Nonparametric estimation of actuarial present value of deferred life annuity.pdf Let x be the age of an individual and at the moment t = 0 payments start. The idea of the r-year deferred life annuity in accordance with [1. P. 174] is this: from the moment t + r = r, an individual starts receiving money once a year, which we take as a monetary unit, and payments are made only during the lifetime of an individual. It is known that the deferred life annuity is associated with the appropriate type of insurance. Thus, the average total cost of the present continuous r-year deferred life annuity is given by the following formula (see [1. P. 184]): 1 - A a (5) =-±-i, r' 5 да where A = j e-t fx (t)dt is the net premium (the expectation of the present value of an insured unit sum for r the deferred life insurance at age x), 5 is a force of interest, f ( t) = f (x +t) is a probability density of future x S (x) lifetime of an individual T = X -x [1. P. 62], f (x) is a probability density of lifetime of an individual X, S(x) = P(X > x) is a survival function. Introduce the random variable , -5T 1 - e x z(x) = ---, Tx > r. (1) о (2) Then, by averaging z(x) (1), we get the formula of the deferred life annuity (see [2-4]): a ©=e( z)=1 f 1 -^(имл r x( ) ( ) si S(x) where E is the symbol of the mathematical expectation, да Ф(x, s, r) = eSx j e~5tdF(t), x+r F(x) = P(X < x) = 1 - S(x) is a distribution function. Note that the whole li fe annuity ax (S) [2] is the special case of the deferred life annuity (2) at r = 0. 1. Construction of the Deferred Annuity Estimator Assume that we have a random sample X1,...,XN of N individuals' lifetimes. Using the empirical survival function 1 N Sn (X) = -11(X< > x), N i=1 where I(A) is the indicator of an event A, obtain the following estimator of (2): Л Ф n (x, 5, r ) Sn( X) £ exp(-5X< )I(Xt > x + r) aN (5)=-r 5 1- (3) 1 -- Sn(x) • N~= V 05x N Фn (x, 5, r) = - £ exp(-5X,< )I(X< > x + r). i=1 2. Bias and Mean Squared Error of the Estimator r^aN(5) Here we will obtain the principal term of the asymptotic MSE and the bias convergence rate of the estimator (3). Introduce the notation according to [5]: t^ = (tlN, t2N ,---,tSN )T is an s-dimensional vector with the components tjN = tjN (x) = tjN(x; X1,...,XN), j = 1, s, x e Ra, Ra is the a-dimensional Euclidean space; H(t):Rs ^R1 is a function, where t = t(x) = (tx(x),...,ts(x))T is an s-dimensional bounded vector function; N (ц, a) is the s-dimensional normally distributed random variable with a mean vector and covariance matrix 8H (z) , j = 1, s; ^ is the symbol of convergence in distri a = o(x); VH(t) = (H1(t),...,Hs(t))T, Hj(t) = bution; || x || is the Euclidean norm of a vector x; ^ is the set of natural numbers. Definition 1. The function H(t): Rs ^ R1 and the sequence {H(tN)} are said to belong to the class N s (t; Y), provided that: 1) there exists an s-neighborhood a = {z:| z -1 s, < = 1,sj, in which the function H (z) and all its partial derivatives up to order v are continuous and bounded; 2) for any values of variables Xj,..., X^ the sequence {H(tN )j is dominated by a numerical sequence C0dY, such that dN Тда, as N ^ да, and 0 < у 0 and S(t) is continuous at a point x, then 1) for the bias of (3), the following relation holds: e( a (s) - a (s)) 2) the MSE of (3) is given by the formula О(N-1); ь( raN (s)) Ф( x,25, r ) -Ф2( x, 5, r)/ S (x) N52 S 2( x) О (N "3/2 ). u2( a(5)) = E( a (5)- a (5))2 = N, + Proof. For the estimator ^ax (5) (3) in the notation of Theorem 1, we have: s = 2; tN = (tin,t2N)t = (*,5,r),SN(x))t; dN = N; t = (tl,t2)1 = (Ф(х,5,r),S(x))T; ( t л 1 - A- V t2 J Ф N (X, 5, r) SN (x) Ф( X, 5, r) S (x) = r\axN (5); H (t) = -5 = Aax (5); H (tN) = - 1- 1 - VH (t) = ( Hi(t), H2 (t) )T = T 1 Ф(х, 5, r) ф 0. 5S (x) 5S 2 (x) The sequence {H(tN)} satisfies the condition 1) of Theorem 1 with C0 = - (1 + e r), у = 0. Indeed. N e Z exp(-5Xt )I(Xt > X + r) 1 + -- N ZI (Xt > X) i=1 Ф N ( X, 5, r ) Ф N (X, 5, r) Sn (X) 1 |H (tN )| = 1 - < < - < SN (X) X + r) i=1 1 X) i=1 V. Further, the function H(t) satisfies the condition 1) in view of t2 = S(x) > 0. Also, this function satisfies the condition 2) due to Lemma 3.1 [6], as for all t е^ such inequalities hold: E{I(X >x)} = S(x) < 1, E{et&ce~tSX It (X > x+r)} < et5*e-S(x + r) = S (x + r) < 1 It is well known that SN (x) is the unbiased and consistent estimator of S(x). Show that ФN (x, 5, r) is the unbiased estimator of Ф^, 5, r) and calculate the variance of ФN (x, 5, r): e5x (N 1 ЕФN (X, 5, r) = - EIZ exp(-5X, )I(X, > x + r) \ = Ф(x, 5,r), (x,5,r) = e- ZD{l(Xt > x + r)e~5X' } = 1 (ф(x,2 5,r)-Ф2(x,5,r)). БФ N Considering that E(tN -1) = 0 and all the conditions of Theorem 1 are fulfilled, in accordance with (4) we get the order of the bias of ,aXN (5): E(HaN(S) - a(b)) -E[VH(t)(tN -1)] ■■O ( N-1). E( AaN(S) - (S)) Find the components of the covariance matrix o( ,ax (5)) = for the statistics ФN (x, 5, r) and Sn (x): = ND {Ф N (x, 5, r)} = Ф( x, 2 5, r) - Ф 2 (x, 5, r); о 22 = ND {Sn (x)} = S (X )(1 - S (x)); °12 = 021 = J11 = N cov(Sn (X), Ф N (X, 5, r)) = N (E {Sn (^Ф n (x, 5, r)} - E {Sn (x)} E {Ф n (x, 5, r)}) = (1 - S (x)^( x, 5, r). Using the previous results on the bias and covariance matrix, we obtain u2(a(5)) = E[VH(t)(tN -1)]2 + O(N-3/2) = H12(t)on + H2(t)022 + 2H1(t)H2(t)o12 + O(n-/2) = Ф( x,2 5, r) -Ф2 (x,5, r)/ S (x) (5) ■ + N52 S2 (x) O ( N-3/2). The proof of Theorem 2 is completed. 3. Asymptotic Normality of the Estimator We need the following two Theorems. Theorem 3 [7, Appendix 5]. If ^£2,...£N,... is a sequence of independent and identically distributed T 1 N I- s-dimensional vectors, E{

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