Mathematics of Life Contingencies MATH 3281 Life annuities contracts Edward Furman Department of Mathematics and Statistics York University February 13, 2012 Edward Furman Mathematics of Life Contingencies MATH 3281 1 / 23
Definition 0.1 (Life annuity.) Life annuity is a series of payments made continuously or at equal intervals while a given life survives. Unlike in the context of life insurances, we have a series of payments. Discrete payments can be made not only at the end of the year, but also at the beginning of the year. Question. Given a life status (u) what is the expected future lifetime? If one dollar is payed at the beginning of every year while (u) is alive, what would the random present value of the contract be? What if the payment was made at the end of the year? Recall that, e.g., K(u) : Ω N u [0, ]. Edward Furman Mathematics of Life Contingencies MATH 3281 2 / 23
Solution. The expected future lifetime is E[K(u)] full years. The random present value if the payments are made at the beginning of each year is a K(u)+1 = 1+v+ +v K(u)+1 1. Also, the random present value of the payments if they are made at the end of each year is a K(u) = v + +v K(u). Once again, recall that we find the price by taking expectation operator (under the identity utility principle). Net premium for an annuity contract Fix a life status (u) (be careful with joint life statuses). The net premium for a continuously payable annuity is E[a T(u) ] = a t tp u µ(u + t)dt = a t d t p u := a u. R u R u Edward Furman Mathematics of Life Contingencies MATH 3281 3 / 23
Net premium for an annuity contract. cont. The net premium for an annuity due payable annually is E[ a K(u)+1 ] = a k+1 kp u q u+k = a k+1 k p u := a u. k N u k N u Last but not least, the net premium for an annuity certain payable annually is E[a K(u) ] = a k kp u q u+k = a k k p u := a u. k N u k N u Example 0.1 Whole life annuity. Let (u) = (x). Then R u = [0, ) and N u = {0, 1,.}. Thus Edward Furman Mathematics of Life Contingencies MATH 3281 4 / 23
Cont. Further And a x = E[ a K(x)+1 ] = a x = E[a K(x) ] = a x = E[a T(x) ] = 0 a k+1 k q x. a k k q x. a t tp x µ(x + t)dt. Recall that. a n = 1 v n, a n = 1 V n, a n = 1 v n, n 0. d i δ Edward Furman Mathematics of Life Contingencies MATH 3281 5 / 23
Proposition 0.1 Fix a life status (u), and let i, d, δ be the effective and discount interests and the force of interest, respectively. Then, a u = 1 A u δ, a u = 1 (1+i)A u, i a u = 1 A u d. Proof. We clearly have that [ ] a u = E [a T(u) = E 1 v T(u) δ ] = 1 A u. δ The same can be done for a u. For the last one [ ] [ ] 1 v K(u) 1 (1+i)v K(u)+1 a u = E = E, i i as required. Edward Furman Mathematics of Life Contingencies MATH 3281 6 / 23
Proposition 0.2 Fix (u). The variance of the random present value can be given by 2 A u (A u ) 2 Var[a T(u) ] = δ 2, and Var[ a 2 A u (A u ) 2 K(u)+1 ] = d 2, as well as Proof. Var[a K(u) ] = (1+i)2(2 A u (A u ) 2) i 2. The proof is straightforward and similar to the proof of the previous proposition. It is thus omitted. Edward Furman Mathematics of Life Contingencies MATH 3281 7 / 23
Proposition 0.3 Consider a whole life annuity due. Then a x = v k kp x = ke x. Proof. We have that a x = = = as required. k a k+1 kp x q x+k = v i P[K(x) = k] i=0 ( ) P[K(x) = k] v i = i=0 k=i P[K(x) > i 1]v i = P[K(x) i]v i i=0 ip x v i, i=0 i=0 Edward Furman Mathematics of Life Contingencies MATH 3281 8 / 23
Proposition 0.4 Let again (u) = (x). Then Here a = 0. Remark. Note that a a = 1/d. Proof. We have that as required. a x = 1+ a x = 1+E x a x+1. k+1p x v k+1 = 1+vp x v k kp x+1, Edward Furman Mathematics of Life Contingencies MATH 3281 9 / 23
Example 0.2 (n year term life annuity.) Let R u = [0, ) and N u = {0, 1,., }. And let (u) = (x : n). In the discrete case, K(x : n) is K(x) if K(x) < n and K(x : n) = n if K(x) n. The above is in fact the number of payments for the discretely payable annuity due. E[ a K(x:n)+1 ] = n 1 a k+1 k q x + a n k q x := a x:n. k=n (Recall the endowment insurance case.) Recall, however that the p.m.f. is kp x q x+k, k n 2 kp x:n q (x:n)+k = n 1p x, k = n 1 0, k n Thus E[ a n 2 K(x:n)+1 ] = a k+1 k q x + a n n 1 p x Edward Furman Mathematics of Life Contingencies MATH 3281 10 / 23
