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OPTIMIZING MEDICAL FLEXIBLE SPENDING ACCOUNTS A Thesis by CASSIDI JACOBS Submitted to the Office of the Graduate Studies of Texas A&M University-Commerce in partial fulfillment of the requirements for the degree of MASTER OF SCIENCE August 2014 OPTIMIZING MEDICAL FLEXIBLE SPENDING ACCOUNTS A Thesis by CASSIDI JACOBS Approved by: Advisor: Thomas Boucher Committee: Tingxiu Wang Charles Dorsett Pamela Webster Head of Department: Tingxiu Wang Dean of the College: Dan Edelman Dean of Graduate Studies: Arlene Horne iii Copyright © 2014 Cassidi Jacobs iv ABSTRACT OPTIMIZING MEDICAL FLEXIBLE SPENDING ACCOUNTS Cassidi Jacobs, MS Texas A&M University-Commerce, 2014 Advisor: Thomas R. Boucher, PhD In this paper we discuss the characteristics of medical flexible spending accounts (FSAs), including their tax benefits and forfeiture risk. We examine previous work that discusses the problem of optimizing FSA contributions to minimize the risk and maximize the benefits, all from the perspective of the utility of money. Finally, we conduct new work in the optimization of FSAs. In particular, we examine the role of risk aversion in FSA optimization. v ACKNOWLEDGMENTS I wish to express my sincere appreciation to the many individuals whose assistance helped make this thesis a reality. The person I owe the most thanks to are Dr. Thomas Boucher, my major adviser for his patience, moral support, and all around mentoring. With equal appreciation, I thank my other committee members: Dr. Pamela Webster, Dr. Charles Dorsett, and Dr. Tingxiu Wang for their encouragement and professional support throughout my academic career. To my parents, Tami and Darren Jacobs, I express my love and appreciation. Thank you for always believing in me, supporting my educational goals, and sacrificing so I would never have to go without. To my Grandparents, Nancy Templeton (Nonnie) and Van Templeton (Papa), for dealing with the tears and shouts of excitement, with every step of my education. Without you I may have never known I was “smart” enough to succeed in college, and I definitely wouldn’t have known the importance of scholarship. To my brother, SR Kenneth Dallas, for your service to our country and for making this a safe place for me to receive an education of such magnitude. To my boyfriend, Simbarashe Mazambani, for always challenging and inspiring me. vi TABLE OF CONTENTS LIST OF TABLES ………………………………………………………..........viii LIST OF FIGURES ………………………………………………….……….....ix CHAPTER 1. INTRODUCTION …………….………..……………………..................1 Flexible Spending Accounts ..……………………………………1 Utility ……………………...…………………………….........….1 2. UTILITY FUNCTIONS ...……………………………………..................3 Risk Aversion …..………………………………..…………..……9 Statement of the Problem ………………………………..………15 3. REVIEW OF RELATED LITERATURE ..……………………………...16 Cuddington (1991) ………..………….…………………………..16 Cardon and Showalter (2003) ..….……………………………….17 Ramsay and Oguledo (2011) …………………………………….17 4. FINDINGS …………………………………...………..………………...19 Scale Families .…………………………………………………..19 Cuddington: Continuation .………………………………………20 Cardon and Showalter: Continuation ..…………………………..20 5. EXAMPLES .……………….………………………………………........38 6. CONCLUSIONS/DISCUSSION ……..…..………….…………………..43 REFERENCES …………………………………………………………...……....44 APPENDICES ………………………………………………………………........45 Appendix vii A. R Code …….……….…..…….…………..…………………......46 VITA …………..…………………..……………………………………………......50 viii LIST OF TABLES TABLE 1. Cuddington …………….…………………………………………………...38 2. Isoelastic ……………………………………………………………………38 3. Hyperbolic …………………………………………………………….……39 4. Logarithmic ………………………………………………………………...40 5. Quadratic ………………………………………………….………..………40 6. Exponential …………………………………………………………………41 7. Negative Exponential ………………………………………………………41 8. Ramsay and Oguledo .……………...………………………………………42 ix LIST OF FIGURES FIGURES 1. Isoelastic Utility Function …..…………………………………..………….3 2. Hyperbolic Utility Function ...…….……………………………………..…5 3. Logarithmic Utility Function ...………………………….…………….…...6 4. Quadratic Utility Function …...…………………………………………….7 5. Exponential Utility Function ……………………….……………………...8 6. Negative Exponential Utility Function …………….……………………...9 7. Arrow-Pratt Risk Aversion Coefficients .………….…………………......14 1 Chapter 1 INTRODUCTION Flexible Spending Accounts A Flexible Spending Account (FSA) is a savings account set up by an employer to pay medical expenses or dependent care (Schweitze, 1996, p. 1079). Employees specify an irrevocable contri-bution before the beginning of the plan year. Currently, the government maximum limit on FSA contributions, set by The Patient Protection and Affordable Care Act, is $2500 (U.S. Department of Health & Human Services [HHS], 2014). One of the benefits of an FSA is that the money put into the account is taken out of the employee’s check before taxes, including federal taxes, state taxes, income taxes, and FICA. The downside of an FSA is that any unused contributions in the FSA are forfeited by the employee to the employer. Thus when the employee is determining the contribution amount they must balance tax savings versus forfeiture risk. To illustrate the tax benefit, consider an employee with a marginal tax rate of 30%. Suppose this employee pays 100 dollars per unit of health care. If the employee had a flexible spending account they