Advanced Microeconomic Theory Jehle and Reny Solution Manual

Solutions Manual for Advanced Microeconomic Theory I Jianfei Shen School of Economics, Shanghai University of Finance and Economics E-mail address: jianfeishen@163.com URL: http://www.shufe.edu.cn Thanks. Contents Part 1. Consumer Theory 1 Chapter 1. Solution for Exercise I 3 1. (JR, 1.4) 3 2. (JR, 1.27) 3 3. (JR, 1.56) 4 4. (JR, 1.24) 5 5. (Homogeneity) 6 Chapter 2. Solution for Exercise II 9 1. (JR, 1.36) 9 2. (JR, 2.10) 9 3. (MWG, 2.F.7) 10 4. (JR, 1.62) 11 5. (JR, 2.9) 11 Part 2. Production and Choice under Uncertainty 13 Chapter 3. Solution for Exercise III 15 1. (JR, 3.36) 15 2. (JR, 3.39) 16 3. (The Allais Paradox) 17 4. (The Simplex) 18 5. (Independence Axiom) 19 Chapter 4. Solution for Exercise IV 21 1. (MWG, 6.C.16) 21 2. (MWG, 6.C.20) 22 3. (Betweenness Axiom) 23 4. (Quadratic v.N-M Utility Function) 24 i Part 1 Consumer Theory CHAPTER 1 Solution for Exercise I 1. (JR, 1.4) Proof. Suppose that x, y, z ∈ RL+, and x � y, y � z. By Definition 1.21, we have (1.1) x � y ⇔ x < y ∧ y � x; and (1.2) y � z ⇔ y < z ∧ z � y. The transitivity implies that x < z. Suppose that z < x. Since y < z [by equation (1.2)], the transitivity then implies that y < x. But this contradicts (1.1). Thus z � x. Hence x � z. Suppose x ∼ y, and y ∼ z. By Definition 1.32, we have (1.3) x ∼ y ⇔ x < y ∧ y < x; and (1.4) y ∼ z ⇔ y < z ∧ z < y. The transitivity implies that x < z and z < x. Hence x ∼ z. � 2. (JR, 1.27) Solution. If x1 < x2, u(x) = max[ax1, ax2] + min[x1, x2] = ax2 + x1; If x2 < x1, u(x) = max[ax1, ax2] + min[x1, x2] = ax1 + x2. Case i: a < p1p2 < 1 a . In this case, x1 = x2 = w p1 + p2 . 1Jehle & Reny, 2001, P6 2Jehle & Reny, 2001, P7 3 4 1. SOLUTION FOR EXERCISE I A x2 x1B C O Figure 2.1. The Indifference Curve Case ii: p1p2 > 1 a . In this case, there is a corner solution x1 = 0, x2 = w p2 . Case iii: p1p2 < a. In this case, there is a corner solution, too x1 = w p1 ,x2 = 0. Case iv: p1p2 = 1 a . In this case, the UMP is x1 = w − p2x2 p1 ,x2 ∈ [ w p1 + p2 , w p2 ] . Case v: p1p2 = a. In this case, the UMP is x1 = w − p2x2 p1 ,x2 ∈ [ 0, w p1 + p2 ] . � 3. (JR, 1.56) Solution. Use the monotone transformation of the given utility function3: v(x) = lnu(x) = n∑ i=1 bi ln(xi − ai). The UMP can written as max n∑ i=1 bi ln(xi − ai); s.t. n∑ i=1 pixi ≤ w. 3Stone, J.E. (1954): Linear Expenditure Systems and Demand Analysis: An Application to the Pattern of British Demand. Economic Journal 64: 511–27 4. (JR, 1.24) 5 Then form the Lagrangian, we obtain L = n∑ i=1 bi ln(xi − ai) + λ ( w − n∑ i=1 pixi ) . The first-order condition of the UMP yields (1.5) ∂L ∂xi = bi xi − ai − λpi = 0, ∀i. Summarize all of the first-order conditions: (1.6) 1 = λ n∑ i=1 pi(xi − ai)⇒ λ = 1∑n i=1 pi(xi − ai) . With (1.5) and (1.6), we have bi = pi(xi − ai)∑n i=1 pi(xi − ai) = pi(xi − ai) w −∑ni=1 piai . So xi = ai + bi (w − ∑n i=1 piai) pi , ∀i. � 4. (JR, 1.24) Solution. A utility function that represents a preference relation < is not unique. For any strictly increasing function: f : R → R, v(x) = f(u(x)) is a new utility function representing the same preferences as u(·)4. a: f(x) = u(x) + [u(x)]2. ∂f(·) ∂u(·) = 1 + 3[u(x)] 2 > 0. So this is a strictly increasing