In Example 20.5 we showed that the Nash equilibrium in this first-price, sealed bid auction was for each participant to adopt a bidding strategy of \(b(v)=[(n-1) / n] v\). The total revenue a seller might expect to receive from such an auction will obviously be \([(n-1) / n] v^{*}-\) where \(v^{*}\) is the expected value of the highest valuation among the \(n\) auction participants. a. Show that if valuations are uniformly distributed over the interval \([0,1],\) the expected value for \(v^{*}\) is \(n /(n+1) .\) Hence expected revenue from the auction is \((n-1) /(n+1)\) Hint: The expected value of the highest bid is given by \\[ E\left(v^{*}\right)=\int_{0}^{1} v f(v) d v \\] where \(f(v)\) is the probability density function of the probability that any particular \(v\) is the maximum among \(n\) bidders. Here \(f(v)=n v^{n-1}\) b. In a famous 1961 article ("Counterspeculation, Auctions, and Competitive Sealed Tenders," Journal of Finance, March \(1961,\) pp. \(8-37\) ) William Vickrey examined second-price sealed bid auctions. In these auctions the highest bidder wins, but pays the price bid by the second highest bidder. Show that the optimal bidding strategy for any participant in such an auction is to bid his or her true valuation: \(b(v)=v\) c. Show that the expected revenue provided by the second-price auction format is identical to that provided by the first-price auction studied in part a (this is Vickrey's "revenue equivalence theorem"). Hint: The probability that any given valuation will be the second highest among \(n\) bidders is given by \(g(v)=(n-1)(1-v) n v^{n-2}\). That is, the probability is given by the probability that any of \((n-1)\) bidders will have a higher valuation \([(n-1)(1-v)]\) times the probability that any of \(n\) bidders will have a valuation exceeding that of \(n-2\) other bidders \(\left[n v^{n-2}\right]\)

Step by step solution

01

Expected value for \(v^{*}\)

We are given the probability density function of the probability that any particular \(v\) is the maximum among \(n\) bidders: \(f(v) = n v^{n-1}\). We have to find the expected value of \(v^{*}\) using the given formula: \[E(v^{*}) = \int_{0}^{1} v f(v) dv\] Substitute \(f(v)\) in the formula: \[E(v^{*}) = \int_{0}^{1} v \cdot n v^{n-1} dv\] Now, integrate the equation: \[E(v^{*}) = n \int_{0}^{1} v^n dv = n \left[\frac{v^{n+1}}{n+1}\right]_0^1 = \frac{n}{n+1}\] So, the expected value for \(v^{*}\) is \(\frac{n}{n+1}\).

02

Expected revenue from the auction

The expected revenue from the auction is given by \([(n-1) / n] v^{*}\). We have found the value of \(v^{*}\) in step 1. Substitute the value to calculate the expected revenue: \[R = \frac{n-1}{n} \cdot \frac{n}{n+1}\] \[R = \frac{n-1}{n+1}\] Hence, the expected revenue from the auction is \(\frac{n-1}{n+1}\). b. Show the optimal bidding strategy for a second-price sealed bid auction

03

Nash Equilibrium in second-price sealed bid auctions

In a second-price sealed bid auction, the highest bidder wins but pays the price bid by the second highest bidder. In this case, there is no incentive for a bidder to place a bid higher than their true valuation. If a bidder bids higher than their true valuation, the bidder might end up paying more than the value of the item. Similarly, if a bidder bids lower than their true valuation, they might miss out on winning the auction. So, the optimal bidding strategy for any participant in a second-price sealed bid auction is to bid their true valuation: \(b(v) = v\). c. Show that the expected revenue of the second-price auction format is identical to the first-price auction

04

Calculate expected revenue in the second-price auction

We are given the probability density function for the second highest among \(n\) bidders: \(g(v)=(n-1)(1-v)n v^{n-2}\). The expected revenue from a second-price auction is given by integrating the product of the valuation and density function: \[R = \int_{0}^{1} v g(v) dv\] Substitute \(g(v)\) in the formula: \[R = \int_{0}^{1} v (n-1)(1-v)n v^{n-2} dv\] Now, integrate the equation: \[R = (n-1)n \int_{0}^{1} v^n (1-v) dv\] We can use integration by parts to solve this integral. Let \(u=v^n\) and \(dv=(1-v)dv\). Then, \(du=nv^{n-1}dv\) and \(v=(1-v)dv\). \[R = (n-1)n \left[ \int u dv - \int v du \right]\] \[R = (n-1)n \left[ \left[v^n(1-v)\right]_0^1 - \int_0^1 nv^{n-1}(1-v) dv \right]\] \[R = (n-1)n \left[ 0 - \int_0^1 nv^n - nv^{n+1} dv \right]\] \[R = (n-1)n \left[ \left[-\frac{v^{n+1}}{n+1}\right]_0^1 - \left[-\frac{v^{n+2}}{(n+1)(n+2)}\right]_0^1 \right]\] \[R = (n-1)n \left[ -\frac{1}{n+1} +\frac{1}{(n+1)(n+2)} \right]\] \[R = (n-1)n \left[ \frac{n+2-(n+1)}{(n+1)(n+2)} \right]\] \[R = \frac{n-1}{n+1}\] Hence, the expected revenue provided by the second-price auction format is identical to that provided by the first-price auction, which is \(\frac{n-1}{n+1}\).

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