CM1 Practise: cm1 handbook
Assurances and Life annuities:
Past exam questions:
Q1
𝑋 = 0 (𝑖𝑓 𝐾𝑥 < 𝑛 + 1)
1 ( 𝑖𝑓 𝐾𝑥 ≥ 𝑛)
𝑌 = 1 ( 𝑖𝑓 𝐾𝑥 < 𝑛 + 1)
=0 𝑖𝑓 𝐾𝑥 ≥ 𝑛 + 1
𝑋𝑌 = 0 (∀ 𝐾𝑥 )
𝐶𝑜𝑣(𝑋, 𝑌) = 𝐸(𝑋𝑌) − 𝐸(𝑋)𝐸(𝑌) = 0 − 𝐴𝑥:𝑛̅1 | 𝐴1𝑥:𝑛̅|
𝑉𝑎𝑟(𝑋 + 𝑌) = 𝑉𝑎𝑟(𝑋) = 𝑉𝑎𝑟(𝑌) − 2𝐶𝑜𝑣(𝑋𝑌)
= (^2𝐴 𝑥:𝑛̅1 | − (𝐴𝑥:𝑛̅1 | )^2 ) + (^2𝐴 1𝑥:𝑛̅| − (𝐴1𝑥:𝑛̅| )^2 ) − 2(𝐴𝑥:𝑛̅1 | 𝐴1𝑥:𝑛̅| )
2𝐴𝑥:𝑛
̅ 1|
=(
1
2
2
+2𝐴𝑥:𝑛̅| ) − (𝐴𝑥:𝑛̅1 | + 𝐴1𝑥:𝑛̅| ) = 2 𝐴𝑥:𝑛̅| + (𝐴𝑥:𝑛̅| ) = 𝑉𝑎𝑟(𝐴𝑥:𝑛̅| )
Q2)
0.5
𝐸(𝑃. 𝑉) = 𝐴30:20
= (𝐴30 − 𝑉 𝑛 𝑛𝑝𝑥 𝐴50 ) ∗ (1.04)0.5 = 0.13271 ∗ (1.04)0.5
̅̅̅̅1 | (1.04)
= 0.013534
Error made -> forgot to multiply (1.04)^0.5 -> conversion from Eoyod to immediate
𝑉𝑎𝑟 = 𝐸(𝑃. 𝑉 2 ) − 𝐸(𝑝. 𝑣)2 = 0.008814 Error made : Forgot to square (1.04)^0.5 in E(P.V^2)
Q4)
3𝑘𝐴40 − 2𝑘𝐴40:20
̅̅̅̅1 | − 20| 𝐴40:201 | = 691.68 − 68.58 − 𝐶 = 508.93
𝐴̅𝑛 = 𝑉 𝑇𝑥 = 𝐴𝑛 (1 + 𝑖)0.5
𝐴𝑛 = 𝑉 (𝐾(𝑥+1) )
𝑃(𝐾𝑥 = 𝑘) = 𝑘|𝑞𝑥
Uncertain Annuities
∞
𝑡
𝑎̈ 𝑛 = 𝑎̈ ̅̅̅̅̅̅̅̅
𝐾𝑥 +1 | = ∑ 𝑉 𝑡𝑝𝑥
0
(1−𝑉 𝐾𝑥+1 )
𝑉𝑎𝑟 (
)=
𝑑
1
(𝑉𝐾𝑥 +1
𝑑2
)
𝑎𝑛 = 𝑎̈ ̅̅̅̅
𝐾𝑥 |=𝑉𝑞𝑥 +(𝑉+𝑉 2 )𝑝𝑥 𝑞𝑥+1 +⋯(𝑎̈ 𝑛
̅ | )𝑛−1𝑝𝑥 (𝑞𝑥+𝑛 )
1 + 𝑎𝑛 = 𝑎̈ 𝑛
\𝑖𝑛𝑓𝑡𝑦
𝑎̅𝑛 = 𝑎̅̅̅̅
𝑇𝑥 | = ∫