function, and it represents thesame <. b: f(x) = u(x) + [u(x)]2. ∂f(·) ∂u(·) = 1 + 2u(x). • ∂f(·)∂u(·) ≤ 0 if u(x) ≤ − 12 . In this case, f(x) does not represents the same <; • ∂f(·)∂u(·) > 0 if u(x) > − 12 . In this case, f(x) represents the same <. c: f(x) = u(x) + ∑n i=1 xi. Not necessarily. f represents the same preferences as u if u(x) is a monotonic transformation of ∑n i=1 xi. Otherwise, they may not represent the same preferences. For example, suppose u(x) = x1. Then u(1, 0, . . . , 0) > u(0, 3, 0, . . . , 0), but f(1, 0, . . . , 0) = 1 + 1 = 2 < 3 = 0 + 3 = f(0, 3, 0, . . . , 0). � 4Let x, y ∈ RL+. Since u(·) represents <, x < y iff u(x) ≥ u(y). Since f(·) is strictly increasing, u(x) ≥ u(y) iff v(x) ≥ v(y). Hence x < y iff v(x) ≥ v(y). Therefore v(·) represents <. This completes the proof. 6 1. SOLUTION FOR EXERCISE I 5. (Homogeneity) Proof. a: We want to show that ∀p ∈ RL++, y ≥ 0, α ≥ 0, and x ∈ RL+, if x is the Marshallian demand at (p, y), then x′ = αx is the Marshallian demand at (p, αy). We’ll complete this proof with two steps: Step I: we prove that the consumption bundle x′ is affordable at (p, αy). Note that p · x ≤ y ⇒ p · αy ≤ αy, so x′ = αx is affordable at (p, αy) indeed. Step II: we prove that the consumption bundle x′ is the Marshallian demand function at (p, αy). Let x′′ ∈ RL+ and p · x′′ ≤ αy. Then p · x ′′ α ≤ y. Hence u ( x′′ α ) < u(x), (by the Strictly quasiconcave) since x is the Marshallian demand function at (p, y). Thus, by the homogeneity of u(·): u ( x′′ α ) = 1 α u (x′′) < u(x)⇒ u(x′′) ≤ αu(x) = u(αx). Hence αx is the Marshiallian demand function at (p, αy), and x(p, αy) = αx(p, y). b: By the result of part (a), we have v(p, αy) = u(x(p, αy)) = u(α · x(p, y)) = α · u(x(p, y)) = α · v(p, y). This proves that the indirect utility function v(·) is homogeneous of degree one in y. c: In this part, we’ll show that ∀x ∈ RL+ and α ≥ 0, we have u(αx) = αu(x). We first prove that x(p, αy) = α · x(p, y), where x(·) is the Marshiallian demand function. Since v(·, ·) is homogeneous of degree one in y: v(p, αy) = αv(p, y), we have ∇pv(p, αy) = α∇pv(p, y), and ∇yv(p, αy) = α · ∇yv(p, αy) = α · ∇yv(p, y)⇒ ∇yv(p, αy) = ∇yv(p, y). Then by Roy’s identity: xi(p, y) = −∂v(p, y)/∂pi ∂v(p, y)/∂y , i = 1, . . . , L, we have x(p, αy) = −∇pv(p, αy)∇yv(p, αy) = −α · ∇pv(p, y) ∇yv(p, y) = α · x(p, y). 5. (HOMOGENEITY) 7 Now, again by the homogeneity of v(·, ·), we have u(α · x(p, y)) = u(x(p, αy)) = v(p, αy) = α · v(p, y) = α · u(x(p, y)). This completes the proof of u(α · x(p, y)) = α · u(x(p, y)). � CHAPTER 2 Solution for Exercise II 1. (JR, 1.36) Solution. a. By definition, e(p, u) , min x∈RL+ p · x s.t. u(x) ≥ u The solution to the EMP is known as the Hicksian demand function: h(p, u). Hence u(x0) ≥ u0, and so e(p, u0) ≤ p · x0, with the equality sign holds when p = p0. b. It follows immediately from part a that f(p) = e(p, u)− p · x0 is maximized at p = p0: max f(p) = 0. c. If f(p) is differentiable at p0, (2.1) ∇pf ( p0 ) = 0. d. If e(p, u) is differentiable in p, then by equation (2.1) we have ∇pf = ∇pe− x0 = 0, so h(p, u) = ∇pe(p, u). � 2. (JR, 2.10) Solution. a. Bundle x0 x1 x2 p0 42 48 40∗ Price p1 33∗ 36 39(∗) p2 52 48∗ 51 9 10 2. SOLUTION FOR EXERCISE II These