TRANSACTIONS JUNE, 1969 AN UPPER BOUND ON THE STOP-LOSS NET PREMIUM--ACTUARIAL NOTE NEWTON L. BOWERS, JR. ABSTRACT



Similar documents
(Quasi-)Newton methods

Chapter 4 Lecture Notes

ALGEBRA 2 CRA 2 REVIEW - Chapters 1-6 Answer Section

MATHEMATICAL METHODS OF STATISTICS

BEST METHODS FOR SOLVING QUADRATIC INEQUALITIES.

Algebra I Vocabulary Cards

Tim Kerins. Leaving Certificate Honours Maths - Algebra. Tim Kerins. the date

4. Continuous Random Variables, the Pareto and Normal Distributions

TOPIC 4: DERIVATIVES

THREE DIMENSIONAL GEOMETRY

Department of Mathematics, Indian Institute of Technology, Kharagpur Assignment 2-3, Probability and Statistics, March Due:-March 25, 2015.

Algebra 1 Course Title

CORRELATED TO THE SOUTH CAROLINA COLLEGE AND CAREER-READY FOUNDATIONS IN ALGEBRA

STAT 830 Convergence in Distribution

A Model of Optimum Tariff in Vehicle Fleet Insurance

Algebra Unpacked Content For the new Common Core standards that will be effective in all North Carolina schools in the school year.

Summary of Formulas and Concepts. Descriptive Statistics (Ch. 1-4)

WHERE DOES THE 10% CONDITION COME FROM?

THE NUMBER OF GRAPHS AND A RANDOM GRAPH WITH A GIVEN DEGREE SEQUENCE. Alexander Barvinok

Rolle s Theorem. q( x) = 1

PRE-CALCULUS GRADE 12

TABLE OF CONTENTS. 4. Daniel Markov 1 173

PUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include

MULTIVARIATE PROBABILITY DISTRIBUTIONS

WARM UP EXERCSE. 2-1 Polynomials and Rational Functions

8 Polynomials Worksheet

Transformations and Expectations of random variables

MATH4427 Notebook 2 Spring MATH4427 Notebook Definitions and Examples Performance Measures for Estimators...

Econ 132 C. Health Insurance: U.S., Risk Pooling, Risk Aversion, Moral Hazard, Rand Study 7

2.1 The Present Value of an Annuity

( ) is proportional to ( 10 + x)!2. Calculate the

Question: What is the probability that a five-card poker hand contains a flush, that is, five cards of the same suit?

(a) We have x = 3 + 2t, y = 2 t, z = 6 so solving for t we get the symmetric equations. x 3 2. = 2 y, z = 6. t 2 2t + 1 = 0,

Vocabulary Words and Definitions for Algebra

Section 3.1 Quadratic Functions and Models

Solution. Let us write s for the policy year. Then the mortality rate during year s is q 30+s 1. q 30+s 1

VERTICES OF GIVEN DEGREE IN SERIES-PARALLEL GRAPHS

Manual for SOA Exam MLC.

Stop Loss Reinsurance

Further Topics in Actuarial Mathematics: Premium Reserves. Matthew Mikola

SOME ASPECTS OF GAMBLING WITH THE KELLY CRITERION. School of Mathematical Sciences. Monash University, Clayton, Victoria, Australia 3168

Eigenvalues, Eigenvectors, Matrix Factoring, and Principal Components

APPLIED MATHEMATICS ADVANCED LEVEL

The Basics of Interest Theory

Math 370/408, Spring 2008 Prof. A.J. Hildebrand. Actuarial Exam Practice Problem Set 2 Solutions

Paper 2 Revision. (compiled in light of the contents of paper1) Higher Tier Edexcel

2010 Solutions. a + b. a + b 1. (a + b)2 + (b a) 2. (b2 + a 2 ) 2 (a 2 b 2 ) 2

Tail inequalities for order statistics of log-concave vectors and applications

4.3 Lagrange Approximation

Math Review. for the Quantitative Reasoning Measure of the GRE revised General Test

Probability and Statistics Prof. Dr. Somesh Kumar Department of Mathematics Indian Institute of Technology, Kharagpur

Exam Introduction Mathematical Finance and Insurance

Lecture 8: More Continuous Random Variables

Definition: Suppose that two random variables, either continuous or discrete, X and Y have joint density

Quantitative Methods for Finance

Math 370, Actuarial Problemsolving Spring 2008 A.J. Hildebrand. Practice Test, 1/28/2008 (with solutions)