𝑉 𝑡 𝑡𝑝𝑥 𝑑𝑡 = (𝑎̈ 𝑥 − 0.5)
0
𝑎̈ 𝑥:𝑛̅| = 𝑎𝑥:𝑛̅| + (1 − 𝑉 𝑛 𝑛𝑝𝑥) = 𝑎̈ ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
min(𝐾𝑥 +1,𝑛)|
𝑎̈ 𝑥:𝑛̅| = 𝑎̅𝑥:𝑛̅| + 0.5(1 − 𝑉 𝑛 𝑛𝑝𝑥)
𝑎̈ 𝑥 ∗ 𝑑 =
1 − 𝐴𝑥
𝑎̅𝑥:𝑛̅| ∗ 𝑑 = 1 − 𝐴̅𝑥:𝑛̅|
Guaranteed:
𝑛
𝑎̈ ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
(𝑥:𝑛̅| )
max(𝐾𝑥 +1,𝑛)| = 𝑎̈ 𝑛̅| + 𝑉 𝑛𝑝𝑥 𝑎̈ 𝑥+𝑛 = 𝑎̈ ̅̅̅̅̅̅̅̅̅
Monthly payments:
(𝑚)
𝑎𝑥
= 𝑎𝑥 +
𝑚+1
2𝑚
(𝑚 )
𝑎̈ 𝑥
= 𝑎̈ 𝑥 −
𝑚−1
2𝑚
𝑚−1
2𝑚
(𝑚)
𝑎̈ 𝑥:𝑛̅| = 𝑎̈ 𝑥:𝑛̅| −
Past Exam Questions:
Q1)
(12)
(12)
10
200𝑘 = 𝑋𝑎̈ 10
10𝑝60 𝑎̈ 70
̅̅̅̅| + 𝑋𝑉
11
24
(12)
10
= 𝑋 (𝑎̈ 10
10𝑝60 (𝑎̈ 70 − ( )))
̅̅̅̅| + 𝑉
(𝟏𝟐)
𝟏𝟏
-> error made 𝑽𝟏𝟎 𝟏𝟎𝒑𝟔𝟎 𝒂̈ 𝟕𝟎 = 𝑽𝟏𝟎 𝟏𝟎𝒑𝟔𝟎 𝒂̈ 𝟕𝟎 ∗ (𝟐𝟒)
Q2)
∞
𝐸(𝑝. 𝑣) = ∫ 𝐸𝑥𝑝(−0.05𝑡 − 0.02𝑡)𝑑𝑡 = 0.07−1 = 13.2857
0
𝑉𝑎𝑟(𝑝. 𝑉) =
1
(𝑉𝑎𝑟(𝑊𝐿𝐴)) =
𝑑2
∞
∞
𝑉𝑎𝑟(𝑊𝐿𝐴) = ∫ 𝐸𝑥𝑝(−0.05𝑡 ∗ 2 − 0.02𝑡) ∗ 0.02𝑑𝑡 − ∫ 𝐸𝑥𝑝(−0.05𝑡 − 0.02𝑡) ∗ 0.02𝑑𝑡 =
0
Var(p.V) = 34.smtn smtn
No errors
Q3)
= 0 𝑖𝑓 𝑇𝑥 ≤ 2
= 5𝑘 ∗ 𝑉 2 ∗ 𝑎̅𝑇63 −2 else
0
= 100 ∗ 5𝑘 ∗ 𝑉 2 2𝑝63 ∗ (𝑎̈65 − 0.5) = 6594347
Error made: Wrong mortality used (Am92)-> silly error
Var(P.V)= 𝐸(𝑃𝑉 2 ) − 𝐸(𝑝. 𝑉)2
iii) Variance of deffered annuity -> unable to solve???
