Formula Sheet for Actuarial Mathematics - Exam MLC - (ASM 2014)

1 Lesson 1 - Probability Review

Lesson 2 – Survival Distributions: Probability Functions

1.1 1.2 1.3 1.4 1.5

𝜇2 = 𝜇2′ − 𝜇 2 𝜇3 = 𝜇3′ − 3𝜇2′ 𝜇 − 2𝜇 3 𝑉𝑎𝑟(𝑋) = 𝐸[𝑋 2 ] − 𝐸[𝑋]2 𝑉𝑎𝑟(𝑎𝑋 + 𝑏𝑌) = 𝑎2 𝑉𝑎𝑟(𝑋) + 2𝑎𝑏𝐶𝑜𝑣(𝑋, 𝑌) + 𝑏 2 𝑉𝑎𝑟(𝑌) 𝑉𝑎𝑟(∑𝑛𝑖=1 𝑋𝑖 ) = 𝑛𝑉𝑎𝑟(𝑋)

1.6

𝑉𝑎𝑟(𝑋̅) = 𝑉𝑎𝑟 (

1.7

Bayes Theorem Pr(𝐵 |𝐴)Pr⁡(𝐴) Pr(𝐴|𝐵) =

∑𝑛 𝑖=1 𝑋𝑖 𝑛

)=

𝑛𝑉𝑎𝑟(𝑋) 𝑛2

=

2.4

𝑡|𝑢𝑞𝑥

2.5

𝑡|𝑢𝑞𝑥 =

Pr⁡(𝐵) 𝑓𝑦 (𝑦|𝑥 )𝑓𝑥 (𝑥)

1.9

Law of Total Probability (Discrete) Pr(𝐴) = ∑𝑖 Pr(𝐴 ∩ 𝐵𝑖 ) = ∑𝑖 Pr(𝐵𝑖 )Pr⁡(𝐴|𝐵𝑖 ) Law of Total Probability (Continuous) Pr(𝐴) = ∫ Pr(𝐴|𝑥) 𝑓(𝑥)𝑑𝑥 Conditional Mean Formula 𝐸𝑋 [𝑋] = 𝐸𝑌 [𝐸𝑋 [𝑋|𝑌]] Double Expectation Formula 𝐸𝑋 [𝑔(𝑋)] = 𝐸𝑌 [𝐸𝑋 [𝑔(𝑋)|𝑌]] Conditional Variance Formula

1.12 1.13

2.3 𝐹𝑥 (𝑡) =

𝑛

𝑓𝑥 (𝑥|𝑦) =

1.11

2.2 𝑆𝑥 (𝑡) =

𝑉𝑎𝑟(𝑥)

1.8

1.10

2.1 𝑆𝑥+𝑡 (𝑢) =

𝑓𝑦 (𝑦)

𝑆𝑥 (𝑡+𝑢)

𝑆𝑥 (𝑡) 𝑆0 (𝑥+𝑡)

𝑆0 (𝑥) 𝐹0 (𝑥+𝑡)−𝐹0 (𝑥) 1−𝐹0 (𝑥)

= 𝑡𝑝𝑥 − 𝑡+𝑢𝑝𝑥 𝑡+𝑢𝑞𝑥

− 𝑡𝑞𝑥

Life Table Functions 𝑡𝑝𝑥

=

𝑡𝑞𝑥

=

𝑡|𝑢𝑞𝑥

𝑡+𝑢𝑝𝑥

𝑙𝑥+𝑡

=

𝑙𝑥 𝑡𝑑𝑥

𝑙𝑥

=

𝑢𝑑𝑥+𝑡

𝑙𝑥

𝑙𝑥 −𝑙𝑥+𝑡 𝑙𝑥

=

𝑙𝑥+𝑡 −𝑙𝑥+𝑡+𝑢 𝑙𝑥

= 𝑡𝑝𝑥 𝑢𝑝𝑥+𝑡

𝑉𝑎𝑟𝑋 (𝑋) = 𝐸𝑌 [𝑉𝑎𝑟𝑋 (𝑋|𝑌)] + 𝑉𝑎𝑟𝑌 (𝐸𝑋 [𝑋|𝑌]) Distribution Bernoulli Binomial Uniform Exponential

Mean 𝑞 𝑚𝑞 𝑎+𝑏 2

𝜃

Variance 𝑞(1 − 𝑞) 𝑚𝑞(1 − 𝑞) (𝑏−𝑎)2 12

𝜃2

Bernoulli Shortcut: If a random variable can only assume two values 𝑎 and 𝑏 with prob 𝑞 and 1 − 𝑞, then 𝑉𝑎𝑟(𝑋) = 𝑞(1 − 𝑞)(𝑏 − 𝑎)2

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

2 Lesson 3 – Survival Distributions: Force of Mortality 3.1 𝜇𝑥+𝑡 =

𝑓𝑥 (𝑡)

𝑆𝑥 (𝑡) 𝑑

3.2 𝜇𝑥+𝑡 = 𝑑𝑡

𝑡𝑞𝑥 ⁡ 𝑡𝑝𝑥

3.3 𝜇𝑥+𝑡 = − 3.4 𝜇𝑥+𝑡 = −

Lesson 4 – Survival Distribution: Mortality 4.1 Gompertz’ Law 𝜇𝑥 = 𝐵𝑐 𝑥 𝑐>1 𝑡𝑝𝑥

= exp⁡(−

𝐵𝑐 𝑥 (𝑐 𝑡 −1)

𝑑 ln(𝑆𝑥 (𝑡))

4.2

𝑑𝑡 𝑑 ln( 𝑡𝑝𝑥 )

4.3 Makeham’s Law

𝑑𝑡

) 𝜇𝑥 = 𝐴 + 𝐵𝑐 𝑥

𝑐>1 A is constant part of force of mortality *Adding A to 𝜇 multiplies 𝑡𝑝𝑥 by e−μt

𝑡

3.5 𝑆𝑥 (𝑡) = exp⁡(− ∫0 𝜇𝑥+𝑠 𝑑𝑠) 𝑡

3.6 𝑡𝑝𝑥 = exp⁡(− ∫0 𝜇𝑥+𝑠 𝑑𝑠) 𝑥+𝑡

3.7 𝑡𝑝𝑥 = exp⁡(− ∫𝑥 𝜇𝑠 𝑑𝑠) 3.8 𝑓𝑥 (𝑡) = 𝑆𝑥 (𝑡)𝜇𝑥+𝑡 = 𝑡𝑝𝑥 𝜇𝑥+𝑡 𝑡+𝑢 3.9 𝑃(𝑡 < 𝑇𝑥 < 𝑡 + 𝑢) = 𝑡|𝑢𝑞𝑥 = ∫𝑡 𝑠 𝑝𝑥 𝜇𝑥+𝑠 𝑑𝑠 𝑡 ∫0 𝑠𝑝𝑥 𝜇𝑥+𝑠 𝑑𝑠

3.10 𝑡𝑞𝑥 = ′ If 𝜇𝑥+𝑠 = 𝜇𝑥+𝑠 + 𝑘 for 0 ≤ 𝑠 ≤ 𝑡 then 𝑡𝑝𝑥′ = 𝑡𝑝𝑥 𝑒 −𝑘𝑡 . If 𝜇𝑥+𝑠 = 𝜇̂ 𝑥+𝑠 + 𝜇̅𝑥+𝑠 for 0 ≤ 𝑠 ≤ 𝑡 then 𝑡𝑝𝑥 = 𝑡𝑝̂𝑥 𝑡𝑝̅𝑥 ′ If 𝜇𝑥+𝑠 = 𝑘𝜇𝑥+𝑠 for 0 ≤ 𝑠 ≤ 𝑡 then 𝑡𝑝𝑥′ = ( 𝑡𝑝𝑥 )

ln⁡(𝑐)

𝑘

4.4

𝑡𝑝𝑥

= exp⁡(−𝐴𝑡 −

𝐵𝑐 𝑥 (𝑐 𝑡 −1) ln⁡(𝑐)

)

Weibull Distribution

𝜇𝑥 = 𝑘𝑥 𝑛 𝑘𝑥 (𝑛+1) 𝑝 = exp⁡ ( − ) 0 𝑥 𝑛+1 Constant Force of Mortality 4.5 𝜇𝑥 = 𝜇 4.6 𝑡𝑝𝑥 = e−μt 4.7 (BLANK) Uniform Distribution 1 4.8 𝜇𝑥 = ⁡⁡⁡⁡0 ≤ 𝑥 ≤ 𝜔 4.9 4.10

𝑡𝑝𝑥

𝜔−𝑥 𝜔−𝑥−𝑡

=

𝑡𝑞𝑥

=

𝜔−𝑥 𝑡 𝜔−𝑥 𝑢

0≤𝑡 ≤𝜔−𝑥 0≤𝑡 ≤𝜔−𝑥

4.11 𝑡|𝑢𝑞𝑥 = 0≤ 𝑡+𝑢 ≤ 𝜔−𝑥 𝜔−𝑥 4.12 (BLANK) Beta Distribution 𝛼 4.13 𝜇𝑥 = 0≤𝑥≤𝜔 4.14

𝑡𝑝𝑥

𝜔−𝑥 𝜔−𝑥−𝑡 𝛼

=(

𝜔−𝑥

) 0≤𝑡 ≤𝜔−𝑥

*The force of mortality is the sum of two uniform forces. 𝑡𝑝𝑥 is the product of uniform probabilities

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

3 Lesson 5 – Survival Distributions: Moments Complete Future Lifetime ∞ 5.1 𝑒̇𝑥 = ∫0 𝑡 𝑡𝑝𝑥 𝜇𝑥+𝑡 𝑑𝑡 ∞

5.2 𝑒̇𝑥 = ∫0 𝑡𝑝𝑥 𝑑𝑡 ∞ 5.3 𝐸[𝑇𝑥2 ] = 2 ∫0 𝑡 𝑡𝑝𝑥 𝑑𝑡 ∞ 2 ∫0 𝑡 𝑡𝑝𝑥

𝑒̇𝑥2

5.4 𝑉𝑎𝑟(𝑇𝑥 ) = 𝑑𝑡 − 5.5 𝑒̇ 𝑥:𝑛| ̅̅̅ = 𝐸[min(𝑇𝑥 , 𝑛)] 𝑛 𝑒̇ 𝑥:𝑛| ̅̅̅ = ∫0 𝑡 𝑡𝑝𝑥 𝜇𝑥+𝑡 𝑑𝑡 + 𝑛 𝑛 𝑝𝑥 𝑛

5.6 𝑒̇ 𝑥:𝑛| ̅̅̅ = ∫0

𝑡𝑝𝑥

𝑑𝑡 𝑛

5.7 𝐸[min(𝑇𝑥 , 𝑛)2 ] = 2 ∫0 𝑡 𝑡𝑝𝑥 𝑑𝑡

e−μ

−μk 5.20 𝑒𝑥 = ∑∞ = ⁡⁡ (CFM) 𝑘=1 e 1−e−μ 5.21 𝑒̇𝑥 = 𝑒𝑥 + 0.5 (UDD) 5.22 𝑒̇ 𝑥:𝑛| ̅̅̅ = 𝑒𝑥:𝑛| ̅̅̅ + 0.5 𝑛𝑞𝑥 (UDD)

*For those surviving n years, min(𝑇𝑥 , 𝑛) = 𝑛 𝑛 *For those not surviving n years, average future lifetime is , since future 2 lifetime is uniform. * 𝑒𝑥 = 𝐸[min(𝐾𝑥 , 𝑛)] * 𝑒̇𝑥 = 𝐸[𝑇𝑥 ] *If curtate,⁡𝑒𝑥+𝑠:𝑎+𝑏| ̅̅̅̅̅̅̅ , 𝑏 < 1, 𝑠 < 1, 𝑎𝜖Ƶ is the same as 𝑒𝑥+𝑠:𝑎| ̅̅̅

Special Mortality Laws 𝜔−𝑥 5.8 𝑒̇𝑥 = 𝐸[𝑇𝑥 ] = (Beta) 𝑒̇𝑥 = 𝐸[𝑇𝑥 ] = 𝑒̇𝑥 = 𝐸[𝑇𝑥 ] =

𝛼+1 𝜔−𝑥 1

2

(UDD)

𝜇 𝛼(𝜔−𝑥)2

5.9 𝑉𝑎𝑟(𝑇𝑥 ) = (𝛼+1)2 𝑉𝑎𝑟(𝑇𝑥 ) = 𝑉𝑎𝑟(𝑇𝑥 ) =

(𝛼+2)

(𝜔−𝑥)2 1

12

𝜇2

(CFM) (Beta) (UDD) (CFM) 𝑛

5.10 𝑒̇ 𝑥:𝑛| ̅̅̅ = 𝑛𝑝𝑥 (𝑛) + 𝑛𝑞𝑥 ( ) (UDD) 𝜔−𝑥−𝑛

𝑛

2

𝑛

(𝑛) + 𝑒̇ 𝑥:𝑛| ( ) (UDD) ̅̅̅ = 𝜔−𝑥 𝜔−𝑥 2 5.11 𝑒̇ 𝑥:1| ̅ = 𝑝𝑥 + 0.5𝑞𝑥 (UDD) Curtate Future Lifetime 5.12 𝑒𝑥 = ∑∞ 𝑘=0 𝑘 𝑘|𝑞𝑥 𝑛−1 5.13 𝑒𝑥:𝑛| ̅̅̅ = ∑𝑘=0 𝑘 𝑘|𝑞𝑥 + 𝑛 𝑛𝑝𝑥 2 5.14 𝐸[𝐾𝑥2 ] = ∑∞ 𝑘=0 𝑘 𝑘|𝑞𝑥 2] 2 2 5.15 𝐸[min(𝐾𝑥 , 𝑛) = ∑𝑛−1 𝑘=0 𝑘 𝑘|𝑞𝑥 + 𝑛 𝑛𝑝𝑥 5.16 𝑒𝑥 = ∑∞ 𝑘=1 𝑘 𝑝𝑥 ∑𝑛𝑘=1 𝑘 𝑝𝑥 5.17 𝑒𝑥:𝑛| = ̅̅̅ 5.18 𝐸[𝐾𝑥2 ] = ∑∞ 𝑘=1(2𝑘 − 1) 𝑘 𝑝𝑥 5.19 𝐸[min(𝐾𝑥 , 𝑛)2 ] = ∑𝑛𝑘=1(2𝑘 − 1) 𝑘 𝑝𝑥

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

4 Lesson 6 – Survival Distributions: Percentiles and Recursions 6.1 𝑒̇𝑥 = 𝑒̇ 𝑥:𝑛| ̅̅̅ + 𝑛𝑝𝑥 𝑒̇𝑥+𝑛 6.2 𝑒𝑥 = 𝑒𝑥:𝑛| ̅̅̅ + 𝑛𝑝𝑥 𝑒𝑥+𝑛 6.3 𝑒𝑥 = 𝑒𝑥:𝑛−1| ̅̅̅̅̅̅̅ + 𝑛𝑝𝑥 (1 + 𝑒𝑥+𝑛 ) 6.4 𝑒𝑥 = 𝑝𝑥 + 𝑝𝑥 𝑒𝑥+1 = 𝑝𝑥 (1 + 𝑒𝑥+1 )⁡⁡⁡𝑛 = 1 6.5 𝑒𝑥:𝑛| ̅̅̅ = 𝑒𝑥:𝑚| ̅̅̅̅ + 𝑚 𝑝𝑥 𝑒𝑥+𝑚:𝑛−𝑚| ̅̅̅̅̅̅̅̅ ⁡⁡⁡⁡𝑚 < 𝑛 6.6 𝑒𝑥:𝑛| ̅̅̅ = 𝑒𝑥:𝑚−1| ̅̅̅̅̅̅̅̅ + 𝑚 𝑝𝑥 (1 + 𝑒𝑥+𝑚:𝑛−𝑚| ̅̅̅̅̅̅̅̅ )⁡⁡⁡⁡𝑚 < 𝑛 6.7 𝑒𝑥:𝑛| ̅̅̅ = 𝑝𝑥 + 𝑝𝑥 𝑒𝑥+1:𝑛−1| ̅̅̅̅̅̅̅ = 𝑝𝑥 (1 + 𝑒𝑥+1:𝑛−1| ̅̅̅̅̅̅̅ )

