1 of 7 ECO 7113 Problem Set 2 Solution 1. Prove "as risky as" is a transitive relation. From page 6.1: "Let x and y be two random variables with ][][yxEE=. We say y is as risky as x for all Uu∈ if )]([)]([yxuEuE≥ for all Uu∈." Assume y is as risky as x and z is as risky as y From the definition above... y is as risky as x means ][][yxEE= and )]([)]([yxuEuE≥ z is as risky as y means ][][zyEE= and )]([)]([zyuEuE≥ Put these together: ][][][zyxEEE== )]([)]([)]([zyxuEuEuE≥≥ Therefore, we have ][][zxEE= and )]([)]([zxuEuE≥ (because numbers are transitive). That means z is as risky as x so "as risky as" is a transitive relation. 2. Prove first order stochastic dominance (FOSD)  second order stochastic dominance (SOSD). FOSD - equivalent definitions (from page 6.6): (a) ε+= xyd, 0≤ε (b) )]([)]([ yuExuE≥ ∀ increasing u (c) )()( tGtF≤ ∀ ],[ bat∈ (where )(tF and )(tG are CDFs of x and y, respectively) SOSD - equivalent definitions (from page 6.7): (a) δε++= xyd, 0≤ε, 0]|[=+εδxE (b) )]([)]([ yuExuE≥ ∀ increasing, concave u (c) [ ]0)()( ≤−TadttGtF ∀ Since the definitions above are equivalent, we just need to show any definition of FOSD implies any definition of SOSD. It seems they are matched up for fairly easy proofs, but (b) is the easiest. FOSD (b)  SOSD (b)... if )]([)]([ yuExuE≥ ∀ increasing u, the condition will also hold for all increasing and concave u.2 of 7 3. Assume {}321,, zzzZ = in the vN-M setup. Now prove Theorem 5.4 in Kreps (for this special case). You may assume his Lemma 5.6 has been proven. (Theorem and Lemma from page 4.4) A1.  ⊆ X x X is a preference relation A2. ∀ p, q, r ∈ X, qp  rqrp)1()1(αααα−+−+ ∀ α ∈ (0,1] A3. rqp  ∃ α, β ∈ (0,1) ∋ rpqrp)1()1(ββαα−+−+ ∀ p, q, r ∈ X Theorem 5.4 - ,X satisfies A1, A2, A3 iff ∃ u : Z → R such that qp ⇔ > )()(iiiizuqzup And )(zu is unique to positive affine transformation Lemma 5.6 - if ,X satisfies A1, A2, A3, then (a) Monotonicity - qp and 10≤<≤βα  qpqp)1()1(ααββ−+−+ (b) Solvability - p q  r and rp  ∃ unique ]1,0[*∈α such that rpq*)1(*~αα−+ (c) qp~ and ]1,0[∈α  rqrp)1()1(αααα−+−+ ∀ X∈r Class adds 2 more: (d) qp and sr and ]1,0[∈α  sqrp)1()1(αααα−+−+ (e) qp~ and ]1,0[∈α  qpp)1(αα−+ Necessary - Kreps didn't bother to cover this (because it's a "straightforward exercise")l Suppose the function u exists such that qp ⇔ )()()()()()(332211332211zuqzuqzuqzupzupzup ++>++ Proof of A1 - need to show that  is asymmetric and negatively transitive to prove it's a preference relation Asymmetric: x  y  not y  x X∈∀yx, Assume x  y That means )()()()()()(332211332211zuyzuyzuyzuxzuxzux ++>++ Since the left side is greater than the right side, it is not possible for the inequality to hold in the reverse in order to get y  x ∴ we have not y  x (i.e.,  is asymmetric) Negatively Transitive: not x  y and not y  z  not x  z Xzyx∈∀,, Assume not x  y and not y  z That means )()()()()()(332211332211zuyzuyzuyzuxzuxzux ++≤++ and )()()()()()(332211332211zuzzuzzuzzuyzuyzuy ++≤++ By transitivity of numbers: )()()()()()(332211332211zuzzuzzuzzuxzuxzux ++≤++ ∴ we have not x  z (i.e.,  is negatively transitive)3 of 7 Proof of A2 - Assume qp That means )()()()()()(332211332211zuqzuqzuqzupzupzup ++>++ Multiply both sides by α ∈ (0,1]: ==>3131)()(iiiiiizuqzupαα Add =−31)()1(iiizurα to both sides: ====−+>−+31313131)()1()()()1()(iiiiiiiiiiiizurzuqzurzupαααα Combine the summations on each side: [ ] [ ] ==−+>−+3131)()1()()1(iiiiiiiizurqzurpαααα That means rqrp)1()1(αααα−+−+ ∴ A2 holds Proof of A3 - Assume rqp That means ===>>313131)()()(iiiiiiiiizurzuqzup We can apply the intermediate value theorem (Kreps p.47) which says that if f is a continuous function which takes on a value 1y at the argument 1x and another value 2y at 2x , then for any value 3y between 1y and 2y there is some argument between 1x and 2x at which the function takes on the value 3y . Modify the IVT for this application by letting: uf= p=1x and ==311)(iiizupy r=2x and ==312)(iiizury q=3x and ==313)(iiizuqy That means there is a "line segment" joining p and r (i.e., convex combination: rp)1(−+ for (0,1)∈) such that some point on the segment is indifferent to q. Suppose that the indifference occurs at the specific value * qrp~)*1(*−+ which means [ ] ===−+3131)()(*)1(*iiiiiiizuqzurp4 of 7 Now choose some )1,0(,∈βα with *>α and *<β That means [ ] ==>−+3131)()()1(iiiiiiizuqzurpαα and [ ] ==<−+3131)()()1(iiiiiiizuqzurpββ Which can be rewritten rpqrp)1()1(ββαα−+−+ ∴ A3 holds Sufficient - this part of the proof is based on Kreps pp.49-50 Now suppose  satisfies A1, A2, and A3 (so that Lemma 5.6 holds) Let {}nzX⊆p be the set of degenerate lotteries; that is, the set of probability measures equivalent to getting each prize (1z , 2z , or 3z ) with certainty; In this example with three prizes, the set will be {}321,,zzzppp where )0,0,1(1=zp , )0,1,0(2=zp , and )1,0,0(3=zp Since {}zp is finite, we can relabel the prizes so that 321 zzzppp  (i.e., 1z is the best prize and 3z is the worst) For X∈p, define α=)(pf , where ppp ~)1(31zzαα−+ ... this is possible by part (b) of Lemma 5.6 We can see from the picture to the right that 1)(1=zf p , *)(2α=zf p , and 0)(3=zf p Now pick some X∈q with qp We can apply part (b) again to say qpqpq ~))(1()(31zzff −+ Since we picked q such that qp, we know 3131))(1()())(1()(zzzzffff pqpqpppp −+−+  From part (a) of Lemma 5.6 and we can conclude that )()(qpff> This argument also works in reverse so we can say that )(⋅f is a representation of  For all X∈qp, and ]1,0[∈α, repeated application of part (c) of Lemma 5.6 gives [][]3131))(1()()1())(1()(~)1(zzzzffff pqpqppppqp −+−+−+−+αααα Therefore, by definition of )(⋅f : )()1()(~))1((qpqpfffαααα−+−+ We want to show ==31)()(iiizupf p, which is done by induction (Kreps p.50) For Zzi∈ , define )()(izifzu p= (i.e., 1)(1=zu , *)(2α=zu , and 0)(3=zu ) 1 0 α* z1 z2 z3 αααα Prize p5 of 7 Suppose the