Random future lifetime LIF E IN S URAN CE P RODUCTS VALUATION IN - PowerPoint PPT Presentation

Random future lifetime LIF E IN S URAN CE P RODUCTS VALUATION IN R Katrien Antonio, Ph.D. Professor, KU Leuven and University of Amsterdam

The random future lifetime ( x ) denotes an individual aged x at this moment, with x ≥ 0 . The random variable T is the future lifetime of ( x ) . x Thus, age at death of ( x ) is x + T . x LIFE INSURANCE PRODUCTS VALUATION IN R

The life table in R Human Mortality Database (HMD, www.mortality.org ). life_table contains the period life table for males in Belgium of 2013. head(life_table, 10) age qx lx dx ex 1 0 0.00381 100000 381 77.95 2 1 0.00047 99619 47 77.24 3 2 0.00019 99572 19 76.28 4 3 0.00015 99553 15 75.30 5 4 0.00013 99538 13 74.31 6 5 0.00010 99525 10 73.32 7 6 0.00011 99514 11 72.32 8 7 0.00008 99504 8 71.33 9 8 0.00011 99496 11 70.34 10 9 0.00008 99485 8 69.34 LIFE INSURANCE PRODUCTS VALUATION IN R

Mortality rates and survival probabilities The one-year probability of dying q = Pr( T ≤ 1). x x q is the mortality rate at age x . x The one-year probability of surviving p = Pr( T > 1). x x Thus, p = 1 − q . x x LIFE INSURANCE PRODUCTS VALUATION IN R

Mortality rates of Belgian sportsmen in R Eden Hazard is a Belgian footballer who plays Eddy Merckx is a Belgian cyclist who won the for Chelsea and is 27 years old. T our de France 5 times and is 72. age <- life_table$age qx[age == 72] qx <- life_table$qx qx[age == 27] 0.02631 0.00062 qx[72 + 1] qx[27 + 1] 0.02631 0.00062 LIFE INSURANCE PRODUCTS VALUATION IN R

Picturing Belgian mortality rates q in R x plot(age, log(qx), main = "Log mortality rates (Belgium, males, 2013)", xlab = "Age x", ylab = expression(paste("Log mortality rate ", log(q[x]))), type = "l") LIFE INSURANCE PRODUCTS VALUATION IN R

LIFE INSURANCE PRODUCTS VALUATION IN R

The life expectancy The (complete) expected future lifetime of ( x ) is E [ T ] x For Eden Hazard who is 27 years old: ex <- life_table$ex ex[27 + 1] 51.74 For Eddy Merckx who is 72 years old: ex[72 + 1] 12.67 LIFE INSURANCE PRODUCTS VALUATION IN R

Picturing the life expectancy in R plot(age, ex, main = "Life expectancy (Belgium, males, 2013)", xlab = "Age x", ylab = expression(paste("Life expectancy E[", T[x], "]")), type = "l") LIFE INSURANCE PRODUCTS VALUATION IN R

Let's practice! LIF E IN S URAN CE P RODUCTS VALUATION IN R

Binomial experiments LIF E IN S URAN CE P RODUCTS VALUATION IN R Roel Verbelen, Ph.D. Statistician, Finity Consulting

The life table in R life_table contains the period life table for # Variables used in this video qx <- life_table$qx males in Belgium of 2013. px <- 1 - qx lx <- life_table$lx head(life_table, 10) dx <- life_table$dx age qx lx dx ex 1 0 0.00381 100000 381 77.95 2 1 0.00047 99619 47 77.24 3 2 0.00019 99572 19 76.28 4 3 0.00015 99553 15 75.30 5 4 0.00013 99538 13 74.31 6 5 0.00010 99525 10 73.32 7 6 0.00011 99514 11 72.32 8 7 0.00008 99504 8 71.33 9 8 0.00011 99496 11 70.34 10 9 0.00008 99485 8 69.34 LIFE INSURANCE PRODUCTS VALUATION IN R

