Actuarial mathematics 2

Actuarial mathematics 2 Life insurance contracts Edward Furman Department of Mathematics and Statistics York University January 3, 212 Edward Furman Actuarial mathematics MATH 328 1 / 45

Definition.1 (Life insurance.) Life insurance is a contract that is designed to reduce the financial impact of an untimely death. The payment is a single one. The payment can be made either upon death or thereafter. Question Given a life status (u), what is its expected future lifetime? If the insurance amount is one dollar: 1.) what is the r.v. representing the payment upon death of (u)? 2.) what is the r.v. representing the payment at the end of the year of death of (u)? Recall that, e.g., T(u) : Ω R u [, ]. Edward Furman Actuarial mathematics MATH 328 2 / 45

Solution The expected lifetime of (u) is E[T(u)]. The payment upon death is v T(u). The payment at the end of the year of death is v K(u)+1. Recall that the price is the expected loss, for identity utility fairness principle (or equivalence principle). Net premium for an insurance contract. The net premium for an immediately payable insurance is E[v T(u) ] for a life status (u). Also, the net premium for an insurance payable at the end of the year of death is E[v K(u)+1 ]. I.e., E[v T(u) ] v t tp u µ(u + t)dt v t d t p u : A u. R u R u Edward Furman Actuarial mathematics MATH 328 3 / 45

Also E[v K(u)+1 ] v k+1 k p u q u+k v k+1 k p u : A u. k R u k R u Example.1 (Whole life insurance.) Let (u) (x). Then R u [, ), and A x : E[v T(u) ] as well as v t tp x µ(x + t)dt, A x : E[v K(u)+1 ] v k+1 kp x q x+k. k Edward Furman Actuarial mathematics MATH 328 4 / 45

Edward Furman Actuarial mathematics MATH 328 5 / 45

Example.2 (n-year term insurance.) Let (u) ( 1 x : n). Thus R u [, n) { }, and as well as Proposition.1 A 1x:n : E[v T(u) ] n v t tp x µ(x + t)dt +, n 1 A 1x:n : E[v K(u)+1 ] v k+1 kp x q x+k. k We have that, for A 1x: (why?), and for all x, A 1x:n vq x + vp x A 1. x+1:n 1 Edward Furman Actuarial mathematics MATH 328 6 / 45

Proof. A 1x:n as required. n 1 n 1 v k+1 kp x q x+k vq x + v k+1 kp x q x+k k k1 n 1 vq x + vp x v k k 1p x+1 q x+k k1 n 2 vq x + vp x v k+1 kp x+1 q x+k+1 k vq x + vp x A 1, x+1:n 1 Edward Furman Actuarial mathematics MATH 328 7 / 45

Corollary.1 We have that A x vq x + vp x A x+1. Proposition.2 Under the UDD assumption for each year of age, we have that A u UDD i δ A u. Proof. We have that Then tp u µ(u + t) UDD q u, t 1, u, 1,... A u 1 v t tp u µ(u + t)dt + 1 v t tp u µ(u + t)dt Edward Furman Actuarial mathematics MATH 328 8 / 45

Proof (cont.) A u UDD 1 1 1 1 1 e δ δ v t tp u µ(u + t)dt + v t tp u µ(u + t)dt + v t+1 t+1p u µ(u + 1+t)dt v t+1 p u tp u+1 µ(u + 1+t)dt v t tp u µ(u + t)dt + vp u v t tp u+1 µ(u + 1+t)dt 1 v t q u dt + vp u A u+1 q u e δt dt + vp u A u+1 q u + vp u A u+1 1 v q u + vp u A u+1. δ Edward Furman Actuarial mathematics MATH 328 9 / 45

Proof (cont.) Thus we have that: A u UDD i δ vq u + vp u A u+1. The domain for the latter relationship is: u, 1,... and A. Further, UDD i ( ) i δ vq u + vp u δ vq u+1 + vp u+1 A u+2 A u i δ vq u + i δ v 2 p u q u+1 + v 2 p u p u+1 A u+2 i ( ) v p u q u+ + v 2 p u q u+1 + v 3 p u p u+1 q u+2 + + δ i v k+1 δ kp u q u+k i δ A u, k which completes the proof. Edward Furman Actuarial mathematics MATH 328 1 / 45