Lesson 7 – Survival Distributions: Fractional Ages Uniform Distribution of Deaths 7.1 𝑙𝑥+𝑠 = (1 − 𝑠)𝑙𝑥 + 𝑠𝑙𝑥+1 = 𝑙𝑥 − 𝑠𝑑𝑥 7.2 𝑠𝑞𝑥 = 𝑠𝑞𝑥 𝑠𝑞𝑥+𝑡

7.3

=

𝑠𝑞𝑥

1−𝑡𝑞𝑥

7.4 𝑠𝑝𝑥 𝜇𝑥+𝑠 = 𝑞𝑥 𝑞 7.5 𝜇𝑥+𝑠 = 𝑥 = 𝑠𝑝𝑥 1

=

𝑠𝑑𝑥

𝑙𝑥 −𝑡𝑑𝑥

𝑙

= 1 − ( 𝑥+𝑠+𝑡) , 0 ≤ 𝑠 + 𝑡 ≤ 1 𝑙𝑥+𝑡

𝑞𝑥 1−𝑠𝑞𝑥

7.6 𝑒̇𝑥 = 𝑒𝑥 + (UDD) 2 Recall: 5.11 (𝑒̇ 𝑥:1| ̅ = 𝑝𝑥 + 0.5𝑞𝑥 ) 7.7 𝑒̇ 𝑥:𝑛| ̅̅̅ = 𝑒𝑥:𝑛| ̅̅̅ + 0.5 𝑛𝑞𝑥 Constant Force of Mortality 7.8 𝑝𝑥 = 𝑒 −𝜇 7.9 𝜇 = −ln⁡(𝑝𝑥 ) 7.10 𝑠𝑝𝑥 = 𝑒 −𝜇𝑠 = (𝑝𝑥 ) 𝑠 7.11 𝑠𝑝𝑥+𝑡 = (𝑝𝑥 ) 𝑠 ⁡⁡⁡0 ≤ 𝑡 ≤ 1 − 𝑠 Table 7.1: Summary of Formulas for Fractional Ages UDD CFM 𝑙𝑥+𝑠 𝑙𝑥 − 𝑠𝑑𝑥 𝑙𝑥 𝑝𝑥𝑠 𝑠𝑞𝑥 1 − 𝑝𝑥𝑠 𝑠𝑞𝑥 1 − 𝑠𝑞𝑥 𝑝𝑥𝑠 𝑠𝑝𝑥 𝑠𝑞 𝑥 1 − 𝑝𝑥𝑠 𝑠𝑞𝑥+𝑡 1 − 𝑡𝑞𝑥 𝑞𝑥 𝜇𝑥+𝑠 − ln(𝑝𝑥 ) 1 − 𝑠𝑞𝑥 𝑞𝑥 −𝑝𝑥𝑠 ln(𝑝𝑥 ) 𝑠𝑝𝑥 𝜇𝑥+𝑠 𝑒̇𝑥 𝑒𝑥 + 0.5 𝑒̇ 𝑥:𝑛| ̅̅̅ 𝑒𝑥:𝑛| ̅̅̅ + 0.5 𝑛𝑞𝑥 𝑒̇ 𝑥:1| 𝑝𝑥 + 0.5𝑞𝑥 ̅

Function



Monica E. Revadulla EXAM MLC – Models for Life Contingencies

𝑒̇ 𝑥:𝑡|̅ = 𝑒̇ 𝑥:𝑘| ̅̅̅ + 𝑘 𝑝𝑥 𝑒̇ 𝑥+𝑘:𝑡−𝑘| ̅̅̅̅̅̅̅

Formula Summary of ASM 2014

5 Lesson 8 – Survival Distributions: Select Mortality  A man whose health was established 5 years ago will have better mortality than a randomly selected man.  A life selected at age 𝑥 can never become a life selected at any higher age. [𝑥] will never become [𝑥 + 1].

Lesson 10 – Insurance: Annual and 1/mthly – Moments ∞ 𝑘+1 𝑘+1 10.1 𝐸[𝑍] = ∑∞ 𝑘=0 𝑏𝑘 𝑣 𝑘 𝑝𝑥 𝑞𝑥+𝑘 𝑘|𝑞𝑥 = ∑𝑘=0 𝑏𝑘 𝑣 ∞ 2 ∞ 2 2] 2(𝑘+1) 2(𝑘+1) 10.2 𝐸[𝑍 = ∑𝑘=0 𝑏𝑘 𝑣 𝑘 𝑝𝑥 𝑞𝑥+𝑘 𝑘|𝑞𝑥 = ∑𝑘=0 𝑏𝑘 𝑣 Table 10.1: Actuarial notation for standard types of insurance Name Present Value of RV Symbol 𝐴𝑥 Whole life 𝑣 𝐾𝑥+1 𝐴𝑥́ :𝑛| Term life 𝑣 𝐾𝑥+1 ⁡⁡⁡⁡𝐾𝑥 < 𝑛 ̅̅̅ { ⁡⁡⁡⁡⁡⁡0⁡⁡⁡⁡⁡⁡𝐾𝑥 ≥ 𝑛 ⁡⁡⁡⁡⁡⁡0⁡⁡⁡⁡⁡⁡𝐾𝑥 < 𝑛 Deferred life 𝑛|𝐴𝑥 { 𝐾𝑥 +1 𝑣 ⁡⁡⁡𝐾𝑥 ≥ 𝑛 Deferred term 0⁡⁡⁡⁡⁡⁡⁡⁡𝐾𝑥 ≤ 𝑛⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡ ̅̅̅̅ 𝑛|𝐴𝑥́ :𝑚| {𝑣 𝐾𝑥+1 ⁡⁡⁡𝑛 < 𝐾𝑥 < 𝑛 + 𝑚 𝑛|𝑚 𝐴𝑥 0⁡⁡⁡⁡⁡⁡⁡𝐾𝑥 ≥ 𝑛 + 𝑚⁡⁡⁡ ⁡0⁡⁡⁡⁡⁡⁡𝐾𝑥 < 𝑛 𝐴𝑥:𝑛| Pure Endowment ́ ̅̅̅ { 𝑛 𝑣 ⁡⁡⁡𝐾𝑥 ≥ 𝑛 𝑡 𝑣 𝑝 ⁡𝑣 𝐾𝑥+1 ⁡⁡⁡⁡⁡⁡𝐾𝑥 < 𝑛 { 𝑣 𝑛 ⁡⁡⁡𝐾𝑥 ≥ 𝑛

Endowment

𝑡 𝑥

𝐴𝑥:𝑛| ̅̅̅

2

10.3 𝑉𝑎𝑟(𝑍) = 2𝐴𝑥:𝑛| ̅̅̅ ⁡ − (𝐴𝑥:𝑛| ̅̅̅ ) 10.4 2𝑖 = 2𝑖 + 𝑖 2 2 𝑑 = 2𝑑 − 𝑑 2 2 𝑣 = 𝑣2 10.5 ⁡ 𝑛|𝐴𝑥 = 𝑛𝐸𝑥 𝐴𝑥+𝑛 10.6 𝐴𝑥́ :𝑛| ̅̅̅ = 𝐴𝑥 − ⁡ 𝑛|𝐴𝑥 = 𝐴𝑥 − 𝑛𝐸𝑥 𝐴𝑥+𝑛 10.7 𝐴𝑥:𝑛| ̅̅̅ = 𝐴𝑥́ :𝑛| ̅̅̅ + 𝑛𝐸𝑥 = 𝐴𝑥 − 𝑛𝐸𝑥 𝐴𝑥+𝑛 + 𝑛𝐸𝑥 𝑞 10.8 𝐴𝑋 = ⁡⁡⁡⁡(𝐶𝐹𝑀) 𝑞+𝑖

𝑉𝑎𝑟(𝑍) =

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

𝑞 𝑞+2𝑖+𝑖 2

−(

𝑞 𝑞+𝑖

)

2

(CFM)

Formula Summary of ASM 2014

6 Table 10.2: EPV for insurance payable at EOY of death Type of Insurance CFM UDD 𝑞 𝑎𝜔−𝑥| ̅̅̅̅̅̅̅̅ Whole Life 𝑞+𝑖 2 𝑞 𝑎𝑛| ̅̅̅ n-year term 𝑛 (1 − (𝑣𝑝) ) 𝑞+𝑖 𝜔−𝑥 𝑛 𝑞 𝑣 𝑎 ̅̅̅̅̅̅̅̅̅̅̅̅̅̅ n-year deferred life 𝑛 𝜔−(𝑥+𝑛)| (𝑣𝑝) 𝑞+𝑖 𝜔−𝑥 (𝑣𝑝)𝑛 n-year pure endowment 𝑣 𝑛 (𝜔 − (𝑥 + 𝑛)) 𝜔−𝑥

Lesson 11 – Insurance: Continuous – Moments – Part I ∞ 11.1 𝐸(𝑍) = 𝐴̅𝑥 = ∫0 𝑣 𝑡 𝑓𝑥 (𝑡)𝑑𝑡 ∞ 11.2 𝐴̅𝑥 = ∫ 𝑒 −𝛿𝑡 𝑡𝑝𝑥 𝜇𝑥+𝑡 𝑑𝑡 0

2 11.3 𝑉𝑎𝑟(𝑍) = 𝐸(𝑍 2 ) − (𝐸(𝑍)) = 2𝐴̅𝑥 − (𝐴̅𝑥 )2 𝜇 11.4 𝑛|𝐴̅𝑥 = 𝑒 −𝑛(𝜇+𝛿) ⁡⁡ 𝜇+𝛿

𝜇 𝐴̅𝑥 = ⁡⁡⁡⁡(𝐶𝐹𝑀) 𝜇+𝛿 𝐴̅𝑥+𝑛 = 𝐴̅𝑥 ⁡⁡⁡(𝐶𝐹𝑀) 11.5 𝐴̅𝑥́ :𝑛| ̅̅̅ = 𝐴̅𝑥 (1 − 𝑛𝐸𝑥 ) =

11.6

̅

̅̅̅̅ 𝑛|𝐴𝑥́ :𝑚|

11.7 𝐴̅𝑥:𝑛| ̅̅̅ = 11.8

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

𝜇 𝜇+𝛿

(1 − 𝑒 −𝑛(𝜇+𝛿) )

𝜇𝑒 = 𝐴̅𝑥 ( 𝑛𝐸𝑥 − 𝑚+𝑛𝐸𝑥 ) = 𝜇 𝜇+𝛿

−𝑛(𝜇+𝛿)

𝜇+𝛿

(1 − 𝑒 −𝑚(𝜇+𝛿) )

(1 − 𝑒 −𝑛(𝜇+𝛿) ) + 𝑒 −𝑛(𝜇+𝛿)

̅ = 𝑛𝐸𝑥 𝐴̅𝑥+𝑛 −𝑛(𝜇+𝛿) 𝐴𝑥:𝑛| ́ = 𝑒 ̅̅̅ 𝑛|𝐴𝑥

Formula Summary of ASM 2014

7 Lesson 12 – Continuous – Moments – Part II

Lesson 13 – Insurance: Probabilities and Percentiles

Table 12.1 EPV for insurance payable at moment of death Type of Insurance CFM UDD 𝜇 𝑎̅𝜔−𝑥| ̅̅̅̅̅̅̅̅ Whole Life 𝜇+𝛿 𝜔−𝑥 𝜇 𝑎̅𝑛| ̅̅̅ n-year term (1 − 𝑒 −𝑛(𝜇+𝛿) ) 𝜇+𝛿 𝜔−𝑥 𝜇 n-year deferred life 𝑒 −𝛿𝑛 𝑎̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅ 𝜔−(𝑥+𝑛)| 𝑒 −𝑛(𝜇+𝛿) 𝜇+𝛿 𝜔−𝑥 n-year pure endowment 𝑒 −𝑛(𝜇+𝛿) 𝑒 −𝛿𝑛 (𝜔 − (𝑥 + 𝑛))

To calculate Pr⁡(𝑍 ≤ 𝑧) for continuous 𝑍, draw a graph of 𝑍 as a function of 𝑇𝑥 . Identify the parts of the graph that are below the horizontal line 𝑍 = 𝑧, and the corresponding 𝑡’s. Then calculate the probability of 𝑇𝑥 being in the range of those 𝑡’s.

𝜔−𝑥 ∞

𝑛!

12.1 Gamma 𝐸𝑃𝑉 = ∫0 𝑡 𝑛 𝑒 −𝑐𝑡 𝑑𝑡 = 𝑛 𝑐 +1 ∞

12.2 If n=1, 𝐸𝑃𝑉 = ∫0 𝑡𝑒 −𝑐𝑡 𝑑𝑡 = ∞

12.3 If n=2, 𝐸𝑃𝑉 = ∫0 𝑡 2 𝑒 −𝑐𝑡 𝑑𝑡 = 𝑢

1

1

𝑐2 2 𝑐3

12.4 ∫0 𝑡𝑒 −𝑐𝑡 𝑑𝑡 = 2 (1 − (1 + 𝑐𝑢)𝑒 −𝑐𝑢 ) 𝑐 ̅ ̅)̅̅̅ 12.5 (𝐼 𝑎 𝑢| =

𝑢 (𝑎̅̅̅̅ 𝑢| −𝑢𝑣 )

𝛿

⁡⁡

𝜇

For CFM, Pr(𝑍 ≤ 𝑧) = 𝑧 𝛿

For discrete 𝑍, identify 𝑇𝑥 and then identify 𝐾𝑥 + 1 corresponding to that 𝑇𝑥 . To calculate percentiles of continuous 𝑍, draw a graph of 𝑍 as a function of 𝑇𝑥 . Identify where the lower parts of the graph are, and how they vary as a function of 𝑇. For example, for whole life, higher 𝑇 leads to lower 𝑍. For 𝑛year deferred whole life, both 𝑇𝑥 < 𝑛 and higher 𝑇𝑥 lead to lower 𝑍. Write an equation for the probability 𝑍 is less than 𝑧 in terms of mortality probabilities expressed in terms of 𝑡. Set it equal to the desired percentile, and solve for 𝑡 or for 𝑒 𝑘𝑡 for any 𝑘. Then solve for 𝑧 (which is often 𝑣 𝑡 )

Variance If 𝑍3 = 𝑍1 + 𝑍2 , 𝑍1 &⁡𝑍2 are mutually exclusive, 𝑉𝑎𝑟(𝑍3 ) = 𝑉𝑎𝑟(𝑍1 ) + 𝑉𝑎𝑟(𝑍2 ) − 2𝐸(𝑍1 )𝐸(𝑍2 )

