A jar has an unknown value V with a flat prior. Each of 10 bidders sees an independent signal uniform on [V - 10, V + 10]. Conditional on your signal being s and on it being the highest of the 10 signals, how far below s does the expected value of V sit?

A jar has an unknown value V with a flat prior. Each of 10 bidders sees an independent signal uniform on [V - 10, V + 10]. Conditional on your signal being s and on it being the highest of the 10 signals, how far below s does the expected value of V sit?

Approach: Condition on V, write the chance every other signal falls below yours, and combine that with your own signal density to get the posterior for the gap between your signal and V.

90/11. The expected value of V sits 90/11 below your signal, about 8.18. Condition on V. Your signal density is flat on [V - 10, V + 10] and each rival signal falls below s with probability (s - V + 10)/20, since the highest signal is the relevant order statistic. Write u = s - V + 10, which lies in [0, 20]. Under a flat prior the posterior density of u is proportional to (u/20)^{n-1} with n = 10, so E[u] = 20n/(n + 1) = 200/11. Since V = s + 10 - u, the conditional expectation is E[V | s and s is highest] = s + 10 - 200/11, which is s - 90/11. Valuing the jar at your signal loses that much on every win, which is the winner's curse, and the bid shading of 10 - 20/(n + 1) grows with the number of bidders.

Follow-up: With 100 bidders instead of 10, how much shading does the same signal structure require?

Key concepts: winner's curse, conditional expectation, order statistics, bid shading.