Q4)
(12)
𝑎̈ 𝑥:1̅| = 𝑎̈ 𝑥:1̅| −
(12)
𝑎̈ 𝑥:1̅| = 1 −
11
(1 − 𝑉 1 ∗ 1𝑝𝑥)
24
11
(1 − 1.06−1 ∗ 0.99) = 09697 = 𝐴
24
1(𝑎𝑝)𝑥 = 0.99 ∗ 0.8 = 𝐵
𝑃. 𝑉 = 240 ∗ 120 ∗ (𝐴 + 𝐴 ∗ 𝐵 ∗ 1.06−1 + 1.06−2 ∗ 𝑏 2 ∗ 𝐴) = 64,386𝑘
Error: Used concept of one year term annuities -> failed to realise 𝒂̈ 𝒙:𝟏̅| = 𝟏
Q5)
(12)
= 5𝑘𝑎̈ ̅5|
(12)
(12)
+ 5 |7𝑘 𝑎̈ 60 − 5| 1𝑘 𝑎60:5̅|
Better method:
(12)
= 5𝑘𝑎̈ 5̅|
(12)
(12)
+ 5 |6𝑘 𝑎̈ 60 + 10| 1𝑘𝑎̈ 60
UDD:𝑡𝑞𝑥 = 𝑡 ∗ 𝑞𝑥
CFM: 𝑡𝑝𝑥 = 𝑝𝑥𝑡
∀ 𝑡 ∈ (0,1)
∀𝑡 ∈ (0,1)
Constant force of mortality
UDD and CFM:
(4)
(4)
Q21) 𝑎73.25 = 𝑎73.25:0.75
̅̅̅̅̅̅| + 1|𝑎73
(4)
0.25
𝑎73.25:0.75
∗ 0.25𝑝73.25 + 𝑉 0.5 ∗ 0.5 𝑝 73.25 + 𝑽𝟎.𝟕𝟓 𝟎. 𝟕𝟓𝒑 𝟕𝟑. 𝟐𝟓 ]
̅̅̅̅̅̅| = [𝑣
𝑝73 =
= [𝑣 0.25 ∗ (𝑝73)0.25 + 𝑉 0.5 ∗ (𝑝 73)0.5 + 𝑉 0.75 𝑝( 73)0.75 ] = 0.729953
(4)
(4)
1|𝑎73 = 𝑉 1 ∗ 𝑝73 ∗ 𝑎74 =
Variable Benefits:
Q4) Overhead expense : expenses independent of the amount of business conducted
Direct expense: expenses that vary with the amount of business -> commission to salesmen,
renewal expense
Note: commissions are also expenses.
Q 19.7)
1
1
(12 )
= 1500 ∗ (
) ∗ (𝑎̈ 67 −
)
12
(1 + 𝑏)
12 @4%
Q13) Explain why an insurance firm sets up reserves for endowment contracts sold:
The expected cost of paying benefits usually increases as the life ages and the probability of a claim
by death increases. In the final year the probability of payment is large, since the payment will be
made if the life survives the term, and for most contracts the probability of survival is large. Level
premiums received in the early years of a contract are more than enough to pay the benefits that fall
due in those early years, but in the later years, and in particular in the last year of an endowment
assurance policy, the premiums are too small to pay for the benefits. It is therefore prudent for the
premiums that are not required in the early years of the contract to be set aside, or reserved, to
fund the shortfall in the later years of the contract. If premiums received that were not required to
pay benefits were spent by the company, perhaps by distributing to shareholders, then later in the
contract the company may not be able to find the money to pay for the excess of the cost of benefits
over the premiums received.
𝟏𝟎𝟎
(𝟓𝟐 )
Q21) 𝑓𝑢𝑛𝑑 = 𝒗𝟒𝟓∗𝟒𝟓𝒑𝟐𝟏 ∗ (𝒂𝟐𝟏:𝟒𝟓
̅̅̅̅| )
Error made: i) do not multiply by 100 -> question asks for value of fund for each student
ii) Use 𝑎̅𝑥:𝑛̅| instead -> weekly payments can be considered as continuous -> easier.
Q22) Errors made: I) Death benefit payble immediately -> silly mistake (used eoyod)
II) IN EXPENSES: 2.5% OF each quarterly premium from the start of 2nd policy year.
Subtract a one year term annuity (to account for start from 2nd year) .
