Problem 3. You are bidding against one other bidder in a first-price sealed-bid

Question

Problem 3. You are bidding against one other bidder in a first-price sealed-bid auction with private values. You believe that the other bidder's valuation is equally likely to lie anywhere in the interval between $O and $500. Your own valuation is $200. Suppose you expect your rival to submit a bid that is exactly one half of its valuation. Thus, you believe that your rival's bids are equally likely to fall anywhere between 0 and $250. Given this, if you submit a bid of Q, the probability that you win the auction is the probability that your bid Q will exceed your rival's bid. It turns out that this probability is equal to Q/250. (Don't worry about where this formula comes from, but you probably should plug in several different values of Q to convince yourself that this makes sense.) Your profit from winning the auction is equal to (200-bid)*probability of winning. Show that your profit maximizing strategy is bidding half of your valuation. a + ???????? 1 ?? 1-

Explanation / Answer

In First-Price sealed bid auction, the bidders submit their respective bids without knowing what the opponents’ bids are.

My valuation is given as $200 while my opponent’s valuation is given as somewhere between $0 and $500. My bid is given as Q and the probability of winning the auction is Q/250. Additionally, the profit function is given as Pr= (200-bid) * (probability of winning)

i.e. Pr = (200-Q)*(Q/250).

Now, the first order condition (f.o.c) for profit maximization requires that the first order derivative is equal to zero , and the second order condition (s.o.c) for profit maximization requires that the second order derivative is negative. Thus,

From the given question,

Pr = 0.8Q – Q2/250

Or dPr/dQ = 0 implies 0.8 –(2Q/250) = 0

Thus, 0.8 = Q/125

Or Q = 100 which is half of the valuation 200.

The s.o.c is also satisfied as d2Pr/dQ2 = -2/125 <0

Now, in order to verify whether this is true for any valuation, let the valuation be assumed a constant value of V.

The profit function then becomes:

Pr = (V-Q)*(Q/250)

Or Pr = VQ/250 – Q2/250

Thus, f.o.c. requires dPr/dQ= 0

Or V/250 – 2Q/250 = 0

Or V/250 = 2Q/250

Or V = 2Q

Or Q = V/2 which is half the valuation.

The s.o.c is also satisfied as d2Pr/dQ2 = -2/125 <0

Thus, it has been proved that Profit will be maximized at a point where the bidding is exactly half of the valuation.

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