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

8 Lesson 14 – Insurance: Recursive Formulas, Varying Insurance Recursive Formulas 14.1 𝐴𝑥 = 𝑣𝑞𝑥 + 𝑣𝑝𝑥 𝐴𝑥+1 14.2 𝐴𝑥:𝑛| ̅̅̅ = 𝑣𝑞𝑥 + 𝑣𝑝𝑥 𝐴𝑥+1:𝑛−1| ̅̅̅̅̅̅̅ 14.3 𝐴𝑥́ :𝑛| = 𝑣𝑞 + 𝑣𝑝 𝐴 ̅̅̅ ̅̅̅̅̅̅̅ ́ :𝑛−1| 𝑥 𝑥 𝑥+1 14.4 𝑛|𝐴𝑥 = 𝑣𝑝𝑥 𝑛−1|𝐴𝑥+1 Applying whole life recursive equation twice: 𝐴𝑥 = 𝑣𝑞𝑥 + 𝑣 2 𝑝𝑥 𝑞𝑥+1 + 𝑣 2 2𝑝𝑥 𝐴𝑥+2 𝜇 ̅ ̅)𝑥 = 14.5 (𝐼 𝐴 (𝜇+𝛿)2 14.6 Continuously whole life insurance (CFM) 2𝜇 𝐸(𝑍 2 ) = (𝜇 + 2𝛿)3 ̅ ̅)𝑥́ :𝑛| ̅ 𝐴̅)𝑥́ :𝑛| 14.7 (𝐼 𝐴 ̅̅̅ + (𝐷 ̅̅̅ = 𝑛𝐴̅𝑥́ :𝑛| ̅̅̅ ̅ ̅ 14.8 (𝐼𝐴)𝑥́ :𝑛| ̅̅̅ + (𝐷𝐴)𝑥́ :𝑛| ̅̅̅ = (𝑛 + 1)𝐴̅𝑥́ :𝑛| ̅̅̅ 14.9 (𝐼𝐴)𝑥́ :𝑛| ̅̅̅ + (𝐷𝐴)𝑥́ :𝑛| ̅̅̅ = (𝑛 + 1)𝐴𝑥́ :𝑛| ̅̅̅ 𝑛 (𝐼𝐴)𝑥́ :𝑛| ̅̅̅ = ∑𝑘=1 𝑘 𝑘−1|𝐴̅𝑥́ :1| ̅ Recursive Formulas for Increasing and Decreasing Insurance 14.10 (𝐼𝐴)𝑥́ :𝑛| ̅̅̅ = 𝐴𝑥́ :𝑛| ̅̅̅ + 𝑣𝑝𝑥 (𝐼𝐴)𝑥+1 ̅̅̅̅̅̅̅ ́ :𝑛−1| 14.11 (𝐼𝐴)𝑥́ :𝑛| = 𝐴 + 𝑣𝑝 (𝐼𝐴𝐴) ̅̅̅ ̅ ̅̅̅̅̅̅̅ ́ :𝑛−1| 𝑥 𝑥́ :1| 𝑥+1 14.12 (𝐷𝐴)𝑥́ :𝑛| ̅̅̅ = 𝑛𝐴𝑥́ :1| ̅ + 𝑣𝑝𝑥 (𝐷𝐴)𝑥+1 ̅̅̅̅̅̅̅ ́ :𝑛−1| 14.13 (𝐷𝐴)𝑥́ :𝑛| ̅̅̅ = 𝐴𝑥́ :𝑛| ̅̅̅ + (𝐷𝐴)𝑥́ :𝑛−1| ̅̅̅̅̅̅̅

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

̅ Lesson 15 – Insurance: Relationships (𝐴𝑥 , 𝐴𝑚 𝑥 , 𝐴𝑥 ) Uniform Distribution of Deaths 𝑖 15.1 𝐴̅𝑥 = ( ) 𝐴𝑥 𝛿

𝑖 15.2 𝐴̅𝑥́ :𝑛| ̅̅̅ = ( ) 𝐴𝑥́ :𝑛| ̅̅̅ 𝛿

15.3

̅ = ( 𝑖 ) 𝑛|𝐴𝑥

𝑛|𝐴𝑥

𝛿

𝑖 15.4 𝐴̅𝑥:𝑛| ̅̅̅ = ( ) 𝐴𝑥́ :𝑛| ̅̅̅ + 𝐴𝑥:𝑛| ́ ̅̅̅ 𝛿

(𝑚)

15.5 𝐴𝑥 15.6

2

=

𝑖 𝑖 (𝑚)

𝐴𝑥

2𝑖+𝑖 𝐴̅𝑥 = 2𝛿

2

2

𝐴𝑥 ⁡

Claims Acceleration Approach 𝐴̅𝑥 = (1 + 𝑖)0.5 𝐴𝑥 0.5 ̅ 𝐴𝑥́ :𝑛| ̅̅̅ = (1 + 𝑖) 𝐴𝑥́ :𝑛| ̅̅̅ 0.5 ̅ 𝑛|𝐴𝑥 = (1 + 𝑖) 𝑛|𝐴𝑥 0.5 𝐴̅𝑥:𝑛| ̅̅̅ = (1 + 𝑖) 𝐴𝑥́ :𝑛| ̅̅̅ + 𝐴𝑥:𝑛| ́ ̅̅̅ (𝑚)

𝑚−1

= (1 + 𝑖) 2𝑚 𝐴𝑥 2 ̅ 𝐴𝑥 = (1 + 𝑖) 2𝐴𝑥 ⁡

𝐴𝑥

Formula Summary of ASM 2014

9 17.12 𝑎̈ 𝑥:𝑛| ̅̅̅ = 𝑎̈ 𝑥 − 𝑛𝐸𝑥 𝑎̈ 𝑥+𝑛 𝑛−1 17.13 𝑎̈ 𝑥:𝑛| ̅̅̅ = ∑𝑘=1 𝑎̈ 𝑘| ̅̅̅ 𝑘−1𝑝𝑥 𝑞𝑥+𝑘−1 + 𝑎̈ 𝑛| ̅̅̅ 𝑛−1𝑝𝑥 𝑛−1 𝑘 ∑ 17.14 𝑎̈ 𝑥:𝑛| = 𝑣 𝑝 ̅̅̅ 𝑘=0 𝑘 𝑥 ∞ 𝑘 ∑ 𝑎̈ = 𝑣 𝑘=𝑛 𝑘 𝑝𝑥 𝑛| 𝑥 17.15 Constant Force of Mortality 1+𝑖 𝑎̈ 𝑥 =

Lesson 17 – Annuities: Discrete, Expectation Annuities-Due Whole Life Annuities 1−𝐴𝑥 17.1 𝑎̈ 𝑥 = 𝑑 17.2 𝐴𝑥 = 1 − 𝑑𝑎̈ 𝑥

𝑞+𝑖

𝑛|𝑎̈ 𝑥 = 𝑛𝐸𝑥 𝑎̈ 𝑥

Temporary Life Annuities 1−𝐴𝑥:𝑛| ̅̅̅

17.3 𝑎̈ 𝑥:𝑛| ̅̅̅ = 𝑑 17.4 𝐴𝑥:𝑛| ̅̅̅ = 1 − 𝑑 𝑎̈ 𝑥:𝑛| ̅̅̅

Annuities-immediate

n-year Deferred Whole Life Annuity 17.5 0⁡⁡⁡𝐾𝑥 ≤ 𝑛 − 1

Whole life annuities 1 − 𝑖𝑎𝑥 𝐴𝑥 = 1+𝑖

17.6 𝑎̈ ̅̅̅̅̅̅̅̅̅ ̅̅̅ = 𝐾𝑥 +1| − 𝑎̈ 𝑛|

𝑣 𝑛 −𝑣 𝐾𝑥 +1 𝑑

⁡⁡⁡𝐾𝑥 ≥ 𝑛

Temporary life annuities 1 = 𝑖𝑎𝑥:𝑛| ̅̅̅ + 𝐴𝑥:𝑛| ̅̅̅ + 𝑖𝐴𝑥́ :𝑛| ̅̅̅ 17.16 𝐴𝑥́ :𝑛| ̅̅̅ = 𝑣𝑎̈ 𝑥:𝑛| ̅̅̅ − 𝑎𝑥:𝑛| ̅̅̅

n-year certain-and-life annuity-due 17.7 𝑎̈ ̅̅̅ 𝑛| =

1−𝑣 𝑛 𝑑

17.8 𝑎̈ ̅̅̅̅̅̅̅̅̅ 𝐾𝑥 +1| =

⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝐾𝑥 ≤ 𝑛 − 1

1−𝑣 𝐾𝑥 +1 𝑑

⁡⁡⁡⁡𝐾𝑥 ≥ 𝑛

Table 17.2: Actuarial notation for standard types of annuity-due Name Whole Life Temporary Life

Annual pmt at k 1⁡⁡⁡⁡⁡0 ≤ 𝑘 ≤ 𝐾𝑥 1⁡⁡⁡0 ≤ 𝑘 ≤ min(𝐾𝑥 , 𝑛 − 1) 0⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑘 > min(𝐾𝑥 , 𝑛 − 1)⁡⁡⁡⁡⁡⁡⁡⁡

Deferred Life Deferred Temporary Life Certainand-life

0⁡⁡⁡0 ≤ 𝑘 < n⁡or⁡k > K x 1⁡⁡⁡𝑛 ≤ 𝑘 ≤ K x ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡ 0⁡⁡⁡0 ≤ 𝑘 < n 1⁡⁡⁡𝑛 ≤ 𝑘 < min⁡(𝑛 + 𝑚, K x + 1) 0⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝐾𝑥 ≥ min⁡(𝑛 + 𝑚, K x + 1)

17.9

1⁡⁡⁡0 ≤ 𝑘 < max⁡(K x + 1, 𝑛) 0⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑘 ≥ max⁡(K x + 1, 𝑛)

PV 𝑎̈ ̅̅̅̅̅̅̅̅̅ 𝐾𝑥 +1| 𝑎̈ ̅̅̅̅̅̅̅̅̅ 𝐾𝑥 +1| ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝐾𝑥 < 𝑛 ⁡𝑎̈ ̅̅̅ 𝑛| ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝐾𝑥 ≥ 𝑛 0⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝐾𝑥 < 𝑛 𝑎̈ ̅̅̅̅̅̅̅̅̅ 𝐾𝑥 +1| − 𝑎̈ ̅̅̅ 𝑛| ⁡⁡⁡⁡𝐾𝑥 ≥ 𝑛 0⁡⁡⁡𝐾𝑥 < 𝑛 𝑎̈ ̅̅̅̅̅̅̅̅̅ 𝐾𝑥 +1| − 𝑎̈ ̅̅̅ 𝑛| ⁡⁡⁡⁡𝑛 ≤ 𝐾𝑥 ≤ 𝑛 + 𝑚 𝑎̈ ̅̅̅̅̅̅̅̅ 𝑛+𝑚| − 𝑎̈ ̅̅̅ 𝑛| ⁡⁡⁡⁡𝐾𝑥 ≥ 𝑛 + 𝑚 𝑎̈ ̅̅̅ 𝑛| ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝐾𝑥 < 𝑛 𝑎̈ ̅̅̅̅̅̅̅̅̅ 𝐾𝑥 +1| ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝐾𝑥 ≥ 𝑛

𝑎̈ 𝑥:𝑛| ̅̅̅̅̅̅̅ ̅̅̅ + 𝑛|𝑎̈ 𝑥 ̅̅̅| = 𝑎̈ 𝑛|

17.10 𝑛|𝑎̈ 𝑥 = 𝑛𝐸𝑥 𝑎̈ 𝑥+𝑛 17.11 𝑎̈ 𝑥 = 𝑎̈ 𝑥:𝑛| ̅̅̅ + 𝑛|𝑎̈ 𝑥

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Symbol 𝑎̈ 𝑥 𝑎̈ 𝑥:𝑛| ̅̅̅ 𝑛|𝑎̈ 𝑥

̅̅̅ 𝑛|𝑎̈ 𝑥:𝑛|

𝑎̈ ̅̅̅̅̅̅̅ ̅̅̅| 𝑥:𝑛|

Certain-and-life annuities 17.17 𝑎̈ 𝑥 = 𝑎𝑥 + 1 17.18 𝑎̈ 𝑥:𝑛| ̅̅̅ = 𝑎𝑥:𝑛−1| ̅̅̅̅̅̅̅ + 1 17.19 𝑎̈ 𝑥:𝑛| ̅̅̅ = 𝑎𝑥:𝑛| ̅̅̅ + 1 − 𝑛𝐸𝑥 17.20 𝑛|𝑎̈ 𝑥 = 𝑛|𝑎𝑥 + 𝑛𝐸𝑥 1/mthly annuities (𝑚)

1−𝐴𝑥

(𝑚)

=

(𝑚)

= 1 − 𝑑 (𝑚) 𝑎̈ 𝑥

(17.1)  𝑎̈ 𝑥

(17.2)  𝐴𝑥

17.21 𝑠̈ 𝑥:𝑛| ̅̅̅ =

𝑑 (𝑚)

(𝑚)

𝑎̈ 𝑥:𝑛| ̅̅̅ 𝑛𝐸𝑥

Official Definition: 𝑎̈ 𝑥 = ∑∞ 𝑛=1 𝑎̈ 𝑛 ( 𝑛−1𝑞𝑥 ) Alternative 𝑛 𝑎̈ 𝑥 = ∑∞ 𝑛=0 𝑣 𝑛𝑝𝑥

Formula Summary of ASM 2014

10 Lesson 18 – Annuities: Continuous, Expectation 18.1 𝑎̅ ̅̅̅̅ 𝑇𝑥 | = Name Whole Life Temporary Life Deferred Life Deferred Temporary Life Certainand-life

𝛿

Annual pmt at k 1⁡⁡⁡⁡𝑡 ≤ 𝑇 1⁡⁡⁡⁡𝑡 ≤ min⁡(𝑇, 𝑛) 0⁡⁡⁡⁡𝑡 > min⁡(𝑇, 𝑛)⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡

PV 𝑎̅ ̅̅̅ 𝑇| 𝑎̅ ̅̅̅ 𝑇| ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑇 ≤ 𝑛 ⁡𝑎̅̅̅̅ 𝑛| ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑇 > 𝑛 0⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑇 ≤ 𝑛 𝑎̅ 𝑇| ̅̅̅̅ ̅̅̅ − 𝑎 𝑛| ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑇 > 𝑛 0⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑇 ≤ 𝑛 𝑎̅ 𝑇| ̅̅̅̅ ̅̅̅ − 𝑎 𝑛| ⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑛 < 𝑇 ≤ 𝑛 + 𝑚 𝑎̅̅̅̅̅̅̅̅̅ ̅̅̅̅ 𝑛+𝑚| − 𝑎 𝑛| ⁡⁡⁡⁡𝑇 > 𝑛 + 𝑚 𝑎̅̅̅̅ 𝑛| ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑇 < 𝑛 𝑎̅̅̅̅̅̅̅̅̅̅ 𝐾𝑥 +1| ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑇 ≥ 𝑛

0⁡⁡⁡⁡𝑡 ≤ n⁡or⁡t > T 1⁡⁡⁡⁡𝑛 < 𝑡 ≤ 𝑇⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡ 0⁡⁡⁡⁡𝑡 ≤ n⁡or⁡t > T 1⁡⁡⁡⁡𝑛 < 𝑡 ≤ 𝑛 + 𝑚⁡𝑜𝑟⁡𝑡 ≤ 𝑇 0⁡⁡⁡⁡𝑇 > 𝑛 + 𝑚 1⁡⁡⁡⁡𝑡 ≤ max⁡(𝑇, 𝑛) 0⁡⁡⁡⁡𝑡 > max⁡(T, 𝑛) 1−𝐴̅