Example 0.2 (cont.) Therefore E[ a K(x:n)+1 ] = = = n 2 a k+1 k q x + n 1 a k+1 k q x + a n k q x k=n 1 a n k q x k=n n 1 a k+1 k q x + a n n p x Also E[a T(x:n) ] = n 0 a t tp x µ(x + t)dt + a n np x. Edward Furman Mathematics of Life Contingencies MATH 3281 11 / 23
Proposition 0.5 For (u) = (x : n), we have that Proof. n 1 a x:n = ke x. a x:n = = = n 1 k i=0 n 1 v i k q x + v i np x i=0 n 1 n 1 n 1 v i k q x + v i np x i=0 k=i i=0 ( n 1 n 1 ) n 1 v i k q x + n p x = ip x v i, i=0 k=i i=0 as required. Edward Furman Mathematics of Life Contingencies MATH 3281 12 / 23
Proposition 0.6 For a general (u) (be careful with the joint life statuses), we have that b a u = a b+1 b+1p u + a a a p u + kp u v k. k=a Proof. Write the annuity as, for a and b being any in {0, 1,., }. a u = b a k+1 k p u. k=a a u = ( a k+1 kp u b+1 a b k+1p u k=a = a b+2 b+1p u + a a+1 ap u + a k+1 ) b k+1p u v k+1. Edward Furman Mathematics of k=a Life Contingencies MATH 3281 13 / 23
Proof. Further a u = a b+2 b+1p u + a a+1 ap u + = a b+1 b+1p u + a a a p u + b+1 k=a+1 b kp u v k k=a kp u v k Edward Furman Mathematics of Life Contingencies MATH 3281 14 / 23
Example 0.3 (Deferred annuities.) We are sometimes interested in an annuity that starts to pay one dollar after n time units only. We thus have, e.g., Also, a x = n = k=n a u a u:n := n a u = a k+1 k q x.ȧ n k=n k q x = ( a k+n+1 a ) n k+n qx = kp u v k. k=n ( a k+1 a ) n k q x k=n ( v n (1 v k+1 ) ) k+n qx d = v n a k+1 n+kp x q x+n+k = n E x a x+n. Edward Furman Mathematics of Life Contingencies MATH 3281 15 / 23
Example (cont.) For the continuous case, we have that (make sure you can prove that), n a x = n E x a x+n, as well as n a x = n te x dt. Note that we can obtain, say the discretely payable deferred whole life annuity as an expectation of the r.v. {.ȧk(x)+1 K a (x) := K(x)+1 0, K(x) < n a K(x)+1 a n, K(x) n. Edward Furman Mathematics of Life Contingencies MATH 3281 16 / 23
Example 0.4 (n year certain and then life annuity) Consider an r.v. K (x) := { Then taking expectation, we have that a x:n = a n n q x + = a n n q x k=n = a n + v n np x + a n, K(x) < n a K(x)+1, K(x) n. a k+1 k q x =.ȧ n nq x a k+1 k p x ( a k+1 kp x n k=n ) k+1p x v k+1 k=n k+1p x v k+1 k=n a n + n a x. In a similar fashion a x:n = a n + n a x = a n + n E x a x+n. Edward Furman Mathematics of Life Contingencies MATH 3281 17 / 23
Proposition 0.7 The variance of the n year certain and whole life annuity is equal to the variance of the n years deferred annuity. Proof. Note that if we denote by K (x) the random present value of the deferred annuity and by k (x) the random present value of the n year certain whole life annuity, then Thus k (x) = a n + K (x). Var[k (x)] = Var[ a n ]+Var[K (x)]+cov[ a n, K (x)], as required. Edward Furman Mathematics of Life Contingencies MATH 3281 18 / 23
Recall. We know that s n := (1+i) n a n, for n 0. The above is an accumulated present value of an annuity at time n instead of time 0. Similarly, we have that s x:n := ( n E x ) 1 a x:n, that represents the actuarial accumulated value at the end of the term of an n-year temporary life annuity of 1 per year payable continuously while (x) survives. The benefit is available at age x + n if x survives till then Edward Furman Mathematics of Life Contingencies MATH 3281 19 / 23
Summary of discrete life annuities of 1 per annum payable at the beginning of each year (due) or at the end of each year (immediate) Edward Furman Mathematics of Life Contingencies MATH 3281 20 / 23
Useful relations between specific discrete annuities and insurances. Edward Furman Mathematics of Life Contingencies MATH 3281 21 / 23
Summary of continuous life annuities of 1 per annum payable continuously. Edward Furman Mathematics of Life Contingencies MATH 3281 22 / 23
Typical distributions for the present value rv T. Edward Furman Mathematics of Life Contingencies MATH 3281 23 / 23