Variances and Relationships Among Present Values of Continuous Whole and Temporary Life Annuities: Practice Problems and Solutions

The following relationship holds with respect to actuarial present values of continuous whole life annuities and of whole life insurance policies.

1 = δā_x + Ā_x

We can also determine the variance of the present value of a continuous whole life annuity:

Var(ā_T¬) = (²Ā_x - (Ā_x) ²)/δ²

If the force of mortality is a constant value μ, then

ā_x = 1/(δ + μ)

Ā_x = μ/(δ + μ)

²Ā_x = μ/(2δ + μ)

Var(ā_T¬) = μ/[(2δ + μ)(δ + μ)²]

Pr(ā_T¬ > ā_x) = (μ/(δ + μ))^μ/δ

If Y is the present value of an n-year temporary life annuity, then

Var(Y) = (2/δ)(ā_x:n¬ - ²ā_x:n¬) - (ā_x:n¬)²

Meaning of Symbols:

Ā_x = the actuarial present value of a whole life insurance policy for life (x).

²Ā_x = the second moment of the actuarial present value of a whole life insurance policy for life (x).

ā_x = the actuarial present value of a continuous whole life annuity for life (x).

ā_T¬ = the present-value random variable for a continuous whole life annuity for an annuitant with a future-lifetime random variable of T.

δ = the annual force of interest.

ā_x:n¬= the actuarial present value of a continuous n-year temporary life annuity for life (x).

_tE_x= the actuarial present value of a t-year pure endowment for life (x).

Source: Bowers, Gerber, et. al. Actuarial Mathematics. 1997. Second Edition. Society of Actuaries: Itasca, Illinois. pp. 136-139

Original Problems and Solutions from The Actuary's Free Study Guide

Problem S3L38-1. A 34-year-old blue pterodactyl can get a whole life insurance policy paying a benefit of 1 at death for 0.56 Golden Hexagons (GH) or a continuous whole life annuity with actuarial present value ā₃₄ = 26. Find the annual force of interest.

Solution S3L38-1. We use the formula 1 = δā_x + Ā_x.

Then 1 - Ā_x = δā_x

(1 - Ā_x)/ā_x = δ

We are given that Ā₃₄ = 0.56

Thus, δ = (1 - 0.56)/26 = δ = about 0.0169230769.

Problem S3L38-2. A whole life insurance policy on the life of a 50-year-old giant mongoose has an actuarial present value of 0.53. Under a force of interest of 0.047, the second moment of the present value of this policy is 0.43. Find the variance of the present value of a continuous whole life annuity for a 50-year-old giant mongoose.

Solution S3L38-2. We use the formula Var(ā_T¬) = (²Ā_x - (Ā_x) ²)/δ².

We are given that Ā₅₀ = 0.53, ²Ā₅₀ = 0.43, and δ = 0.047.

Thus, Var(ā_T¬) = (0.43 - (0.53)²)/0.047² = Var(ā_T¬) = about 67.4966048.

Problem S3L38-3. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. The annual force of interest in Triceratopsland is currently 0.06. Charlie the Triceratops has a continuous whole life annuity whose actuarial present value is denoted by ā_x. What is the variance of the present value of this annuity?

Solution S3L38-3. The lifetimes of triceratopses are exponentially distributed, which means that triceratopses exhibit a constant force of mortality - in this case, 0.34. We can thus use the formula Var(ā_T¬) = μ/[(2δ + μ)(δ + μ)²] for μ = 0.34 and δ = 0.06. Thus,

Var(ā_T¬) = 0.34/[(2*0.06 + 0.34)(0.34 + 0.06)²] = 0.34/0.0736 = Var(ā_T¬) = about 4.619565217.

Problem S3L38-4. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. The annual force of interest in Triceratopsland is currently 0.06. Charlie the Triceratops has a continuous whole life annuity whose actuarial present value is denoted by ā_x = 2.5. What is the probability that the actuarial present value of this annuity will exceed 2.5?

Solution S3L38-4.

Since triceratopses have a constant force of mortality, we can use the formula

Pr(ā_T¬ > ā_x) = (μ/(δ + μ))^μ/δ with μ = 0.34 and δ = 0.06.

Thus, Pr(ā_T¬ > 2.5) = (0.34/(0.06 + 0.34))^0.34/0.06 = (0.85)^17/3 =

Pr(ā_T¬ > 2.5) = about 0.3981443701.

Problem S3L38-5. The life of a triceratops has the following survival function associated with it: s(x) = e^-0.34x. The annual force of interest in Triceratopsland is currently 0.06. Desiderius the Triceratops has a continuous 4-year temporary life annuity whose actuarial present value is ā_x:4¬. What is the variance of the present value of Desiderius's annuity?

Solution S3L38-5. We use the formula

Var(Y) = (2/δ)(ā_x:n¬ - ²ā_x:n¬) - (ā_x:n¬)²

We can find ā_x:4¬. We use the formula ā_x:n¬ = ₀ⁿ∫v^t*_tp_x*dt. Since the lifetimes of triceratopses are exponentially distributed, _tp_x = e^-0.34t for all x. Moreover, v^t = e^-0.06t. Thus,

ā_x:4¬ = ₀⁴∫e^-0.06t*e^-0.34t*dt = ₀⁴∫e^-0.4t*dt = (-5/2)e^-0.4t│₀⁴ = (5/2)(1 - e^-1.6) = ā_x:4¬ = about 1.99525871.

To find ²ā_x:4¬, we can use the Rule of Moments and apply the resulting formula

²ā_x:n¬ = ₀ⁿ∫v^2t*_tp_x*dt.

Thus, ²ā_x:4¬ = ₀⁴∫e^-0.12t*e^-0.34t*dt = ₀⁴∫e^-0.46t*dt = (-100/46)e^-0.46t│₀⁴ = (100/46)(1 - e^-1.84) = about 1.828657769.

Hence, Var(Y) = (2/0.06)(1.99525871 - 1.828657769) - (1.99525871)²

Var(Y) = about 1.572307369.

See other sections of The Actuary's Free Study Guide for Exam 3L.