𝑥 18.2 𝑎̅𝑥 = 𝛿 18.3 𝐴̅𝑥 = 1 − 𝛿𝑎̅𝑥

18.4 𝑎̅𝑥 = 18.5 𝑎̅𝑥 = 18.6 𝑎̅𝑥 =

18.10

̅𝑥 𝑛|𝑎

=

=

𝛿 𝛿 1−𝐴̅𝑥:𝑛| ̅̅̅

̅𝑥 𝑛|𝑎

1−𝐴̅𝑥 𝛿

−

̅𝑥 𝑛|𝑎 ̅𝑥:𝑛| ̅̅̅ 𝑛|𝑎 𝑎̅̅̅̅̅̅̅̅ ̅̅̅| 𝑥:𝑛|

Whole Life and Temporary Life ∞

19.1 𝐸[𝑎̅2̅̅̅̅ ̅2𝑇| ̅̅̅ 𝑡𝑝𝑥 µ𝑥+𝑡 𝑑𝑡 𝑇𝑥 | ] = ∫0 𝑎 ∞ 1−𝑣 𝑡

19.2 𝐸[𝑎̅ 2̅̅̅̅ 𝑇𝑥 | ] = ∫0 ( 19.3 𝑉𝑎𝑟(𝑎̅ ̅̅̅ 𝑇𝑥 | ) =

𝛿2 2 2 ̅ 𝐴𝑥:𝑛| ̅̅̅ −(𝐴̅𝑥:𝑛| ̅̅̅ ) 𝛿2

19.4 𝑉𝑎𝑟(𝑌) = 19.5 2𝐴̅𝑥 = 1 − (2𝛿) 2𝑎̅𝑥

2(𝑎̅ − 2𝑎̅ )

𝑥 19.6 𝑉𝑎𝑟(𝑌) = 𝑥 − (𝑎̅𝑥 )2 𝛿 2 19.7 2𝐴̅𝑥:𝑛| ̅𝑥:𝑛| ̅̅̅ = 1 − (2𝛿) 𝑎 ̅̅̅

𝛿

= 𝑛𝐸𝑥 𝑎̅𝑥 =

2̅ 2(𝑎̅𝑥:𝑛| ̅̅̅ − 𝑎 ̅̅̅ ) 𝑥:𝑛| 𝛿

19.9 𝑉𝑎𝑟(𝑎̈ ̅̅̅̅̅̅̅̅ 𝐾𝑥 +1| ) = 2

𝜇+𝛿

=

− (𝑎̅𝑥:𝑛| ̅̅̅ )

2

⁡⁡⁡(𝐶𝐹𝑀)

𝐴̅𝑥:𝑛| ̅̅̅ −𝐴̅𝑥

𝑒 −𝑛(𝜇+𝛿) 𝜇+𝛿⁡

18.11 𝑎̅𝑥:𝑛| ̅𝑥 (1 − 𝑛𝐸𝑥 ) = ̅̅̅ = 𝑎

⁡⁡⁡(𝐶𝐹𝑀) 𝜇+𝛿

𝐴𝑥 −(𝐴𝑥 )2 𝑑2

𝐴𝑥:𝑛| ̅̅̅ −(𝐴𝑥:𝑛| ̅̅̅ )

2

2(𝑎̈ 𝑥 − 2𝑎̈ 𝑥 ) 𝑑

+ 2𝑎̈ 𝑥 − (𝑎̈ 𝑥 )2

Other Annuities 2 19.13 𝐸[𝑌𝑥̈ 2 ] = ∑∞ ̅̅̅ 𝑘−1|𝑞𝑥 𝑘=1 𝑎̈ 𝑘|

𝛿

1−𝑒 −𝑛(𝜇+𝛿)

2

19.10 𝑉𝑎𝑟(𝑌) = 𝑑2 19.11 2𝐴̅𝑥 = 1 − 2𝑑 2𝑎̈ 𝑥 = 1 − (2𝑑 − 𝑑 2 ) 2𝑎̈ 𝑥 19.12 𝑉𝑎𝑟(𝑌) =

1−𝐴̅𝑥:𝑛| ̅̅̅

2

) 𝑡𝑝𝑥 µ𝑥+𝑡 𝑑𝑡 𝛿 2 ̅ 𝐴𝑥 −(𝐴̅𝑥 )2

Note: 2𝑎̅𝑥 is 1st moment at twice FOI

∞ ∫0 𝑎̅𝑡|̅ 𝑡𝑝𝑥 𝜇𝑥+𝑡 𝑑𝑡 ∞ ∫0 𝑣𝑡 𝑡𝑝𝑥 𝑑𝑡 𝜇 𝛿 1−𝜇+𝛿 1 𝜇+𝛿

=

Symbol 𝑎̅𝑥 𝑎̅𝑥:𝑛| ̅̅̅

19.8 𝑉𝑎𝑟(𝑌) =

18.7 𝑎̅𝑥:𝑛| ̅̅̅ = 𝛿 18.8 𝐴̅𝑥:𝑛| ̅𝑥:𝑛| ̅̅̅ = 1 − 𝛿𝑎 ̅̅̅ 18.9

Lesson 19 – Variance

1−𝑣 𝑇𝑥

⁡⁡⁡(𝐶𝐹𝑀)

2 2 2 𝑛−1 2 ̈ 2̅̅̅ ] = ∑𝑛𝑘=1 𝑎̈ ̅̅̅ 19.14 𝐸[𝑌𝑥:𝑛| ̅̅̅ 𝑘−1|𝑞𝑥 + 𝑛−1𝑝𝑥 𝑎̈ ̅̅̅ 𝑘| 𝑘−1|𝑞𝑥 + 𝑛 𝑝𝑥 𝑎̈ ̅̅̅ 𝑛| = ∑𝑘=1 𝑎̈ 𝑘| 𝑛|

CFM: 𝑎̅𝑥 = 𝑎̅𝑥+𝑛 Relationships: 𝑎̅𝑥 = 𝑎̅𝑥:𝑛| ̅𝑥+𝑛 ̅̅̅ + 𝑛𝐸𝑥 𝑎 ̅ ̅̅̅ ̅𝑥 𝑎̅̅̅̅̅̅̅̅ ̅̅̅| = 𝑎 𝑛| + 𝑛|𝑎 𝑥:𝑛|

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

11 Lesson 20 – Annuities: Probabilities and Percentiles For the continuous whole life annuity PVRV Y, the relationship of 𝐹𝑌 (𝑦)to 𝐹𝑥 (𝑡) as follows: 𝐹𝑌 (𝑦) = Pr⁡(𝑌 ≤ 𝑦) 1 − 𝑣 𝑇𝑥 ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡= Pr⁡( ≤ 𝑦) 𝛿 𝑇𝑥 ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡= Pr⁡(𝑣 ≥ 1 − 𝛿𝑦) ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡= Pr⁡(𝑇𝑥 ln 𝑣 ≥ ln(1 − 𝛿𝑦)) ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡= Pr⁡(−𝑇𝑥 𝛿 ≥ ln(1 − 𝛿𝑦)) ln(1 − 𝛿𝑦) ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡= Pr (𝑇𝑥 ≤ ) 𝛿 ln(1 − 𝛿𝑦) ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡= 𝐹𝑥 (− ( )) 𝛿 To calculate a probability for an annuity, calculate the 𝑡 for which 𝑎̅𝑡 has the desired property. Then calculate the probability 𝑡 is in that range. To calculate a percentile of an annuity, calculate the percentile of 𝑇𝑥 , then calculate 𝑎̅ ̅̅̅ 𝑇𝑥 | Some adjustments may be needed for discrete annuities or non-whole-life annuities, as discussed in the lesson. If forces of mortality and interest are constant, then the probability that the present value of payments on a continuous whole life annuity will be greater than its expected present value is 𝜇

Pr(𝑎̅ ̅̅̅ 𝑇𝑥 |

𝜇 𝛿 > 𝑎̅𝑥 ) = ( ) 𝜇+𝛿

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Lesson 21 – Annuities: Varying, Recursive Formulas Whole Life 21.1 𝑎̈ 𝑥 = 𝑣𝑝𝑥 𝑎̈ 𝑥+1 + 1 (𝑚)

21.2 𝑎̈ 𝑥

1

= 𝑣 𝑚 1 𝑝𝑥 𝑎̈ 𝑚

(𝑚)

1 𝑚

𝑥+

+

1 𝑚

𝑎𝑥 = 𝑣𝑝𝑥 𝑎𝑥+1 + 𝑣𝑝𝑥 𝑎̅𝑥 = 𝑣𝑝𝑥 𝑎̅𝑥+1 + 𝑎̅𝑥:1| ̅ Temporary 𝑎̈ 𝑥:𝑛| ̅̅̅ = 𝑣𝑝𝑥 𝑎̈ 𝑥+1:𝑛−1| ̅̅̅̅̅̅̅ + 1 𝑎𝑥:𝑛| = 𝑣𝑝 𝑎 ̅̅̅ ̅̅̅̅̅̅̅ + 𝑣𝑝𝑥 𝑥 𝑥+1:𝑛−1| 𝑎̅𝑥:𝑛| ̅𝑥+1:𝑛−1| ̅𝑥:1| ̅̅̅ = 𝑣𝑝𝑥 𝑎 ̅̅̅̅̅̅̅ + 𝑎 ̅ Deferred life 𝑛|𝑎̈ 𝑥 = 𝑣𝑝𝑥 𝑛−1|𝑎̈ 𝑥+1 𝑛|𝑎𝑥 = 𝑣𝑝𝑥 𝑛−1|𝑎𝑥+1 ̅𝑥 = 𝑣𝑝𝑥 𝑛−1|𝑎̅𝑥+1 𝑛|𝑎 n-year certain and life 𝑎̈ ̅̅̅̅̅̅ ̅̅̅ = 1 + 𝑣𝑞𝑥 𝑎̈ ̅̅̅̅̅̅̅ ̅̅̅̅̅̅̅ 𝑛−1| + 𝑣𝑝𝑥 𝑎̈ ̅̅̅̅̅̅̅̅̅̅̅̅̅̅ 𝑥:𝑛| 𝑥+1:𝑛−1| 𝑎̅̅̅̅̅̅ = 𝑣 + 𝑣𝑞 𝑎 + 𝑣𝑝 𝑎 ̅̅̅̅̅̅̅ ̅̅̅̅̅̅̅̅̅̅̅̅̅̅ ̅̅̅ ̅̅̅̅̅̅̅ 𝑥 𝑥 𝑛−1| 𝑥:𝑛| 𝑥+1:𝑛−1| 𝑎̅̅̅̅̅̅̅ = 𝑎 ̅ + 𝑣𝑞 𝑎 ̅ + 𝑣𝑝 𝑎 ̅ ̅ ̅̅̅̅̅̅̅ ̅̅̅ ̅̅̅̅̅̅̅ 𝑥 𝑛−1| 𝑥 ̅̅̅̅̅̅̅̅̅̅̅̅̅̅ 1| 𝑥:𝑛| 𝑥+1:𝑛−1| Increasing/decreasing annuities (𝐼𝑎̈ )𝑥 = 𝑎𝑥 + 𝑣𝑝𝑥 (𝐼𝑎̈ )𝑥+1 ̅ ̅)𝑥 = 1 2 ⁡⁡⁡⁡(𝐶𝐹𝑀) (𝐼 𝑎 (𝛿+μ) ̅ ̅)𝑥:𝑛| ̅ 𝑎̅)𝑥:𝑛| (𝐼 𝑎 ̅𝑥:𝑛| ̅̅̅ + ⁡ (𝐷 ̅̅̅ = 𝑛𝑎 ̅̅̅ (𝐼𝑎)𝑥:𝑛| ̅̅̅ + ⁡ (𝐷𝑎)𝑥:𝑛| ̅̅̅ = (𝑛 + 1)𝑎𝑥:𝑛| ̅̅̅

Formula Summary of ASM 2014

12

Lesson 22 – Annuities: 1/m-thly Payments 𝑚−2 (𝑚) 22.1 𝑎̈ 𝑥 ≈ 𝑎̈ 𝑥 − (𝑚)

𝑎𝑥

≈ 𝑎𝑥 −

Lesson 24 – Premiums: Net Premiums for Discrete Insurances – Part I Future Loss = PV(Benefits) – PV(Gross Premiums) Equivalence Principle: EPV(Premiums) = EPV(Payments)  E[FutureLoss]=0 Net Premium  E[PVFB] = E[PVFP]

2𝑚 𝑚−2 2𝑚

Uniform Distribution of Deaths (UDD) 𝑖𝑑 𝛼(𝑚) = (𝑚) (𝑚)⁡ 𝑖

𝛽(𝑚) =

22.2 22.3 22.4

If P = net premium A = EPV of Benefits a = EPV of Annuity 𝐴 Equivalence Principle: 𝑃𝑎 = 𝐴 → 𝑃 =

𝑑 𝑖−𝑖 (𝑚)

𝑖 (𝑚) 𝑑 (𝑚) (𝑚) 𝑎̈ 𝑥 = 𝛼(𝑚)𝑎̈ 𝑥 − 𝛽(𝑚) (𝑚) 𝑎̈ 𝑥:𝑛| ̅̅̅ − 𝛽(𝑚)(1 − 𝑛𝐸𝑥 ) ̅̅̅ = 𝛼(𝑚)𝑎̈ 𝑥:𝑛| (𝑚) = 𝛼(𝑚) 𝑛|𝑎̈ 𝑥 − 𝛽(𝑚) 𝑛𝐸𝑥 𝑛|𝑎̈ 𝑥

𝑎

P Whole Life

Woolhouse’ Formula (𝑚)

22.5 𝑎̈ 𝑥

≈ 𝑎̈ 𝑥 − 1

𝑚−1 2𝑚

−

𝑚2 −1 12𝑚2

n-year term

(µ𝑥 + 𝛿)

22.6 µ𝑥 ≈ − (ln 𝑝𝑥−1 + ln 𝑝𝑥 ) (𝑚)

2

22.7 𝑎̈ 𝑥:𝑛| ̅̅̅ − ̅̅̅ ≈ 𝑎̈ 𝑥:𝑛| (𝑚)

≈

(1 − 𝑛𝐸𝑥 ) −

2𝑚 𝑚−1

𝑛|𝑎̈ 𝑥 − 2𝑚 𝑛𝐸𝑥 1 1 22.9 𝑎̅𝑥 ≈ 𝑎̈ 𝑥 − − (µ𝑥 + 𝛿) 2 12 1 1 22.10 𝑒̇𝑥 ≈ 𝑒𝑥 + − µ𝑥 2 12

22.8

𝑛|𝑎̈ 𝑥

𝑚−1

−

n-year deferred

𝑚2 −1

(µ𝑥+𝑛 + 𝛿)) 2 (µ𝑥 + 𝛿 − 𝑛𝐸𝑥

12𝑚 𝑚2 −1 𝐸 (µ 12𝑚2 𝑛 𝑥 𝑥+𝑛

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

+ 𝛿)

𝑛|𝐴𝑥

𝑎̈ 𝑥:𝑛| ̅̅̅ 𝑛|𝐴𝑥

𝑎̈ 𝑥

𝐴𝑥 𝑎̈ 𝑥 𝐴𝑥́ :𝑛| ̅̅̅ 𝑎̈ 𝑥:𝑛| ̅̅̅ P payable during deferral P payable for life

Formula Summary of ASM 2014

13

Lesson 25: Premiums: Net Premiums for Discrete Insurances – Part II

Lesson 27 – Premiums: Net Premiums for Fully Continuous Insurances Whole life

Fully discrete whole life 𝐴 1−𝑑𝑎̈ 𝑥 25.1 𝑃𝑥 = 𝑥 = =

27.1 𝑃 =

25.2 𝑃𝑥 =

𝑎̈ 𝑥 𝐴𝑥 𝑎̈ 𝑥

=

𝑎̈ 𝑥 𝐴𝑥

1−𝐴𝑥 = 𝑑

1

𝑎̈ 𝑥 𝑑𝐴𝑥

−𝑑

0𝐿

=

𝑥 )𝛿

1 𝑎̅𝑥

=

−𝛿

𝛿𝐴̅𝑥 1−𝐴̅𝑥

1−𝐴𝑥

n-year endowment 1 27.3 𝑃 = ̅ − 𝛿⁡

𝑎̈ 𝑥:𝑛| ̅̅̅ 𝑑𝐴𝑥:𝑛| ̅̅̅

27.4 ⁡𝑃 =

𝑎𝑥:𝑛| ̅̅̅ 𝛿𝐴̅𝑥:𝑛| ̅̅̅

1−𝐴̅𝑥:𝑛| ̅̅̅

⁡

1−𝐴𝑥:𝑛| ̅̅̅

Continuous Whole Life 𝜋 𝜋 27.5 0𝐿 = 𝑣 𝑇𝑥 (𝑏 + ) −

Net Premiums Whole Life and n-year term: (constant 𝑞𝑥 )⁡ 25.5 𝑃𝑥 = 𝑃𝑥:𝑛| ̅̅̅ = 𝑣𝑞𝑥 Future Loss at issue Fully discrete whole life 25.6