consumption bundles are plausible in that the WARP is satisfied. In partic- ular, we note p1x1 = 36 > 33 = p1x0, p0x0 = 42 < 48 = p0x1, so (2.2) x1Rx0. p0x0 = 42 > 40 = p0x2, p2x2 = 50 < 52 = p2x0, so (2.3) x0Rx2. p2x2 = 50 > 48 = p2x1, p1x1 = 36 < 39 = p1x2, so (2.4) x2Rx1. b. Now, from (2.2) and (2.3) we have x1Rx0, x0Rx2; but from (2.4), something utterly revolting occurs: x2Rx1. So the revealed preference R is intransitivity. � 3. (MWG, 2.F.7) Proof. If the Walrasian demand function x(p, w) satisfies Walras’ law, then for all p and w: (2.5) L∑ `=1 p` ∂x`(p, w) ∂pk + xk(p, w) = 0, ∀ k = 1, . . . , L, or (2.6) p ·Dpx(p, w) + x(p, w)′ = 0′ and (2.7) L∑ `=1 p` ∂x`(p, w) ∂w = 1, or (2.8) p ·Dwx(p, w) = 1. If the Walrasian demand function x(p, w) is h.o.d. 0, then for all p and w: (2.9) L∑ k=1 ∂x`(p, w) ∂pk pk + ∂x`(p, w) ∂w w = 0, ∀ ` = 1, . . . , L, or (2.10) Dpx(p, w)p+Dwx(p, w)w = 0. 5. (JR, 2.9) 11 By the Slutsky Equation: (2.11) ∂h`(p, u) ∂pk = ∂x`(p, w) ∂pk + ∂x`(p, w) ∂w xk(p, w), ∀ `, k or equivalently, (2.12) Dph(p, u) = Dpx(p, w) +Dwx(p, w)x(p, w)′, where Dph(p, u) is equal to the matrix S(p, w) = s11(p, w) · · · s1L(p, w)... . . . ... sL1(p, w) · · · sLL(p, w)  with s`k(p, w) = ∂x`(p, w) ∂pk + ∂x`(p, w) ∂w xk(p, w). So we have p · S(p, w) = p ·Dpx(p, w) + p ·Dwx(p, w)x(p, w)′ [from (2.12)] = p ·Dpx(p, w) + x(p, w)′ [from (2.8)] = 0′, [from (2.6)] (2.13) and S(p, w)p = Dpx(p, w)p+Dwx(p, w)x(p, w)′p [from (2.12)] = Dpx(p, w)p+Dwx(p, w)w [from Walras’ law] = 0. [from (2.10)] (2.14) � 4. (JR, 1.62) Solution. Since p · S(p, w) = 0, S(p, w)p = 0 and by symmetric, we have a = −8, b = 2, p = 32. � 5. (JR, 2.9) Proof. a. In the case of two goods, equation (2.13) can be written as p · S(p, w) = [p1 p2] [s11(p, w) s12(p, w)s21(p, w) s22(p, w) ] = [ p1s11(p, w) + p2s21(p, w) p1s12(p, w) + p2s22(p, w) ] = [ 0 0 ] . (2.15) From equation (2.15) we have s21(p, w) = −p1 p2 s11(p, w), 12 2. SOLUTION FOR EXERCISE II and s12(p, w) = −p2 p1 s22(p, w). And the equation (2.14) becomes S(p, w)p = [ s11(p, w) s12(p, w) s21(p, w) s22(p, w) ] [ p1 p2 ] = [ p1s11(p, w) + p2s12(p, w) p1s21(p, w) + p2s22(p, w) ] = [ 0 0 ] . (2.16) From equation (2.16) we have s12(p, w) = −p1 p2 s11(p, w), and s21(p, w) = −p2 p1 s22(p, w). Thus s12(p, w) = s21(p, w). � Part 2 Production and Choice under Uncertainty CHAPTER 3 Solution for Exercise III 1. (JR, 3.36) Proof. We can get some intuition from the following figure 1. x2 x1x01 tx 0 1 x02 tx02 A B O y0 y1 Figure 1.1. Homothetic production function Now we give the regular proof. The cost function for all homothetic production function can be written (3.1) c(w, y) = h(y)φ(w), where φ(w) is linear homogeneous. 1 Then using the Shephard’s Lemma2, we have xi(w, y) = ∂c(w, y) ∂wi = h(y)φi(w), xj(w, y) = ∂c(w, y) ∂wj = h(y)φj(w). Hence (3.2) xi(w, y) xj(w, y) = φi(w) φj(w) , ψ(w), that is, ∂ (xi(w, y)/xj(w, y)) ∂y = 0. 1Any cost function is linear homogeneous in the factor prices. 