would pay 1000 dollars for 10 units of health care in pre-tax dollars. Without an FSA, they could buy 10 units of health care with 1000 after-tax dollars, which is equivalent to 1000/(1−0.3)=$1428.57 in pre-tax dollars. So without an FSA they actually paid 142.87 dollars per unit of health care. With a flexible spending account, the employee saves 42.87 dollars per unit or 428.57 dollars total. Now consider the risk of forfeiture. If the employee contributes 2000 pre-tax dollars to their flexible spending account and buys 15 units of healthcare at a price of 100 dollars per unit, then the employee only used 1500 dollars. Thus, they forfeited 500 dollars to their employer. Therefore, their health care actually cost them 133.34 pre-tax dollars. Utility Utility is a concept in economics to describe the subjective value of money. Mathematically, utility is described by utility functions u(x), x > 0 which must follow two rules; the first derivative 2 must be greater than zero, u′(x) > 0, and the second derivative is less than or equal to zero, u′′(x) ≤ 0, which means that the function must be positive, increasing, and concave down. Examples of functions that follow these rules are the line u(x) = x and u(x) = √x. The theory of utility speaks not of the dollar value of money, but the personal value of money. As one’s wealth increases the personal value of money per dollar usually decreases. For example if a person makes $1000 a month and a unit of health care cost them $100 their cost of health care is more concerning to them than to one who makes $4000 a month. It is important to note that the first derivative of the utility function, u′(x), is called the marginal utility, i.e. the rate of change of the the utility function. The second derivative, u′′(x), is the rate of change of the marginal utility. The marginal utility from a conceptual stand point, in reference to health care, describes the increased satisfaction one gets from buying an additional unit of health care, since u” < 0 the increase of u is decreasing. There are many utility functions that follow this model. This leads to the Principle Of Maximum Expected Utility; rational investor act to maximize their expected utility of wealth. Formally, for each investment I in a set of competing feasible in-vestment alternatives F, let X(I) be the random variable giving the ending value of the investment for the time period in question. Then a rational investor with utility function U faces the optimiza-tion problem of finding an investment Iopt ∈ F for which: E[U(X(Iopt))] = MaxI∈FE[U(X(I))] (Norstaad, 1991, p. 2). Note that we take the expectation, E[·], because X is random. 3 Chapter 2 UTILITY FUNCTIONS The following show the utility functions the author has chosen to use for the remainder of the paper. Isoelastic or Power Utility Function: u(x) = x1−g 1−g ; 0 < g < 1, x > 0 (1) The figure below shows the graph of the Isoelastic utility function for various values of g. When 0 2 4 6 8 10 0 1 2 3 4 5 6 x u(x) gamma=.25 gamma=.5 gamma=.75 Figure 1: Isoelastic utility function for g =0.25, 0.5, 0.75. the first and second derivative of u(x) are taken the two requirements of a utility function are met: u′(x) = x−g > 0; u′(x) > 0 u′′(x) = −gx−g−1 < 0; u′′(x) < 0. 4 Hyperbolic Utility Function: u(x) = 1−g g ( ax 1−g +b )g , b > 0, x > 0. (2) Then, u′(x) = (1−g) ( ax 1−g +b )g−1( a 1−g ) = a ( ax 1−g +b )g−1 > 0, x > 0, and u′′(x) = a(g−1) ( a2 1−g +b )g−2( a 1−g ) = a2 ( ax 1−g +b )g−2 < 0, x > 0. 5 The figure that follows shows the graph of the Hyperbolic utility function for various values of g. Let the values a = 2 and b = 3 chosen for simplicity. 0 2 4 6 8 10 2 3 4 5 6 x u(x) gamma=.25 gamma=.5 gamma=.75 Figure 2: Hyperbolic utility function for g =0.25, 0.5, 0.75 and a = 2, b = 3. Logarithmic Utility Function: u(x) = ln(x), x > 0 (3) Note the requirements of a utility function are satisfied, since u′(x) = 1 x > 0 and u′′(x) = − 1 x2 < 0. 6 The figure below shows the graph of the Logarithmic utility function, because this utility function does not include any parameters, such as g, we have only one curve to plot. 0 2 4 6 8 10 −2 −1 0 1 2 x u(x) Figure 3: Logarithmic utility function. Quadratic Utility Function: u(x) = x−gx2, g > 0, x > 0. (4) Note the requirements of a utility function are satisfied, since u′(x) = 1−2gx > 0 and u′′(x) = −2g < 0. 7 The figure below shows the graph of the quadratic utility function for various values of g. Note g must be small for the quadratic function and over average range of x: x < 1 2g . 0 2 4 6 8 10 0 2 4 6 8 x u(x) gamma=.002 gamma=.02 gamma=.05 Figure 4: Quadratic utility function for g =0.002, 0.02, 0.5. Exponential Utility Function: u(x) = 1−e−gx, x > 0 (5) Note the requirements of a utility function are satisfied, since u′(x) = ge−gx > 0 and u′′(x) = −g2e−gx < 0. 