𝑎̅𝑥 𝐴̅𝑥

27.2 𝑃 = (1−𝐴̅

Fully discrete endowment 1 25.3 𝑃𝑥:𝑛| −𝑑 ̅̅̅ = 25.4 𝑃𝑥:𝑛| ̅̅̅ =

1−𝛿𝑎̅𝑥

𝑑

𝑑

𝐾𝑥 +1 = 𝑏𝑣 𝐾𝑥+1 − 𝜋(𝑎̈ ̅̅̅̅̅̅̅̅ − 𝐾𝑥 +1| ) = 𝑏𝑣 𝜋

𝜋

𝑑

𝑑

= 𝑣 𝐾𝑥+1 (𝑏 + ) −

𝜋

𝜋

𝑑

𝑑

25.7 𝐸[ 0𝐿] = 𝑏𝐴𝑥 − 𝜋𝑎̈ 𝑥 = 𝐴𝑥 (𝑏 + ) −

𝑑

𝜋 𝜋 27.6 𝐸[ 0𝐿] = 𝐴̅𝑥 (𝑏 + ) − 𝑑

𝜋(1−𝑣 𝐾𝑥 +1 ) 𝑑

n-year term insurance: 0𝐿

𝜋(1−𝑣 min⁡(𝐾𝑥 +1,𝑛) )

=𝑍− 𝑑 𝑏𝑣 𝐾𝑥 +1 ⁡⁡⁡𝐾𝑥 < 𝑛 𝑍={ 0⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝐾𝑥 ≥ 𝑛⁡

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

14

Lesson 28 – Premiums: Gross Reserves To calculate gross premium G by equivalence principle, equate G times an annuity-due for the premium payment period with the sum of 1. An insurance for the face amount plus settlement expenses 2. G times an annuity-due for the premium payment period of renewal percent of premium expense, plus the excess of the first year percentage over the renewal percentage 3. An annuity-due for the coverage period of the renewal per-policy and per-1000 expenses, plus the excess of first year over renewal expenses

Lesson 29 – Premiums: Variance of Future Loss, Discrete Net Future Loss Whole Life – 𝜋 is gross premium

𝜋 2

29.1 𝑉𝑎𝑟( 0𝐿) = ( 2𝐴𝑥 − (𝐴𝑥 )2 ) (𝑏 + ) 29.2 𝑉𝑎𝑟( 0𝐿) = 𝑏 2 (

2

𝐴𝑥 −(𝐴𝑥 )2

(1−𝐴𝑥 )2

𝑑

)

Endowment - 𝜋 is gross premium

2

𝜋 2

29.3 𝑉𝑎𝑟( 0𝐿) = ( 2𝐴𝑥:𝑛| ̅̅̅ − (𝐴𝑥:𝑛| ̅̅̅ ) ) (𝑏 + ) 29.4 𝑉𝑎𝑟( 0𝐿) = 𝑏 2 (

2

𝑑

2

𝐴𝑥:𝑛| ̅̅̅ −(𝐴𝑥:𝑛| ̅̅̅ ) (1−𝐴𝑥:𝑛| ̅̅̅ )

2

)⁡⁡(𝐸𝑃)

Whole Life – constant 𝑞𝑥 29.5 𝑉𝑎𝑟( 0𝐿) =

𝑞(1−𝑞) 𝑞+ 2𝑖

Gross Future Loss – Whole Life 𝐾𝑥 +1 + (𝑒𝑓 − 𝑒) − (𝐺 − 𝑒)⁡𝑎̈ 𝐾𝑥+1 0𝐿 = (𝑏 + 𝐸)𝑣 29.6 𝑉𝑎𝑟( 0𝐿) =( 2𝐴𝑥 − 𝐴2𝑥 )(𝑏 + 𝐸 + Where ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡b = Face⁡Amount ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡E = Settlement⁡Expenses⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡ ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡e = level⁡renewal⁡expenses

𝐺−𝑒 2 ) ⁡ 𝑑

For two fully discrete whole life or endowment insurance, one with 𝑏’ units and premium 𝜋′, and the other with 𝑏 units and premium 𝜋, the relative variance of net future loss of the first to the second is 𝑏′𝑑 + ⁡ 𝜋 ′ 2 ( ) 𝑏𝑑 + 𝜋

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

15 Lesson 30 – Premiums: Variance of Future Loss, Continuous Let 0𝐿 be net future loss. The following are for fully continuous whole life and endowment insurance with premium 𝜋 and face amount 𝑏. For whole ̅. life, drop 𝑛| 2 𝜋 30.1 𝑉𝑎𝑟( 0𝐿) = ( 2𝐴̅𝑥 − (𝐴̅𝑥 ) ) (𝑏 + ) 2

30.2 𝑉𝑎𝑟( 0𝐿) = 𝑏 (

2 2 ̅ 𝐴𝑥 −(𝐴̅𝑥 )

(1−𝐴̅𝑥 )

2

2

𝑑

) ⁡⁡𝜋 is net premium 2

30.4 𝑉𝑎𝑟( 0𝐿) = 𝑏 2 ( 30.5 𝑉𝑎𝑟( 0𝐿) =

𝜇 2𝛿+𝜇

𝑑

)⁡(𝐸𝑃)

⁡⁡⁡(𝐶𝐹𝑀⁡𝑊ℎ𝑜𝑙𝑒⁡𝐿𝑖𝑓𝑒)

Let 0𝐿 be gross future loss. The following is for fully continuous whole life and endowment insurance with premium 𝜋 and face amount 𝑏, and excess ̅. first year expenses payable at issue. For whole life, drop 𝑛| 2

𝐺−𝑒 30.6 𝑉𝑎𝑟( 0𝐿) = ( 2𝐴̅𝑥:𝑛| ) ̅̅̅ − (𝐴̅𝑥:𝑛| ̅̅̅ ) ) (𝑏 + 𝐸 +

2

𝛿

For two whole life or endowment insurance, one with 𝑏’ units and gross premium 𝜋′, and the other with 𝑏 units and premium 𝜋, the relative variance of future loss of the first to the second is (

𝑏′𝛿 + ⁡ 𝜋 ′ 2 ) 𝑏𝛿 + 𝜋

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

For level benefit or decreasing benefit insurance, the loss at issue decreases with time for whole life, endowment, and term insurances. To calculate the probability that the loss at issue is less than something, calculate the probability that survival time is greater than something. For level benefit or decreasing benefit deferred insurance, the loss at issue decreases during the deferral period, then jumps at the end of the deferral period and declines thereafter.

𝜋 2

30.3 𝑉𝑎𝑟( 0𝐿) = ( 2𝐴̅𝑥:𝑛| ̅̅̅ − (𝐴̅𝑥:𝑛| ̅̅̅ ) ) (𝑏 + ) 2 2 ̅ 𝐴𝑥:𝑛| ̅̅̅ −(𝐴̅𝑥:𝑛| ̅̅̅ ) 2 (1−𝐴̅𝑥:𝑛| ̅̅̅ )

Lesson 31 – Premiums: Probabilities & Percentiles of Future Loss

To calculate the probability that the loss at issue is greater than a positive number, calculate the probability that survival time is less than something minus the probability that survival time is less than the deferral period. To calculate the probability that the loss at issue is greater than a negative number, calculate the probability that survival time is less than something that is less than the deferral period, and add that to the probability that survival time is less than something that is greater than the deferral period minus the probability that survival time is less than the deferral period. For a deferred annuity with a single premium, the loss at issue is a negative constant during the deferral period. If regular premiums are payable during the deferral period, the loss at issue decreases until the end of the deferral period increases thereafter.

Formula Summary of ASM 2014

16 Lesson 32 – Premium: Special Topics

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Lesson 34 – Reserves: Prospective Net Premium Reserve

Formula Summary of ASM 2014

17 Lesson 35 – Reserves: Gross Premium Reserve & Expense Reserve

Lesson 36 – Reserves: Retrospective Formula

The gross premium reserve at time t is calculated as expected present value of future benefits and expenses minus future gross premiums at time t, given that the policy is in force.

Retrospective Formula – net premium reserve equals the accumulated value of net premiums minus the accumulated cost of insurance. Whole Life

The gross premium used in the calculation does not have to be determined by the equivalence principle. Even if it is, it may use different assumptions than used in calculating the gross premium reserve. If the gross premium is calculated using the equivalence principle on the same basis as the reserve, then the expense loading is the excess of the gross premium over the net premium. The expense reserve at time t is the expected present value of future expenses minus the expected present value of future expense loadings.

𝑡𝑉

=

𝑃𝑎̈ 𝑥:𝑡̅| − 𝐴𝑥́ :𝑡̅|

𝑡𝐸𝑥 𝑃𝑎̈ 𝑥:𝑡̅| - EPV at issue of premiums through t; 𝐴𝑥́ :𝑡̅| - EPV at issue of term insurance through time t

𝑃𝑥 = 𝑃𝑥́ :𝑛| ̅̅̅ + 𝑃𝑥:𝑛| ́ 𝑛𝑉 ̅̅̅ Premium Difference Formula 𝐴𝑡 – EPV of insurance at duration t 𝑎𝑡 – EPV of annuity at duration t 𝑃𝑡 – annual net premium at duration t 𝑡𝑉

= 𝐴𝑡 − 𝑃0 𝑎𝑡 → 𝑎𝑡 (𝑃𝑡 − 𝑃0 )

Paid up Insurance Formula 𝑃0 ) 𝑡𝑉 = 𝐴𝑡 (1 − 𝑃𝑡

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

18 Lesson 37 – Reserves: Special Formulas for Whole Life and Endowment Insurance 37.1 Annuity-ratio formula Endowment ̅̅̅̅̅̅ | − 𝑃𝑥:𝑛̅| 𝑎̈ 𝑥+𝑘:𝑛−𝑘 ̅̅̅̅̅̅ | 𝑘 𝑉 = 𝐴𝑥+𝑘:𝑛−𝑘 = 1 − 𝑑𝑎̈ 𝑥+𝑘:𝑛−𝑘 ̅̅̅̅̅̅ | − (

1

𝑎̈ 𝑥:𝑛 ̅|

= 1 − 𝑑𝑎̈ 𝑥+𝑘:𝑛−𝑘 ̅̅̅̅̅̅ | − =1− Whole Life

𝑎̈ 𝑥+𝑘:𝑛−𝑘 ̅̅̅̅̅̅̅|

𝑎̈ 𝑥+𝑘:𝑛−𝑘 ̅̅̅̅̅̅ | 𝑎̈ 𝑥:𝑛̅|

𝑎̈ 𝑥:𝑛 ̅|

𝑎̈ 𝑥+𝑘 𝑘𝑉 = 1 − 𝑎̈ 𝑥

37.2 Insurance-ratio formula Endowment 𝑘𝑉

=1−

=1− =

𝑎̈ 𝑥+𝑘:𝑛−𝑘 ̅̅̅̅̅̅̅| 𝑎̈ 𝑥:𝑛 ̅|

1−𝐴𝑥+𝑘:𝑛−𝑘 ̅̅̅̅̅̅̅|

1−𝐴𝑥:𝑛 ̅| 𝐴𝑥+𝑘:𝑛−𝑘 ̅̅̅̅̅̅̅| −𝐴𝑥:𝑛 ̅| 1−𝐴𝑥:𝑛 ̅|

− 𝑑)𝑎̈ 𝑥+𝑘:𝑛−𝑘 ̅̅̅̅̅̅ |

⁡⁡

+ 𝑑𝑎̈ 𝑥+𝑘:𝑛−𝑘 ̅̅̅̅̅̅ |

Lesson 38 – Reserves: Variance of Loss 38.1 Continuous Whole Life or Endowment

𝑃 2 𝑉𝑎𝑟( 𝑡𝐿|𝑇𝑥 ≥ 𝑡) = 𝑉𝑎𝑟(𝑍) (𝑏 + ) 𝛿 38.2 Continuous Premium Continuous Whole Life (EP) 2 2 ̅ 𝐴𝑥+𝑡 − (𝐴̅𝑥+𝑡 ) ⁡ 𝑉𝑎𝑟( 𝑡𝐿|𝑇𝑥 ≥ 𝑡) = (1 − 𝐴̅𝑥 )2 38.3 Annual Premium Annual Whole Life, integral k (EP)

2

2

𝐴𝑥+𝑘 − (𝐴𝑥+𝑘 ) ⁡ (1 − 𝐴𝑥 )2 38.4 Continuous Premium Continuous Endowment (EP) 𝑉𝑎𝑟( 𝑘𝐿|𝐾𝑥 ≥ 𝑘) = 2

𝑉𝑎𝑟( 𝑡𝐿|𝑇𝑥 ≥ 𝑡) =

(1 − 𝐴̅𝑥:𝑛| ̅̅̅ ) 38.5 Annual Premium Annual Endowment 2

𝑉𝑎𝑟( 𝑘𝐿|𝐾𝑥 ≥ 𝑘) =

2

𝐴̅𝑥+𝑡:𝑛−𝑡| ̅̅̅̅̅̅̅ − (𝐴̅𝑥+𝑡:𝑛−𝑡| ̅̅̅̅̅̅̅ ) ⁡ 2

2

𝐴𝑥+𝑘:𝑛−𝑘| ̅̅̅̅̅̅̅ − (𝐴𝑥+𝑘:𝑛−𝑘| ̅̅̅̅̅̅̅ ) ⁡ 2

(1 − 𝐴𝑥:𝑛| ̅̅̅ ) 38.6 Gross Future Loss of Whole Life and Endowment 2

𝑉𝑎𝑟( 𝐾𝐿𝑔 |𝑇𝑥 > 𝑘) =( 2𝐴𝑥+𝑘:𝑛−𝑘| ̅̅̅̅̅̅̅ − (𝐴𝑥+𝑘:𝑛−𝑘| ̅̅̅̅̅̅̅ ) )(𝑏 + 𝐸 +

𝐺−𝑒 2 ) 𝑑

Whole Life 𝐴𝑥+𝑘 −𝐴𝑥 ⁡⁡ 𝑘𝑉 = 1−𝐴𝑥

37.3 Premium-Ratio Formula Endowment 𝑘𝑉

=

𝑃𝑥+𝑘:𝑛−𝑘 ̅̅̅̅̅̅ | − 𝑃𝑥:𝑛̅| ⁡⁡ 𝑃𝑥+𝑘:𝑛−𝑘 ̅̅̅̅̅̅ | + 𝑑

Whole Life 𝑘𝑉

=

𝑃𝑥+𝑘 − 𝑃𝑥 ⁡⁡ 𝑃𝑥+𝑘 + 𝑑

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

19 Lesson 39: Reserves: Recursive Formulas General Recursion formulas 39.1

𝑘𝑉

=

(𝑘−1𝑉 +𝜋𝑘−1 )(1+𝑖)−𝑏𝑘 𝑞𝑘+1 𝑝𝑥+𝑘−1

39.2 ( 𝑘−1𝑉 + 𝜋𝑘−1 )(1 + 𝑖) = 𝑞𝑥+𝑘−1 (𝑏𝑘 − 𝑘𝑉 ) + 𝑘𝑉 1) For paid up insurance, the net premium reserve is the net single premium 2) For term insurance, the net premium reserve at expiry is 0 3) For endowment insurance, the net premium reserve right before maturity is the endowment benefit. 4) Deferred annuities and insurances have no benefit during the deferral period, so omit the 𝑏𝑘 𝑞𝑥+𝑘−1 term in equation (39.1) for recursions during the deferral period.