Random variables, probability distributions, binomial random variable

SOLVING QUADRATIC EQUATIONS - COMPARE THE FACTORING ac METHOD AND THE NEW DIAGONAL SUM METHOD By Nghi H. Nguyen

vector calculus 2 Learning outcomes

Lecture 13 Linear quadratic Lyapunov theory

Institute of Actuaries of India Subject CT3 Probability and Mathematical Statistics

Lecture 6: Discrete & Continuous Probability and Random Variables

CITY UNIVERSITY LONDON. BEng Degree in Computer Systems Engineering Part II BSc Degree in Computer Systems Engineering Part III PART 2 EXAMINATION

MATH 21. College Algebra 1 Lecture Notes

Higher Education Math Placement

Chapter 5 Discrete Probability Distribution. Learning objectives

Math 370/408, Spring 2008 Prof. A.J. Hildebrand. Actuarial Exam Practice Problem Set 5 Solutions

1 if 1 x 0 1 if 0 x 1

3.2 The Factor Theorem and The Remainder Theorem

6.4 Logarithmic Equations and Inequalities

minimal polyonomial Example

H/wk 13, Solutions to selected problems

Section 13.5 Equations of Lines and Planes

3. Let A and B be two n n orthogonal matrices. Then prove that AB and BA are both orthogonal matrices. Prove a similar result for unitary matrices.

**BEGINNING OF EXAMINATION**

5. Continuous Random Variables

UNIT I: RANDOM VARIABLES PART- A -TWO MARKS

The Heat Equation. Lectures INF2320 p. 1/88

( 1) = 9 = 3 We would like to make the length 6. The only vectors in the same direction as v are those

3.6 The Real Zeros of a Polynomial Function

Modern Actuarial Risk Theory

ALGEBRA 2: 4.1 Graph Quadratic Functions in Standard Form

South Carolina College- and Career-Ready (SCCCR) Pre-Calculus

THE DYING FIBONACCI TREE. 1. Introduction. Consider a tree with two types of nodes, say A and B, and the following properties:

Undergraduate Notes in Mathematics. Arkansas Tech University Department of Mathematics

7.1 Graphs of Quadratic Functions in Vertex Form

CA200 Quantitative Analysis for Business Decisions. File name: CA200_Section_04A_StatisticsIntroduction

6 EXTENDING ALGEBRA. 6.0 Introduction. 6.1 The cubic equation. Objectives

24. The Branch and Bound Method

ab = c a If the coefficients a,b and c are real then either α and β are real or α and β are complex conjugates

5.3 Improper Integrals Involving Rational and Exponential Functions

Approximation of Aggregate Losses Using Simulation

Polynomial Operations and Factoring

Discrete Mathematics and Probability Theory Fall 2009 Satish Rao, David Tse Note 13. Random Variables: Distribution and Expectation

1 Formulating The Low Degree Testing Problem

Homework # 3 Solutions

ALGEBRA 2/TRIGONOMETRY

MA107 Precalculus Algebra Exam 2 Review Solutions

QUADRATIC EQUATIONS AND FUNCTIONS

A note on exact moments of order statistics from exponentiated log-logistic distribution

Transcription:

TRANSACTIONS OF SOCIETY OF ACTUARIES 1969 VOL. 21 PT. 1 NO. 60 VOL. XXI, PART I M~TINO No. 60 TRANSACTIONS JUNE, 1969 AN UPPER BOUND ON THE STOP-LOSS NET PREMIUM--ACTUARIAL NOTE NEWTON L. BOWERS, JR. ABSTRACT One type of reinsurance contract which has attracted some attention recently is stop-loss reinsurance. In its simplest form, the reinsurer agrees to pay all losses of the insurer in excess of an agreed limit. This Note concerns itself with a simple upper bound on this premium. The bound depends only on the mean and variance of the distribution of total claims. After a proof that is strongly reminiscent of that of the Chebyshev inequality, tables are given comparing the upper bound with the net premium calculated under certain distribution assumptions and with net premiums calculated as part of the work of Bohman and Esscher. The Note is concluded with a very brief review of related inequalities which hold for stop-loss premiums where claim-limit maximums and coinsurance features are included. S ~.V'm~AL recent papers [1-5] have discussed the approximation of the stop-loss net premium. This Note concerns itself with a simple upper bound on this premium. The upper bound is expressed in the following theorem. We note that, if X is the random variable of total claims, then II(z) is the net premium for stop-loss reinsurance for losses in excess of an amount z. Tm~o~r: Let X be a random variable with mean t~, variance o ~, and distribution function F(x). Then if z = t~ + Ka, we have II(z) = f(x - z)df(x) < ~ 1 (1) Z'K + v'l + K 2 Pgoo : Let 2_z= x_~_ Ke and let h(x) = a[x - (# + b~)] 2. g(x) = x > z x<z 211

