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Rubinstein2005-page119

Rubinstein2005-page119 - on a pair of lotteries p and q one...

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October 21, 2005 12:18 master Sheet number 117 Page number 101 Risk Aversion 101 1. Assume he possesses wealth x 0 . 2. Offer him a lottery that will reduce his wealth to 0 with prob- ability 1 / 2 and will increase his wealth to x 1 with probability 1 / 2 so that u ( x 0 ) < [ u ( 0 ) + u ( x 1 ) ] / 2. By the unboundedness of u , there exists such an x 1 . 3. If he loses, you are happy. If he is lucky, a moment before you give him x 1 , offer him a lottery that will give him x 2 with probability 1 / 2 and 0 otherwise, where x 2 is such that u ( x 1 ) < [ u ( 0 ) + u ( x 2 ) ] / 2. 4. And so on . . . Our decision maker will find himself with wealth 0 with probabil- ity 1! First-Order Stochastic Domination We say that p first-order stochastically dominates q (written as pD 1 q ) if p q for any on L ( Z ) satisfying vNM assumptions as well as mono- tonicity in money. That is, pD 1 q if Eu ( p ) Eu ( q ) for all increasing u . Obviously, pD 1 q if the entire support of p is to the right of the entire support of q . But, we are interested in a more interesting condition
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Unformatted text preview: on a pair of lotteries p and q , one that will be not only sufFcient, but also necessary for p to Frst-order stochastically dominate q . ±or any lottery p and a number x , deFne G ( p , x ) = ∑ z ≥ x p ( z ) (the probability that the lottery p yields a prize at least as high as x ). Denote by F ( p , x ) the cumulative distribution function of p , that is, F ( p , x ) = Probability { z | z < x } . Claim: pD 1 q iff for all x , G ( p , x ) ≥ G ( q , x ) (alternatively, pD 1 q iff for all x , F ( p , x ) ≤ F ( q , x ) ). (See Fg. 9.1.) Proof: Let x < x 1 < x 2 < . . . < x K be the prizes in the union of the supports of p and q . ±irst, note the following alternative expression for Eu ( p ) : Eu ( p ) = X k ≥ p ( x k ) u ( x k ) = u ( x ) + X k ≥ 1 G ( p , x k )( u ( x k ) − u ( x k − 1 ))....
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