Lesson 40 – Reserves: Modified Reserves 1) For the full preliminary term method, the reserve at the end of the first year is 0. Thereafter, the reserve is the net premium reserve for an otherwise similar policy (with the same maturity date as the original policy) issued one year later. 2) For any modified reserve method, the expected present value of the valuation premiums must equal the expected present value of the benefits.

If FA + net premium reserve is paid at the end of the year of death: 39.3 𝑗𝑉 = ( 𝑗−1𝑉 + 𝜋𝑗−1 )(1 + 𝑖) − 𝑞𝑥+𝑗−1 𝐹𝐴 𝑘 𝑘−𝑗 39.4 𝑘𝑉 = 𝜋𝑠̈ 𝑘| ̅̅̅ − 𝐹𝐴 ∑𝑗=1 𝑞𝑥+𝑗−1 (1 + 𝑖) Deferred annuities and insurances 39.5 𝑘𝑉 = 𝜋𝑠̈ 𝑘| ̅̅̅ Gross Premium Reserve 39.6

𝑘𝑉

=

(𝑘−1𝑉 +𝐺𝑘−1 −𝑒𝑘−1 )(1+𝑖)−𝑞𝑥+𝑘−1 (𝑏𝑘 +𝐸𝑘 ) 1−𝑞𝑥+𝑘−1

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

20 Lesson 41: Reserves – Other Topics Valuation between premium dates

𝑘+𝑠𝑉

=

( 𝑘𝑉 +𝑃𝑘+1 )(1+𝑖)𝑠 −𝑏𝑘+1 𝑠𝑞𝑥+𝑘 𝑣 1−𝑠 𝑠𝑝𝑥+𝑘

Exact Formula with UDD: 41.2

𝑘+𝑠𝑉

=

+EPV of future net premiums, altered contract = EPV (Future Benefits, altered contract)

Reduced paid up face amount 𝑏 for whole life nonforfeiture option:

Exact Formula 41.1

𝑡𝐶𝑉

( 𝑘𝑉 +𝑃𝑘+1 )(1+𝑖)𝑠 −𝑠𝑏𝑘+1 𝑞𝑥+𝑘 𝑣 1−𝑠

41.9

𝑡𝑊𝑥

=

𝑡𝐶𝑉𝑥

𝐴𝑥+𝑡

Extended term duration n for whole life nonforfeiture option: 41.10 𝑡𝐶𝑉𝑥 = 𝐴𝑥+𝑡 ̅̅̅ ́ :𝑛|

1−𝑠𝑞𝑥+𝑘

Traditional approximation (linear interpolation): 41.3 𝑘+𝑠𝑉 = (1 − 𝑠)( 𝑘𝑉 + 𝑃𝑘+1 ) + 𝑠 𝑘+1𝑉

Pure endowment PE for extended term option on endowment insurance: 41.11 𝑡𝐶𝑉𝑥 = 𝐴𝑥+𝑡 ̅̅̅̅̅̅̅ + 𝑃𝐸⁡ 𝑛−𝑡𝐸𝑥+𝑡 ́ :𝑛−𝑡|

Thiele’s differential equation 𝑑 41.4 𝑡𝑉 = 𝛿𝑡 𝑡𝑉 + 𝐺𝑡 − 𝑒𝑡 − (𝑏𝑡 + 𝐸𝑡 − 𝑡𝑉 )𝜇[𝑥]+𝑡 𝑑𝑡

Numerical solutions with Euler’s method: Using derivatives at the lower end of each interval to go from 𝑡 + ℎ to 𝑡 41.5 𝑡+ℎ𝑉 − 𝑡𝑉 ≈ ℎ(𝛿𝑡 𝑡𝑉 + 𝐺𝑡 − 𝑒𝑡 − (𝑏𝑡 + 𝐸𝑡 + 𝑡𝑉 )𝜇[𝑥]+𝑡 ) 41.6

𝑡𝑉

=

𝑡+ℎ𝑉 −ℎ(𝐺𝑡 −𝑒𝑡 −(𝑏𝑡 +𝐸𝑡 )𝜇[𝑥]+𝑡 )

1+ℎ(𝜇[𝑥]+𝑡 +𝛿)

Using derivatives at the upper end of each interval to go from 𝑡 to 𝑡 + ℎ 41.7 41.8

𝑡𝑉

− 𝑡−ℎ𝑉 ≈ ℎ(𝛿𝑡 𝑡𝑉 + 𝐺𝑡 − 𝑒𝑡 − (𝑏𝑡 + 𝐸𝑡 + 𝑡𝑉 )𝜇[𝑥]+𝑡 )

𝑡−ℎ𝑉

= 𝑡𝑉 (1 − ℎ(𝜇[𝑥]+𝑡 + 𝛿)) + ℎ(−𝐺𝑡 + 𝑒𝑡 + (𝑏𝑡 + 𝐸𝑡 )𝜇[𝑥]+𝑡 )

Policy alterations Equivalence principle formula, with 𝑡𝐶𝑉 the value transferred: 𝑡𝐶𝑉 +EPV of future gross premiums, altered contract = EPV (Future Benefits and expenses, altered contract)

Equivalence principle formula on net basis:

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

21 Lesson 43 – Markov Chains: Discrete – Probabilities Single Life Mortality Insurance/Annuity model Alive (0)  Dead (1) Multiple decrements and pensions Double Indemnity model Alive (0)  Accidental Death (1) Alive (0)  Death from other causes (2) Pension model Active (0)  Terminated (1) Active (0)  Disabled (2) Active (0)  Retired (3) Active (0)  Dead (4) Permanent Disability Permanent disability model Healthy (0)  Disabled (1) Disabled (1)  Dead (2) Healthy (0)  Dead (2)

Continuing Care Retirement Community Common Shock model Both Alive (0)  Husband Alive (1) Both Alive (0)  Wife Alive (2) Both Alive (0)  Neither Alive (3) Husband Alive (1)  Neither Alive (3) Wife Alive (2)  Neither Alive (3) General Independent Living Unit (0)  Temporary Health Care (1) Independent Living Unit (0)  Permanent Health Care (2) Independent Living Unit (0)  Gone (3) Temporary Health Care (1)  Independent Living Unit (0) Temporary Health Care (1)  Gone (3) Temporary Health Care (1)  Permanent Health Care (2) Permanent Health Care (2)  Gone (3) Chapman – Kolmogorov Equations 𝑛 𝑖𝑗

𝑚𝑗

𝑖𝑚 𝑘 𝑝𝑥 = ∑ 𝑙 𝑝𝑥 𝑘−𝑙 𝑝𝑥+𝑙 𝑚=1

Disability income Disability income model Healthy (0)  Sick (1) Sick (1)  Healthy (0) Healthy (0)  Dead (2) Sick (1)  Dead (2) Multiple Lives Multiple lives model Both Alive (0)  Husband Alive (1) Both Alive (0)  Wife Alive (2) Husband Alive (1)  Neither Alive (3) Wife Alive (2)  Neither Alive (3)

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

22 Lesson 44 – Markov Chains: Continuous – Probabilities

Probability of moving two states with constant forces of transition:

Assumptions for Markov Chain models: 1) Probabilities of transitions are independent of length of time in state (Markov property) 2) Probability of 2 or more transitions in time ℎ is 𝑜(ℎ). 𝑖𝑗 3) 𝑡𝑝𝑥 is differentiable function of 𝑡 for all 𝑖, 𝑗.

44.10

44.1 44.2 44.3 44.4

𝑖𝑗 𝜇𝑥

= lim

ℎ→0

𝑖𝑗 ℎ𝑝𝑥 ⁡

ℎ

= 𝜇01 𝜇12 (

0∙ 𝑡

𝑒 −𝜇

(𝜇1∙ −𝜇0∙ )(𝜇2∙ −𝜇0∙ )

+

1∙ 𝑡

𝑒 −𝜇

(𝜇0∙ −𝜇1∙ )(𝜇2∙ −𝜇1∙ )

+

2∙ 𝑡

𝑒 −𝜇

(𝜇0∙ −𝜇2∙ )(𝜇1∙ −𝜇2∙ )

)

Kolmogorov’s forward equations: 𝑑 𝑖𝑗 𝑖𝑗 𝑗𝑘 𝑛 𝑖𝑘 𝑘𝑗 44.11 𝑡𝑝𝑥 = ∑𝑘=0( 𝑡𝑝𝑥 𝜇𝑥+𝑡 − 𝑡𝑝𝑥 𝜇𝑥+𝑡 ) 𝑑𝑡

𝑖𝑗

Definition of 𝜇𝑥

02 𝑡𝑝𝑥

𝑘≠𝑗

Euler approximate solution to Kolmogorov’s forward equations 𝑘𝑗 𝑖𝑗 𝑗𝑘 44.12 𝑡+ℎ𝑝𝑥𝑖𝑗 ≈ 𝑡𝑝𝑥𝑖𝑗 + ℎ ∑𝑛𝑘=0( 𝑡𝑝𝑥𝑖𝑘 𝜇𝑥+𝑡 − 𝑡𝑝𝑥 𝜇𝑥+𝑡 )

⁡

𝑘≠𝑗

𝑖𝑗 𝑖𝑗 ℎ 𝑝𝑥 ⁡ = ℎ𝜇𝑥 + 𝑜(ℎ) ̅ 𝑖𝑖 𝑖𝑖 ℎ 𝑝𝑥 ⁡ = ℎ 𝑝𝑥 ⁡ + 𝑜(ℎ) 𝑖𝑗 ̅ 𝑖𝑖 ℎ 𝑝𝑥 ⁡ = 1 − ℎ ∑𝑗≠1 𝜇𝑥

+ 𝑜(ℎ)

Probability of staying in any state continuously: 𝑡 𝑖𝑗 44.5 𝑡𝑝𝑥𝑖𝑖̅ = exp⁡(− ∫0 ∑𝑗≠1 𝜇𝑥+𝑠 𝑑𝑠) Probability of at least one direct transition from state 0 to state 1 by time t 1 ̅̅̅̅ 01 44.6 ∫0 𝑡𝑝𝑥00 𝜇𝑥+𝑡 𝑑𝑡 For permanent disability model: 𝑡 ̅̅̅̅ 01 ̅̅̅̅ 11 44.7 𝑡𝑝𝑥01 = ∫0 𝑠𝑝𝑥00 𝜇𝑥+𝑠 𝑡−𝑠𝑝𝑥+𝑠 𝑑𝑠 For permanent disability model with constant forces of transition: 44.8

01 𝑡𝑝0

={

𝜇 01 𝑒 −𝜇

12 𝑡

01 −𝜇 12 𝑡

𝜇 𝑒

(

01 +𝜇02 −𝜇12 )𝑡

(1−𝑒 −(𝜇

𝜇 01 +𝜇 02 −𝜇 12

)

)⁡⁡⁡⁡⁡𝜇 01 + 𝜇 02 ≠ 𝜇12

⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝜇

01

+𝜇

02

=𝜇

⁡

12

Probability of moving to next state with constant forces of transition: 44.9

01 𝑡𝑝0

= 𝜇 01 (

0∙ 𝑒 −𝜇 𝑡

𝜇 1∙−𝜇 0∙

+

1∙ 𝑒 −𝜇 𝑡

𝜇 0∙−𝜇 1∙

)

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

23 Lesson 45 – Markov Chains: Premiums and Reserves ∞ 45.1 𝑎̅𝑥𝑖𝑗 = ∫0 𝑒 −𝛿𝑡 𝑡𝑝𝑥𝑖𝑗 𝑑𝑡 𝑖𝑗 𝑘 45.2 𝑎̈ 𝑥𝑖𝑗 = ∑∞ 𝑘=0 𝑣 𝑘𝑝𝑥 ∞ 𝑖𝑗 −𝛿𝑡 𝑖𝑘 𝑘𝑗 45.3 𝐴̅𝑥 = ∫0 ∑𝑘≠𝑗 𝑒 𝑡𝑝𝑥 𝜇𝑥+𝑡 𝑑𝑡 45.4 45.5

𝑑 𝑉 (𝑖) 𝑑𝑡 𝑡 𝑡−ℎ𝑉

(𝑖)

(𝑖𝑗)

(𝑖)

= 𝛿𝑡 𝑡𝑉 (𝑖) − 𝐵𝑡 − ∑𝑛𝑗=0 𝜇𝑖𝑗𝑥+𝑡 (𝑏𝑡

≈ 𝑡𝑉 (𝑖) (1 − 𝛿𝑡 ℎ) +

𝑗≠𝑖 (𝑖) ℎ𝐵𝑡 +

𝑖𝑗

Lesson 46 – Multiple Decrement Models: Probabilities

+ 𝑡𝑉 (𝑗) − (𝑖𝑗)

ℎ ∑𝑛𝑗=0 𝜇𝑥+𝑡 (𝑏𝑡 𝑗≠𝑖

𝑡𝑉

(𝑖)

)

+ 𝑡𝑉 (𝑗) − 𝑡𝑉 (𝑖) )

Correspondence of Markov Chain notation with multiple decrement notation Markov Chain Notation Multiple Decrement Notation 0𝑗 (𝑗) 𝑡𝑝𝑥 𝑡𝑞𝑥 (𝜏) 0∙ 𝑡𝑝𝑥 𝑡𝑞𝑥 (𝜏) 00 𝑝 𝑡 𝑥 𝑡𝑝𝑥 Probability Formulas (𝜏) (𝑗) 46.1 𝑡𝑞𝑥 = ∑𝑛𝑗=1 𝑡𝑞𝑥 46.2 46.3 46.4 46.5

(𝜏) (𝑗) 𝑛 𝑡𝑝𝑥 = 1 − ∑𝑗=1 𝑡𝑞𝑥 (𝑗) (𝜏) (𝑗) 𝑘|𝑞𝑥 = 𝑘 𝑝𝑥 𝑞𝑥+𝑘 (𝑗) (𝜏) (𝑗) 𝑡−1 𝑡𝑞𝑥 = ∑𝑘=0 𝑘 𝑝𝑥 𝑞𝑥+𝑘 (𝑗) (𝜏) (𝑗) (𝜏) (𝑗) 𝑡+𝑢−1 𝑘 𝑝𝑥 𝑞𝑥+𝑘 𝑡|𝑢𝑞𝑥 = 𝑡𝑝𝑥 𝑢𝑞𝑥+𝑡 = ∑𝑘=𝑡

Life Table Formulas (𝑗) (𝜏) 𝑑𝑥 = ∑𝑚 𝑗=1 𝑑𝑥 (𝜏)

(𝜏)

(𝜏)

𝑙𝑥+1 = 𝑙𝑥 − 𝑑𝑥 (𝜏) (𝜏) (𝜏) ⁡𝑙𝑥+𝑘 = 𝑙𝑥 𝑘𝑝𝑥 ⁡ (𝑗) (𝜏) (𝜏) (𝑗) (𝜏) (𝑗) 𝑑𝑥+𝑘 = 𝑙𝑥 𝑘 𝑝𝑥 𝑞𝑥+𝑘 = 𝑙𝑥 𝑘|𝑞𝑥 ⁡⁡ (𝜏)