Example.3 (Pure endowment insurance.) Let (u) (x : 1 n). Then it only makes sense to speak of the discrete case. The r.v. representing the future payment is v K(x: 1 n), Thus A x: 1 n : v n kp x q x+k v n P[T(x) n] v n np x kn Example.4 (General endowment insurance.) Let (u) (x : n). Then by definition of T(x : n) we have that Note that A x:n : n v t tp x µ(x + t)dt + v n np x A 1x:n + A. x: n 1 v T(x:n) v T(1 x:n) + v T(x: 1 n). Edward Furman Actuarial mathematics MATH 328 11 / 45

Example.5 (Example.4 (cont.)) Do we have that Also, Note: T(x : n) T( 1 x : n)+t(x : 1 n)? A x:n : E[v K(1 x:n)+1 + v K(x: 1 n) ], i.e., n 1 A x:n v k+1 kp x q x+k + v n np x A 1x:n + A. x: n 1 k Of course, we have that, lim A 1 A x. n x:n Edward Furman Actuarial mathematics MATH 328 12 / 45

Proposition.3 The variance of the r.v. representing the future payment due to death of (u) is Var[v T(u) ] 2 A u (A u ) 2, where 2 A u is an insurance payable using δ 2δ. Proof. We have that E[(v T(u) ) 2 ] e δ2t tp u µ(u + t)dt e δ t t p u µ(u + t)dt, R u R u that is an insurance payable with a new force of interest. This completes the proof. Edward Furman Actuarial mathematics MATH 328 13 / 45

Corollary.2 The variance of the r.v. representing the present value of due to the pure endowment insurance is Var[v T(x: 1 n) ] v 2n np x ( n q x ). Proof. Note that for the pure endowment insurance we have that the second moment is E[v 2T(x: n) 1 ] v 2n np x. Thus the variance is v 2n np x (v n np x ) 2, as required. Edward Furman Actuarial mathematics MATH 328 14 / 45

Proposition.4 The variance of the general endowment insurance is Var[v T(x:n ] ( ) 2 2 A 1x:n A 1x:n + v 2n n p x n q x 2A 1x:n v n np x. Proof. Note that Var[v T(x:n ] Var[v T(1 x:n + v T(x: 1n ] Var[v T(1 x:n ]+Var[v T(x: 1n ]+2Cov[v T(1 x:n, v T(x: n) 1 ] Var[v T(1 x:n ]+Var[v T(x: 1n ] 2E[v T(1 x:n ]]E[v T(x: 1n ], which completes the proof. Edward Furman Actuarial mathematics MATH 328 15 / 45

Insurances mentioned above can be used as building blocks to create other forms of insurances. Example.6 (Deferred insurance.) An m year deferred n years term insurance provides for a benefit following the death of the insured only if the insured dies at least m years following policy issue and before the end of the policy. Thus we have that m na x : A 1u:m+n A 1u:m The continuous counterpart is then: m na u : A 1u:m+n A 1u:m m+n 1 km m+n m v k+1 kp u q u+k. v t tp u µ(u + t)dt. Edward Furman Actuarial mathematics MATH 328 16 / 45

Example.7 (Deferred whole life insurance.) Take n in the previous example, and get m A u A u A 1u:m v k+1 kp u q u+k, km with the continuous counterpart given then by: m A u A u A 1u:m m v t tp u µ(u + t)dt. Edward Furman Actuarial mathematics MATH 328 17 / 45

Note, that we have been interested in calculating expectations of transformed future life time r.v. s. These r.v. s can be more general. Example.8 (Annually increasing whole life insurance.) Consider the r.v. (K(u)+1)v K(u)+1. The present value is (IA) u : E[K(u)+1)v K(u)+1 ] (k + 1)v k+1 kp u q u+k. k Example.9 (Continuously increasing whole life insurance.) Consider the r.v. T(u)v T(u), then (IA) u : E[T(u)v T(u) ] tv t tp u µ(u + t)dt. Edward Furman Actuarial mathematics MATH 328 18 / 45