8 The figure below shows the graph of the exponential utility function for various values of g. 0 2 4 6 8 10 0.0 0.2 0.4 0.6 0.8 1.0 x u(x) gamma=.25 gamma=.5 gamma=.75 Figure 5: Exponential utility function for g =0.25, 0.5, 0.75. Negative Exponential Utility Function: u(x) = −e−gx, x > 0. (6) Note the requirements of a utility function are satisfied, since u′(x) = ge−gx > 0 and u′′(x) = −g2ex < 0 9 The figure below shows the graph of the negative exponential utility function for various values of g. 0 2 4 6 8 10 −1.0 −0.8 −0.6 −0.4 −0.2 0.0 x u(x) gamma=.25 gamma=.5 gamma=.75 Figure 6: Negative Exponential utility function for g =0.25, 0.5, 0.75. Now that we have introduced the various types of utility functions we will be using we must study the reason for having such functions in the first place, risk aversion. Risk Aversion Risk aversion refers to the behavioral tendency of preferring a less risky alternative to a choice with greater uncertainty when in fact the more risky alternative could be more profitable in the long run. Risk aversion has been used to explain how consumers make choices in a variety of financially risky decisions, from investments to gambling (Cather, 2010, p. 131). The reason for this is that higher utility dollars (small x) are valued more than lower utility dollars (large x). Mathematically, risk aversion occurs when the first derivative of the utility function is greater than zero; u′(x) > 0, and the second derivative is less than zero, u′′(x) < 0. Since u′′(x) is the rate of change in marginal utility, u′(x), greater risk aversion means less gain in utility. From previous information we know that the second derivative can also be equal to zero, u′′(x) = 0, this case is 10 referred to as risk neutral (Cardon and Showalter, 2003, p. 45). For this paper we will investigate risk aversion when applied to flexible spending accounts. FSAs have a forfeiture risk as we discussed previously, and the tax risk incurred by not putting enough money into your FSA; both result in a form of loss. Risk aversion applied to this situation is the idea that you want to put the optimal amount of money M∗ into your FSA to minimize both types of risk. When considering risk aversion in a financial setting we often use the Arrow-Pratt absolute risk aversion coefficient (Norstaad, 1999, p.21): A(x) = −u′′(x) u′(x) . (7) Interpret as normalized rate of change in marginal utility. The larger it is the slower u(x) increases. The utility functions we discussed earlier can be grouped by four types of risk aversion: Hy-perbolic absolute risk aversion (HARA), Decreasing absolute risk aversion (DARA), Increasing absolute risk aversion (IARA), and Constant absolute risk aversion (CARA). The names refer to the behavior of A(x) (Moss, 2010, p. 1). The Isoelastic Utility Function we discussed previously, u(x) = x1−g 1−g ; 0 < g < 1, x > 0. with u′(x) = x−g and u′′(x) = −gx−g−1 is a special type of HARA. The Arrow-Pratt absolute risk aversion coefficient for the Isoelastic utility function is A(x) = −−gx−g−1 x−g 11 = g x , (8) which is a Hyperbolic function. The general form of an HARA is as follows: u(x) = 1−g g ( ax 1−g +b )g , b > 0. So that, u′(x) = (1−g) ( ax 1−g +b )g−1( a 1−g ) = a ( ax 1−g +b )g−1 and u′′(x) = a(g−1) ( a2 1−g +b )g−2( a 1−g ) = a2 ( ax 1−g +b )g−2 . The Arrow-Pratt absolute risk aversion coefficient for the HARA Hyperbolic utility function is A(x) = − [ −a2( ax 1−g +b)g−2 a( ax 1−g +b)g−1 ] = a(1−g) ax+b(1−g) , (9) which is a Hyperbolic function. HARAs are a special form of utility function with decreasing absolute risk aversion (DARA). DARAs occur when A(x) is a decreasing function. DARA utility functions would reflect how the majority of people feel about money. The more money they have, the less they need, and the less value an increase in wealth has to them. Experimental and empirical evidence is mostly consistent 12 with decreasing absolute risk aversion (Grover, 2010, p. 1). Another example of a DARA is the Logarithmic utility function u(x) = ln(x), x > 0, where u′(x) = 1 x and u′′(x) = − 1 x2 The Arrow-Pratt absolute risk aversion coefficient for the DARA Logarithmic utility function is A(x) = −(− 1 x2 ) 1 x = ( 1 x2 )(x 1 ) = 1 x . (10) This is a decreasing function. Increasing absolute risk aversion (IARA) occurs when A(x) is increasing; people with this form of utility function are less common; they may have a spending problem or gambling addiction. Money increases in value to them as they gain more and decrease the more they lose. An example of a IARA is the Quadratic Utility Function. u(x) = x−gx2, g > 0, x > 0 with u′(x) = 1−2gx and u′′(x) = −2g. 13 The Arrow-Pratt risk aversion coefficient for the IARA Quadratic utility function is A(x) = 2g (1−2gx) (11) is increasing if x < 1 2g . The CARA utility functions we have discussed are different from the other forms of absolute risk aversion because they exhibit constant absolute risk aversion where the Arrow-Pratt absolute risk aversion coefficient is constant with respect to x. Exponential utility is of the form u(x) = 1−e−gx where u′(x) = ge−gx and u′′(x) = −g2e−gx The Arrow-Pratt absolute risk aversion coefficient of the CARA Exponential utility function is A(x) = −(−g2e−gx) ge−gx = g. (12) Another example of CARAs is the Negative Exponential Utility Function, where a < 0: u(x) = −e−gx where u′(x) = ge−gx 14 and u′′(x) = −g2ex The Arrow-Pratt absolute risk aversion coefficient of the CARA Negative Exponential utility func-tion is A(x) = [ −ge−gx ge−gx ] = g. (13) The following figure shows the graph of each of the Arrow-Pratt absolute risk aversion coeffi-cients we have discussed above. Let g = 0.5. 