Total annuity – one year term
ii) 𝐴160:5̅| @0% = ∑40 𝑡|𝑞60 = ∑
𝒍
𝒅
𝒕|𝒒𝒙 = 𝒍𝒙 ∗ 𝒍 𝒙+𝒕 =
𝒙
𝒙+𝒕
𝒅𝟔𝟎+𝒕
𝑳𝟔𝟎
𝒅𝒙+𝒕
𝒍𝒙
Calculated reserves using wrong basis
Bonuses were certain -> to be included in prospective reserve calculation
Q23) Error : P.V given represents an increasing term assurance.
Q24) Error: silly mistakes in reading question
iii) Error: Mistook age of policy holder. (wrote as 49) -> age!= policy year
Age 40 at 1J 2000 => Age = 50 at 31 Dec 2009
Q28) i) net premium ignores all bonuses -> given to confuse
ii) a super compound bonus increases more gradually than a simple compound or simple
bonus -> allows life office to retain surplus for longer for the same bonus payments.
Q31) Note renewal expense -> from start of second and subsequent payments->
= 𝑎̈ 𝑥:𝑛̅| − 1 -> this removes the first payment -> continue to use this instead of arrears
If using arrears -> need to remove last payment
Error : 75 pa from 2nd year + 5pa from 3rd year
= 65(𝑎̈ 𝑥:𝑛̅| − 1) + 5(I𝑎̈ 𝑥:𝑛̅| − 1)
Note: since its an increasing annuity, 2nd payment is 10 not 5
Q34) Describe use of terminal bonus:
Distributes surplus available to policy based on asset share.
Distributing available surplus as a terminal bonus -> delays distribution of surplus-> firm can
invest in more long-term assets.
Q49) Error:
ii) calculation of gross prem prospective reserve -> forgot to include bonuses vested till date.
 Took V_7 instead of V_17
Project Appraisal:
Q17) Errors made: time of sale is calculated from time 0, not from completion of development
(𝟒)
ii) Rental income increases by 50k annually -> 𝑰𝒂̈ 𝒏̅|
question doesn’t mention that it increases in the BOY -> this would be more complex
 Can just account for a 50k increase over 4 quarters
Q22) Errors made: silly -> discounted inflow to 0 but discounted outflows to t=5
 Conecptualy okay -> need to be faster
Q26) Error made : Time between 01 Jan 2001 – 30 June 2004 -> 3.5 years (I took 4.5)
 Unecesarry balancing for compound increase
 Arrear w/ 10% compound increase -> balance denom
 Adv -> no balancing in denom required
Bond Valuation:
JOINT LIFE and Reversionary Benefits:
Q39)
𝑃. 𝑉 = 𝑎̅̅̅̅̅̅̅̅̅̅
𝑇_(𝑥𝑦)
|
Q40) joint life annuity of 1 pa payable in advance, as long as both lives aged 60,50 are alive or a
maximum of 20 years
= 𝑎̈ 60:50 − 𝑣 20 ∗ 20𝑝60 ∗ 20𝑝50 ∗ 𝑎̈ 80:70 = 12.747
(12)
Q41) 20𝑘 (𝑎̈ 5̅|
(12)
(12)
(12)
+ 𝑉 5 ∗ 5𝑝65 ∗ 𝑎̈ 70 ) + 10𝑘 ∗ 𝑉 5 ( 5𝑞65 ∗ 5𝑝62 ∗ 𝑎̈ 67 + 5𝑝65: 62 ∗ 𝑎̈ (70|67) )
Error made: The first 5 yrs are guaranteed-> reversionary bit starts only after 5 years.
Q46) Benefits -> reversionary annuity payble on death of 65m to 60 f.
Premiums -> “paid monthly until the annuity commences (male dies) or risk ceases (female
dies-> no one receives money).
Error made: Premiums taken as simple annuity contingent on 65(m).
Premium-> Joint life annuity -> ceases on death of either life.
Q47) Error: didn’t notice it was continuous-> used d instead of 𝛿
1
𝑉𝑎𝑟(𝑔(𝑇)) = (𝐴𝑥𝑦 @𝑖 2 + 2𝑖 + 𝐴2𝑥𝑦 ) 𝛿 2
Q48)