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Slides16-2010

# Slides16-2010 - Problems with Expected Utility Charles B...

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Problems with Expected Utility Charles B. Moss Outline Allais’ Paradox Common Consequence Effect Problems with Expected Utility: Lecture XVI Charles B. Moss September 28, 2010 Charles B. Moss Problems with Expected Utility

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Problems with Expected Utility Charles B. Moss Outline Allais’ Paradox Common Consequence Effect 1 Allais’ Paradox 2 Common Consequence Effect Charles B. Moss Problems with Expected Utility
Problems with Expected Utility Charles B. Moss Outline Allais’ Paradox Common Consequence Effect Allais’ Paradox Machina starts by describing how the expected utility formulation is linear in probabilities. We start by assuming that there are three possible outcomes for a random variable { x 1 , x 2 , x 3 } such that x 1 < x 2 < x 3 with probabilities P = ( p 1 , p 2 , p 3 ). Given this formulation 3 i =1 p i = 1 p 2 = 1 p 1 p 2 (1) Which implies U ( x ) = p 1 U ( x 1 ) + p 2 U ( x 2 ) + p 3 U ( x 3 ) = p 1 U ( x 1 ) + (1 p 1 p 3 ) U ( x 2 ) + p 3 U ( x 3 ) (2) Charles B. Moss Problems with Expected Utility

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Problems with Expected Utility Charles B. Moss Outline Allais’ Paradox Common Consequence Effect Continued Holding the payoffs in Equation 3.52 constant and differentiating with respect to the levels of probability yields dU ( x ) = U ( x 1 ) dp 1 + ( dp 1 dp 3 ) U ( x 2 ) + U ( x 3 ) dp 3 = 0 [ U ( x 1 ) U ( x 2 )] dp 1 = [ U ( x 2 ) + U ( x 3 )] dp 3 [ U ( x 2 ) U ( x 1 )] [ U ( x 3 ) U ( x 2 )] = dp 3 dp 1 (3) which given that x 1 < x 2 < x 3 and if U ( x ) > 0 implies that any increase in p 3 relative to p 1 that holds the total probability constant (i.e., the probabilities still sum to one) yields more expected utility.
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