Ch9 An extra detail on second degree PD

Ch9 An extra detail on second degree PD - ) ≥ P L-P H ⇒...

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Since a lot of students asked why the following is true, I will explain it in detail here, for your interest. This will not be tested! The process of deriving the optimal pairs will not be tested. The application of it to examples will. We will talk about examples in lecture on Nov 13. Lemma 1. If constraint 1 binds (i.e., holds with equality), then constraint 2 is satisfied. Proof. First, we show that at the seller’s optimal choice, it must be that q H q L . Suppose not, that is, q H - q L < 0. Constraint 1 binds implies θ H = P H - P L q H - q L The second constraint implies: θ L ( q L - q H
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Unformatted text preview: ) ≥ P L-P H ⇒ θ L ≥ P L-P H q L-q H = P H-P L q H-q L = θ H ⇒ θ L ≥ θ H This is a contradiction to the assumption that θ H &gt; θ L . Therefore, q H ≥ q L when the seller is optimizing. This implies q H ( θ H-θ L ) ≥ q L ( θ H-θ L ). Then a binding constraint 1 implies: θ H q H-P H = θ H q L-P L ⇒ θ H q H-P H-q H ( θ H-θ L ) ≤ θ H q L-P L-q L ( θ H-θ L ) ⇒ θ L q H-P H ≤ θ L q L-P L The last expression is exactly constraint 2. Therefore, constraint 2 is implied. 1...
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