Tutorial 2 Flashcards
(7 cards)
Question 1
Which of the following is not correct for the variance of whole life annuity payable annually in arrears?
a. var [1−𝑣^(𝐾,𝑥)/𝑖]
b. (1/𝑖^2)var[𝑣^(𝐾,𝑥) +1 / 𝑣^2)]
c. (1/𝑖^2.𝑣^2)var[𝑣^(𝐾,𝑥) +1]
d. (1/𝑑^2)[^2Ax – (Ax)^2]
b. (1/𝑖^2)var[𝑣^(𝐾,𝑥) +1 / 𝑣^2)]
Question 2
Which of the following is true for a temporary annuity (age x, duration n) payable annually in advance ?
I. ἂ,𝑥 - 𝑎,𝑥+n(𝐷,𝑥+𝑡)/(𝐷,𝑥)
II. 1 - d𝐴,x:n¬
III. ἂ,𝑥:𝑛¬(𝑚) + ([𝑚−1]/2𝑚)(1-v^(n)npx)
a) I & II
b) I & III
c) II & III
d) I, II & III
b) ἂ,𝑥:n¬ = (1 −𝐴,𝑥:𝑛¬)/d
Question 3
Rank the following annuity functions in order from
highest to lowest:
ἂ,𝑥, a,𝑥, ᾱ,𝑥 and ἂ,𝑥(𝑚)
ἂ,𝑥
ἂ,𝑥(𝑚)
ᾱ,𝑥
a,𝑥
ἂ,𝑥(𝑚) ≈ ἂ,𝑥 - [𝑚−1]/2𝑚
ᾱ,𝑥 ≈ ἂ,𝑥 - 1/2
a,𝑥 = ἂ,𝑥 - 1
Question 4
A life currently aged 60 is entitled to an annual
pension of £15,000 pa, calculate the EPV if the pension is:
- Payable annually in advance
- Payable annually in arrears
- Payable continuously
- Payable annually in arrears on retirement from age 65
- Payable annually in advance for 10 years only
Basis AM92 Ultimate @ 4% pa
- Payable annually in advance
15000 ἂ,60 = 15000 x 14.134 = £212,010 - Payable annually in arrears
15000 a,60 = 15000 x (14.134 - 1) = £197,010 - Payable continuously
15000 ᾱ,60 = 15000 x (14.134 - 1/2) = £204,510 - Payable annually in arrears on retirement from age 65
5|a60 = v^5 5p60 a,65 = (1.04^-5)(8821.2612/9287.2164)(12.276 - 1)
= 15000 5|a,60 = 15000 x 8.80306 = £132,046 - Payable annually in advance for 10 years only
ἂ,60:10¬ = ἂ,60 – v^10 10p60 ἂ,70
= 14.134 – (1.04-10)(8054.0544/9287.2164) x 10.375 = 8.0557
15000 ἂ,60:10 = 15000 x 8.0557 = £120,836
Question 5
A whole life annuity of £20,000 pa is payable to a
female currently aged 60. Calculate the EPV and
standard deviation, if the annuity is payable annually
in arrears.
Basis PFA92C20 @ 4% pa
EPV = 20000a,60 = 20000(16.652-1) = £313,040
SD = {(20000^2/d^2).[^2A,60 – (A,60)^2]}^0.5
Use A,x = 1 - dἂ,x —— d = 0.04/1.04 = 3.846%
A,60 = (1 - 0.03846 x 16.652) = 0.35956
SD = {(20000^2/0.03846^2).[0.15015 – 0.35956^2]}^0.5 = £75,118
Question 6
Calculate the following:
Basis PFA92C20 @ 4% pa
- 100,000Ȃ,65
- 50,000Ȃ,50:15¬
• 100,000Ȃ,65
100,000(1 - δᾱ,65) = 100000[1-ln1.04x(14.871 - 0.5)] = £43,636
• 50,000Ȃ,50:15¬
- Ȃ,50:15¬ = 1 - δᾱ,50:15¬
- ᾱ,50:15¬ ≈ ἂ,50:15¬ - ½(1 – v^15 15p50)
ἂ,50:15¬ = ἂ,50 - v^15 15p50 ἂ,65
= 19.539 – (1.04^-15)(9703.708/9952.697)x14.871 = 11.4882
ᾱ,50:15¬ = 11.4882 – 0.5[1 - (1.04^-15)(9703.708/9952.697)] = 11.2589
• 50000Ȃ,50:15 =50000(1-ln1.04 x 11.2589) = 27,921
Question 7
Calculate the expected present value of the following:
• A temporary annuity of £5,000 pa payable to a male
currently aged 55 for a period of 15 years
• An endowment assurance with a sum assured of £90,000 with a period of 15 years
• A term assurance with a sum assured £90,000 with a period of 15 years
• A whole life assurance deferred for 15 years with a sum assured £90,000.
• Basis PMA92C20 @ 4% pa
• Assume the annuity is payable annually in advance and any death benefits are paid at the end of the year
• EPV of a temporary annuity of £5,000 pa payable to a male currently aged 55 for a period of 15 years
• Basis PMA92C20 @ 4% pa
• annually in advance
• ἂ,55:15 = ἂ,55 – v^(15) 15p55 ἂ,70
• = 17.364 - (1.04^-15)(9238.134/9904.805)x11.562=11.3761
• = 5000 x 11.3761 = £56,881
• Expected present value of an endowment assurance
with a sum assured of £90,000 with a period of 15 years
• Basis PMA92C20 @ 4% pa
• death benefits are paid at the end of the year
• 𝐴55:15¬= 1 - dἂ,55:15¬ with d = 0.04/1.04 = 0.03846
• 90000𝐴55:15¬ = 90000(1-0.03846 x 11.3761)
• = £50,623
• Expected present value of a term assurance with a sum
assured £90,000 with a period of 15 years
• Basis PMA92C20 @ 4% pa
• death benefits are paid at the end of the year
• 𝐴55:15¬ = 𝐴1^55:15¬ + 𝐴55:15^1¬
• 90000𝐴1^55:15¬ = 50,623 - 90000v^15 15p55
• = 50623 – 90000(1.04^-15)(9238.134/9904.805)
• = £4,013
• Expected present value of a whole life assurance
deferred for 15 years with a sum assured £90,000
• Basis PMA92C20 @ 4% pa
• death benefits are paid at the end of the year.
•15|A55 = A55 - 𝐴1^55:15¬
• A55 = 1 - dἂ,55 = 1- 0.03846 x 17.364 = 0.33218
• 90000 15|A55 = 90000 x 0.33218 - 4,013
• £25,883