212 AN UPPER BOUND ON THE STOP-LOSS NET PILEMILrM Y gcx /Z-I- bo- /z-l- No- If a and b are chosen so that g(x) ~_ h(x) for all x, as in the above diagram, then E[g(X)] ~ E[h(X)].But we observe that while co E[g(X)] = f (x -- z)df(x) = II(z), $ E[h(X)] = af (x -- u -- ba)2df(x) = aa2(1 -b b2), --03 so that II(z) < a~(1 + b2). Now, for g(x) < h(x) to hold for all x, it is sufficient to require that (1) b < K and that (2) g(x) = h(x) at exactly one x, where x > z, that is, the graph of g(x) is tangent to the graph of h(x) at one point, with x > z in addition to x = u q- ba. The equation g(x) = h(x) has a single solution if the discriminant of the quadratic h(x) -- g(x) is zero. This leads to the condition that so that 1 a-- 4o~(K-- b)' ~(z) < 'l+b 2 4 K- b" (2) This inequality holds for all b < K. To make the inequality as sharp as possible, we choose b to minimize the right side of formula (2). This turns out to be equivalent to solving b ~ -- 2bK- 1 -- 0, subject to b < K. The solution is b = K--,/1 + K 2. (3)

AN UPPER BOUND ON TSE STOP-LOSS NET PRE~ 213 Substituting this into formula (2) gives our result that O" n(z) <~ K + Ci+/~",. To show that the inequality developed is the "best possible" involving just the first two moments, we demonstrate a distribution of X for which equality holds. Consider a particular value of K. Assume X is a discrete random variable taking on just two values, the two values where g(x) = h(x). From formula (3) we have tt + ba = tt + a(k -- 1Vq---+--~) as one of the two values of x, where g(x) = It(x). The other is the point where if(x) = 1. This can be shown to be x = tt + a(k + Vrl + KS). If we assign probabilities as and Pr [X = g + ~(K -- ~/1 + K2)] = Pr [X = ~ + ~(K + C1 + K2)I = (K -[- ~ q-- K2) 2 1+ (K+ a/l+k') 2 1 1 + (K + ~/i + K~) 2' it can be verified that E[X] = ~ and Var [X] = e2 as needed. Further n(z) = f (x - z)df(x) or, equivalently, in the discrete case It n(~) = ~E(~,- z)p(x = x,) =i~>s ([. + ~(K + ~/1 + Ks)] -- (. + ~K)} i + (K +,,/i + K~)'- 1 2 K+V,1 +K ~' as given in the theorem. Therefore for any value of K we can demonstrate a distribution of X for which equality holds, and thus the inequality as stated cannot be improved without additional information regarding the distribution of X. We now compare this upper bound with the net premium under certain distribution assumptions on X, the amount of total claims (see Table 1). These distributions are illustrative of the effect of using different distributions to evaluate stop-loss premiums. The normal distri-

214 AN UPPER BOUN'D ON TKE STOP-LOSS NET PREMIUM bution is a symmetric distribution, while the other two are positively skewed. The Pareto distribution chosen has finite variance but infinite third moment [f(x) = 3x -~ for x > 1]. No suggestion that the Pareto distribution, in particular, be used to estimate stop-loss premiums is intended. For a final comparison, we show results for part of the study by Bohman and Esscher [3] (see Table 2). Stop-loss net premiums are compared with the corresponding bound developed above. The particular case used is for 100 expected claims where the number of claims follows a particular (k = 20) member of the negative binomial family of distributions. The individual claim-size distribution is what the authors labeled as Life Insurance B, being based on data from a Swedish life company between 1957 and 1961. The claim amounts were scaled so that the expected size of a single claim would be one. Thus, for this example, E(X) = 100. The authors indicate that Var (X) = (67.947) 5. The disparity between the upper bound and the stop-loss premium given by Bohman and Esscher is not too excessive for stop-loss limits not more than 3 standard deviations above the mean. TABLE 1 STOP-LOSS NET PREMIUM (As Proportion of Standard Deviation) Stop-Loss Upper Normal Pearson III Pareto Level (K) Bound Distribution ~8 m o~ Distribution 0.0 0.5 1.0 115 2.0 2.5 3.0 O. 5000.3090 2071 1514.1180 0963 0.0811 0.3989.1978 0833 0293 0085.0020 0.0004 0.3907.2184 1165.0598 0297.0144 0.0068 O. 2566 1545 1031 0737 0553.0430 0.0344 TABLE 2 Stop-Loss Limit to0... 0 t67.9... 1.~35.9... 2 ;03,8... 3 ;71.8... 4 i07.7... 6 K Stop-Loss Pre- I mium, Bobmsn-[ ' Esscber [ Upper Bound 20.99 33.97 8.42 14.07 4. 680 8.020 3.035 5.513 1.740 4 182 0.1741 I 2.812

