14127lec6_bndrat

14127lec6_bndrat - 14.127 Lecture 6 Xavier Gabaix 0.0.1...

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14.127 Lecture 6 Xavier Gabaix March 11, 2004
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0.0.1 Shrouded attributes. A continuation Rational guys U i = q p + max ( V p,V e ) + σε i ¯ = q p + V min ( p ,e ) + σε i = U i + σε i Rational demand for good 1 D 1 = P U 1 > max U i = P ¯ i =2 ,...,n U 1 + σε 1 > max U i i =2 ,...,n We look for symmetric equilibrium with ( p i ,p i ) = ( p ,p ) for i = 2 , ..., n .
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Profits π 1 = D 1 p + p 1 p e Assume c = 0 and c = 0 Call α fraction of rational guys. Then π = α p + p 1 p e D ( A ) + (1 α ) p + p 1 p V D ( p + p ) where A = p min ( p,e ) + p + min ( p ,e )
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The slope of profit ∂π = αD ( A ) α p + p 1 p e D ( A ) + (1 α ) D ( A ) (1 α ) p + p 1 p ∂p = αD (0) α p + p 1 p e D (0) + (1 α ) D (0) (1 α ) p + p 1 p V = D (0) p + pQ D (0) where Q denotes the fraction of consumers that buy the add-on. Q = α 1 p e + (1 α ) 1 p V
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Hence D (0) p + pQ = D (0) = µ thus the total industry profits do not depend by the add-on structure ( α,V,e ). Proposition. An optimum p ∈ { e,V } . Proof. If e <p < V , then it is better to set p = V . If 0 p < e , then it is better to increase p to e and lower p by e p . Then rational demand stay the same, and irrational demand increases. QED
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So, p e and min ( e,p ) = e , and π 1 = D ( p + p ) α p + p 1 p e + (1 α ) ( p + p ) = D ( p + p ) p + p α 1 p e + 1 α �� = D ( p + p ) [ p + F ] If p = V then F = V (1 α ) . if p = e then F = e . Consequently: p = V iff V (1 α ) >e , or in other words, when e α<α = 1 V p = e iff e α>α = 1 V
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Take α<α . Then p = µ pQ + c = µ (1 α ) + c So if µ = 0 then p<c and the base good is a loss leader.
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1 Advertising. Does it solve the problem? Furthermore, manufacturers in a competitive equipment market have incentives to avoid even this inefficiency by providing informa- tion to consumers. A manufacturer could capture profits by raising its [base-good] prices above market levels (i.e., closer to cost), lowering
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