A binomial experiment: surviving one year Focus on ℓ in life_table . x lx[0 + 1] 1e+05 LIFE INSURANCE PRODUCTS VALUATION IN R

A binomial experiment: surviving one year The number of survivors up to age x + 1 follows a BIN( ℓ , p ). x x lx[72 + 1] 73977 px[72+ 1] 0.97369 rbinom(n = 1, size = lx[72 + 1], prob = px[72 + 1]) 72022 LIFE INSURANCE PRODUCTS VALUATION IN R

A binomial experiment: surviving one year Now in a vectorized way! sims <- rbinom(n = length(lx), size = lx, prob = px) head(sims) 99637 99567 99553 99546 99525 99515 LIFE INSURANCE PRODUCTS VALUATION IN R

A binomial experiment: surviving k years The number of 1-year survivors follows a BIN( ℓ , p ). x x Expected value: ℓ = ℓ ⋅ p . x +1 x x The number of k -year survivors follows a BIN( ℓ , p ). x k x Expected value: ℓ = ℓ ⋅ p . x + k x k x Thus: ℓ x + k p = . k x ℓ x LIFE INSURANCE PRODUCTS VALUATION IN R

A binomial experiment: the number of deaths The number of deaths follows a BIN( ℓ , q ). x x Expected value: d x = ℓ ⋅ q x x = ℓ ⋅ (1 − p ) x x = ℓ − ℓ . x +1 x dx[72 + 1] 1946 lx[72 + 1] - lx[73 + 1] 1946 LIFE INSURANCE PRODUCTS VALUATION IN R

Survival probabilities in R ℓ 70 = Compute p . 5 65 ℓ 65 # Probability that (65) survives 5 more years lx[age == 70] / lx[age == 65] 0.9143957 # Alternatively lx[70 + 1] / lx[65 + 1] 0.9143957 LIFE INSURANCE PRODUCTS VALUATION IN R

Picturing survival probabilities in R # probability that (65) survives to age 65 + k k <- 0:45 plot(k, lx[65 + k + 1] / lx[65 + 1], pch = 20, xlab = "k", ylab = expression(paste(""[k], "p"[65]))) LIFE INSURANCE PRODUCTS VALUATION IN R

Let's practice! LIF E IN S URAN CE P RODUCTS VALUATION IN R

Calculating probabilities LIF E IN S URAN CE P RODUCTS VALUATION IN R Katrien Antonio, Ph.D. Professor, KU Leuven and University of Amsterdam

From one-year to multi-year survival probabilities LIFE INSURANCE PRODUCTS VALUATION IN R

From one-year to multi-year survival probabilities LIFE INSURANCE PRODUCTS VALUATION IN R

From one-year to multi-year survival probabilities LIFE INSURANCE PRODUCTS VALUATION IN R

From one-year to multi-year survival probabilities LIFE INSURANCE PRODUCTS VALUATION IN R

The multiplication rule Rewriting the survival probabilities : p = p ⋅ p . t + u x + u x u x t LIFE INSURANCE PRODUCTS VALUATION IN R

The multiplication rule Rewriting the survival probabilities : p = p ⋅ p . t + u x + u x u x t With k an integer we obtain: x = p ⋅ p p k −1 x +1 k x = p ⋅ p ⋯ p x +1 x + k −1 x k −1 ∏ = p x + l l =0 which is a product of one-year survival probabilities. LIFE INSURANCE PRODUCTS VALUATION IN R

Calculating survival probabilities in R ℓ 70 = p ⋅ p ⋅ p ⋅ p ⋅ p = Compute p . Compute p . 5 65 65 66 67 68 69 5 65 ℓ 65 # One-year survival probabilities # Alternatively (difference due to px <- 1 - life_table$qx rounding) px[(65 + 1):(69 + 1)] lx[70 + 1] / lx[65 + 1] 0.98491 0.98320 0.98295 0.98091 0.97935 0.9143957 # Probability that (65) survives 5 more years prod(px[(65 + 1):(69 + 1)]) 0.9144015 LIFE INSURANCE PRODUCTS VALUATION IN R

