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Variances And Relationships Among Present Values Of Continuous Whole And Temporary Life Annuities: Practice Problems And Solutions

Variances and Relationships Among Current Values of Steady Whole and Temporary Life Annuities: Apply Issues and Options
The Actuary's Free Study Information for Exam 3L - Section 38
G. Stolyarov II
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This section of pattern issues and solutions is a part of The Actuary's Free Examine Information for Examination 3L, authored by Mr. Stolyarov. That is Part 38 of the Research Guide. See an index of all sections by following the link in this paragraph.

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

1 = δāx + Āx

We can additionally decide the variance of the current value of a continuous whole life annuity:

Var(āT¬) = (2Āx - (Āx) 2)/δ2

If the power of mortality is a constant worth μ, then

āx = 1/(δ + μ)

Āx = μ/(δ + μ)

2Āx = μ/(2δ + μ)

Var(āT¬) = μ/(2δ + μ)(δ + μ)2

Pr(āT¬ > āx) = (μ/(δ + μ))μ/δ

If Y is the present worth of an n-yr short-term life annuity, then

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

Meaning of Symbols:

Āx = the actuarial present value of an entire life insurance coverage for all times (x).

2Āx = the second moment of the actuarial present worth of an entire life insurance coverage coverage for life (x).

āx = the actuarial present value of a steady entire life annuity for life (x).

āT¬ = the current-worth random variable for a steady whole life annuity for an annuitant with a future-lifetime random variable of T.

δ = the annual force of interest.

āx:n¬= the actuarial current worth of a continuous n-12 months short-term life annuity for all times (x).

tEx= the actuarial current 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

Unique Issues and Solutions from The Actuary's Free Study Information

Drawback S3L38-1. A 34-12 months-previous blue pterodactyl can get an entire life insurance coverage paying a benefit of 1 at death for 0.fifty six Golden Hexagons (GH) or a steady whole life annuity with actuarial current value ā34 = 26. Discover the annual drive of interest.

Answer S3L38-1. We use the formula 1 = δāx + Āx.

Then 1 - Āx = δāx

(1 - Āx)/āx = δ

We're provided that Ā34 = 0.56

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

Downside S3L38-2. An entire life insurance coverage coverage on the life of a 50-year-old big mongoose has an actuarial present worth of 0.53. Beneath a pressure of interest of 0.047, the second second of the current worth of this policy is 0.43. Discover the variance of the current value of a continuous entire life annuity for a 50-year-outdated big mongoose.

Resolution S3L38-2. We use the method Var(āT¬) = (2Āx - (Āx) 2)/δ2.

We're given that Ā50 = 0.fifty three, 2Ā50 = 0.43, and δ = 0.047.

Thus, Var(āT¬) = (0.forty three - (0.fifty three)2)/0.0472 = Var(āT¬) = about 67.4966048.

Downside S3L38-3. The life of a triceratops has the next survival operate associated with it: s(x) = e-0.34x. The annual drive of interest in Triceratopsland is at the moment 0.06. Charlie the Triceratops has a steady entire life annuity whose actuarial present value is denoted by āx. What's the variance of the current worth of this annuity?

Answer S3L38-3. The lifetimes of triceratopses are exponentially distributed, which implies that triceratopses exhibit a continuing drive of mortality - on this case, 0.34. We will thus use the formula Var(āT¬) = μ/(2δ + μ)(δ + μ)2 for μ = 0.34 and δ = 0.06. Thus,

Var(āT¬) = 0.34/(2*0.06 + 0.34)(0.34 + 0.06)2 = 0.34/0.0736 = Var(āT¬) = about 4.619565217.

Problem S3L38-4. The lifetime of a triceratops has the next survival operate related to it: s(x) = e-0.34x. The annual pressure of curiosity in Triceratopsland is at the moment 0.06. Charlie the Triceratops has a continuous entire life annuity whose actuarial present value is denoted by āx = 2.5. What's the probability that the actuarial present worth of this annuity will exceed 2.5?

Solution S3L38-4.

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

Pr(āT¬ > āx) = (μ/(δ + μ))μ/δ with μ = 0.34 and δ = 0.06.

Thus, Pr(āT¬ > 2.5) = (0.34/(0.06 + 0.34))0.34/0.06 = (0.85)17/three =

Pr(āT¬ > 2.5) = about 0.3981443701.

Downside S3L38-5. The life of a triceratops has the next survival perform associated with it: s(x) = e-0.34x. The annual force of curiosity in Triceratopsland is at present 0.06. Desiderius the Triceratops has a steady 4-12 months short-term life annuity whose actuarial current worth is āx:four¬. What's the variance of the current value of Desiderius's annuity?

Answer S3L38-5. We use the method

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

We are able to discover āx:four¬. We use the components āx:n¬ = 0n∫vt*tpx*dt. Because the lifetimes of triceratopses are exponentially distributed, tpx = e-0.34t for all x. Moreover, vt = e-0.06t. Thus,

āx:four¬ = 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) = āx:4¬ = about 1.99525871.

To find 2āx:four¬, we are able to use the Rule of Moments and apply the resulting formula

2āx:n¬ = 0n∫v2t*tpx*dt.

Thus, 2āx:four¬ = 04∫e-0.12t*e-0.34t*dt = 04∫e-0.46t*dt = (-100/forty six)e-0.46t│04 = (a hundred/46)(1 - e-1.84) = about 1.828657769.

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

Var(Y) = about 1.572307369.

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