2JR, Theorem 3.2, p.129 15 16 3. SOLUTION FOR EXERCISE III � 2. (JR, 3.39) Proof. a: If the production function is CRS, then c(w, y) = n∑ i=1 wixi(w, y) = n∑ i=1 λfi xi(w, y) [ From the F.O.C. ] = λ n∑ i=1 xi(w, y) fi = λy [ From the Euler Equation ] = ∂c(w, y) ∂y y [ By the interpretation of λ ] or (3.3) ∂c(w, y) c(w, y) = ∂y y . Solve the partial differential equation (3.3), we get (3.4) c(w, y) = yφ(w), where φ(w) is a function of factor prices w only. b: If c(w, y) = yφ(w), then from the F.O.C. of the cost minimization problem that for ∀ x(w, y) > 0 (3.5) wi = λfi, and (3.6) λ = ∂c(w, y) ∂y . Combine equation (3.5) and (3.6), we obtain wi = φ(w)fi, or xifi = wixi φ(w) . Hence, for all xi(w, y) n∑ i=1 xifi = n∑ i=1 wixi φ(w) = 1 φ(w) n∑ i=1 wixi = c(w, y) φ(w) = yφ(w) φ(w) = f(x). 3. (THE ALLAIS PARADOX) 17 It follows from the Euler’s Theorem3 that (3.7) f(tx) = tf(x), that is, the production function is CRS. � 3. (The Allais Paradox) Proof. Defining X = {x1, x2, x3} = {$ 0; $ 1, 000, 000; $ 5, 000, 000}, these four gambles are seen form a parallelogram in the (p1, p3) triangle, as in the following figure. p3 p1 (a) p3 p1 1 1 p3 p1 p3 p1 (b) 1 1g1 g2 g3 g4 g1 g2 g3 g4 Figure 3.1. Indifference curves and the Allais Paradox The Independence Axiom is in fact equivalent to linearity in the probabilities. IA implies that: (1) Indifference curves are straight lines: if, ∀ g, g′ ∈ G , we have g ∼ g′ implies αg + (1− α)g = g ∼ αg′ + (1− α)g, ∀ α ∈ (0, 1). (2) Indifference curves are parallel lines: if, ∀ g, g′, g′′ ∈ G , we have g ∼ g′, the IA implies that αg + (1− α)g′′ ∼ αg′ + (1− α)g′′, ∀ α ∈ (0, 1). 3Euler’s Theorem: f(x) is homogeneous of degree r iff rf(x) = n X i=1 xifi. 18 3. SOLUTION FOR EXERCISE III Now consider the conditions in the Allais Paradox. A preference for g1 in the first pair gambles would indicate that the individual’s indifference curves were relatively steep, and hence a preference for g4 in the second pair.4 If, on the contrary, g1 is preferred in the first pair, and g3 in the second, which implies that indifference curves are parallel but rather fan out, as in figure (b). Now we turn to do the job follows another way. We can rewrite the gambles as g1 = (0.10 ◦ 1, 0.01 ◦ 1, 0.89 ◦ 1); g2 = (0.10 ◦ 5, 0.01 ◦ 0, 0.89 ◦ 1); g3 = (0.10 ◦ 5, 0.01 ◦ 0, 0.89 ◦ 0); g4 = (0.10 ◦ 1, 0.01 ◦ 1, 0.89 ◦ 0). Consider the following three gambles g5 = Get 1 with certainty; g6 = ( 0.10 0.11 ◦ 5, 0.01 0.11 ◦ 0 ) ; g7 = Get 0 with certainty. By the Completeness Axiom, we know either g5 < g6 or g6 < g5. (1) If g5 < g6: by the IA, we have 0.11g5 + 0.89g5 < 0.11g6 + 0.89g5, or g1 < g2; and 0.11g5 + 0.89g7 < 0.11g6 + 0.89g7, or g4 < g3. (2) If g6 < g5: we can do the job with the same logic as part (1), and get g2 < g1, and g3 < g4. � 4. (The Simplex) Solution. The slope: Keeping the level of v.N.M utility constant p1u(x1) + p2u(x2) + (1− p1 − p2)u(x3) = Const., and varying p1 and p2 alone, one has, locally, u(x1) + dp2 dp1 u(x2) + (−1− dp2 dp1 )u(x3) = 0, or dp2 dp1 = u(x3)− u(x1) u(x2)− u(x3) ≥ 0, 4In the alternative case of relatively flat indifference curves, the gambles g2 and g3 would be preferred. 