0.0 0.5 1.0 1.5 2.0 0 5 10 15 20 25 x Arrow−Pratt Isoelastic Hyperbolic Logarithmic Quadratic Exponential Negative Exponential Figure 7: Arrow-Pratt Absolute Risk Aversion Coefficients, where g = 0.5 15 Statement of the Problem This leads to the problem at hand: deriving the optimal FSA contribution level M∗, given a tax rate t, income Y, and healthcare expense X. The author will find the employee’s contribution M∗ to their flexible spending account while simultaneously maximizing pre-tax advantage, minimizing forfeiture risk, and maximizing expected after-tax utility E[u(YD)], of disposable income YD, where YD = (Y −M)(1−t)−max(X −M,0) (14) Note that (Y −M)(1−t) gives the after tax and after FSA value of income Y minus FSA contri-bution M and max(X −M,0) gives the out of pocket expense value of health care expense greater than M. 16 Chapter 3 REVIEW OF RELATED LITERATURE Previous work on the problem above can be credited to Cuddington (1991), Cardon and Showalter (2003), Ramsay and Oguledo (2011). Cuddington (1991) Cuddington (1991) focused on after-tax, after-healthcare disposable income YD using the utility function u(x)=x. Because healthcare expense x is not predictable and can be modelled as a random variable, so can YD, so he took the expectation of the disposable income. To optimize he sought to maximize the principle of expected utility by taking the derivative of E[YD] with respect to the contribution M, set it equal to zero and solved for M∗, the optimal contribution. Where fx(x) is the chosen density function for health care expense x and x ≥ 0: E[YD] = (Y −M)(1−t)−∫ ¥ M (x−M) fx(x)dx =Y(1−t)−M(1−t)−∫ ¥ M x fx(x)dx+M∫ ¥ M fx(x)dx (15) Note that by using integration by parts we get d dM [ −∫ ¥ M x fx(x)dx+M∫ ¥ M fx(x)dx ] = −Mfx(M)+∫ ¥ M fx(x)dx+Mfx(M) = 0+P(X > M), so that the amount M∗ which yields maximum utility is determined by d dM E[YD] = −(1−t)+P(X > M) = 0 ⇒t = P(X ≤ M∗) (16) In other words, Cuddington found that the optimal contribution M∗ occurs when the probability of 17 healthcare expense X being less than the contribution M∗ equals the tax rate t i.e. forfeiture risk equals tax rate. Cardon and Showalter(2003) Cardon and Showalter (2003) used equation (15) for the expected value of the disposable in-come YD that Cuddington used, but replaced u(x) = x with a generic utility function u, and added a fixed, known healthcare expense upper limit L to get the following. Where P(X ≤ L) = 1 and fx(x) is defined on [0,L], L > M. E[u(YD)] = ∫ M 0 u((Y −M)(1−t)) fx(x)dx +∫ L M u(Y(1−t)+tM−X) fx(x)dx (17) where fx(x) is the probability density function (PDF) of X. Cardon and Showalter found that in the case of risk neutrality, that is when the first derivative is greater than zero, u′(x) > 0, and the second derivative is equal to zero; u′′(x) = 0, they arrived at the same result as Cuddington, the optimal contribution takes place when the tax rate equals the probability that the loss is less than the contribution M∗: t = P(X < M∗) As for the risk adverse case, which is when the first derivative is greater than zero; u′(x) > 0; and the second derivative is less than zero; u′′(x)<0, the optimal contribution occurs when Arrow-Pratt absolute risk aversion coefficient is applied. Explicitly, M∗ is such that u′′((Y −M∗)(1−t)) u′((Y −M∗)(1−t)) = (t −P(X < M∗)) t(P(X > M∗))E[X −M∗ X > M∗] . (18) Ramsay and Oguledo (2011) Ramsay and Oguledo (2011) retained the utility function u, the cap L, and added a predictable lower limit on healthcare expense x∗; fx(x) is defined on [x∗,L]. Ramsay and Oguledo chose to add 18 in the lower limit because they believe some floor on healthcare expenses can be predicted with certainty, such as eye glasses, monthly prescriptions, and yearly checkups. With this addition, we get the following equation for Ramsay and Oguledo: E[u(YD)] = E[u((Y −M)(1−t)−max(X −M,0))], dE[u(YD)] dM = 0 (19) Ramsay and Oguledo found that if a person has stable health and a reasonably low tax rate the optimal contribution should be set equal to the predictable expenses. M∗ = x∗ (20) 19 Chapter 4 FINDINGS The author continued the work of the three authors above by solving each equation (16), (18) and (20), given a probability distribution for health care expense X, for the optimal contribution M∗, excluding Ramsay and Oguledos result because it was already in the form of M∗ = x∗. Scale Families For this paper we will use the Exponential loss distribution fX(x) = 1 be−x/b; x > 0, b > 0. We choose the Exponential loss distribution because it is a Scale Family. A scale family: x ∼ fX(x), Y = sx, has fY (y) = 1 s fX( y s). We choose to use a loss distribution that is a Scale Family because a scale family allows the equations to be valid regardless of the currency and/or inflation. Below the author proved that the Exponential loss distribution to be a Scale Family. Recall the Exponential Loss Distribution is fX(x) = 1 b e−x/b; x > 0, b > 0. (21) Lemma 0.1 Exponential distribution is a scale family. Proof: X ∼ 1 b e−x/b; x > 0 Y = sx fY (y) = 1 s ( 1 b e−x/sb ) ; x > 0, s > 0 = 1 s fx( x s ) a scale family. Thus the Exponential distribution is a scale family. 