Example.1 (Annually increasing continuous whole life insurance.) Let the r.v. of interest be T(u)+1 V T(u). Then (IA) u : E[ T(u)+1 V T(u) ] t + 1 v t tp u µ(u + t)dt. Example.11 (m-thly per year increasing continuous whole life insurance.) Let the r.v. of interest be T(u)m+1 m v T(u). Then: [ ] T(u)m+1 (I (m) A) u : E v T(u) m Remark. tm+1 m v t tp u µ(u+t). Of course, for m, the just mentioned insurance reduces to the continuously increasing one. Edward Furman Actuarial mathematics MATH 328 19 / 45

Proposition.5 We have that (IA) u s A u ds. Proof. (IA) u ( t s s A u ds, ) ds v t tp u µ(u + t)dt v t tp u µ(u + t)dtds as required. Edward Furman Actuarial mathematics MATH 328 2 / 45

Figure: Auxiliary plot. Note: For the totally discrete counterpart, i.e., for (IA) u, we should again have (why?) (IA) u j A u A u + 1 A u + 2 A u +... j Edward Furman Actuarial mathematics MATH 328 21 / 45

Benefits must not be increasing. Example.12 (Annually decreasing n-year term insurance.) Let the r.v. of interest be (n K(u))v K(1 u:n+1. Then n 1 (DA) 1u:n : E[(n K(u))v K(1 u:n+1 ] (n k)v k+1 kp u q u+k. k Proposition.6 We have that n 1 (DA) 1u:n A 1u:n j. j Edward Furman Actuarial mathematics MATH 328 22 / 45

Proof. Note that Then (DA) 1u:n n k n k 1 j (1). n 1 (n k)v k+1 kp u q u+k k n 1 n k 1 k j n 1 n j 1 (1)v k+1 kp u q u+k n 1 (1)v k+1 kp u q u+k A 1u:n j, j k j as needed. Edward Furman Actuarial mathematics MATH 328 23 / 45

Figure: Annually decreasing 8-year term insurance. Edward Furman Actuarial mathematics MATH 328 24 / 45

Proposition.7 Let the actuarial present value r.v. be generally given by Z b(k(u)+1)v T(u). And assume the UDD approximation for each integer u. Then E[b(K(u)+1)v T(u) ] UDD i δ E[b(K(u)+1)v K(u)+1 ] for any life status (u). Proof (cont.) Recall that T K + J with J U(, 1) because of the UDD. We have also proven that under the UDD K and J are independent. Namely P[K k, J j] k p u q u+k j P[K k]p[j j]. Edward Furman Actuarial mathematics MATH 328 25 / 45

Proof (cont.) Also where E[b(K(u)+1)v T(u) ] E[b(K(u)+1)v K(u)+1+J(u) 1 ] E[b(K(u)+1)v K(u)+1 v J(u) 1 ] UDD E[b(K(u)+1)v K(u)+1 ]E[v J(u) 1 ] E[b(K(u)+1)v K(u)+1 ]E[(1+i) 1 J(u) ], E[(1+i) 1 J(u) ] 1 (1+i) 1 s ds 1 d(1+i) 1 s ln(1+i), which is E[(1+i) 1 J(u) ] i/δ. Edward Furman Actuarial mathematics MATH 328 26 / 45

Corollary.3 Thus, we easily have that with b(k + 1) 1. And also with b(k + 1) K + 1. A x UDD i δ A x, (IA) x UDD i δ (IA) x, Question: What about(ia) x? Edward Furman Actuarial mathematics MATH 328 27 / 45

Proposition.8 Under the UDD, we have that (IA) x UDD i δ ( (IA) x ( 1+i i 1 δ ) A x ). Proof. The r.v. corresponding to the price above is Z Tv T (K + J)v K+J (K + 1)v K+J +(J 1)v K+J (K + 1)v K+J (1 J)v K+1 v J 1 (K + 1)v K+1 (1+i) 1 J (1 J)v K+1 (1+i) 1 J. Also, note that 1 J U(, 1). Indeed P[1 J j] P[J 1 j] j. Recall that J and K are independent because of the UDD assumption and take expectations Edward Furman Actuarial mathematics MATH 328 28 / 45

