance of annuity-due

Var(Y) = Var(Z) ÷ d² = (²A_x − (A_x)²) ÷ d²

Continuous: Var(Ȳ) = (²Ā_x − (Ā_x)²) ÷ δ².

Pure endowment variance

Var(Z) = v^(2n) × np_x × (1 − np_x)

Z is v^n with probability np_x, otherwise 0.

Annuity-immediate link

a_x = ä_x − 1

For a whole life annuity paid in arrears.

How to solve Present Value Random Variables and Actuarial Notation questions

1. 1Identify the benefit: when it is paid (end of year of death, moment of death, on survival) and how much.
2. 2Identify the life variable: use K_x for discrete payments and T_x for continuous payments.
3. 3Write the present value variable as a function of K_x or T_x. For term and endowment contracts, define Z on each range of K_x. Include the zero case.
4. 4Write the probabilities for each outcome, such as k|q_x = kp_x × q_(x+k), or np_x for survival to n.
5. 5Find the mean: E[Z] = Σ (value × probability), or the equivalent integral. Add up each contract's part separately.
6. 6For the variance, find E[Z²] by replacing v with v² in every term. Then use Var(Z) = E[Z²] − (E[Z])².
7. 7For annuities, express Y using Z, for example Y = (1 − Z) ÷ d, and use the linear link to get mean and variance. Multiply by the benefit or payment size (S for the mean, S² for the variance).
8. 8Check: the answer must be positive, the mean of an assurance must be less than 1 per unit benefit, and the variance must not be negative.

Quickest way: Link everything to Z

Common mistakes in Present Value Random Variables and Actuarial Notation

• Using v^(K_x) instead of v^(K_x+1) for a whole life assurance paid at the end of the year of death.

  K_x counts completed years lived, and students forget that payment is made at the end of the year of death, which is one year later.

  Fix: Say it in words: if K_x = 0, death is in year 1 and payment is at time 1, so the factor is v¹. Always use K_x + 1 for end-of-year payment.

• Writing E[Z²] as (A_x)², giving a variance of zero.

  Students confuse the square of the mean with the mean of the square.

  Fix: E[Z²] = ²A_x, which is A_x calculated with v². Never square the number A_x to get the second moment.

• Forgetting the zero outcome in term assurance or pure endowment.

  Students write Z = v^(K_x+1) for all K_x and sum to infinity.

  Fix: Write Z piece by piece. A term assurance is v^(K_x+1) when K_x < n, and 0 otherwise. The sum only goes to n−1.

• Using Var(Y) = Var(Z) instead of Var(Z) ÷ d².

  Students copy the variance of the assurance and forget that Y = (1 − Z) ÷ d is a scaled variable.

  Fix: If Y = (1 − Z) ÷ d, then Var(Y) = Var(Z) ÷ d². Constants like 1 drop out, and the divisor d is squared.

• Mixing discrete and continuous notation, for example using δ with ä_x or d with ā_x.

  The bar and the symbol for the interest rate look like small details.

  Fix: Discrete annuity-due: divide by d. Continuous: divide by δ. Check that the bar on the assurance matches the bar on the annuity.

• Treating a_x (immediate) as ä_x (due), or omitting the −1 in a_x = ä_x − 1.

  The two symbols differ only by a double dot.

  Fix: Annuity-due pays at time 0, so it is worth 1 more than the immediate annuity. Write down the first payment time before you pick the symbol.

Worked examples