5. (INDEPENDENCE AXIOM) 19 since x2 > x3 > x1. Direction where the utility increasing: Since upward movements in the triangle increase p2 at the expense of p3 (i.e. shift probability from the outcome x3 up to x2) and leftward movements reduce p1 to the benefit of p3 (shift probability from x1 to x3), these movements (and more generally, all northwest movements) lead to stochastically dominating gambles and would accordingly be preferred. p2 p1 1 1 Figure 4.1. Expected utility indifference curves in the sim- plex diagram � 5. (Independence Axiom) Proof. Let us assume without loss of generality that the elements of A have been indexed so that a1 < a2 < · · · < an. Now, let gk, 0 ≤ k ≤ n, be the gamble that yields outcome k with probability one: gk = (0 ◦ a1, 0 ◦ a2, . . . , 1 ◦ ak, . . . , 0 ◦ an). Then a1 < gk < an, since all of them can be identified with sure outcomes. Let g = (p1 ◦ a1, . . . , pn ◦ an) be any gamble in G , then g = n∑ k=1 pkgk. If there is only one k, s.th. pk = 1, that is, pj = 0,∀ j 6= n, there is nothing to prove. So let N = #{(1, . . . , n) : pk 6= 0, 0 ≤ k ≤ n} > 1 20 3. SOLUTION FOR EXERCISE III and suppose the proposition a1 < N −1∑ k=1 pkgk < an is true for N − 1. By the definition of a compound gamble, N∑ k=1 pkgk = (1− pN ) N −1∑ k=1 pk 1− pN gk + pN gN . By the induction hypothesis, a1 < (1− pN ) N −1∑ k=1 pk 1− pN gk < an. Hence, by the independence axiom, we have (1− pN )a1+ pN gN < (1− pN ) N −1∑ k=1 pk 1− pN gk + pN gN < (1− pN )an+ pN gN . Applying the axiom once again, we obtain a1 = (1− pN )a1 + pN a1 < (1− pN )a1 + pN gN ; (1− pN )an + pN gN < (1− pN )an + pN an = an. Hence, by the transitivity, a1 < g < an,∀ g ∈ G . � CHAPTER 4 Solution for Exercise IV 1. (MWG, 6.C.16) Solution. The maximum amount the person is willing to buy the gamble: (4.1) 0.5u(w −Rb − y) + 0.5u(w −Rb + x) = u(w), where Rb is the maximal buying price. w − y w w + x A u(w) B u(·) w Rb Rb Figure 1.1. The maximum amount the person is willing to buy the gamble The minimum amount the person is willing to sell the gamble: (4.2) 0.5u(w − y) + 0.5u(w + x) = u(w +Rs), where Rs is the minimal selling price. In general, these two prices are different. However, if u(·) is CARA, then they are the same. In face, equation (4.1) and (4.2) can be restated as (4.3) CEw−Rb = w, (4.4) CEw = w +Rs, 21 22 4. SOLUTION FOR EXERCISE IV w − y w w + x C u(w) D u(·) ww +Rs Rs E[u(·)] Figure 1.2. The minimum amount the person is willing to sell the gamble where CEw−Rb and CEw are certainty equivalence for (4.1) and (4.2), respectively. The CARA implies that1 (w −Rb)− CEw−Rb = w − CEw, thus Rb = Rs. � 2. (MWG, 6.C.20) Proof. u(CE) = 0.5u(x+ ε) + 0.5u(x− ε), where CE is the Certainty Equivalent. Hence (4.5) u′(CE) ∂CE ∂ε = 0.5u′(x+ ε)− 0.5u′(x− ε), (4.6) u′′(CE) ( ∂CE ∂ε )2 + u′(CE) ∂2CE ∂ε2 = 0.5u′′(x+ ε) + 0.5u′′(x− ε). Thus lim ε↓0 ∂2CE ∂ε2 = lim ε↓0 0.5u′′(x+ ε) + 0.5u′′(x− ε)− u′′(CE) (∂CE∂ε )2 u′(CE) = u′′(x) u′(x) = −Ra(x), 1See MWG (1995) Section 6.C. 3. (BETWEENNESS AXIOM) 23 since lim ε↓0 ∂CE ∂ε = lim ε↓0 0.5u′(x+ ε)− 0.5u′(x− ε) u′(CE) = 0, and lim ε↓0 u(CE) = u(x). � 3. (Betweenness Axiom) Solution. The Betweenness Axiom2 only requires that indifference sets be convex, i.e., if an individual is indifferent between two lotteries, then any probability mixture of these two is equally good: if g ∼ g′, then λg + (1− λ)g′ ∼ g, ∀λ ∈ [0, 1]. Essentially, the betweennss axiom is a substantially weaker version of the contro- versial independence axiom. 