Proof (cont.) E[Z] UDD E[(K + 1)v K+1 (1+i) 1 J ] E[v K+1 ]E[(1 J)(1+i) 1 J ] i δ (IA) x A x E[(1 J)(1+i) 1 J ] i δ (IA) x A x E[(J)(1+i) J ] i 1 δ (IA) x A x j d(1+i)j ln(1+i) ( ) i δ (IA) A 1 x x jd(1+i) j ln(1+i) ( i ) δ (IA) A 1 x x j(1+i) j 1 ln(1+i) (1+i) j dj A x i δ (IA) x ln(1+i) ( (1+i) i ln(1+i) ). Edward Furman Actuarial mathematics MATH 328 29 / 45

An insurance can be payable m-thly. Example.13 (m-thly payable whole life insurance.) The price of an m-thly payable whole life insurance is A (m) x : (v 1/m ) k+1 k k m p x 1 q m x+ k. m Note that this is not the expectation of the transformed K(x), but rather an expectation of a transformation of K (m) (x) K(x)m+J(x), where this time J(x) counts the number of total m-thly periods (x) was alive. Thus A (m) x : E[(v 1/m ) Km+J+1 ] E[(v) K+(J+1)/m ] m 1 v k+(j+1)/m P[K k, J j]. k j Edward Furman Actuarial mathematics MATH 328 3 / 45

Proposition.9 We have that under the UDD for integer ages, A (m) x where i (m) m((1+i) 1/m 1). Proof. We have that UDD i i (m) A x, E[v K+(J+1)/m ] m 1 k j v k+(j+1)/m kp x j/m 1/m q x+k A (m) x. In addition, under the UDD j/m 1/mq x+k (j+1)/m q x+k (j)/m q x+k UDD j + 1 m q x+k j m q x+k 1 m q x+k. Edward Furman Actuarial mathematics MATH 328 31 / 45

Proof (cont.) Thus A (m) x UDD k m 1 v k+1 j v (j+1)/m 1 kp x 1 m q x+k m 1 v k+1 kp x q x+k (1+i) 1 (j+1)/m 1 m k k k j m 1 v k+1 kp x q x+k (1+i) (v (1/m) ) j+1 1 m j v k+1 kp x q x+k (1+i) 1 1/m 1 v(m 1+1)/m v m 1 v 1/m k v k+1 kp x q x+k (1+i) 1 1 v m v 1/m (1 v 1/m ). Moreover Edward Furman Actuarial mathematics MATH 328 32 / 45

Proof. A (m) x UDD v k+1 kp x q x+k (1+i) 1 1 v m(1+i) 1/m 1 k v k+1 (1 v)(1+i) kp x q x+k i (m) k v k+1 i kp x q x+k i (m) i i (m) A x, k which completes the proof. Edward Furman Actuarial mathematics MATH 328 33 / 45

Figure: Insurances payed at the end of the year of death. Edward Furman Actuarial mathematics MATH 328 34 / 45

Figure: Immediately payed insurances. Edward Furman Actuarial mathematics MATH 328 35 / 45

Proposition.1 We have that as well as A x:y + A x:y A x + A y, A x:y + A x:y A x + A y. Proof. Because of the definition of say T(x : y) and T(x : y), we have that v T(x:y) + v T(x:y) v T(x) + v T(y). Taking expectations throughout then completes the proof. Example.14 An insurance that pays one dollar upon the death of the first of (x) and (y) is A x:y : E[v T(x:y) ] v t tp x:y µ((x : y)+t)dt. Edward Furman Actuarial mathematics MATH 328 36 / 45

Example.15 An insurance that pays one dollar upon the death of the last one of (x) and (y) is A x:y : E[v T(x:y) ] Example.16 v t tp x:y µ((x : y)+t)dt. An insurance that pays one dollar upon the death of (x) if he/she dies first is A 1x:y. Of course it is an expectation of v T(1 x:y). The latter is v T(x) if T(x) < T(y) and v if T(x) T(y). Thus A 1x:y : E[v T(1 x:y) ] v t f T(x),T(y) (t, s)dsdt t Edward Furman Actuarial mathematics MATH 328 37 / 45