AN UPPER BOUND ON THE STOP-LOSS NET PREM~ 215 The usual stop-loss reinsurance contract covers total claims in excess of a lower limit but has a claim-limit maximum. Often a coinsurance feature is also included--that is, the reinsurer pays only a percentage of claims in excess of the lower limit up to the maximum. Let us study what changes result in the inequality presented above when maximumlimit and coinsurance features are included. We will indicate what the statement of the inequality becomes with these changes and hint at the proof of these new statements. For this purpose, let K be the lower limit in standard measure, that is, the lower limit is g -4- Kg; let g + K'~ be the upper limit beyond which no additional reinsurance payments are made; and let 100 c be the percentage of claims paid in excess of the lower limit. CASE 1. Coinsurance, but no maximum claim limit. In this case it is easy to see that ca n(z) < 2(K + %/1-4- K 2) This benefit is just c times the benefit with no coinsurance, so that the upper bound is c times the upper bound established in the theorem. Further, the example following the theorem can be adjusted to show this result is "best possible." For the two other cases we shall assume that coinsurance is included. When coinsurance is not present, c is replaced by 1. CASE 2. Maximum claim limit with K' > K + v'l + K S, The bound developed in Case 1, ca/2(k + ~ + KS), is clearly an upper bound for the modified benefit since a benefit with a maximum claim limit is less expensive than one with no maximum. Again the example following the proof of the theorem shows that this bound is "best possible," involving only the mean and variance when the upper limit, ;t + K'a, is greater than g + (K + ~ K9 ~. CASE 3. Maximum claim limit with K' _< K + ~/1 + K 2. The inequality can in this case be improved to be n(z) < c~(k' -- K) i + (K') 2 " The proof for this result follows lines similar to that of the main theorem. However, the condition the parabola h(x) must satisfy is that it go through the point [g + K'o',c,;(K' -- K)] rather than be tangent to the graph of g(x) at one point x > z. This leads to the changed form of the inequality. The author would like to thank the reviewer of the paper for the sug-

216 AN UPPER BOUND ON" THE STOP-LOSS NET PRE]~I'U~ gestion to include a discussion of coinsurance and maximum benefit limits and to indicate more refined bounds in these cases. BIBLIOGRAPHY 1. B~T~Tr, D. K. "Excess Ratio Distribution in Risk Theory," TSA, XVII (1965), 435. 2. BOWERs, N. L., JR. "Expansion of Probability Density Functions as a Sum of Gamma Densities with Applications in Risk Theory," TSA, XVIII (1966), 125. 3. BOH~A~, H., AND ESSCHER, F. "Studies in Risk Theory with Numerical Illustrations concerning Distribution Functions and Stop-Loss Premiums," Skandina~isk Aktuarietidskrifl, XLVI (1963), 173, and X_LVII (1964), 1. 4. Fr.Ay, H. L., AND KABAK, I. W. "Frequency Formulas for Determining Stop-Loss Reinsurance Premiums," Journal of Risk and Insurance, XXXV (1968), 371. 5. KA~N, P. M. "An Introduction to Collective Risk Theory and Its Applications to Stop-Loss Reinsurance," TSA, XIV (1962), 400. ADDENDUM I have received a considerably more elegant proof of the theorem of this note from Hilary Seal, and I would like to present it as a very interesting addition to the paper. Let and o o I = f(x- z)df(x) = gf(y- K)dFo(y) z K K J = f(x -- z)df(x) = cr f(y -- K)dFo(y), 0 -~1 where Fo(y) is the equiwdent of F(x) when y = (x -- #)/or and where z = ~ -t- Kcr. Then while I -b J = ~ f(y -~/~ -- K)dFo(y) = --~K, o K I -- J = ~f(y -- K)dFo(y) -- ~ f(y -- K)dFo(y) K --~/~ = fly - K[dFo(y). -~/~

Hence AN UPPER BOUND ON THE STOP-LOSS NET PREMILr~ 217 L--~/a J < ~ fdfo(y) f(y -- K)*dFo(y) oo = ~2 f(/ _ 2yK + K2)dFo(y) = ~(1 + K9 The above inequality is an application of the well-known Cauch) Schwarz inequality. Thus I -- J < rx/1 -b K 2, and since I -b J = --~rk, we have i<~(vq+ K2 ~( 1 ) --K)=~ vq+k~+k