20 Cuddington: Continuation First the author explicitly solved Cuddington’s result (16) for M∗ on the case where X is an exponential random variable. Since b is unknown the best unbiased estimator is ˆb = ¯ x, estimated using past medical expenses. Lemma 0.2 Assume loss X is exponential ( ¯ x). The optimal contribution M∗ that maximizes E[YD] in (16), yields M∗ = −ln(1−t) ¯ x. Proof: Cuddington showed M∗ defined implicitly by P(X < M∗) = t. If X is exponential ( ¯ x), then the cumulative distribution function (CDF) is Fx(x) = 1−e−x/ ¯ x and so 1−e−M∗/ ¯ x = t −e−M∗/ ¯ x = t −1 e−M∗/ ¯ x = 1−t ln(e−M∗/ ¯ x) = ln(1−t) −M∗ ¯ x = ln(1−t) −M∗ = ln(1−t) ¯ x M∗ = −ln(1−t) ¯ x ⋄ Cardon and Showalter: Continuation The author explicitly solved (18) Cardon and Showalter’s risk averse case, u′′((Y −M∗)(1−t)) u′((Y −M∗)(1−t)) = (t −P(X < M∗)) t(P(X > M∗))E[X −M∗ X > M∗] , 21 for M∗ using the exponential loss distribution and all six utility functions discussed previously. We will first simplify the right hand side of Cardon and Showlater’s risk averse case. Again let X be an exponential random variable, let ˆb = ¯ x. Lemma 0.3 Assume loss X is exponential ( ¯ x). The optimal contributionM∗ that maximizes E[u(YD)] in (18), yields (t −P(X < M∗)) t(P(X > M∗))E[X −M∗ X > M∗] = [t + −M∗ ¯ x ] t[1+ −M∗ ¯ x (−2M∗+ ¯ x)] Proof: We simplified E[(X −M∗)I(X>M∗)] as follows ∫ ¥ M∗ (x−M∗) 1 ¯ x e−x/ ¯ xdx =∫ ¥ M∗ x 1 ¯ x e−x/ ¯ xdx−∫ ¥ M∗ M∗ 1 ¯ x e−x/ ¯ xdx The author then worked each integral separately. For ∫ ¥ M∗ x 1 ¯ x e−x/ ¯ xdx we used integration by parts ∫ udv = uv−∫ vdu, where u = X, du = dx, v = −e−X/ ¯ x, and dv = 1 ¯ x e−X/ ¯ xdx to get x(−e−X/ ¯ x)−∫ ¥ M∗ −e−x/ ¯ xdx = x(−e−X/ ¯ x)−[ ¯ xe−x/ ¯ x] ¥ M∗ = x(−e−x/ ¯ x)+ ¯ xe−M∗/ ¯ x, 22 and for ∫ ¥ M∗ M∗ 1 ¯ x e−X/ ¯ xdx =M∗ ∫ ¥ M∗ 1 ¯ x e−X/ ¯ xdx = M∗P(X > M∗) = M∗1−(1−e−M∗/ ¯ x) = M∗e−M∗/ ¯ x. Thus ∫ ¥ M∗ X 1 ¯ x e−X/ ¯ xdx−∫ ¥ M∗ M∗ 1 ¯ x e−X/ ¯ xdx simplifies to X(−e−X/ ¯ x)+ ¯ xe−M∗/ ¯ x−M∗e−M∗/ ¯ x. Next we put the simplification back into the original equation to get [t −P(X < M∗)] t[e−M∗/ ¯ x(−2M∗+ ¯ x)] The author decided to simplify e−M∗/ ¯ x by use of the Taylor Series first order approximation: ex = 1+x+ x2 2! + x3 3! +...+ xn n! = S¥k =0 xk k! e−M∗/ ¯ x ≈ 1+ −M∗ ¯ x . Now we see that t −P(X < M∗) = t −[1−e−M∗/ ¯ x ≈ t −[1−(1+ −M∗ ¯ x )] = t − M∗ ¯ x . 23 This last simplification is put into the equation to give [t + −M∗ ¯ x ] t[(1+ −M∗ ¯ x )(−2M∗+ ¯ x)] . ⋄ Now our equation is u′′((Y −M∗)(1−t)) u′((Y −M∗)(1−t)) = [t + −M∗ ¯ x ] t[1+ −M∗ ¯ x (−2M∗+ ¯ x)] . Next the author simplified the left hand side, by substituting in the various utility functions discussed previously. We can see that the left hand side is the Arrow-Pratt coefficient u′′(x) u′(x) where x = (Y −M∗)(1−t). Lemma 0.4 (I) Using the Isoelastic Utility Function, u(x) = x1−g 1−g : u′′((Y −M∗)(1−t)) u′((Y −M∗)(1−t)) = ((Y −M∗)(1−t))1−g 1−g . (II) Using the Hyperbolic Utility Function, u(x) = 1−g g ( ax 1−g +b )g : u′′((Y −M∗)(1−t)) u′((Y −M∗)(1−t)) = a−ga a((Y −M∗)(1−t))+b(1−g) . (III) Using the Logarithmic Utility Function, u(x) = ln(x) : u′′((Y −M∗)(1−t)) u′((Y −M∗)(1−t)) = 1 (Y −M∗)(1−t) . 24 (IV) Using the Quadratic Utility Function, u(x) = x−gx2 : u′′((Y −M∗)(1−t)) u′((Y −M∗)(1−t)) = 2g [1−2g((Y −M∗)(1−t))] . (V) Using the Exponential Utility Function, u(x) = 1−e−gx : u′′((Y −M∗)(1−t)) u′((Y −M∗)(1−t)) = g. (VI) Using the Negative Exponential Utility Function, u(x) = −e−gx : u′′((Y −M∗)(1−t)) u′((Y −M∗)(1−t)) = −g. Proof: (I) If u′(x) = 1−g, u′′(x) = x1−g and x = (Y −M∗)(1−t), then u′′((Y −M∗)(1−t)) u′((Y −M∗)(1−t)) = ((Y −M∗)(1−t))1−g 1−g . (II) If u′(x) = a( ax 1−g +b)g−1, u′′(x) = a2( ax 1−g )1−g and x = (Y −M∗)(1−t), then u′′((Y −M∗)(1−t)) u′((Y −M∗)(1−t)) = a2( a((Y−M∗)(1−t)) 1−g )1−g a( a((Y−M∗)(1−t)) 1−g +b)g−1 = a(1−g) a((Y −M∗)(1−t))+b(1−g) . (III) If u′(x) = 1 x , u′′(x) = −1 x2 and x = (Y −M∗)(1−t), then u′′((Y −M∗)(1−t)) u′((Y −M∗)(1−t)) = −1 ((Y−M∗)(1−t))2 1 (Y−M∗)(1−t) = 1 (Y −M∗)(1−t) . 25 (IV) If u′(x) = 1−2gx, u′′(x) = −2g and x = (Y −M∗)(1−t), then u′′((Y −M∗)(1−t)) u′((Y −M∗)(1−t)) = 2g (1−2g(Y −M∗)(1−t)) . (V) If u′(x) = ge−gx, u′′(x) = −g2e−gx and x = (Y −M∗)(1−t), then u′′((Y −M∗)(1−t)) u′((Y −M∗)(1−t)) = −(−g2e−g(Y−M∗)(1−t)) ge−g(Y−M∗)(1−t) = g. (VI) If u′(x) = ge−gx, u′′(x) = g2e−gx and x = (Y −M∗)(1−t), then u′′((Y −M∗)(1−t)) u′((Y −M∗)(1−t)) = −(g2e−g(Y−M∗)(1−t)) ge−g(Y−M∗)(1−t) = −g. ⋄ Now we have six equations to solve for M∗, (I) ((Y −M∗)(1−t))1−g 1−g = [t + −M∗ ¯ x ] t[(1+ −M∗ ¯ x )(−2M∗+ ¯ x)] (II) a(1−g) a((Y −M∗)(1−t))+b(1−g) = [t + −M∗ ¯ x ] t[(1+ −M∗ ¯ x )(−2M∗+ ¯ x)] (III) 1 (Y −M∗)(1−t) = [t + −M∗ ¯ x ] t[(1+ −M∗ ¯ x )(−2M∗+ ¯ x)] (IV) 2g (1−2g(Y −M∗)(1−t)) = [t + −M∗ ¯ x ] t[(1+ −M∗ ¯ x )(−2M∗+ ¯ x)] 26 (V) g = [t + −M∗ ¯ x ] t[(1+ −M∗ ¯ x )(−2M∗+ ¯ x)] (VI) −g = [t + −M∗ ¯ x ] t[(1+ −M∗ ¯ x )(−2M∗+ ¯ x)] To finish this problem we will solve each of the above equations explicitly for M∗. Lemma 0.5 (I) Suppose X is Exponential ( ¯ x), then ((Y −M∗)(1−t))1−g 1−g = [t + −M∗ ¯ x ] t[(1+ −M∗ ¯ x )(−2M∗+ ¯ x)] , when solved for M∗ can be expressed as the solution to a quadratic function, M∗ = −b± √b2−4ac 2a where a = (g−gt −2t) b = (−gt ¯ x+gt2 ¯ x−gY +gYt +3t ¯ x) c = (Ygt ¯ x−Ygt2 ¯ x−t ¯ x2). (II) Suppose X is Exponential ( ¯ x), then a(1−g) a((Y −M∗)(1−t))+b(1−g) = [t + −M∗ ¯ x ] t[(1+ −M∗ ¯ x )(−2M∗+ ¯ x)] , when solved for M∗ can be expressed as the solution to a quadratic function, M∗ = −b± √b2−4ac 2a 27 where a = (a−ta−2+2g) b = (t2a ¯ x−at ¯ x−Ya+Yta−b+gb+2t ¯ x+ ¯ x−2t ¯ xg− ¯ xg) c = (Yat ¯ x−Yt2a ¯ x+bt ¯ x−gbt ¯ x− ¯ x2tg). (III) Suppose X is Exponential ( ¯ x), then 1 (Y −M∗)(1−t) = [t + −M∗ ¯ x ] t[(1+ −M∗ ¯ x )(−2M∗+ ¯ x)] , when solved for M∗ can be expressed as the solution to a quadratic function, M∗ = −b± √b2−4ac 2a where a = (3− ¯ x−t) b = (t2 ¯ x−t ¯ x−Y +Yt +2t ¯ x) c = (Yt ¯ x−Yt2 ¯ x− ¯ x2). (IV) Suppose X is Exponential ( ¯ x), then 2g (1−2g(Y −M∗)(1−t)) = [t + −M∗ ¯ x ] t[(1+ −M∗ ¯ x )(−2M∗+ ¯ x)] , when solved for M∗ can be expressed as the solution to a quadratic function, M∗ = −b± √b2−4ac 2a 28 where a = (1−t +4tg) b = (Yt −Y −t ¯ x+2 ¯ xg+t2 ¯ x+4t ¯ xg−1) c = (Yt ¯ x−2Y ¯ xg−Yt2 ¯ x+2Yt ¯ xg−2t ¯ x2g+t ¯ x). (V) Suppose X is Exponential ( ¯ x), then g = [t + −M∗ ¯ x ] t[(1+ −M∗ ¯ x )(−2M∗+ ¯ x)] , when solved for M∗ can be expressed as the solution to a quadratic function, M∗ = −b± √b2−4ac 2a where a = (2tg) b = (−2 ¯ xtg) c = ( ¯ x2tg−t ¯ x). (VI) Suppose X is Exponential ( ¯ x), then −g = [t + −M∗ ¯ x ] t[(1+ −M∗ ¯ x )(−2M∗+ ¯ x)] , when solved for M∗ can be expressed as the solution to a quadratic function, M∗ = −b± √b2−4ac 2a 29 where a = (2tg) b = (−4 ¯ xtg) c = ( ¯ x2tg+t ¯ x). Proof: (I) (gt −g+g(1+ −M∗ ¯ x ))(Y −Yt −M∗+M∗t) = t(1+ −M∗ ¯ x )(−2M∗+ ¯ x) (gt −g+g− M∗g ¯ x )(Y −Yt −M∗+M∗t) = (t − M∗t ¯ x )(−2M∗+ ¯ x) (gt − M∗g ¯ x )(Y −Yt −M∗+M∗t) = −2M∗t +t ¯ x+ 2M∗2 t ¯ x −M∗t ⇒Ygt −Ygt2−M∗gt +M∗gt2− M∗gY ¯ x + M∗gYt ¯ x + M∗2 gt ¯ x − M∗2 gt ¯ x = −2M∗t +t ¯ x+ 2M∗2 t ¯ x −M∗t ⇒Ygt ¯ x−Ygt2 ¯ x−M∗gt ¯ x+M∗gt2 ¯ x−M∗gY +M∗gYt +M∗2 g−M∗2 gt = −2M∗t ¯ x+t ¯ x2+2M∗2 t −M∗t ¯ x ⇒Ygt ¯ x−Ygt2 ¯ x−M∗gt ¯ x+M∗gt2 ¯ x−M∗gY +M∗gYt +M∗2 g−M∗2 gt +2M∗t ¯ x−t ¯ x2−2M∗2 t +M∗t ¯ x = 0 30 The author then rearranged the problem by the powers of M∗ M∗2 g−M∗2 gt −M∗2 2t −M∗gt ¯ x+M∗gt2 ¯ x−M∗gY +M∗gYt +M∗2t ¯ x+M∗t ¯ x +Ygt ¯ x−Ygt2 ¯ x−t ¯ x2 = 0 Then the author factored out the powers of M∗ M∗2 (g−gt −2t)+M∗(−gt ¯ x+gt2 ¯ x−gY +gYt +3t ¯ x)+(Ygt ¯ x−Ygt2 ¯ x−t ¯ x2) = 0 Now that we have a function in the form of ax2+bx+c = 0 we can use the quadratic equation to solve for M∗ M∗ = −b± √b2−4ac 2a (22) where a = (g−gt −2t) (23) b = (−gt ¯ x+gt2 ¯ x−gY +gYt +3t ¯ x) (24) c = (Ygt ¯ x−Ygt2 ¯ x−t ¯ x2) (25) (II) 31 a(1−g) a((Y −M∗)(1−t))+b(1−g) = [t + −M∗ ¯ x ] t[1+ −M∗ ¯ x (−2M∗+ ¯ x)] (a−ag) a((Y −M∗)(1−t))+(b−bg) = [t + −M∗ ¯ x ] t[1+ −M∗ ¯ x (−2M∗+ ¯ x)] a((Y −M∗)(1−t))+(b−bg) (a−ag) = t[1+ −M∗ ¯ x (−2M∗+ ¯ x)] [t + −M∗ ¯ x ] a((Y −M∗)(1−t))+(b−bg) = t[1+ −M∗ ¯ x (−2M∗+ ¯ x)](a−ag) [t + −M∗ ¯ x ] a((Y −M∗)(1−t)) = t[1+ −M∗ ¯ x (−2M∗+ ¯ x)](a−ag) [t + −M∗ ¯ x ] −(b−bg) (Y −M∗)(1−t) = t[1+ −M∗ ¯ x (−2M∗+ ¯ x)](a−ag) a[t + −M∗ ¯ x ] − b−bg a (Y −M∗)(1−t) = t[1+ −M∗ ¯ x (−2M∗+ ¯ x)](1−g) [t + −M∗ ¯ x ] − b−bg a Y −Yt −M∗+M∗t = t(−2M∗+ ¯ x+ 2(M∗)2 ¯ x − M∗ ¯ x ¯ x )(1−g) [t + −M∗ ¯ x ] − b−bg a Y −Yt −M∗+M∗t + b a − bg a = (−2t ¯ xM∗+ ¯ x2t +2(M∗)2−M∗ ¯ x)(1−g) t ¯ x−M∗ Ya−Yta−M∗a+M∗ta+b−bg = (−2t ¯ xM∗+ ¯ x2t +2(M∗)2−M∗ ¯ x)(1−g) t ¯ x−M∗ (t ¯ x−M∗)(Ya−Yta−M∗a+M∗ta+b−bg) = (−2t ¯ xM∗+ ¯ x2t +2(M∗)2−M∗ ¯ x)(1−g) ⇒Yat ¯ x−Yt2a ¯ x−M∗at ¯ x+M∗t2a ¯ x+bt ¯ x−gbt ¯ x−YaM∗+YtaM∗+(M∗)2a−(M∗)2ta−bM∗+gbM∗ = −2t ¯ xM∗+ ¯ x2t +2(M∗)2−M∗ ¯ x+2t ¯ xM∗g− ¯ x2tg−2(M∗)2g+M∗ ¯ xg ⇒Yat ¯ x−Yt2a ¯ x−M∗at ¯ x+M∗t2a ¯ x+bt ¯ x−gbt ¯ x−YaM∗+YtaM∗+(M∗)2a−(M∗)2ta−bM∗+gbM∗ +2t ¯ xM∗− ¯ x2t −2(M∗)2+M∗ ¯ x−2t ¯ xM∗g+ ¯ x2tg+2(M∗)2g−M∗ ¯ xg = 0 32 The author then rearranged the problem by the powers of M∗ and factored out M∗ ⇒(a−ta−2+2g)(M∗)2 +(t2a ¯ x−at ¯ x−Ya+Yta−b+gb+2t ¯ x+ ¯ x−2t ¯ xg− ¯ xg)M∗ +(Yat ¯ x−Yt2a ¯ x+bt ¯ x−gbt ¯ x− ¯ x2t + ¯ x2tg) = 0 Now that we have a function in the form of ax2+bx+c = 0 we can use the quadratic equation to solve for M∗ M∗ = −b± √b2−4ac 2a where a = (a−ta−2+2g) b = (t2a ¯ x−at ¯ x−Ya+Yta−b+gb+2t ¯ x+ ¯ x−2t ¯ xg− ¯ xg) c = (Yat ¯ x−Yt2a ¯ x+bt ¯ x−gbt ¯ x− ¯ x2tg) (III) 33 1 (Y −M∗)(1−t) = [t + −M∗ ¯ x ] t[1+ −M∗ ¯ x (−2M∗+ ¯ x)] (Y −M∗)(1−t) = t[1+ −M∗ ¯ x (−2M∗+ ¯ x)] [t + −M∗ ¯ x ] Y −Yt −M∗+M∗t = −2tm∗+ ¯ xt + 2(M∗)2 ¯ x −M∗ [t + −M∗ ¯ x ] Y −Yt −M∗+M∗t = −2tm∗ ¯ x+ ¯ x2t +2(M∗)2−M∗ ¯ x t ¯ x−M∗ (Y −Yt −M∗+M∗t)(t ¯ x−M∗) = −2tm∗ ¯ x+ ¯ x2t +2(M∗)2−M∗ ¯ x Yt ¯X −Yt2 ¯ x−M∗t ¯ x+M∗t2 ¯ x−M∗Y +YtM∗+(M∗)2−(M∗)2t = −2tm∗ ¯ x+ ¯ x2t +2(M∗)2−M∗ ¯ x Yt ¯X −Yt2 ¯ x−M∗t ¯ x+M∗t2 ¯ x−M∗Y +YtM∗+(M∗)2−(M∗)2t +2tm∗ ¯ x− ¯ x2t −2(M∗)2+M∗ ¯ x = 0 The author then rearranged the problem by the powers of M∗ and factored out M∗ ⇒(2− ¯ x+1−t)(M∗)2 +(t2 ¯ x−t ¯ x−Y +Yt +2t ¯ x)M∗ +(Yt ¯ x−Yt2 ¯ x− ¯ x2) = 0 Now that we have a function in the form of ax2+bx+c = 0 we can use the quadratic equation to solve for M∗ M∗ = −b± √b2−4ac 2a 34 where a = (3− ¯ x−t) b = (t2 ¯ x−t ¯ x−Y +Yt +2t ¯ x) c = (Yt ¯ x−Yt2 ¯ x− ¯ x2) (IV) 2g (1−2g(Y −M∗)(1−t)) = [t + −M∗ ¯ x ] t[1+ −M∗ ¯ x (−2M∗+ ¯ x)] (1−2g(Y −M∗)(1−t)) 2g = t[1+ −M∗ ¯ x (−2M∗+ ¯ x)] [t + −M∗ ¯ x ] (1−2g(Y −M∗)(1−t)) = t[1+ −M∗ ¯ x (−2M∗+ ¯ x)](2g) [t + −M∗ ¯ x ] 2g(Y −M∗)(1−t) = t[1+ −M∗ ¯ x (−2M∗+ ¯ x)](2g) [t + −M∗ ¯ x ] −1 (Y −M∗)(1−t) = t[1+ −M∗ ¯ x (−2M∗+ ¯ x)](2g) [t + −M∗ ¯ x (2g)] − 1 2g Y −Yt −M∗+M∗t = 2tg(−2M∗+ ¯ x+ 2(M∗)2 ¯ x −M∗)−(t + −M∗ ¯ x ) (t + −M∗ ¯ x )−2g Y −Yt −M∗+M∗t = −4tM∗g+2t ¯ xg+ 4tg(M∗)2 ¯ x −2tM∗g−t + M∗ ¯ x t − M∗ ¯ x −2g Y −Yt −M∗+M∗t = −4tM∗ ¯ xg+2t ¯ x2g+4tg(M∗)2−2tM∗ ¯ xg−t ¯ x+M∗ t ¯ x−M∗−2 ¯ xg (Y −Yt −M∗+M∗t)(t ¯ x−M∗−2 ¯ xg) = −4tM∗ ¯ xg+2t ¯ x2g+4tg(M∗)2−2tM∗ ¯ xg−t ¯ x+M∗ 35 ⇒Yt ¯ x−YM∗−2Y ¯ xg−Yt2 ¯ x+YtM∗+2Yt ¯ xg−t ¯ xM∗+(M∗)2+2M∗ ¯ xg+t2 ¯ xM∗−t(M∗)2−2tM∗ ¯ xg = −4tM∗ ¯ xg+2t ¯ x2g+4tg(M∗)2−2tM∗ ¯ xg−t ¯ x+M∗ ⇒Yt ¯ x−YM∗−2Y ¯ xg−Yt2 ¯ x+YtM∗+2Yt ¯ xg−t ¯ xM∗+(M∗)2+2M∗ ¯ xg+t2 ¯ xM∗−t(M∗)2 −2tM∗ ¯ xg+4tM∗ ¯ xg−2t ¯ x2g−4t(M∗)2g+2tM∗ ¯ xg+t ¯ x−M∗ = 0 The author then rearranged the problem by the powers of M∗ and factored out M∗ ⇒(1−t +4tg)(M∗)2 +(Yt −Y −t ¯ x+2 ¯ xg+t2 ¯ x+4t ¯ xg−1)M∗ +(Yt ¯ x−2Y ¯ xg−Yt2 ¯ x+2Yt ¯ xg−2t ¯ x2g+t ¯ x) = 0 Now that we have a function in the form of ax2+bx+c = 0 we can use the quadratic equation to solve for M∗ M∗ = −b± √b2−4ac 2a where a = (1−t +4tg) b = (Yt −Y −t ¯ x+2 ¯ xg+t2 ¯ x+4t ¯ xg−1) c = (Yt ¯ x−2Y ¯ xg−Yt2 ¯ x+2Yt ¯ xg−2t ¯ x2g+t ¯ x) (V) 36 g = [t + −M∗ ¯ x ] t[1+ −M∗ ¯ x (−2M∗+ ¯ x)] g = [t + −M∗ ¯ x ] t(−2M∗+ ¯ x+ 2(M∗)2 ¯ x + −M∗ ¯ x ¯ x ) g = [t ¯ x−M∗] (−2M∗t ¯ x+ ¯ x2t +2(M∗)2t −M∗ ¯ xt) g = [t ¯ x−M∗] ((2t)(M∗)2−(3 ¯ xt)M∗+( ¯ x2t)) (2t)(M∗)2−(3 ¯ xt)M∗+( ¯ x2t) = [t ¯ x−M∗] g (2t)(M∗)2−(3 ¯ xt)M∗+( ¯ x2t) = t ¯ x g − M∗ g (2t)(M∗)2−(3 ¯ xt)M∗+( ¯ x2t)− t ¯ x g + M∗ g = 0 (2tg)(M∗)2−(3 ¯ xtg)M∗+( ¯ x2tg)−t ¯ x+M∗ = 0 (2tg)(M∗)2−(2 ¯ xtg)M∗+( ¯ x2tg)−t ¯ x = 0 Now that we have a function in the form of ax2+bx+c = 0 we can use the quadratic equation to solve for M∗ M∗ = −b± √b2−4ac 2a where a = (2tg) b = (−2 ¯ xtg) c = ( ¯ x2tg−t ¯ x) (VI) 37 −g = [t + −M∗ ¯ x ] t[1+ −M∗ ¯ x (−2M∗+ ¯ x)] −g = [t + −M∗ ¯ x ] t(−2M∗+ ¯ x+ 2(M∗)2 ¯ x + −M∗ ¯ x ¯ x ) −g = [t ¯ x−M∗] (−2M∗t ¯ x+ ¯ x2t +2(M∗)2t −M∗ ¯ xt) −g = [t ¯ x−M∗] ((2t)(M∗)2−(3 ¯ xt)M∗+( ¯ x2t)) (2t)(M∗)2−(3 ¯ xt)M∗+( ¯ x2t) = [t ¯ x−M∗] −g (2t)(M∗)2−(3 ¯ xt)M∗+( ¯ x2t) = −t ¯ x g + M∗ g (2t)(M∗)2−(3 ¯ xt)M∗+( ¯ x2t)+ t ¯ x g − M∗ g = 0 (2tg)(M∗)2−(3 ¯ xtg)M∗+( ¯ x2tg)+t ¯ x−M∗ = 0 (2tg)(M∗)2−(4 ¯ xtg)M∗+( ¯ x2tg)+t ¯ x = 0 Now that we have a function in the form of ax2+bx+c = 0 we can use the quadratic equation to solve for M∗ M∗ = −b± √b2−4ac 2a where a = (2tg) b = (−4 ¯ xtg) c = ( ¯ x2tg+t ¯ x) ⋄ The author created code in the computer program R to solve this complex equation for M∗, given t, ¯ x, and g. The code is in appendix A. 38 Chapter 5 EXAMPLES After the above results were found the author calculated various examples for the equations resulting from Lemma 0.3, Lemma 0.4, and equation (16). Each are displayed in the tables below: Lemma 0.3: Table 1: Various examples of Cuddington’s result solved for M∗. Let t = Let mean ( ¯ x =) M∗ = −ln(1−t) ¯ x M∗ = .15 1000 M∗ = −ln(1−.15)1000 M∗ = $162.52 .30 1000 M∗ = −ln(1−.30)1000 M∗ = $356.67 .30 2000 M∗ = −ln(1−.30)2000 M∗ = $713.35 Lemma 0.4: (I) M∗ = −b± √b2−4ac 2a where a = (g−gt −2t) b = (−gt ¯ x+gt2 ¯ x−gY +gYt +3t ¯ x) c = (Ygt ¯ x−Ygt2 ¯ x−t ¯ x2) Table 2: Examples of Cardon and Showalter’s result solved for M∗ with the Isoelastic Utility Function. Let t = Let Y = Let g = Let mean ¯ x = M∗ = .15 50,000 .5 1000 M∗ = $145.71 .30 50,000 .5 1000 M∗ = $295.02 .15 50,000 .5 2000 M∗ = $282.50 (II) M∗ = −b± √b2−4ac 2a 39 where a = (a′−ta′−2+2g) b = (t2a′ ¯ x−a′t ¯ x−Ya′+Yta′−b+gb′+2t ¯ x+ ¯ x−2t ¯ xg− ¯ xg) c = (Ya′t ¯ x−Yt2a′ ¯ x+b′t ¯ x−gb′t ¯ x− ¯ x2tg) For simplicity we let a′ = 1 and b′ = 0 making the above a = (1−t −2+2g) b = (t2 ¯ x−t ¯ x−Y +Yt +2t ¯ x+ ¯ x−2t ¯ xg− ¯ xg) c = (Yt ¯ x−Yt2 ¯ x− ¯ x2tg) Table 3: Examples of Cardon and Showalter’s result solved for M∗ with the Hyperbolic Utility Function. Let t = Let Y = Let g = Let mean ¯ x = M∗ = .15 50,000 .5 1000 M∗ = $150.18 .30 50,000 .5 1000 M∗ = $330.51 .15 50,000 .5 2000 M∗ = $300.74 (III) M∗ = −b± √b2−4ac 2a where a = (3− ¯ x−t) b = (t2 ¯ x−t ¯ x−Y +Yt +2t ¯ x) c = (Yt ¯ x−Yt2 ¯ x− ¯ x2) 40 Table 4: Examples of Cardon and Showalter’s result solved for M∗ with the Logarithmic Utility Function. Let t = Let Y = Let mean ¯ x = M∗ = .15 50,000 1000 M∗ = $150.37 .30 50,000 1000 M∗ = $301.02 .15 50,000 2000 M∗ = $301.49 (IV) M∗ = −b± √b2−4ac 2a where a = (1−t +4tg) b = (Yt −Y −t ¯ x+2 ¯ xg+t2 ¯ x+4t ¯ xg−1) c = (Yt ¯ x−2Y ¯ xg−Yt2 ¯ x+2Yt ¯ xg−2t ¯ x2g+t ¯ x) Table 5: Examples of Cardon and Showalter’s result solved for M∗ with the Quadratic Utility Function. Let t = Let Y = Let g = Let mean ¯ x = M∗ = .15 50,000 .5 1000 M∗ = $149.63 .30 50,000 .5 1000 M∗ = $298.97 .15 50,000 .5 2000 M∗ = $298.52 (V) M∗ = −b± √b2−4ac 2a 41 where a = (2tg) b = (−2 ¯ xtg) c = ( ¯ x2tg−t ¯ x) Table 6: Examples of Cardon and Showalter’s result solved for M∗ with the Exponential Utility Function. Let t = Let g = Let mean ¯ x = M∗ = .15 .5 1000 M∗ = $494.34 .30 .5 1000 M∗ = $497.67 .15 .5 2000 M∗ = $994.34 (VI) M∗ = −b± √b2−4ac 2a where a = (2tg) b = (−4 ¯ xtg) c = ( ¯ x2tg+t ¯ x) Table 7: Examples of Cardon and Showalter’s result solved for M∗ with the Negative Exponential Utility Function. Let t = Let g = Let mean ¯ x = M∗ = .15 .02 1000 M∗ = $110.03 .30 .02 1000 M∗ = $166.97 .15 .02 2000 M∗ = $176.29 42 Equation (16): Table 8: Various examples of Ramsay and Oguledo’s result solved for M∗. Let x∗ = M∗ = x∗ 700 M∗ = 700 1000 M∗ = 1000 2500 M∗ = 2500 43 Chapter 6 CONCLUSIONS/DISCUSSION In conclusion, determining the optimal contribution to an FSA depends on utility, risk aver-sion, income, tax rate, and previous medical history. The author found the following results. The Taylor serice first approximation was more than sufficient. This is determined when you compare the results in the tables above to the porpotion given by M∗ ¯ x . The results in the tables also showed for Isoelastic, Hyperbolic, and Quadratic utility functions it is clear the risk tolerance parame-ter, g, does not have a large effect on maximum contribution. Also a proportional increase in tax bracket and/or previous medical expenses have proportional effects on maximum contribution. For the Logarithmic utility function the risk tolerance parameter, g, is not a factor, and a proportional