Cumulative product of survival probabilities in R for k = 1,2,3,4,5 . for k = 0,1,2,3,4,5 . Compute p Compute p 65 65 k k # One-year survival probabilities # Multi-year survival probabilities of (65 px[(65 + 1):(69 + 1)] c(1, cumprod(px[(65 + 1):(69 + 1)])) 1.0000000 0.9849100 0.9683635 0.9518529 0.98491 0.98320 0.98295 0.98091 0.97935 0.9336820 0.9144015 # Multi-year survival probabilities of (65 cumprod(px[(65 + 1):(69 + 1)]) 0.9849100 0.9683635 0.9518529 0.9336820 0.9144015 LIFE INSURANCE PRODUCTS VALUATION IN R

A deferred mortality probability Focus on a speci�c deferred mortality probability : ( x ) survives k whole years, but dies before reaching age x + k + 1 : k ∣ x = p ⋅ q . q x + k k x LIFE INSURANCE PRODUCTS VALUATION IN R

A deferred mortality probability in R = p ⋅ q Compute q . 5∣ 65 5 65 70 # 5-year deferred mortality probability of (65) prod(px[(65 + 1):(69 + 1)]) * qx[70 + 1] 0.02086664 d 70 = Compute q . 5∣ 65 ℓ 65 # Alternatively (difference due to rounding) dx[70 + 1] / lx[65 + 1] 0.02086817 LIFE INSURANCE PRODUCTS VALUATION IN R

Deferred mortality probabilities in R = p ⋅ q for k = 0,1,2,… Compute q k ∣ 65 65 65+ k k # Survival probabilities of (65) kpx <- c(1, cumprod(px[(65 + 1):(length(px) - 1)])) head(kpx) 1.0000000 0.9849100 0.9683635 0.9518529 0.9336820 0.9144015 # Deferred mortality probabilities of (65) kqx <- kpx * qx[(65 + 1):length(qx)] head(kpx) 0.01509000 0.01654649 0.01651060 0.01817087 0.01928053 0.02086664 LIFE INSURANCE PRODUCTS VALUATION IN R

Let's practice! LIF E IN S URAN CE P RODUCTS VALUATION IN R

Calculating life expectancies LIF E IN S URAN CE P RODUCTS VALUATION IN R Roel Verbelen, Ph.D. Statistician, Finity Consulting

The curtate future lifetime K = ⌊ T ⌋ , the number of whole years lived Compute x x Pr ( K = 5) = p ⋅ q = p − p by ( x ) in the future. . 65 5 65 70 5 65 6 65 Pr ( K = k ) = Pr ( k ≤ T < k + 1) # 5-year deferred mortality probability of (65) x x = p ⋅ q prod(px[(65 + 1):(69 + 1)]) * qx[70 + 1] x + k k x = p − p , k +1 k x x 0.02086664 # Alternatively prod(px[(65 + 1):(69 + 1)]) - prod(px[(65 + 1):(70 + 1)]) 0.02086664 LIFE INSURANCE PRODUCTS VALUATION IN R

The curtate life expectancy The expected value of K is called the curtate life expectancy : x ∞ ∑ E [ K ] x = k ⋅ Pr ( K = k ) x k =0 ∞ ∑ = k ⋅ ( p − p ) k +1 k x x k =0 = … ∞ ∑ = p . k x k =1 LIFE INSURANCE PRODUCTS VALUATION IN R

The life expectancy of a superhero Mr. Incredible is 35 years old and lives in Belgium. As an independent superhero he needs to take care of his �nancial planning . What is a good estimate of his curtate future lifetime ? Can you help? LIFE INSURANCE PRODUCTS VALUATION IN R