Ex. (cont.) The latter expression is rewritten as ind A 1x:y : E[v T(1 x:y ] v t f T(x),T(y) (t, s)dsdt v t f T(y) T(x) (s t)ds f T(x) (t)dt t v t f T(y) (s)ds f T(x) (t)dt t v t F T(y) (t) f T(x) (t)dt v t tp y tp x µ(x + t)dt, t i.e., if (x) dies at any time t when (y) is alive, then v t dollars are payed. Check at home that if δ, then A 1x:y q 1x:y. Edward Furman Actuarial mathematics MATH 328 38 / 45

Example.17 An insurance payable upon death of (y) if it precedes the death of (x) is A x: 2 y : E[v T(x: 2 y) ] Proposition.11 t v t f T(x),T(y) (s, t)dsdt. Under independence of the future lifetimes of (x) and (y), we have that ind A 2 A y A 1. x: y x: y Proof. By definition A 2 x: y t v t f T(x) T(y) (s t)ds f T(y) (t)dt. Edward Furman Actuarial mathematics MATH 328 39 / 45

Proof. By independence and in actuarial notation A x: 2 y v t tq x tp y µ(y + t)dt v t (1 t p x ) tp y µ(y + t)dt, as required. Remark. The result holds for dependent future lifetimes too. Prove at home by looking at the corresponding r.v. s. Proposition.12 Under independence of the future lifetimes of (x) and (y), we have that ind A 2 s Ay sp x µ(x + s)ds. x: y Edward Furman Actuarial mathematics MATH 328 4 / 45

Proof. From the previous proposition, changing the order of integration and by substitution u t s, ind A x: 2 y v t tq x tp y µ(y + t)dt t v t sp x µ(x + s)ds tp y µ(y + t)dt s v t sp x µ(x + s) tp y µ(y + t)dtds v u+s sp x µ(x + s) u+s p y µ(y + u + s)duds v s sp y s p x µ(x + s) v s sp y s p x µ(x + s)a y+s ds. v u up y+s µ(y + u + s)duds Also, we can write the deferred insurance as Edward Furman Actuarial mathematics MATH 328 41 / 45

Proof. cont. s A y A y A 1y:s v u+s u+sp y µ(y + u + s)du v s sp y v u up y+s µ(y + u + s)du v s sp y A y+s. This completes the proof. Remark. We will often use the notation s E x : v s sp x. This is refereed to as the stochastic discount factor. Edward Furman Actuarial mathematics MATH 328 42 / 45

Example.18 Assume m sources of decrement and a whole life insurance contract due to each one. Also, let b(x + t) (j), j 1,...,m be the payment due to the decrement j. Then the overall price is A m j1 b(x + t) (j) v t tp τ x µ(j) (x + t)dt. Example.19 Let b(x + t) (1) t and b(x + t) (2) for all t >. Assume UDD for each year of death. Then A 1 k tv t tp τ x µ(1) (x + t)dt k k k+1 (k + s)v k+s k+sp τ xµ (1) (x + k + s)ds tv t tp τ x µ(1) (x + t)dt Edward Furman Actuarial mathematics MATH 328 43 / 45

Example.2 (Example. cont.) Then by the UDD and UDD A k k v k kpx τ k tp x µ(x + t) UDD q x, 1 v k+1 kp τ x q (1) x+k v k+1 kp τ x q(1) x+k v k+1 kpx τ q(1) k x+k (k + s)v s sp τ x+k µ(1) (x + k + s)ds 1 1 (k + s)v s 1 ds (k + s)(1+i) 1 s ds i (k + 1δ δ 1i ). Edward Furman Actuarial mathematics MATH 328 44 / 45

Remark Note that if b(x + k + s) (j) is more cumbersome than (k + s) than using, e.g., the midpoint rule 1 b(x + k + s) (j) (1+i) 1 s ds b(x + k + 1/2) (j) (1+i) 1 1/2. Edward Furman Actuarial mathematics MATH 328 45 / 45