Econ139 Lecture 08 Spring 2021 PDF

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ECON 139 Notes - Lecture 8 Students: Helton Arata-Suzuki, Kuitai Wang, Virginia Bergamasco February 2021

Expected Utility Theorem The preference relation  within the space of simple lotteries L can be represented by the following function with linear probabilities: U: L→R Within L, L  L’ if and only if U(L)  U(L’), in such a way that: U(L) = p1 u(x1 ) + ... + pN u(xN ) U(L′ ) = p′1 u(x1 ) + ... + p′N u(xN ) Under the four assumptions delineated below: 1) Completeness/Rationality: There must exist a rational preference relation  in the defined space of simple lotteries L. This assumes that the individual has well-defined preferences in L, and can always choose between them. For each (L,L’), LL’ or L’L. 2) Continuity: The preference relation  must be continuous. This implies that if there are 3 lotteries (L, L’ and L”), in which there’s a preference L  L’ and L’  L”, then there ought to be a combination of L and L” that the individual regards as equivalent to lottery L’. In such a way that: For any L, L’ and L” ∈ L where L  L’  L” there must exist an α ∈ [0, 1] such that: L′ ∼ αL + (1 − α)L” 3) Independence: The preference relation  in L is such that for all L, L’ and L” within L and α in [0,1] two preference relations will stay the same regardless of a third element. Let three lotteries, L, L’ and L”, such that α ∈ [0,1], if: αL + (1 − α)L”  αL′ + (1 − α)L” then the third choice L” is irrelevant with regard to the preference relation between L and L’ which will remain the same regardless of the value of L”. 4) Best and worst: There’s a best and a worst lotteries: L and L respectively. Observation: A degenerate lottery of L, denoted L”, pays XN with probability 1 and any other payoffs with probability 0, so that: U(Ln )=u(Xn ), where u:W ⊆R→R, and U :L→R Proof using a von Neumann-Morgenstern Utility Function Steps to construct a von Neumann-Morgenstern (VNM) utility function that represents the preference L L’ given that U(L)  U(L”): Step 1: By completeness and continuity, there exists a α and β ∈ [0,1] such that: L ∼ α + (1 − α)L ′

L ∼ β + (1 − β)L Step 2: Be cognizant that L  L’ if and only if α ≥ β

1

We start with the reverse direction: suppose α ≥ β If α = β, then we must have L ∼ αL + (1 − α)L ∼ βL + (1 − β)L ∼ L′ for transativity, it implies L ∼ L′ If α > β and let γ =

α−β 1−β

∈ (0, 1) :

from step one we know L ∼ αL + (1 − α)L

= γL + (1 − γ)(βL + (1 − β)L) ≻ γ(βL + (1 − β )L) + (1 − γ)(βL + (1 − β )L) = βL + (1 − β )L ∼ L′

(1)

this implies L ≻ L′ for the equation (1) In fact we know that L ≻ βL +(1−β)L, β ∈ (0, 1); for indipendence, with (1−γ)(βL +(1−β)L) as third lottery we obtain that Now we prove the first direction: suppose L  L′ . we will argue by contradiction: suppose β > α : this implies L′ ≻ L, that contradicts the supposition. Therefore we must have α ≥ β . Step 3: Define U such that U (L) = α and U (L′ ) = β Step 4: by completeness and continuity, ∃ scalars γ1, ...., γα ∈ (0, 1) such that L1 ∼ γ1L + (1 − γ1)L . . . N

L

∼ γN L + (1 − γN )L

Step 5: observe that for any lottery L = (p1, ...pN) we can write so, as a linear combination of degenerate lotteries L1, L2,..,LN

1

N

2

      0 0 1  0  1  0            0  + p2 0 + ... + pN 0 L = p1  . . .       . . . 1 0 0

Corollary

10000

15000

0.1

0.09 B

0.9

0.91 0

0

10000

15000

1

0.9

C

D 0

0.1 0

0

: 0....