3 The axioms 1-4, 6, and the betweenness axiom means that the indifference curves are straight lines can be established in the same way as in Chapter 3, exercise (3). Note that we do not use the independence axiom in that exercise, in fact, betweenness axiom is suffices. Note also that these straight indifference curves need not be parallel, because the betweenness axiom imposes restrictions only on straight indifference curves and nothing on the relative positions of different indifference curves. p2 p1 1 1 Figure 3.1. Betweenness axiom means the indifference curves are straight lines, but need not be parallel. � 2See Dekel, E. (1986) for further discussion. 3See MWG Exercise 6.B.1A. 24 4. SOLUTION FOR EXERCISE IV 4. (Quadratic v.N-M Utility Function) a. Solution. The restrictions are u′(w) > 0, u′′(w) < 0, so b > −4cw¯; c < 0. � b. Solution. Ra(w) = − 4c b+ 4cw , so ∂Ra(w) ∂w = 16c2 (b+ 4cw)2 > 0. This means that the quadric utility functions are unsatisfactory. Not only do they imply that utility reaches a maximum, they also entail that the absolute degree of risk aversion is increasing in wealth, approaching infinity as utility approaches its maximum. Consequently, one is led to the absurd result that the willingness to gamble for a bet of fixed size should decrease as wealth is increased. � c. Proof. Before going through the proof, it is worthwhile to consider the intu- ition of the representation. The expected utility hypothesis suggests that prefer- ences toward gambles can be represented by the expected value of a v.N-M utility function E [ u(w) ] , where w is a random variable that represents the income from an uncertain gamble. Expected utility in general depends on the form of the function u(·) and on the distribution of w. Suppose the distribution of w can be completely characterized by a vector of parameters α. In particular, let w be distributed on the real line with a P.D.F. f(w,α). Then4 E [ u(w) ] = ∫ u(w)f(w,α) dw. The integral on the right-hand side of this equation is a function of α.5 If we let this integral be represented by u(α), then u(α) = E [ u(w) ] is a valid representation of preferences. 4From this subsection through the end of the chapter, we focus on continuous monetary variable for convenience. 5It is not a function of w since w is just the variable of integration. 4. (QUADRATIC V.N-M UTILITY FUNCTION) 25 Many problems in the economics of uncertainty are related to the trade-off between the average level of income and its degree of riskiness. Since the mean is a summary measure of average and the variance is a summary measure of risk, it will be particularly convenient to represent preferences by a function of the mean and variance of the income distribution. Unfortunately, this is not always possible, because in general the mean and variance do not completely determine the distri- bution of a random variable. There are many income streams that have the same mean and variance but different probability distributions. The expected utility associated with these income streams are different. Although u(α) is a valid repre- sentation of preferences, the vector α generally contains more than two parameters. Thus a utility function that depends only on mean and variance can be at best be viewed as an approximation to expected utility. There are some special cases, however, when a function involv