(𝜏)

𝑑𝑥+𝑘 = 𝑙𝑥

(𝜏) (𝜏) 𝑘 𝑝𝑥 𝑞𝑥+𝑘

(𝜏)

= 𝑙𝑥

(𝜏) 𝑘|𝑞𝑥

Expected present value of discrete life insurance (𝑗) (𝑗) (𝜏) 𝑚 𝑘 46.6 𝐴 = ∑∞ 𝑘=1 𝑣 𝑘−1𝑝𝑥 ∑𝑗=1 𝑞𝑥+𝑘−1 𝑏𝑘

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

24 Lesson 47 – Multiple Decrement Models: Forces of Decrement 𝑡 (𝑗) (𝜏) (𝑗) 47.1 𝑡𝑞𝑥 = ∫0 𝑠𝑝𝑥 𝜇𝑥+𝑠 𝑑𝑠 (𝑗)

𝑑

(𝑗) 𝑡𝑞𝑥 (𝜏) 𝑡𝑝𝑥

47.2 𝜇𝑥+𝑡 = 𝑑𝑡 47.3 47.4 47.5 47.6

𝑡 (𝜏) (𝜏) 𝑡𝑝𝑥 = exp⁡(− ∫0 𝜇𝑥+𝑠 𝑑𝑠) (𝑗) (𝜏) 𝜇𝑥+𝑡 = ∑𝑛𝑗=1 ⁡𝜇𝑥+𝑡 𝜕(𝐹(𝑡,𝑗)−𝐹(𝑡,𝑗−1)) (𝜏) (𝑗) 𝑓𝑇,𝐽 (𝑡, 𝑗) = = 𝑡𝑝𝑥 𝜇𝑥+𝑡 𝜕𝑡 ∞ (𝑗) 𝑓𝐽 (𝑗) = ∫0 𝑓(𝑡, 𝑗)𝑑𝑡 = ∞𝑞𝑥 (𝜏) (𝜏) 𝑓𝑇 (𝑡) = ∑𝑚 𝑗=1 𝑓(𝑡, 𝑗) = 𝑡𝑝𝑥 𝜇𝑥+𝑡

Fractional Ages Uniform distribution of decrement between integral ages in multiple decrement table: (𝑗) (𝑗) 𝑠𝑞𝑥 = 𝑠𝑞𝑥 ⁡⁡⁡0 ≤ 𝑠 ≤ 1 Constant Force of decrement between integral ages (𝑗) 𝑞𝑥 (𝑗) (𝜏) 𝑠 𝑞 = (1 − (𝑝𝑥 ) )⁡ 𝑠 𝑥 (𝜏) 𝑞𝑥

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Lesson 48 – Multiple Decrement Models: Associated Single Decrement Tables General Formulas and methods: 𝑡 (𝑗) ′(𝑗) 48.1 𝑡𝑝𝑥 = exp⁡(− ∫0 𝜇𝑥+𝑠 𝑑𝑠) 48.2

′(𝑗) 𝑡𝑞𝑥

𝑡

′(𝑗) (𝑗)

= ∫0 𝑠𝑝𝑥 𝜇𝑥+𝑠 𝑑𝑠

(𝜏) ′(𝑗) 48.3 ∏𝑗 𝑡𝑝𝑥 = 𝑡𝑝𝑥

From 46.1 (𝜏) (𝑗) 𝑛 𝑡𝑞𝑥 = ∑𝑗=1 𝑡𝑞𝑥 To go from multiple-decrement probabilities to associated singledecrement probabilities: (𝜏) 1. Calculate 𝑡𝑝𝑥 (j) 2. Calculate µx+t using equation (47.2) ′(𝑗) 3. Calculate 𝑡𝑝𝑥 using equation (48.1) 4. As a check, if you are calculating all of the rates, you can use (𝜏) equation (48.3) and see if you can reproduce 𝑡𝑝𝑥 To go from associated single-decrement probabilities to multipledecrement probabilities: (j) 1. Calculate µx for all j’s using one of the single-decrement formulas (𝜏) ′(𝑗) 2. Calculate 𝑡𝑝𝑥 by multiplying the 𝑡𝑝𝑥 ’s together, or by (j) summing µ ’s and exponentiating. (𝜏) 3. Calculate 𝑡𝑞𝑥 using equation (47.1)

Formula Summary of ASM 2014

25 Lesson 49 – Multiple Decrement Models: Relations between Multiple Decrement Rates and Associated Single Decrement Rates 49.1 49.2

(𝜏)

′(𝑗) 𝑠𝑝𝑥 (𝑗) 𝑞𝑥

(𝑗)

= ( 𝑠𝑝𝑥 )𝑞𝑥

=

(𝜏)

/𝑞𝑥

⁡

Lesson 50 – Multiple Decrements: Discrete Decrements

0≤𝑠≤1

′(𝑗)

(𝜏) ln 𝑝 𝑞𝑥 ( 𝑥(𝜏) ) ln 𝑝𝑥

49.3

(1) 𝑡𝑞𝑥

= 𝑞𝑥′(1) (1 −

49.4

(1) 𝑡𝑞𝑥

= 𝑞𝑥′(1) (1 −

′(2)

𝑞𝑥

)

2 ′(2) ′(3) 𝑞𝑥 +𝑞𝑥 2

+

′(2) ′(3) 𝑞𝑥

𝑞𝑥

3

)

Table 49.2: Summary of formulas for going between multipledecrement and associated single-decrement tables Assumptions 0 < 𝑠 + 𝑡 ≤ 1 Decrements are uniformly distributed in multiple decrement table There are two decrements uniformly distributed in the associated singledecrement tables There are three decrements uniformly distributed in the associated singledecrement tables

Formula ′(𝑗) 𝑠𝑝𝑥+𝑡

(𝜏)

(𝑗)

= ( 𝑠𝑝𝑥+𝑡 )𝑞𝑥

(𝜏)

/𝑞𝑥

′(𝑗)

(𝑗) (𝜏) ln 𝑝𝑥 𝑞𝑥 = 𝑞𝑥 ( ) (𝜏) ln 𝑝𝑥

′(2)

𝑞𝑥 ) 2 𝑠 (1) ′(1) ′(2) (1 − ( ) 𝑞𝑥 ) 𝑠𝑞𝑥 = 𝑠𝑞𝑥 2 ′(2) ′(3) ′(2) ′(3) 𝑞𝑥 + 𝑞𝑥 𝑞𝑥 𝑞𝑥 (1) ′(1) 𝑞𝑥 = 𝑞𝑥 (1 − + ) 2 3 2 3 𝑠 𝑠 (1) ′(1) ′(2) ′(3) ′(2) ′(3) (𝑞 + 𝑞𝑥 ) + (𝑞𝑥 𝑞𝑥 )) 𝑠𝑞𝑥 = 𝑞𝑥 (𝑠 − 2 𝑥 3 (1)

′(1)

𝑞𝑥 = 𝑞𝑥 (1 −

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

26 Lesson 51 – Multiple Decrements: Continuous Insurances

Lesson 53 – Multiple Lives: Joint Life Probabilities (𝑗)

If 𝑍 is the benefit random variable for an insurance paying 𝑏𝑡 at time 𝑡 upon occurrence of decrement 𝑗, then 51.1 51.2

53.1

𝑡𝑝𝑥𝑦

+ 𝑡𝑞𝑥𝑦 = 1 𝑡|𝑢𝑞𝑥𝑦 = 𝑡𝑝𝑥𝑦 𝑢𝑞𝑥+𝑡:𝑦+𝑡 = 𝑡𝑝𝑥𝑦 − 𝑡+𝑢𝑝𝑥𝑦 𝑡+𝑢𝑝𝑥𝑦

∞ (𝑗) (𝑗) (𝜏) 𝐸[𝑍] = ∫0 𝑣 𝑡 𝑡𝑝𝑥 ∑𝑚 𝑗=1 𝜇𝑥+𝑡 𝑏𝑡 𝑑𝑡 ∞ (𝑗) (𝑗) 2 (𝜏) 𝐸[𝑍 2 ] = ∫0 𝑣 2𝑡 𝑡𝑝𝑥 ∑𝑚 𝑗=1 𝜇𝑥+𝑡 (𝑏𝑡 ) ⁡ 𝑑𝑡

53.2

∗

(2)

Special formula for additional EPV of additional benefit 𝑏 = 𝑏 − 𝑏 (1) paid for the first 𝑛 years on secondary decrement with constant force 𝜇 (2) : 𝑏 ∗ 𝜇 (2) 𝑎̅𝑥:𝑛| ̅̅̅

𝑡𝑞𝑥𝑦

𝑡

0∙ 00 = 𝑡𝑝𝑥𝑦 = exp (− ∫0 𝜇𝑥+𝑡:𝑦+𝑡 𝑑𝑠)

= 1 − (1 − 𝑡𝑞𝑥 )(1 − 𝑡𝑞𝑦 ) = 𝑡𝑞𝑥 + 𝑡𝑞𝑦 − 𝑡𝑞𝑥 𝑡𝑞𝑦

If lives are independent, then 𝑡𝑝𝑥𝑦 = 𝑡𝑝𝑥 𝑡𝑝𝑦 𝜇𝑥+𝑡:𝑦+𝑡 = 𝜇𝑥+𝑡 + 𝜇𝑦+𝑡 𝑡

𝑡𝑝𝑥𝑦

= exp (− ∫ (𝜇𝑥+𝑡 + 𝜇𝑦+𝑡 ) 𝑑𝑠) 0

In particular: For CFM: 𝜇𝑥𝑦 = 𝜇𝑥 + 𝜇𝑦 For beta with parameters 𝛼𝑥 ,⁡𝛼𝑦 and 𝜔𝑥 − 𝑥 = 𝜔𝑦 − 𝑦, 𝜇𝑥+𝑡:𝑦+𝑡 is beta with parameters 𝛼𝑥 + 𝛼𝑦 and 𝜔𝑥 − 𝑥

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

27 Lesson 54 – Multiple Lives: Last Survivor Probabilities 54.1 𝑇𝑥 + 𝑇𝑦 = 𝑇𝑥𝑦 + 𝑇𝑥𝑦 ̅̅̅̅ 54.2 𝑡𝑝𝑥𝑦 = 𝑝 + 𝑝 ̅̅̅̅ 𝑡 𝑥 𝑡 𝑦 − 𝑡 𝑝𝑥𝑦

Lesson 55 – Multiple Lives: Moments Expected Value Formulas: ∞ 55.1 𝑒̇𝑥𝑦 = ∫0 𝑡𝑝𝑥𝑦 𝑑𝑡 ∞

𝑒̇𝑥𝑦 ̅̅̅̅ = ∫0

̅̅̅̅ 𝑑𝑡 𝑡𝑝𝑥𝑦 𝑛 ∫0 𝑡𝑝𝑥𝑦 𝑑𝑡

If independent, 54.3 𝑡𝑝̅̅̅̅ 𝑥𝑦 = 𝑡𝑝𝑥 + 𝑡𝑝𝑦 − 𝑡 𝑝𝑥 𝑡𝑝𝑦 54.4 𝑡|𝑢𝑞𝑥𝑦 ̅̅̅̅ = 𝑡𝑝𝑥𝑦 ̅̅̅̅ − 𝑡+𝑢𝑝𝑥𝑦 ̅̅̅̅

𝑒̇ 𝑥𝑦:𝑛| ̅̅̅ = 55.2 𝑒̇𝑥𝑦 ̅̅̅̅ = 𝑒̇𝑥 + 𝑒̇𝑦 − 𝑒̇𝑥𝑦

If independent,

For two independent uniform lives with 𝑎 = 𝜔𝑥 − 𝑥 ≤ 𝑏 = 𝜔𝑦 − 𝑦:

54.5 𝜇𝑥𝑦 ̅̅̅̅ (𝑡) =

𝑡𝑞𝑥 𝑡𝑝𝑦 𝜇𝑦+𝑡 + 𝑡𝑞𝑦 𝑡𝑝𝑥 𝜇𝑥+𝑡 ̅̅̅̅ 𝑡𝑝𝑥𝑦

𝑎

𝑎2

2

6𝑏

55.3 𝑒̇𝑥𝑦 = −

Variance and Covariance Formulas: ∞

2

55.4 𝑉𝑎𝑟(𝑇𝑥𝑦 ) = 2 ∫0 𝑡 𝑡𝑝𝑥𝑦 𝑑𝑡 − (𝑒̇𝑥𝑦 ) ⁡⁡ 𝐶𝑜𝑣(𝑇𝑥𝑦 , 𝑇𝑥𝑦 ̅̅̅̅ ) = 𝐶𝑜𝑣(𝑇𝑥 , 𝑇𝑦 ) + (𝑒̇𝑥 − 𝑒̇𝑥𝑦 )(𝑒̇ 𝑦 − 𝑒̇ 𝑥𝑦 ) If independent, 55.5 𝐶𝑜𝑣(𝑇𝑥𝑦 , 𝑇𝑥𝑦 ̅̅̅̅ ) = (𝑒̇ 𝑥 − 𝑒̇𝑥𝑦 )(𝑒̇𝑦 − 𝑒̇𝑥𝑦 )

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

28 Lesson 56 – Multiple Lives: Contingent Probabilities

Lesson 57 – Multiple Lives: Common Shock

Relationships for probabilities (double bar = 2) 56.1 ∞𝑞𝑥́ 𝑦 = ∞𝑞𝑥𝑦̿ 56.2 ∞𝑞𝑥́ 𝑦 + ∞𝑞𝑥𝑦́ = 1 56.3 ∞𝑞𝑥̿ 𝑦 + ∞𝑞𝑥𝑦̿ = 1 56.4 𝑛𝑞𝑥́ 𝑦 = 𝑛𝑞𝑥𝑦̿ + 𝑛𝑞𝑥 𝑛𝑞𝑦 56.5 𝑛𝑞𝑥́ 𝑦 + 𝑛𝑞𝑥𝑦́ = 𝑛𝑞𝑥𝑦 56.6 𝑛𝑞𝑥̿ 𝑦 + 𝑛𝑞𝑥𝑦̿ = 𝑛𝑞𝑥𝑦 ̅̅̅̅ 56.7 𝑛𝑞𝑥 = 𝑛𝑞𝑥́ 𝑦 + 𝑛𝑞𝑥̿ 𝑦

Both Alive (0)  Husband Alive (1) Both Alive (0)  Wife Alive (2) Both Alive (0)  Neither Alive (3) Husband Alive (1)  Neither Alive (3) Wife Alive (2)  Neither Alive (3) 𝜇 03 is constant 02 03 𝜇13 𝑥+𝑡 = 𝜇𝑥+𝑡:𝑦+𝑡 + 𝜇 01 23 𝜇𝑥+𝑡 = 𝜇𝑥+𝑡:𝑦+𝑡 + 𝜇 03

Probabilities when 𝜇𝑦+𝑡 = 𝑐𝜇𝑥+𝑡 for 0 ≤ 𝑡 ≤ 𝑛 56.8

𝑛𝑞𝑥́ 𝑦

=

1 𝑞 1+𝑐 𝑛 𝑥𝑦

Formula for contingent survival probabilities for uniform mortality Let (x) and (y) have lifetimes with uniform mortality with 𝜔𝑥 and 𝜔𝑦 respectively. Set 𝑎 = 𝜔𝑥 − 𝑥 and 𝑏 = 𝜔𝑦 − 𝑦. Then 𝑛 𝑎