increase in tax bracket and/or previous medical expenses have proportional effects on maximum contribution. For the Exponential utility function income, Y, is not a factor. Also a change in tax bracket does not have a large impact, while a change in previous medical expenses does. Fi-nally the Negative Exponential utility function income, Y, is not a factor, and the risk tolerance parameter, g, must be very small to get proper results. Also a proportional increase in tax bracket and/or previous medical expenses have proportional effects on maximum contribution. Note that Isoelastic, Hyperbolic, Logarithmic, Quadratic utility functions are similar to one another, while Exponential and Negative Exponential are different. Depending on you and your healthcare, one of the results from above should be able to help you determine what formula works best for you and your family. I believe future research can be done by considering other loss distributions, exploring the idea of co-insurance, and documenting the effect the Affordable Health Care Act will have on FSAs in the coming years. 44 REFERENCES Cardon, J.H. and Showalter, M.H. (2003). “Flexible Spending Accounts as Insurance”. The Journal of Risk and Insurance, Vol.70,No.1, pgs.43-51 Cather, David A. "A Gentle Introduction To Risk Aversion And Utility Theory." Risk Management & Insurance Review 13.1 (2010): 127-145. Business Source Complete. Web. 29 Aug. 2013. Cuddington, J.T. (1999). “Optimal Annual Contributions to Flexible Spending Accounts: a Rule-of-Thumb”. Economics Letters. 62, 59-61 Grover, Ekta. (2010). “Measures of Risk Aversion.” Retrieved from http://thelittlescientists.files.wordpress.com/2010/10/macro-key-notes-2.pdf Moss, Charles B. (2010, October 05). “Utility functions, risk aversion coeficients and.” Retrieved from http://ricardo.ifas.ufl.edu/aeb6182.risk/Slides18-2010.pdf Norstaad, John. (1991, March 29). “An Introduction to Utility Theory.” Ramsay, C.M. and Oguledo, V.I. (2011). “Optimum Allocations to Health Care Flexible Spending Accounts.” North American Actuarial Journal, Vol. 13, No.3, pgs.448-467 Schweitze, M. and Asch, D.A. (1996). “The Role of Employee Flexible Spending Accounts in HealthCare Financing.” American Journal of Public Health, Vol. 86, No. 8. pgs. 1079-1081 Vanden, J. M. (2012), “General Properties of Isoelastic Utility Economics.” Mathematical Finance. doi: 10.1111/mafi.12010 45 APPENDICES 46 APPENDIX A R CODE 47 R CODE CalcMiso.r calc.m.iso <- function(x,y,t,g){ # x = mean healthcare expenses # y = income # t = tax rate # g = utility function coefficient a <- g - g*t - 2*t b <- - g*t*x + g*t^2*x - g*y + g*y*t + 3*t*x c <- y*g*t*x - y*g*t^2*x - t*x^2 m1 <-(-b - sqrt(b^2 - 4*a*c))/(2*a) m2 <-(-b + sqrt(b^2 - 4*a*c))/(2*a) return(list(m1,m2)) } calc.m.iso(1000,50000,0.2,0.5) CalcMhyp.r calc.m.hyp <- function(x,y,t,g){ # x = mean healthcare expenses # y = income # t = tax rate # g = utility function coefficient a <- -1 +3*t -2*g*t b <- g*t*x -t^2*x +y -t*y c <- t*x -g*t*x -t*x*y +t^2*x*y m1 <-(-b - sqrt(b^2 - 4*a*c))/(2*a) m2 <-(-b + sqrt(b^2 - 4*a*c))/(2*a) return(list(m1,m2)) } calc.m.hyp(1000,50000,0.2,0.5) 48 R CODE CalcMlog.r calc.m.log <- function(x,y,t){ # x = mean healthcare expenses # y = income # t = tax rate a <- -1 +3*t b <- -t^2*x +y -t*y c <- t*x -t*x*y +t^2*x*y m1 <-(-b - sqrt(b^2 - 4*a*c))/(2*a) m2 <-(-b + sqrt(b^2 - 4*a*c))/(2*a) return(list(m1,m2)) } calc.m.log(1000,50000,0.2) CalcMquad.r calc.m.quad <- function(x,y,t,g){ # x = mean healthcare expenses # y = income # t = tax rate # g = utility function coefficient a <- -2*g -2*g*t b <- -1 +4*g*t*x -2*g*t^2*x +2*g*y -2*g*t*y c <- t*x -2*g*t*x -2*g*t*x*y +2*g*t^2*x*y m1 <-(-b - sqrt(b^2 - 4*a*c))/(2*a) m2 <-(-b + sqrt(b^2 - 4*a*c))/(2*a) return(list(m1,m2)) } calc.m.quad(1000,50000,0.2,0.5) 49 R CODE CalcMexpo.r calc.m.expo <- function(x,t,g){ # x = mean healthcare expenses # t = tax rate # g = utility function coefficient a <- 2*t*g b <- 1-g*t*x c <- -t*x +g*t*x m1 <-(-b - sqrt(b^2 - 4*a*c))/(2*a) m2 <-(-b + sqrt(b^2 - 4*a*c))/(2*a) return(list(m1,m2)) } calc.m.expo(1000,0.2,0.5) CalcMnegexpo.r calc.m.negexpo <- function(x,t,g){ # x = mean healthcare expenses # t = tax rate # g = utility function coefficient a <- 2*t*g b <- 1 +g*t*x c <- -t*x -g*t*x m1 <-(-b - sqrt(b^2 - 4*a*c))/(2*a) m2 <-(-b + sqrt(b^2 - 4*a*c))/(2*a) return(list(m1,m2)) } calc.m.negexpo(1000,0.2,0.5) 50 VITA Cassidi Jacobs was born in Atlanta, TX, in 1991. She graduated Valedictorian from Avinger High School in 2009. In her efforts to continue her education she applied for and received nine scholarships, including two national scholarships, totaling $13,000. She then attended Paris Junior College, meeting the qualifications of an Associate’s degree in mathematics and theatre, graduating in May of 2011. From there she transferred to Texas A&M University – Commerce where it took her a year and a half to finish her Bachelors of Science in mathematics with a minor in theatre. After her graduation in December 2012 Cassidi began the course work for a Master’s of Science in mathematics. While she worked on this endeavor Cassidi was a graduate assistant teaching (GAT) for the mathematics department. Cassidi loves teaching at the university level, and will begin to teach at Northeast Texas Community College Fall 2014. 2313 CR 1597 Avinger TX, 75630 Email:Calford3@live.com |