56.9

𝑛𝑞𝑥́ 𝑦

−

= 1− {

𝑏 2𝑎

𝑛2

2𝑎𝑏 𝑎

2𝑏

⁡⁡⁡𝑛 ≤ min(𝑎, 𝑏)⁡⁡⁡⁡

⁡⁡⁡⁡⁡⁡⁡𝑛 ≥ 𝑎⁡𝑎𝑛𝑑⁡𝑏 ≥ 𝑎⁡⁡⁡⁡

⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑛 ≥ 𝑎⁡𝑎𝑛𝑑⁡𝑏 ≤ 𝑎

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

29 Lesson 58 – Multiple Lives: Insurances

Lesson 59 – Multiple Lives: Annuities

58.1 𝑎̅𝑥 + 𝑎̅𝑦 = 𝑎̅𝑥𝑦 + 𝑎̅𝑥𝑦 ̅̅̅̅ 58.2 𝐴̅𝑥 + 𝐴̅𝑦 = 𝐴̅𝑥𝑦 + 𝐴̅𝑥𝑦 ̅̅̅̅ 58.3 𝐴̅𝑥́ 𝑦 + 𝐴̅𝑥𝑦́ = 𝐴̅𝑥𝑦 𝐴̅𝑥̿ 𝑦 + 𝐴̅𝑥𝑦̿ = 𝐴̅𝑥𝑦 ̅̅̅̅ 𝐴̅𝑥́ 𝑦 + 𝐴̅𝑥̿ 𝑦 = 𝐴̅𝑥 ̅ ̅ 𝐴̅𝑥́ 𝑦 − 𝐴̅𝑥𝑦̿ = 𝐴̅𝑥 − 𝐴̅𝑥𝑦 ̅̅̅̅ = 𝐴𝑥𝑦 − 𝐴𝑦

59.1 𝑎𝑥|𝑦 = 𝑎𝑦 − 𝑎𝑥𝑦

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

To convert last-survivor annuities to joint annuities, use a formula of the following form: 𝑎̅𝑥𝑦 ̅𝑥 + 𝑎̅𝑦 − 𝑎̅𝑥𝑦 ̅̅̅̅ = 𝑎 Techniques for evaluating two-life annuities: 1. Two-annuity technique – Calculate the annuities for each life, ignoring the other life. Compare the amount paid in the joint status to the amounts paid by the individual annuities, and add the excess (or subtract the deficiency) from the sum of the individual annuities 2. Three-annuity technique – Calculate the annuities for each life when exactly one is alive and the joint status annuity, and add these three annuities up. 3. Reversionary annuity technique – If an annuity is paid to (𝑧2 ) after (𝑧1 ) expires, where (𝑧1 ) and (𝑧2 ) can be any combination of lives and certain statuses, then 𝑎𝑧1|𝑧2 = 𝑎𝑧2 − 𝑎𝑧1:𝑧2 𝑎̈ 𝑧1|𝑧2 = 𝑎̈ 𝑧2 − 𝑎̈ 𝑧1:𝑧2 𝑎̅𝑧1|𝑧2 = 𝑎̅𝑧2 − 𝑎̅𝑧1:𝑧2

Formula Summary of ASM 2014

30 Lesson 61 – Pension Mathematics

Lesson 62 – Interest rate risk: Replicating Cash Flows

Salary rate function 𝑠̅𝑥 is an index of the instantaneously earned salary.

Let 𝑣(𝑡) be the present value of 1 paid at time 𝑡. Then

Salary Scale 𝑠𝑥 is an index of the salary earned during the year of age (𝑥, 𝑥 + 1].⁡ The salary rate earned at age 𝑥 is approximated by 𝑠𝑥 − 0.5 k-year final average salary is the average of the salaries earned in the last 𝑘 years before retirement. Career average salary is the average of salaries earned during the entire career.

The spot rate 𝑦𝑡 is the annual interest rate on a 𝑡-year zero-coupon bond: 1 (1 + 𝑦𝑡 )𝑡 = 𝑣(𝑡) The forward rate 𝑓(𝑡, 𝑇) is the rate for a 𝑇 − 𝑡 year zero-coupon bond issued at time 𝑡. 𝑣(𝑡) (1 + 𝑦𝑇 )𝑇 𝑇−𝑡 (1 + 𝑓(𝑡, 𝑇)) = = 𝑣(𝑇) (1 + 𝑦𝑡 )𝑡

Replacement ratio is the ratio of the pension paid in the first year of retirement over the salary earned in the last year before retirement. Service table is a table showing, year by year, the number of decrements from each of four causes: death, withdrawal, disability retirement, age retirement. Accrual rate in a defined benefit plan is the percentage of the benefit basis that is paid annually as a pension for each year of service. For example, if the benefit basis is a 3-year final average, the accrual rate is 1.5%, and a retiree had 40 years of service, then the retiree’s benefit would be 60% of 3-year final average. Salary scale is defined from salary rate as follows: 1 61.1 𝑠𝑥 = ∫0 𝑠̅𝑥+𝑡 𝑑𝑡 Salary rate is approximated from salary scale as follows: 61.2 𝑠̅𝑥 = 𝑠𝑥−0.5

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

31 Lesson 63 – Interest rate risk: Diversifiable and non-diversifiable risk

Lesson 64 – Profit Tests: Asset Shares

If 𝐼 is some situation (such as interest rate or mortality rate) which is the same for all policyholders, then 𝑝 𝐸[𝑉𝑎𝑟(𝐿 |𝐼 )] 𝐿𝑝 63.1 𝑉𝑎𝑟 ( ) = 𝑉𝑎𝑟(𝐸[𝐿𝑡 |𝐼]) +

Recursive Formula: (𝑑) (𝑤) ( 𝑘−1𝐴𝑆 + 𝐺(1 − 𝑐𝑘−1 ) − 𝑒𝑘−1 )(1 + 𝑖) − 𝑏𝑘 𝑞𝑥+𝑘−1 − 𝑘𝐶𝑉 𝑞𝑥+𝑘−1 ⁡ 𝐴𝑆 = 𝑘 (𝑑) (𝑤) 1 − 𝑞𝑥+𝑘−1 − 𝑞𝑥+𝑘−1

The first summand is non-diversifiable risk. The second summand is diversifiable risk.

Accumulating gross premiums - The accumulated value of unit of gross premium after 𝑛 years is 𝑐𝑟 − 𝑐𝑓 (1 − 𝑐𝑟 )𝑠̈ 𝑥:𝑛| ̅̅̅ + 𝑛𝐸𝑥 ⁡⁡ Where 𝑐𝑓 is the percentage of first year premium expense and 𝑐𝑟 is the percentage of renewal premium expense.

𝑛

𝑛

q

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

32 Lesson 65 – Profit Tests: Profit for Traditional Products

Lesson 66 – Profit Tests: Participating Insurance

Precontract expenses – Expenses incurred initially before issue of the contract. Placed at the end of year 0 in profit tests.

If dividends are paid as cash, subtract them from the profit. No other change is needed to the profit test.

Profit (𝑑) (𝑑) 65.1 𝑃𝑟𝑘 = ( 𝑘−1𝑉 + 𝐺𝑘−1 − 𝑒𝑘−1 )(1 + 𝑖) − 𝑞𝑥+𝑘−1 (𝑏𝑘 + 𝐸𝑘 ) − (𝑤) (𝜏) 𝑞 (𝑤) (𝐶𝑉𝑘 + 𝐸𝑘 ) − 𝑝𝑥+𝑘−1 𝑘𝑉

If dividends are used to buy paid-up insurance and are paid in cash to deaths and withdrawals, divide the dividend by the net single premium for the insurance to compute the amount of paid-up insurance purchased. When dividends are used to buy paid-up insurance, the paid-up insurance is called a “reversionary bonus”.

Change in Reserve (𝜏) 65.2 𝛥 𝑘𝑉 = (1 + 𝑖) 𝑘−1𝑉 − 𝑝𝑥+𝑘−1 𝑘𝑉 Profit Vector – Vector of net profits per policy in force at the beginning of each year. Denoted by 𝑃𝑟𝑘 . Profit Signature – Vector of net profits per policy issued. In singledecrement model, equals 𝑝𝑥+𝑘−1 𝑃𝑟𝑘 . Denoted by 𝛱𝑘 .

If dividends are used to buy paid-up insurance but not paid to all deaths and withdrawals, increase the reversionary bonus by dividing the probability of receiving it. Reversionary bonuses increase the reserves, death benefits, and surrender benefits. Higher reserves lead to higher interest.

Profit Measures IRR – Internal rate of return. The rate at which the expected present value of the profit signature is 0. NPV – Net present value of profit signature components discounted at risk discount rate 𝑟, which is also called the ℎ𝑢𝑟𝑑𝑙𝑒⁡𝑟𝑎𝑡𝑒 𝑗 65.3 𝑁𝑃𝑉 = ⁡ ∑∞ 𝑗=1 𝛱𝑗 𝑣𝑟

The bonus rate is the bonus divided by the face amount related to the bonus. There are three types of reversionary bonus: Simple: the bonus is applied only to the original face amount. Compound: the bonus is applied to the original face amount + cumulative reversionary bonuses through the end of the previous year. Super-compound: separate bonus rates are computed for the original face amount and for the cumulative reversionary bonuses.

NPV (k) – Partial NPV. Net present value of profit signature components up to and including time 𝑘 discounted at the hurdle rate 𝑟. 𝑗 65.4 𝑁𝑃𝑉⁡(𝑘) = ⁡ ∑𝑘𝑗=1 𝛱𝑗 𝑣𝑟 Profit margin – Ratio of NPV to present value of premiums computed at the hurdle rate. DPP – Discounted payback period Zeroization of reserves means setting the reserves so that the profit is 0 in each year except the first.

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

33 Lesson 67 – Profit tests: Universal life Type B 67.1 𝐴𝑉𝑡 = (𝐴𝑉𝑡−1 + 𝑃𝑡 − 𝑒𝑡 − 𝑣𝑞 𝑞𝑥+𝑡−1 𝐹𝐴)(1 + 𝑖) 67.2 𝐴𝑉𝑡 = (𝐴𝑉𝑡−1 + 𝑃𝑡 − 𝑒𝑡 )(1 + 𝑖) − 𝑞𝑥+𝑡−1 𝐹𝐴 if 𝑖𝑞 = 𝑖 67.3 𝐶𝑂𝐼𝑡 = 𝑣𝑞 𝑞𝑥+𝑡−1 𝐹𝐴 Type A 67.4 𝐴𝑉𝑡 = (𝐴𝑉𝑡−1 + 𝑃𝑡 − 𝑒𝑡 − 𝐶𝑂𝐼𝑡 )(1 + 𝑖) 67.5 (𝐴𝑉𝑡−1 + 𝑃𝑡 − 𝑒𝑡 )(1 + 𝑖) − 𝑞𝑥+𝑡−1 𝐹𝐴 𝐴𝑉𝑡 = 1 − 𝑞𝑥+𝑡−1 67.6 if 𝑖𝑞 = 𝑖 (𝐴𝑉𝑡−1 + 𝑃𝑡 − 𝑒𝑡 − 𝑣𝑞 𝑞𝑥+𝑡−1 𝐹𝐴)(1 + 𝑖) 𝐴𝑉𝑡 = 1 − 𝑣𝑞 𝑞𝑥+𝑡−1 (1 + 𝑖) 67.7 𝑣𝑞 𝑞𝑥+𝑡−1 (𝐹𝐴 − (𝐴𝑉𝑡−1 + 𝑃𝑡 − 𝑒𝑡 )(1 + 𝑖)) 𝐶𝑂𝐼𝑡 = 1 − 𝑣𝑞 𝑞𝑥+𝑡−1 (1 + 𝑖) 67.8 𝑣𝑞𝑥+𝑡−1 (𝐹𝐴 − (𝐴𝑉𝑡−1 + 𝑃𝑡 − 𝑒𝑡 )(1 + 𝑖)) 𝐶𝑂𝐼𝑡 = 1 − 𝑞𝑥+𝑡−1 If corridor applies 67.10 (𝐴𝑉𝑡−1 + 𝑃𝑡 − 𝑒𝑡 )(1 + 𝑖) 𝐴𝑉𝑡 = 1 − 𝑣𝑞 𝑞𝑥+𝑡−1 (1 + 𝑖)(𝛾 − 1) 67.9 𝑖𝑞 = 𝑖 (𝐴𝑉𝑡−1 + 𝑃𝑡 − 𝑒𝑡 )(1 + 𝑖) 𝐴𝑉𝑡 = 1 − 𝑞𝑥+𝑡−1 (1 + 𝑖)(𝛾 − 1) 67.11 𝐶𝑂𝐼𝑡 =

Simplified version of account formulas Let the accumulated fund be (𝐴𝑉𝑡−1 + 𝑃𝑡 − 𝑒𝑡 )(1 + 𝑖). Then: For Type A, this amount pays 𝐴𝑉𝑡 to those who survive and the face amount to those who die. For Type B, this amount pays 𝐴𝑉𝑡 to everybody plus the face amount to those who die. If the corridor applies, this amount pays the 𝐴𝑉𝑡 to everybody plus (𝛾 − 1)𝐴𝑉𝑡 to those who die, where 𝛾 is the corridor factor.

𝑣𝑞 𝑞𝑥+𝑡−1 (𝛾 − 1)(𝐴𝑉𝑡−1 + 𝑃𝑡 − 𝑒𝑡 )(1 + 𝑖) 1 − 𝑣𝑞 𝑞𝑥+𝑡−1 (1 + 𝑖)(𝛾𝑡 − 1)

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014

34 Lesson 68: Profit Tests: Gain by Source 68.1 𝑃𝑟𝑘 (𝑑) (𝑑) (𝑤) (𝑤) ( 𝑘−1𝑉 + 𝐺𝑘−1 − 𝑒𝑘−1 )(1 + 𝑖) − 𝑞𝑥+𝑘−1 (𝑏𝑘 + 𝐸𝑘 ) − 𝑞𝑥+𝑘−1 (𝐶𝑉𝑘 + 𝐸𝑘 ) = ⁡⁡ (𝜏) 𝑝𝑥+𝑘−1 Total profit formula: ( 𝑘−1𝑉 + 𝐺𝑘−1 − 𝑒𝑘−1 )(1 + 𝑖) − 𝑞𝑥+𝑘−1 (𝑏𝑘 + 𝐸𝑘 ) − 𝑝𝑥+𝑘−1 𝑘𝑉 Total gain is the excess of actual profit over expected profit. Components of gain: (In each component, primed are actual, starred are assumed) Interest (𝑖 ′ − 𝑖 ∗ )( 𝑘−1𝑉 + 𝐺𝑘 − 𝑒𝑘 ) (𝑒𝑘∗ − 𝑒𝑘′ )(1 + 𝑖) + 𝑞𝑥+𝑘−1 (𝐸𝑘∗ − 𝐸𝑘′ ) Expense: ∗ ′ Mortality: (𝑞𝑥+𝑘−1 − 𝑞𝑥+𝑘−1 )(𝑏𝑘 + 𝐸𝑘 − 𝑘𝑉 ) Lapse:

(𝑤)∗

(𝑤)′

(𝑤)

(𝑞𝑥+𝑘−1 − 𝑞𝑥+𝑘−1 )(⁡ 𝑘𝐶𝑉 + 𝐸𝑘

− 𝑘𝑉 )

When computing the components of gain, assumptions should be changed sequentially. In the above order, actual interest would be used in expense and actual settlement expense would be used in mortality, but assumed annual expense would be used in interest. Different orders are possible.

Monica E. Revadulla EXAM MLC – Models for Life Contingencies

Formula Summary of ASM 2014